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# Talk:Conjectures/List

A000002 Conjecture: Taking the sequence in word lengths of 10, for example, batch 1-10, 11-20, etc... then there can only be 4, 5 or 6 1's in each batch. - _Jon Perry_, Sep 26 2012 A000006 Conjecture: No two successive terms in the sequence differ by more than 1. Proof of this would prove the converse of the theorem that every prime is surrounded by two consecutive squares, namely |sqrt(p)|^2 and (|sqrt(p)|+1)^2. - _Cino Hilliard_, Jan 22 2003 A000040 Conjecture: a(n) = (6*f(n)+(-1)^f(n)-3)/2, n>2, where f(n) = floor(ithprime(n)/3)+1. See A181709. - _Gary Detlefs_, Dec 12 2011 A000041 Conjecture: starting with offset 1 represents the numbers of ordered compositions of n using the signed (++--++...) terms of A001318 starting (1, 2, -5, -7, 12, 15,...). - _Gary W. Adamson_, Apr 04 2013 (this is true by the pentagonal number theorem, _Joerg Arndt_, Apr 08 2013) A000108 Conjecture: For any positive integer n, the polynomial sum(k=0..n, a(k)*x^k) is irreducible over the field of rational numbers. [From _Zhi-Wei Sun_, Mar 23 2013] A000120 Conjecture: The sequence where a(n) is the sum of digits of (n written in base b), can be written as a triangle T(r,k) in which all positive terms in column k are equal. Row 0 is a(0)=0 and row r lists a(b^(r-1)) .. a(b^r - 1), for r>=1 and b>=2. [From _Omar E. Pol_, Feb 20 2010] A000126 Conjecture. Let S(1)={1} and, for n>1, let S(n) be the smallest set containing x+1 and 2x+1 for each element x in S(n-1). Then a(n) is the sum of the elements in S(n). (See A122554 for a sequence defined in this way.) - _John W. Layman_, Nov 21 2007 A000172 Conjecture: a(n)== 2 (mod n^3) iff n is prime. - Gary Detlefs, Mar 22 2013 A000215 Conjecture: let the smallest prime factor of Fermat number F(n) be P(F(n)). If F(n) is composite, then P(F(n)) < 3*2^(2^n/2 - n - 2). - _Arkadiusz Wesolowski_, Aug 10 2012 A000217 Conjecture: For n>0, there is always a prime between A000217(n) and A000217(n+1). Sequence A065383 has the first 1000 of these primes. - _Ivan N. Ianakiev_, Mar 11 2013 A000350 Conjecture: Other than 1 and 5, there is no n such that Fibonacci(n) in binary ends with n in binary. The conjecture holds up to n=50000. - _Ralf Stephan_, Aug 21 2006 A000350 Conjecture is true. It is easy to prove (by induction on k) that if F(n) ends with n in binary, then n == 0, 1, or 5 modulo 3*2^k for any positive integer k, i.e., n must simply be equal to 0, 1, or 5. [From _Max Alekseyev_, Jul 03 2009] A000392 Conjecture. Let S(1)={1} and, for n>1, let S(n) be the smallest set containing x, 2x and 3x for each element x in S(n-1). Then a(n) is the sum of the elements in S(n). (It is easy to prove that the number of elements in S(n) is the n-th triangular number given by A001952.) See A122554 for a sequence defined in this way. - _John W. Layman_, Nov 21 2007 A000984 Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = sum_{k=0}^n(m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - _Zhi-Wei Sun_, Mar 23 2013 A001110 Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - _Ivan N. Ianakiev_, Sep 19 2012 A001122 Conjecture : sequence contains infinitely many pairs of twin primes. - _Benoit Cloitre_, May 08 2003 A001359 Conjecture: For any integers n >= m > 0, there are infinitely many integers b > a(n) such that the number sum_{k=m}^n a(k)*b^(n-k) (i.e., (a(m), ..., a(n)) in base b) is prime; moreover, when m = 1 there is such an integer b < (n+6)^2. [_Zhi-Wei Sun_, Mar 26 2013] A001554 Conjectures for o.g.f.s for this type of sequences appear in the PhD thesis by _Simon Plouffe_. See A001552 for the reference. These conjectures are proved in a link given in A196837. [Wolfdieter Lang, Oct 15 2011] A001555 Conjectures for o.g.f.s for this type of sequences appear in the PhD thesis by _Simon Plouffe_. See A001552 for the reference. These conjectures are proved in a link given in A196837. [Wolfdieter Lang, Oct 15 2011] A001556 Conjectures for o.g.f.s for this type of sequences appear in the PhD thesis by _Simon Plouffe_. See A001552 for the reference. These conjectures are proved in the link given in A196837. [Wolfdieter Lang, Oct 15, 2011] A001557 Conjectures for o.g.f.s for this type of sequences appear in the PhD thesis by _Simon Plouffe_. See A001552 for the reference. These conjectures are proved in the link given in A196837. - _Wolfdieter Lang_, Oct 15 2011 A001576 Conjecture: Let n>1, if a(n)= 1^n+2^n+4^n is a prime number then n is the form 3^h. Example, for h=1, n=3, a(n)= 1^3+2^3+4^3=73 (prime); h=2, n=9, a(n)= 1^9 + 2^9 + 4^9 = 262657 (prime); for h=3, n=27, a(n) is not prime. [From _Vincenzo Librandi_, Aug 03 2010] A001599 Conjecture: Every harmonic number is practical (A005153). I've verified this refinement of Ore's conjecture for all terms less than 10^14. - _Jaycob Coleman_, Oct 12 2013 A002093 Conjecture: (a) Every highly abundant number >10 is practical (A005153). (b) For every integer k there exists A such that k divides a(n) for all n>A. Daniel Fischer proved that every highly abundant number greater than 3, 20, 630 is divisible by 2, 6, 12 respectively. The first conjecture has been verified for the first 10000 terms. - _Jaycob Coleman_, Oct 16 2013 A002383 Conjecture: the set of these numbers, except 3, is the intersection of sets A085104 and A059055. See A225148. - _Thomas Ordowski_, May 02 2013 A002620 Conjectured size of the smallest critical set in a Latin square of order n (true for n <= 8). - Richard Bean (rwb(AT)eskimo.com), Jun 12 2003 and Nov 18 2003 A002878 Conjecture: for n>0, a(n)=sqrt(Fibonacci(4n+3)+sum_{k=2..2n}Fibonacci(2k)). [_Alex Ratushnyak_, May 06 2012] A002893 Conjecture: For each n=1,2,3,... the polynomial g_n(x) = sum_{k=0}^n binomial(n,k)^2*binomial(2k,k)*x^k is irreducible over the field of rational numbers. [_Zhi-Wei Sun_, Mar 21 2013] A002895 Conjecture: Let D(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,n. Then the number D(n)/12^n is always a positive odd integer. - _Zhi-Wei Sun_, Aug 14 2013. A003116 Conjecture: a(n) is the number of compositions p(1) + p(2) + ...p(m) = n with p(i)-p(i-1) <= 1, see example; cf. A034297. - _Vladeta Jovovic_, Feb 09 2004 A003273 Conjectured asymptotics (based on random matrix theory) on p. 453 of Cohen's book. [From _Steven Finch_, Apr 23 2009] A003630 Conjecture: Primes congruent to (5, 7) mod 12. - _Vincenzo Librandi_, Aug 06 2012 A004050 Conjecture: Only 5, 11, 17, 35, and 259 can be expressed in two different ways. See A085634. - _Robert G. Wilson v_, Sep 19 2012 A005117 Conjecture: For each n=2,3,... there are infinitely many integers b > a(n) such that sum_{k=1}^n a(k)*b^(k-1) is prime, and the smallest such an integer b does not exceed (n+3)*(n+4). [_Zhi-Wei Sun_, Mar 26, 2013] A005153 Conjecture: The sequence a(n)^(1/n) (n=3,4,...) is strictly decreasing to the limit 1. - _Zhi-Wei Sun_, Jan 12 2013 A005165 Conjecture: for n > 2, smallest prime divisor of a(n) > n. - _Gerald McGarvey_, Jun 19 2004. Rebuttal: This is not true - Zivkovic (Math. Comp. 68 (1999), pp. 403-409) has demonstrated that 3612703 divides A_{n} for all n >= 3612702. - Paul Jobling, Oct 18 2004. A005234 Conjecture: if p# + 1 is a prime number, then the next prime is less than p# + exp(1)*p. - _Arkadiusz Wesolowski_, Feb 20 2013 A005234 Conjecture: if p# + 1 is a prime, then the next prime is less than p# + p^2. - _Thomas Ordowski_, Apr 07 2013 A005251 Conjecture: a(n+1) + |A078065(n)| = 2*A005314(n+1). - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Dec 21 2004 A005258 Conjecture: For each n=1,2,3,... the polynomial a_n(x) = sum_{k=0}^n C(n,k)^2*C(n+k,k)*x^k is irreducible over the field of rational numbers.[_Zhi-Wei Sun_, Mar 21 2013] A005259 Conjecture: For each n=1,2,3,... the Apery polynomial A_n(x) = sum_{k=0}^n binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. [_Zhi-Wei Sun_, Mar 21 2013] A005329 Conjecture: this sequence is the inverse binomial transform of A075272 or, equivalently, the inverse binomial transform of the BinomialMean transform of A075271. - _John W. Layman_, Sep 12 2002 A005376 Conjecture: a(n) is approximately c*n, where c is the real root of x^5+x-1 = 0, c=0.754877666246692760049508896... - _Benoit Cloitre_, Nov 05 2002 A005408 Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q>0)); the top row of its coach is comprised of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - _Gary W. Adamson_, Sep 09 2012 A005430 Conjecture: the terms of the inverse binomial transform are 2*A132894(n). - _R. J. Mathar_, Oct 21 2012 A005809 Conjecture: a(n)==3 (mod n^3) iff n is an odd prime. - _Gary Detlefs_, Mar 23 2013 A005810 Conjecture: a(n) == 4 (mod n^3) iff n is prime. - _Gary Detlefs_, Apr 03 2013 A006010 Conjectures: If n is odd then a(n) = (1/8) * (n^4 + 2*n^3 + 2*n^2 + 2*n + 1) = Det(Transpose[M]*M) where M is the 2 X 3 matrix whose rows are [(n-1)/2, (n-1)/2], [(n-1)/2 + 1, 0] and [(n-1)/2 + 1, (n-1)/2 + 1]. If n is odd then a(n) = (1/8) * (n^4 + 2*n^3 + 2*n^2) = Det(Transpose[M]*M) where M is the 2x3 matrix whose rows are [n/2, 0], [n/2, n/2] and [n/2 + 1, 0]. - _Gerald McGarvey_, Oct 30 2007 A006077 Conjecture: Let W(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,n. If n == 1 (mod 3) then W(n) = 0. When n == 0 or 2 (mod 3), W(n)*(-1)^(floor[(n+1)/3])/6^n is always a positive odd integer. - _Zhi-Wei Sun_, Aug 21 2013 A006077 Conjecture: Let p == 1 (mod 3) be a prime, and write 4*p = x^2 + 27*y^2 with x, y integers and x == 1 (mod 3). Then W(p-1) == (-1)^{(p+1)/2}*(x-p/x) (mod p^2), where W(n) is defined as the above. - _Zhi-Wei Sun_, Aug 23 2013 A006318 Conjecture: For each n > 2, the polynomial sum_{k = 0}^n a(k)*x^{n-k} is irreducible modulo some prime p < n*(n+1). - _Zhi-Wei Sun_, Apr 07 2013 A006990 Conjecture: for n > 2, n! - a(n) is 1 or a prime i.e. a non-composite number. E.g. 3! - 5 = 1, 5!- 113 = 7 etc. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Mar 19 2002 A007059 Conjecture: for n>0 a(n+1) is the number of "numbral" divisors of (4^n-1)/3 = A002450(n) (see A048888 for the definition of numbral arithmetic). This has been verified computationally up to n=15. - _John W. Layman_, Dec 18 2001. This conjecture follows immediately from Proposition 2.3 of Frosini and Rinaldi. - _N. J. A. Sloane_, Apr 29 2011. A007185 Conjecture: For any m coprime to 10 and for any k, the density of n such that a(n) == k (mod m) is 1/m. - _Eric M. Schmidt_, Aug 01 2012 A007356 Conjecture: for n > 26069, a(n) = n + 3715. [_Charles R Greathouse IV_, Sep 28 2011] A007413 Conjecture. The sequence is formed by the numbers of 1's between every pair of consecutive 2's in A076826. - Vladimir Shevelev, May 31 2009 A007489 Conjectured to be the length of the shortest word over {1,...,n} that contains each of the n! permutations as a factor [see Johnston]. - _N. J. A. Sloane_, May 25 2013 A007645 Conjecture: this sequence is Union(A002383,A162471). [From _Daniel Tisdale_, Jul 04 2009] A007669 Conjecture (V. Kotesovec, 2010): A007669(n) = A178717(n-1) - 1 [From _Vaclav Kotesovec_, Sep 02 2010] A007675 Conjecture: for every prime p, the numbers p#-1, p#, p#+1 are squarefree, where primorial p# = product of all primes <= p. - _Thomas Ordowski_, Apr 21 2013 A007691 Conjecture: If n is such that 2^n-1 is in A066175 then a(n) is a triangular number. - _Ivan N. Ianakiev_, Aug 26 2013 A007691 Conjecture: Every multiply-perfect number is practical (A005153). I've verified this conjecture for the first 5261 terms with abundancy > 2 using Achim Flammenkamp's data. The even perfect numbers are easily shown to be practical, but every practical number > 1 is even, so a weak form says every even multiply-perfect number is practical. - _Jaycob Coleman_, Oct 15 2013 A007850 Conjecture: Giuga numbers are the solution of the differential equation n'=n+1, being n' the arithmetic derivative of n. [From _Paolo P. Lava_, Nov 16 2009]. A007933 Conjecture: a(n) ~ 3n/2. - _Charles R Greathouse IV_, Sep 19 2012 A007934 Conjecture: a(n) ~ 3n/2. - _Charles R Greathouse IV_, Sep 19 2012 A007935 Conjecture: a(n) ~ 3n/2. - _Charles R Greathouse IV_, Sep 19 2012 A008347 Conjecture: For any m = 1, 2, 3, ... and r = 0, ..., m - 1, there are infinitely many positive integers n with a(n) == r (mod m). [_Zhi-Wei Sun_, Feb. 27, 2013] A008347 Conjecture: For any n > 9 we have a(n+1) < a(n-1)^(1+2/(n+2)). (This yields an upper bound for prime(n+1) - prime(n) in terms of prime(1), ..., prime(n-1). The conjecture has been verified for n up to 10^8.) -- _Zhi-Wei Sun_, June 9, 2013. A008364 Conjecture: these are numbers n such that (sum(k^4, k=1..n) mod n = 0) and (sum(k^6, k=1..n) mod n =0). [_Gary Detlefs_, Dec 20 2011] A008364 Conjecture: these are numbers n such that (n^6 mod 210=1) or (n^6 mod 210=169). [_Gary Detlefs_, Dec 30 2011] A008365 Conjecture: Numbers n such that n^24 is congruent to {1,421,631,841} mod 2310. [From Gary Detlefs, Dec 30 2011] A008407 Conjecture: (i) The sequence a(n)^(1/n) (n=3,4,...) is strictly decreasing (to the limit 1). (ii) We have 0 < a(n)/n - H_n < (gamma + 2)/(log n) for all n > 4, where H_n denotes the harmonic number 1+1/2+1/3+...+1/n, and gamma refers to the Euler constant 0.5772... [The second inequality has been verified for n = 5, 6, ..., 5000.] - _Zhi-Wei Sun_, Jun 28 2013. A008407 Conjecture: For any integer n > 2, there is 1 < k < n such that 2*n - a(k)- 1 and 2*n - a(k) + 1 are twin primes. Also, every n = 3, 4, ... can be written as p + a(k)/2 with p a prime and k an integer greater than one. - _Zhi-Wei Sun_, Jun 29-30 2013. A008486 Conjecture from Dmitry Kamenetsky, Jun 29 2008: This is also the maximum number of edges possible in a planar simple graph with n+2 vertices. A008578 Conjecture: the sequence contains exactly those n such that sigma(n) > n*BigOmega(n). - _Irina Gerasimova_, Jun 08 2013 A008676 Conjecture: a(n) = Floor[2*(n + 3)/3] - Floor[3*(n + 3)/5]. - _John W. Layman_, Sep 23 2009 A008683 Conjecture: a(n) = determinant of Redheffer matrix A143104 where T(n,n)=0. Verified for 50 first terms. - Mats O. Granvik, Jul 25 2008 A008683 Conjecture: Consider the table A051731 and treat 1 as a divisor. Move the value in the lower right corner vertically to a divisor position in the transpose of the table and you will find that the determinant is the Moebius function. The number of permutation matrices that contribute to the Moebius function appears to be A074206. - Mats Granvik, Dec 08 2008 A008749 Conjecture: For n >= 1, A067628(a(n+2)) appears for the first time in A067628. Equivalently, A067628(a(n+2)) is the first T such that the minimal perimeter of polyiamonds of T triangles is a(n+2). - Winston C. Yang (winston(AT)cs.wisc.edu), Feb 05 2002 A010048 Conjecture: polynomials with (positive) Fibonomial coefficients are reducible iff n odd >1. - Ralf Stephan, Oct 29 2004 A011754 Conjecture: a(n)/n tends to log(3)/(2*log(2)) = 0.792481250... - _Ed Pegg Jr_, Dec 05 2002 A014688 Conjecture: this sequence contains an infinite number of primes (A061068), yet contains arbitrarily long "prime deserts" such as the 11 composites in A014688 between a(6) = 19 and a(18) = 79 and the 17 composites in A014688 between a(48) = 271 and a(66) = 383. - _Jonathan Vos Post_, Nov 22 2004 A016090 Conjecture: For any m coprime to 10 and for any k, the density of n such that a(n) == k (mod m) is 1/m. - _Eric M. Schmidt_, Aug 01 2012 A018800 Conjecture: If a(n) = (n concatenated with k) then k < n. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), May 01 2002 A019469 Conjecture: sequence is union of powers of two >1 (A000079) and 3 * A096304. A020697 Conjecture: all numbers of the form 3*2^n are in this sequence - _J. Lowell_, Mar 12 2008. However, this is false: 3*2^12 = 12288 is the smallest counterexample to above conjecture. - Klaus Brockhaus, Jul 24 2008 A022344 Conjecture: Every pair of Fibonacci sequences, F1 and F2, appear in rows n and m of Wythoff's Array, respectively and have respective characteristics a(n) and a(m). Also, there is a third Fibonacci sequence F3, defined by F3(i) = F1(i) * F2(j+1) - F1(i+1)*F2(j) where j is held constant. The sequence F3 appears in row p of Wythoff's array and has the characteristic a(p) = a(n)*a(m). - _Kenneth J Ramsey_, Feb 11 2007 A022544 Conjecture: if n is not the sum of 2 squares sigma(n)==0 mod 4 (the converse does not hold). - _Benoit Cloitre_, May 19 2002 A023402 Conjecture: a(n) = 3rd divisor of smallest number having exactly n+2 divisors, a(n)=T(n+2,3), T as defined in A081532. - _Reinhard Zumkeller_, Jul 18 2003 A023855 Conjecture: Antidiagonal sums of triangle A075462. - _L. Edson Jeffery_, Jan 20 2012 A024187 Conjecture: essentially the same as A001711. - _Ralf Stephan_, Dec 30 2004 A024812 Conjecture (verified for k<=10^6 by _M. F. Hasler_): A024812(k) = (A024813(k)+k-1)/2, k=1,2,.... - _L. Edson Jeffery_, Mar 21 2013 A024813 Conjecture (verified for m < 10^6 by _M. F. Hasler_): A024813(n) = 2*A024812(n)-n+1, n=1,2,.... - _L. Edson Jeffery_, Mar 21 2013 A025060 Conjecture: If i, j and k are allowed to be negative, but not zero, and are still distinct, then the sequence is all the integers. - _Jon Perry_, Apr 21 2013 A025163 Conjecture: (-n+1)*a(n) +2*(2*n-1)*a(n-1) -8*n*a(n-2)=0. - _R. J. Mathar_, Feb 05 2013 A025191 Conjectures: a(n) = A027914(n)-A027914(n-1) = 1/2*(A081673(n)-A081673(n-1). A025416 Conjecture : The sequence never becomes monotonic increasing. - _Jon Perry_, Nov 03 2012 A026002 Conjecture: define an infinite array to have m(n,1) = m(1,n) = n*(n-1)+1 in the first row and column, and m(i,j) = m(i,j-1) + m(i-1,j-1) + m(i-1,j); then m(n,n) = a(n). - _J. M. Bergot_, Apr 24 2013 A027687 Conjecture: A010888(a(n)) divides a(n). Tested for n up to 36 incl. - _Ivan N. Ianakiev_, Oct 31 2013 A027697 Conjecture: a(n) < A027699(n) except for n = 2; verified up to n=5*10^7. Moreover, I conjecture that A027699(n) - a(n) tends to infinity. - _Vladimir Shevelev_ A027833 Conjecture: a(n) < log(A014574(n))^2. [Thomas Ordowski, Jul 21 2012] A028723 Conjectured to be crossing number of complete graph K_n, see A000241. A029766 Conjecture: for n>0, a(n) = n! * A040039(n-1). - _Vaclav Kotesovec_, Sep 30 2013 A029795 Conjecture: there exists some m and N for which a(n) = m + n for all n >= N. [_Charles R Greathouse IV_, Jun 28 2011] A029797 Conjecture: there exists some m and N for which a(n) = m + n for all n >= N. [_Charles R Greathouse IV_, Jun 28 2011] A029858 Conjectured to be the number of integers from 0 to 10^(n-1) - 1 that lack 0, 1, 2, 3, 4, 5 and 6 as a digit. - _Alexandre Wajnberg_, Apr 25 2005. This is easily verified to be true. - Renzo Benedetti, Sep 25 2008 A029905 Conjectured to be complementary to h(n), see A029902. A030179 Conjectured to be crossing number of complete bipartite graph K_{n,n}. Known to be true for n <= 7. A030210 Conjecture: |a(p)| < 2*p^(3/2) for p prime. - _Michael Somos_, Oct 31 2005 A032594 Conjecture: sequence is finite. [_Charles R Greathouse IV_, Mar 22, 2011] A033932 Conjecture: No term is a composite number. a(n)is a prime > 3*prime(k) where prime(k) < n = < prime(k+1). - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 07 2004 A034175 Conjectured to be a permutation of the nonnegative integers. A034496 Conjecture: No primes in this sequence (checked for first 10000 terms). [From _Artur Jasinski_, Sep 23 2008] A034693 Conjecture: for every n > 1 there exists a number k < n such that n*k + 1 is a prime - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 17 2001 A035312 Conjectured to form a permutation of the positive integers. A035313 Conjecture: lim(n->infinity, a(n+1)/a(n)) = 2. - _David W. Wilson_, Feb 26 2012 A036263 Conjecture: |a(1)|+|a(2)|+..+|a(n)| ~ prime(n). [Thomas Ordowski, Jul 21 2012] A036284 Conjecture: For n>=1, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x^3 + 1) ^ A000225(n-1). If this holds, then for n>=1, a(n) = A048720bi(A136380(n),A048723bi(9,A000225(n-1))). Conjecture 2: there is also one extra (x^1 + 1) factor present, see A136384. A036496 Conjecture: a(n) is half the minimal perimeter of a polyhex of n hexagons. - Winston C. Yang (winston(AT)cs.wisc.edu), Apr 06 2002. This conjecture follows from the Brunvoll et al. reference. - _Sascha Kurz_, Mar 17 2008 A036840 Conjecture: a(n) is never 0; i.e. the sequence {T(n,k)} is eventually periodic for every n. A036845 Conjecture: The sequence {T(n,k)} is eventually periodic for every n, so a(n) can be computed in finite time. A036845 Conjecture: a(n) -> infinity as n -> infinity. A037084 Conjecture : for any x, the iterated process "x ->3x-1" if x is odd or "x ->x/2" if x is even leads to one of the following three cycles: (1, 2), (5, 14, 7, 20, 10), (41, 122, 61, 182, 91, 272, 136, 68, 34, 17, 50, 25, 74, 37, 110, 55, 164, 82). - _Benoit Cloitre_, May 14 2002 A037096 Conjecture: For n>=3, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x + 1) ^ A055010(n-1). If this holds, then for n>=3, a(n) = A048720bi(A136386(n),A048723bi(3,A055010(n-1))). A037097 Conjecture: For n>=3, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x + 1) ^ A000225(n-1) (= A051179(n-2)). If this holds, then for n>=3, a(n) = A048720bi(A136386(n),A048723bi(3,A000225(n-1))) = A048720bi(A136386(n),A051179(n-2)). A037223 Conjecture: a(n) = Prod(i, 1<=i<=n and phi(i)<=floor(i/2)). - _Enrique Pérez Herrero_, May 31 2012. This conjecture is WRONG, counterexample is n=105. [From _Vaclav Kotesovec_, Sep 07 2012] A039004 Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n). - _Gerald McGarvey_, Nov 18 2007 A039824 Conjecture: for n>6, a(n) = n^2 - 3. - _Ralf Stephan_, Mar 07 2004 A046060 Conjectured finite and probably these are the only terms; cf. Flammenkamp's link. [_Georgi Guninski_, Jul 25 2012] A046061 Conjectured finite and probably these are the only terms; cf. Flammenkamp's link. [_Georgi Guninski_, Jul 25 2012] A046869 Conjecture: primes p(n) such that 2*p(n) >= p(n-1) + p(n+1). [Thomas Ordowski, Jul 25 2012] A047211 Conjecture: n such that the characteristic polynomial of M(n) is irreducible over the rationals where M(n) is an n x n matrix with ones on the skew diagonal and below it and the skew line two positions above it and otherwise zeros; see example for one such matrix. Tested up to n=177. [_Joerg Arndt_, Aug 10 2011] A047926 Conjecture: Number of representations of 3^(2n) as a sum a^2 + b^2 + c^2 with 0 < a <= b <= c. That is, a(1)=3 because 3^2 = 1^2+2^2+2^2, a(2)=3 because 3^4 = 1^2+4^2+8^2 = 3^2+6^2+6^2 = 4^2+4^2+7^2. - Zak Seidov, Mar 01 2012 A048963 Conjecture: regarded as a decimal fraction, this number is normal in base 10. - _Franklin T. Adams-Watters_, Aug 20 2012 A049048 Conjecture: If c is composite then n! == 1 (mod c) for at most one n >= 2. - David Wilson, Oct 06 2009 A049094 Conjecture: 2^n-1 is squarefree iff gcd(n,2^n-1)=1. If true the conjecture would imply that Mersenne numbers (cf. A001348) are squarefree. - _Vladeta Jovovic_, Apr 12 2002. But the conjecture is not true: counterexamples are n = 364 and n = 1755, i.e. gcd(364,2^364-1) = 1 and (2^364-1) mod 1093^2 = 0 and gcd(1755,2^1755-1) = 1 and (2^1755-1) mod 3511^2 = 0, cf. A001220. - _Vladeta Jovovic_, Nov 01 2005 A049096 Conjecture: lim n -> infinity a(n)/n = C exists and 4<C<9/2. There seems to be a sequence of primes p such that p^2 never divides numbers of the form 2^x+1: the first few are 5,7,23,31. - _Benoit Cloitre_, Aug 20 2002 A049591 Conjecture: start from any initial value f(1)>=2 and define f(n) to be the largest prime factor of f(1)+f(2)+...+f(n-1) then f(n)=n/2+O(log(n)) and there are infinitely primes p such that f(2p)=p. Conjecture: current sequence gives primes satisfying f(2p)=p when f(1)=3. - _Benoit Cloitre_, Jun 04 2003 A050295 Conjecture: for k=1,2,3,..., a(6k+1)=2a(6k) and a(6k+5)=2a(6k+4) (these relations hold through a(35)). - _John W. Layman_, Jun 22 2002 A051250 Conjecture: the sequence is finite and 60 is the largest term, empirically verified up to 10^7; A051451 Conjecture: For every n > 2, there exists a twin prime pair [p, p+2] with p < a(n) such that the "next" twin prime pair > a(n) is [a(n)+p, a(n)+p+2]. Example: For n=6 we can take p=11, because the next twin prime pair > a(6) = 420 is [431, 433], which is [420+11, 420+13]. This has been verified for 2 < n <= 200. - _Mike Winkler_, Sep 12 2013 A051827 Conjectured o(10^n) A051924 Conjecture: a(n) mod n^2 = n+2 iff n is an odd prime. - _Gary Detlefs_, Feb 19 2013 A053029 Conjecture: m is on this list iff m is an odd number all of whose factors are on this list or m is twice such an odd number A053031 Conjecture: m is on this list iff m is an odd number all of whose factors are on this list or m is 2 or 4 times such an odd number A053175 Conjecture: Let P(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,n. Then P(n)/2^(n*(n+3)) is a positive odd integer. - _Zhi-Wei Sun_, Aug 14 2013 A053319 Conjecture: a(n) < log(A014574(n))^3 for n > 2. [Thomas Ordowski, Jul 21 2012] A053697 Conjecture. For any positive integer a(1), the sequence generated according to the above rule eventually cycles through the forms a(k)=[1][4^a][3][(84)^b],..., a(k+6)=[1][4^a][3][(84)^(b+1)], or through a(k)=[1][5^a][4][(84)^b],..., a(k+6)=[1][5^a][4][(84)^(b+1)], for nonnegative integers a and b. The sequence listed above, with a(1)=1, is an example of the first type. A054096 Conjecture: sequence is A006183 shifted right. - _Ralf Stephan_, Jan 15 2004 A054120 Conjecture: T(n,k) = T(n-1,k-1) + 2*T(n-2,k-1) + T(n-1,k) (except for T(0,0) = 1 and T(2,1) = 3 and assuming that T(n,k) = 0 for elements outside the triangular array). - _Gerald McGarvey_, Sep 20 2007 A054120 Conjecture: T(n,k) = A081577(n,k) - A081577(n-2,k-1). (A081577 is Pascal-(1,2,1) array). - _Gerald McGarvey_, Sep 20 2007 A054271 Conjecture: every other even integer appears in the sequence an infinite number of times. (end) A054465 Conjecture : the number of k>0 such that a(k)<=n is asymptotic to C*log(n) C>0 A054656 Conjecture: if n >= 23 then a(n)=2 if both (n-6) and (n-4) are prime, a(n)=1 if one of (n-6), (n-4) or (n-1) is prime, a(n)=0 otherwise A054770 Conjecture: this is the sequence of numbers for which the base phi representation includes phi itself, where phi = (1 + sqrt(5))/2 = the golden ratio. Example: let r = phi; then 6 = r^3 + r + r^(-4). [_Clark Kimberling_, Oct 17 2012] A054979 Conjecture: Every e-perfect number is divisible by 36. - _Jon Perry_, Nov 13 2012 A055170 Conjecture: this sequence is a permutation of the nonnegative integers. A055173 Conjecture: this sequence is a permutation of the positive integers. A055174 Conjecture: every positive integer occurs. A055176 Conjecture: this sequence is a permutation of the positive integers. A055177 Conjecture: every positive integer occurs. A055179 Conjecture: this sequence is a permutation of the positive integers. A055180 Conjecture: every positive integer occurs. A055182 Conjecture: this sequence is a permutation of the positive integers. A055183 Conjecture: every positive integer occurs. A055185 Conjecture: this sequence is a permutation of the positive integers. A055187 Conjectures: limit as n goes to infinity of max {a(k) : 1<=k<=n}/sqrt(n) = 2; A055234 Conjecture: For each n, a(n) > 0. - _Farideh Firoozbakht_, Sep 12 2004 A055272 Conjecture in "Introduction a la theorie des nombres" by d'Armel Mercier and J. M. Deconinck: this is the period length of the fraction 1/7^n. For example 1/7^2=0.0204081632653061224489795918367346938775510204....with a period of 42 digits =6*7=a(2). The period of 1/7^3 has exactly 294=a(3) digits. - _Benoit Cloitre_, Feb 02 2002 A055487 Conjecture: Unless n!+1 is prime (i.e., n in A002981), a(n)=pq where p is the least prime > sqrt(n!) such that (p-1) | n! and q=n!/(p-1)+1 is prime. A055778 Conjecture: For all n, A007895(n) <= A055778(n). There is equality at 1, 7, 18, 19, 47, 48, 54, 123, 124, 130, 141, 142, 322, 323, 329, 340, 341, 369, 370, 376, 843, 844, 850, 861, 862, 890, 891, 897, 966, 967, 973, 984, 985, 2207, 2208, 2214, 2225, 2226, 2254, 2255, 2261, 2330, 2331, 2337, 2348, 2349, 2529, 2530, 2536, 2547, 2548, 2576, 2577, 2583, ... - Dale Gerdemann at Sun Apr 01 17:09:19 EDT 2012 A056142 Conjecture: if n has 3 or more digits, a(n) is a palindrome only if all the digits of n are the same. It is easy to see that any palindrome can have at most 2 distinct digits: matching digits from the initial n in the concatenation matches each digit after the second with an earlier digit. - _Franklin T. Adams-Watters_, Sep 07 2006 A057083 Conjecture: Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n)-A057682(n)*M+z(n)*M^2, where z(0)=z(1)=0 and, apparently, z(n+2)=a(n). - Stanislav Sykora, Jun 10 2012 A057194 Conjecture: a(n) < A216151(n) for all n > 1. - _Jon Perry_, Sep 12 2012 A057368 Conjecture: a(n)>0 for all n>0. - _Franklin T. Adams-Watters_, May 05 2006 A057659 Conjectured to be a finite sequence. A057702 Conjectured to be complete. A057746 Conjectured to be complete. A057748 Conjecture: this sequence is always positive (with n>1). A057767 Conjecture: this sequence is always positive. A057768 Conjecture: this sequence is always positive (with n>2). A057856 Conjecture: For all pairs of relative prime numbers (x, y) there exists at least one number n=2^m and one prime number p such p=x^n+y^n. This sequence show one case of this conjecture where y=x+1. - _Tomas Xordan_, Jun 02 2007 A058188 Conjecture: if prime(n)>=127, there is always at least one prime between prime(n) and prime(n) + sqrt(prime(n)). Easily checked for prime(n)<1.1e15 in existing maximal gap tables A058347 Conjecture: T(n,k) = T(k,n). A058927 Conjecture: S(x)=sum[n=0,1,..., ((2*n+1)^(n-1)/(n!*2^n))*x^(2*n+1)] = (A052750(n)/A000165(n))*x^(2*n+1). Letting D_n be the set of divisors of n! and d_n=max(k in D_n : k | (2*n+1)^(n-1)), then a(n)=A052750(n)/d_n, with the next few terms, for n=12,...,20, being: A058928 Conjecture: S(x)=sum[n=0,1,..., ((2*n+1)^(n-1)/(n!*2^n))*x^(2*n+1)] = (A052750(n)/A000165(n))*x^(2*n+1). Letting D_n be the set of divisors of n! and d_n=max(k in D_n : k | (2*n+1)^(n-1)), then a(n)=A000165(n)/d_n, with the next few terms, for n=12,...,20, being: A059169 Conjecture: this is 0 followed by A026922. - _R. J. Mathar_, Oct 05 2008 A059297 Conjecture: the matrix inverse is A137452. - _R. J. Mathar_, Mar 12 2013 A059324 Conjecture : There exists no pair of primes (p,q>p^2) such that q-p^2 = 6*n-4 (see A138479) - Philippe LALLOUET (philip.lallouet(AT)orange.fr), Mar 20 2008 A059391 Conjecture: k(n)= n-th prime (conjectured by Naohiro Nomoto). The next new term is bigger than 5*10^7. A059456 Conjecture: From the sequence of prime numbers, let 2 and remove the first data iteration of 2*p+1; leave 3 and remove the prime data by the iteration 2*p+1 and we get the sequence. Example for p=2, remove(5,11,23,47); p=3, remove(7); p=13, p=17, p=19, p=23, remove(47); and so on. [From _Vincenzo Librandi_, Aug 07 2010] A059970 Conjectures: (1) Nim-Factorial(2^n-1)=1 (verified for n=1,2,3,...,16). (2) Nim-Factorial(2^n+2^(n-1)-1)=2 (verified for n=1,2,3,...,15). A060199 Conjecture: There are always more than 3 primes between two consecutive nonzero cubes. - _Cino Hilliard_, Jan 05 2003 A060275 Conjecture: we always have gcd(i,j)=1. A060298 Conjecture: Lim_n->inf. of a(2n)/10^n=1-1/sqrt(10) and Lim_n->inf. of a(2n-1)/10^n=1/sqrt(10)-1/10. A060318 Conjecture: all odd Catalan numbers have smallest factor 3, except cat[3] has smallest divisor 5 and cat[31] and cat[255] have smallest divisor 7 (checked up to cat[ -1+2^2048 ]). A060464 Conjecture: n is a sum of three cubes iff n is in this sequence. A061002 Conjecture: If p is the n-th prime and H(n) is the n-th harmonic number, then denominator(H(p)/H(p-1))/numerator(H(p-1)/p^2) = p^3. A193758(p)/a(n) = p^3, p > 3. - _Gary Detlefs_, Feb 20 2013 A061262 Conjecture: for n large enough, 1<a(n)/n^2<2 - _Benoit Cloitre_, May 10 2003 A061357 Conjecture: for n>=4 a(n)>0. - Benoit Cloitre, Apr 29 2003 A061357 Conjectures from _Rick L. Shepherd_, Jun 24 2003: 1) For each integer N>=1 there exists a positive integer m(N) such that for n>=m(N) a(n)>a(N). (After the first m(N)-1 terms, a(N) does not reappear). In particular, for N=1 (or 2 or 3), m(N)=4 and a(N)=0, giving Benoit Cloitre's conjecture. (cont.) A062368 Conjecture: this is the third inverse Mobius transform of the sequence 4^A001221(n). - _R. J. Mathar_, Aug 09 2012 A063108 Conjecture: no matter what the starting term is, the sequence eventually joins this one. This should be true in any base - base 2, for example, is trivial. A063170 Conjectures: The exponent in the power of 2 in the prime factorization of a(n) (its 2-adic valuation) equals 1 if n is odd and equals n - A000120(n) if n is even. - _Gerald McGarvey_, Nov 17 2007, Jun 29 2012 A063224 Conjecture: a(n) = 2 * A004523(n), n > 0. - _Wesley Ivan Hurt_, Sep 17 2013 A063289 Conjecture: Except for the first term,a(n)=9*n-a(n-1)-18 (with a(2)=2)[From Vincenzo Librandi, Dec 05 2010] A063289 Conjecture: a(n) = 9*n/4+(-1)^n/4-45/4, n>=3 with first differences in A010710 [R. J. Mathar, Dec 06 2010] A063305 Conjecture: a(n) = 18n - 41 for n > 2. - Peter Luschny, Mar 04 2012 A063321 Conjecture: a(n) = 13*(n + (n-1)mod 2) + 14*(n + n mod 2) - 72 for n > 2. - Peter Luschny, Mar 05 2012 A063337 Conjecture: a(n) = 72*n - 155 for n > 2. - Peter Luschny, Mar 05 2012 A063793 Conjecture: a(n) > 0 for n > 0 A064016 Conjecture: It would appear that the sum from 1 to n approaches the limit 0.1960364*n^2 = (1/2 - 3/Pi^2)*n^2. A064413 Conjecture: If a(n) =/= p, then almost everywhere a(n) > n. - _Tomasz Ordowski_, Jan 23 2009 A064413 Conjecture: lim #(a_n > n) / n = 1, i.e. #(a_n > n) ~ n. - _Tomasz Ordowski_, Jan 23 2009 A064413 Conjecture: A term p^2, p a prime, is immediately preceded by p*(p+1) and followed by p*(p+2) - Vladimir Baltic, Oct 03, 2001. This is false, for example the sequence contains the 3 terms p*(p+2), p^2, p*(p+3) for p = 157. - Eric Rains A064537 Conjectured to be a permutation of the natural numbers. A064796 Conjecture: a(n) = (n+1)(n+3)/4 for odd n, a(n) = (n)(n+4)/4 for even n. - _Jud McCranie_, Oct 25 2001 A065056 Conjecture : there is a constant c=1.9... such that a(n)/n^c is bounded. - _Benoit Cloitre_, Nov 03 2002 A065075 Conjecture: every non-multiple of 3 does appear in the sequence. - _Franklin T. Adams-Watters_, Jun 29 2009. See A230289. - _N. J. A. Sloane_, Oct 17 2013 A065129 Conjecture: A003285(m) = even or A004613, if m is divisible by A003285(m). A065204 Conjecture: A064413(a(n))=A002110(n) (primorial numbers), i.e. a(n) sets positions of primorial numbers in A064413. A065255 Conjecture: lim n -> infinity b(k)/k = 10. A065558 Conjecture: a(2n) = 2*A001615(n). - _Ralf Stephan_, Mar 26 2004 A065560 Conjecture : a(n)+n > prime(n) A065796 Conjecture; there are an infinite number of values which do not appear in this sequence (in the signed version, of course). The first example appears to be 2. _Sean A. Irvine_ has checked this up to 10^9. - _Robert G. Wilson v_, Dec 10 2001 A065806 Conjectures: -1 < a(n) < n+2; If a(n) = n+1 then a(n+1) = 0; All integers >1 occur in the sequence; For each c in [0,1] there exists a subsequence a(i_j) with (a(i_j)+1)/(i_j+3)-> c, j->infinity. A065907 Conjecture: no integer occurs more than three times in this sequence. Confirmed for the first 2399 terms of A007522 (primes < 100000). There are integers which do occur thrice, e.g. 221, 1159. A065908 Conjecture: no integer occurs more than three times in this sequence. Confirmed for the first 2399 terms of A007522 (primes < 100000). In this section, there are no integers which do occur thrice. A065909 Conjecture: no integer occurs more than three time in this sequence. Confirmed for the first 1182 terms of A014754 (primes < 100000). There are integers which do occur thrice, e.g. 6624. Moreover, no integer is first, second, third or fourth solution for more than three primes. Confirmed for the first 2399 terms of A007522 and the first 1182 terms of A014754 (primes < 100000). A065910 Conjecture: no integer occurs more than three time in this sequence. Confirmed for the first 1182 terms of A014754 (primes < 100000). In this section, there are no integers which do occur thrice. Moreover, no integer is first, second, third or fourth solution for more than three primes. Confirmed for the first 2399 terms of A007522 and the first 1182 terms of A014754 (primes < 100000). A065911 Conjecture: no integer occurs more than three time in this sequence. Confirmed for the first 1182 terms of A014754 (primes < 100000). In this section, there are no integers which do occur thrice. Moreover, no integer is first, second, third or fourth solution for more than three primes. Confirmed for the first 2399 terms of A007522 and the first 1182 terms of A014754 (primes < 100000). A065912 Conjecture: no integer occurs more than three time in this sequence. Confirmed for the first 1182 terms of A014754 (primes < 100000). In this section, there are no integers which do occur thrice. Moreover, no integer is first, second, third or fourth solution for more than three primes. Confirmed for the first 2399 terms of A007522 and the first 1182 terms of A014754 (primes < 100000). A066437 Conjecture: a(n) is always finite; i.e. the sequence {T(n,k)} is eventually periodic for every n. A066450 Conjectures: a(n) shows the same asymptotic behavior as n^2. For infinitely many n, a(n)=n^2-n-1. Again, it is an open question, if the values of the sequence really lead to infinitely many 'reverse and add!' steps or not. Is the sequence always positive? A066774 Conjecture: the set of k such that a(k+1)-a(k)=2 is infinite. A066852 Conjecture: There are an infinite number of odd values in A066820. A066861 Conjecture: a(n) = n*ln(n) asymptotically. A066881 Conjecture: all natural numbers 1,2,3,..n will eventually occur among the integer values of sigma(phi(n))/sigma(n). A067126 Conjecture: 4 and 9 are the only composite terms. A067129 Conjecture : sequence gives odd values only A067274 Conjecture: The difference a(n)-a(n-1) is 6 if and only if n is a prime number. This has been checked up to about n=300 and may be easy to prove. A067348 Conjecture: sequence contains most of 2*A000384(k). Exceptions are k= 8, 18, 20, 23, 35, ... - _Ralf Stephan_, Mar 15 2004 A067723 Conjecture: There are no terms with an even number of digits. - Klaus Brockhaus A067730 Conjecture: sequence contains odd values only - _Benoit Cloitre_, Feb 18 2002 A067745 Conjecture: odd part of 3n-2. -- Ralf Stephan, Nov 18 2010 A067793 Conjecture: Odd composite n such that (n^2 + 8) mod 3 = 0. (All primes > 3 meet this criterion). - _Gary Detlefs_, May 03 2012 A067864 Conjecture: the sequence is finite, since (sum of the digits of 6^n)/n->log_10(6)*4.5~3.50168 as n->00 (this is also a conjecture). - _Robert Gerbicz_, May 08 2008 A068638 Conjecture: 25 is the largest odd term of this sequence. A068696 Conjecture: a(2) is the only 0 in this sequence. A068704 Conjecture: For every n there exists a k which yields such a prime. A068890 Conjecture: the sequence is finite. A068891 Conjecture: the sequence is finite. A068991 Conjecture : if n is in the sequence and n is squarefree then the denominator of the 2n-th Bernoulli's number contains n. E.g. 2310 is squarefree, is in the sequence and A002445(2310)=744535159372016163713900138929458330 is divisible by 2310. A069003 Conjecture: a(n) does not exceed 4*sqrt(n+1) for any positive integer n. --_Zhi-Wei Sun_, Apr 15 2013. A069004 Conjecture: a(n)>0 for all n>1. - Entries checked by _Franklin T. Adams-Watters_, May 05 2006 A069187 Conjecture: sequence is A071253 minus those entries of A071253 that have their index in A049532, i.e. a(n) is of form n^2*(n^2+1) for all n not in A049532. - Ralf Stephan, Aug 18 2004 A069191 Conjecture: Let M(n) be the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is prime or not. For each n = 24, 25, ..., the characteristic polynomial of M(n) is irreducible over the field of rational numbers, and the n eigenvalues can be listed as lambda(1), ..., lambda(n) such that lambda(1) > -lambda(2) > lambda(3) > -lambda(4) > .... > (-1)^n*lambda(n) > 0. - _Zhi-Wei Sun_, Aug 25 2013 A069202 Conjecture: More generally let a(1)=x a(2)=y be 2 distinct positive integers then for any x,y >0 lim n -> infinity ln(a(n))/n = 1/4 A069803 Conjectured to be complete. A069804 Conjectured to be complete. A069837 Conjecture: For every n there exists an n-digit prime which comprises of digits 2,3,5 and 7. i.e. no prime > 7 is required in this concatenation. i.e. a(n) of A069637 contains exactly n digits. This is a looser conjecture than the one by _Patrick De Geest_ in A036937. A069869 Conjecture: a(n) = 0 only for n = 1 or n = 3k with k>1. A069883 Conjecture: sequence is bounded. A069884 Conjecture: sequence is bounded. A069941 Conjecture : if n>=2 there are at least 3 primes p such that n!<=p<=n!+n^2 (or stronger : for n>1 a(n)>ln(n)). This is stronger than the conjecture described in A037151(n). Because if n!+k is prime, k composite, k=A*B, where A and B must have, each one, at least one prime factor>n (if not : A=q*A' q<=n then n!+k is divisible by q), hence k>n^2. Also stronger (but more restrictive) than the Schinzel conjecture : "for m large enough there's at least one prime p such that m<=p<=m+ln(m)^2" since n^2<ln(n!)^2 for n>5. A070077 Conjecture: for any squarefree k exists m such that a(m)=k; see A070078. A070219 Conjectures: There is at least one number k with n<k<2n such that the concatenation n and n+k is a prime. For all m there is at least one number k with m*n < k < (m+1)*n such that the concatenation n and n+k is a prime. A070226 Conjecture : If A000005(n) divides A007947(n) for some n, then A007947(n)=A000005(n). [From _Ctibor O. Zizka_, Feb 05 2009] A070255 Conjecture: sequence is finite. A070831 Conjecture: The sequence is unbounded. A070961 Conjecture: lim (+) n ->infinity a(n)/n = 1 and lim (-) n ->infinity a(n)/n = 0,7.... A070965 Conjecture: all integers are present - Edwin Clark Aug 20 2004 A071010 Conjecture : if n is not the sum of 2 squares sigma(n)==0 mod 4 (converse is not true : if sigma(n)==0 mod 4, n is sometimes the sum of 2 squares : sigma(65)=84==0 mod 4 but 65=49+16 is a sum of 2 squares) A071071 Conjectured by J. Shallit to be complete. A071089 Conjecture: Every non-negative integer can appear in the sequence at most finitely many times. - _Thomas Ordowski_, Jul 22 2013 A071241 Conjecture: 101 is the largest prime member, the only other primes being 2 and 11. A071256 Conjecture: lim sup n ->infinity a(n)/n^2 exists = C, where 0<C<1. - _Benoit Cloitre_, May 23 2002 A071524 Conjecture: a(n) = 0 for no n > 28. - _Zhi-Wei Sun_, Aug 26 2013 A071532 Conjecture: asymptotically, a(n) ~ C * Log(n)^2 with C = 1.4..... A071537 Conjecture: this is a permutation of the nonprimes A018252. A071558 Conjecture: a(n) < sqrt(n)*log(n) for all n > 17261. This has been verified for n up to 3*10^7. It implies the inequality a(n) < n for each n > 127. - _Zhi-Wei Sun_, Jan 07 2013 A071561 Conjecture: the lim n-> oo, the number of n's which have no middle divisor / n -> > 3/4. A071622 Conjecture : for n>2300, a(n) > ln(n)^2 A071636 Conjecture: there are no more terms. A071681 Conjecture: a(n)>0 for n>2. A071823 Conjecture : for n > 1000, a(n) - n/2 > sqrt(n). A071824 Conjecture : for n > 1000 a(n) - n/2 < - sqrt(n) ( if b(n) denotes the number of x with largest prime factor of the form 4k+3 less than or equal to n, it is conjectured that if n > 1000 b(n)- n/2 > sqrt(n) ) A072341 Conjecture: a(n) is less than or equal to n for all n. A072342 Conjecture: a(n) is less than or equal to n for all n. A072532 Conjecture: If a(n) = k*a(n-1)-1 then k < a(n-1). A072592 Conjecture: this is exactly the sequence whose terms are twice those of A009003. (This has been verified for all terms<=500.) Compare A009003. - _John W. Layman_, Mar 12 2008 A072893 Conjectures: (1) Lim n ->infinity a(n)/n = C = 3.5.... (2) For any n > 2, a(n+1)-a(n) = 2, 3, 5, 7, 8 or 10 only. First differences a(n+1)-a(n) for n>2 are 3, 2, 3, 2, 5, 3, 2, 5, 3, 2, 10, 3, 5, 2, 3, 2, 8... (3) If x is not in this sequence, for k large enough, c(k)= -1 or c(k) reaches one of the two cycles {3, 1, 2, -1} or {0, -1, 1, 0, 1, -1}. A072894 Conjectures : (1) a(n+1)-a(n) = 0 or 1; (2) lim n ->infinity a(n)/n = 2/3; (3) 1/2 < (3a(n)-2n)/Log(n) <3/2 for any n > 1000. Does lim n -> infinity (3a(n)-2n)/Log(n) = 1 ? A072901 Conjecture: Numbers n such that n=p*p_1*p_2*,.,* p_n and p is not repeat. Example: 6=2*3; 15=3*5; 30=2*3*5; 154=2*7*11; 195=3*5*13 [From _Vincenzo Librandi_, Aug 08 2010] A072939 Conjecture: a(n) = A036554(n)+1. - _Vladeta Jovovic_, Apr 01 2003 A072989 Conjecture: limit of a(n)/n is zero. A073008 Conjectured to be equal to (4/153)*(1+2*sqrt(2))*sqrt(51) A073117 Conjecture (seems provable): More generally let a and b(1) be integers. If b(n+1) = b(n)+ b(n) (mod(n+a)) there is an integer x(a,b(1)) such that b(n+1) = b(n)+x(a,b(1)) for n sufficiently large. We have x(0,1) = x(1,1) = x(2,1) = 97, x(3,1) = 1, x(4,1) = 3, x(5,1) = 3, x(6,1) = 6 ...x(97,1) = 43, x(0,11) = 2 etc. - _Benoit Cloitre_, Aug 20 2002 A073271 Conjecture: a(n) is even except for a(1)=3, a(3)=7 and a(8)=23 (A073272(n)=0). A073310 Conjecture: a(n) < 2n. See A073316 for a generalization for all positive even numbers less than 2n. A073316 Conjecture: a(n) < 2n. Note that the truth of this conjecture implies that for any pair of positive even numbers e1 < e2 <= 2n, there is a positive odd number d < 2n such that e1+d and e2+d are primes. Note that this conjecture can also be stated with odd and even swapped: for any pair of positive odd numbers d1 < d2 < 2n, there is a positive even number e <= 2n such that e+d1 and e+d2 are primes. Also note that proving this conjecture would prove the twin primes conjecture. A073503 Conjectures: 2n > a(n) or 2n < a(n) for infinitely many values of n and abs(a(n)-2n) < sqrt(n) for n > 45. a(n)=2n for n = 318, 338, 350, 488, 490, 492, 494,... A073602 Conjecture: Every prime belongs to this sequence. A073641 Conjecture: every prime besides 5 is in this list. - Gabriel Cunningham (gcasey(AT)mit.edu), Apr 11 2003 A073652 Conjecture: Every prime is a member. A073653 Conjecture: Every odd prime eventually appears; a(n) ~ prime(n). A073842 Conjectured to be a rearrangement of the natural numbers. A074057 Conjecture : a(n)=0 if and only if n is a Fermat prime A074100 Conjecture: the sequence is finite. A074482 Conjecture: a(n) is defined for all n (as well as A074483); A074483 Conjecture: a(n) is defined for all n (as well as A074482); A074766 Conjecture: p(2n)-2p(n) < 2n for all n. A074964 Conjecture: subsequence of A066522, implying finiteness. [_Reinhard Zumkeller_, Nov 14 2011] A075000 Conjecture: For every n there exists a nonzero a(n). A075001 Conjecture: For every n there exists a k. A075002 Conjecture: For every n there exists a nonzero k such that a(n) = k. A075092 Conjecture: a(n) nonnegative. A075343 Conjecture: there exists a prime between consecutive terms. A075344 Conjecture: a(n) > 0 for all n. A075362 Conjecture that antidiagonal sums are A023855. - _L. Edson Jeffery_, Jan 20 2012 A075568 Conjectured to consist of consecutive primes after 53. (See also comments in A123055.) A075594 Conjecture: a(n+r)= A075593(n) for n > k for some k and r. What is the value of k and r? A075650 Conjecture: a(n) == 0 (mod n). A075864 Conjecture: Number of Dyck n-paths with all ascent lengths being a power of 2. [_David Scambler_, May 07 2012] A075885 Conjecture: limit a(n)^(1/n) = L where L = 2.200161058099... is the geometric mean of Luroth expansions, where log(L) = Sum_{n>=1} log(n)/(n*(n+1)) = 0.7885305659115... (cf. A085361). A076043 Conjecture: every squarefree number except 1 is a member. A076044 Conjecture: for every m greater than a(n), there are more than n primes between m and m+sqrt(m); true if a(n) less than 1000000. A076070 Conjecture: No entry is zero. (At least one multiple of n can be formed by using the digits of the next n numbers). A076072 Conjecture: No entry is zero. (At least one multiple of n can be formed by using the digits of the next n numbers). A076157 Conjecture: a(3*2^n) = -1 + 2^[(n+1)((n+2)!) ]. - _Ralf Stephan_, May 17 2005 A076502 Conjecture : a(n) - c*n is bounded and a(n)=floor(c*n) + 0 + 1 or +2 A076635 Conjecture : lim k->infinity (b(k)-k)/ln(k) = f(n), constant depending on n. f(n) seems erratic: f(2)=2.9..., f(3)=2.5..., f(4)=3.2..., f(5)=2.25..., f(6)=4.0... A076874 Conjecture: For n>=7, a(n)-2 is the maximum number of steps in a 2D self-avoiding random walk trapped after n steps having only 2 choices for the next step. a(n)>=A077484(n)+2 A076973 Conjecture: start from any initial value a(1)=m>=2 and define a(n) to be the largest prime factor of a(1)+a(2)+...+a(n-1); then a(n)=n/2+O(log(n)) and there are infinitely primes p such that a(2p)=p. - _Benoit Cloitre_, Jun 04 2003 A077128 Conjecture : every member is a prime. A077183 Conjecture: a(n) > 0 for all n > 3, since prime(1) = 2 and prime(3) = 5 are the only primes whose multiples cannot end in 1. - _Ryan Propper_, Jul 29 2005 A077202 Conjecture: Every odd prime occurs in this sequence infinitely many times. A077220 Conjectured to be a permutation of the natural numbers (cf. A099130). The first few fixed points are: 1, 2, 19, 92, 220, 467, 556, 616, 690, 842. A077259 Conjecture: a(0)=0, a(1)=2, a(2)=6, a(3)=44, a(n)=18a(n-2)-a(n-4)+8 [From Robert Phillips (bobanne(AT)bellsouth.net), Sep 01 2008] A077361 Conjecture: no entry is zero for n>2. A077362 Conjecture: no entry is zero for n>2. A077371 Conjecture: The sequence is finite. A077373 Conjecture: sequence is finite. Is there any term > 89 in this sequence? A077391 Conjecture: No term is zero. A077510 Conjecture: for k > 5, prime(n) = < k < prime(n+1) <= k + Pi(k), i.e. the smallest prime greater than k is <= k +Pi(k). Equality holds for k = 7. A077594 Conjecture 1: For each k > 0 the trajectory of k eventually leads to a term in the trajectory of some j which belongs to A063048, i.e. whose trajectory (presumably) never leads to a palindrome. Conjecture 2: There is no k > 0 such that the trajectory of k contains more than twelve palindromes, i.e. a(n) = -1 for n > 12. A077602 Conjecture: M(n,1) ~ A077596(n) * sqrt(Pi*n/2), where A077596(n) is the largest coefficient of the n-th Moebius polynomial, M(n,x). A077653 Conjecture : let z(1)=x; z(2)=y; z(3)= z; z(n)=abs(z(n-1)-z(n-2)-z(n-3)) if z(n) is unbounded (i.e. x,y,z are such that z(n) doesn't reach a cycle of length 2), then there are 2 integers n(x,y,z) and w(x,y,z) such that M(n) = floor(sqrt(n+w(x,y,z))) for n>n(,x,y,z) where M(n) = Max ( a(k) : 1<=k<=n ). As example : w(1,2,2)=9 n(1,2,2)=4; w(1,2,4)=29 n(1,2,4)=4; w(1,2,8)=157 n(1,2,8)=9 A077712 Conjecture: Terms contain only two types of digits i.e. 0 and 1. A077739 Conjecture: a(n) = 0 if and only if n is 11, 17 or 19. A077740 Conjecture: a(n) = 0 if and only if n is 11, 17 or 19. - _Ray Chandler_, Sep 05 2003 A077754 Conjecture: zero occurs only for indices which are multiples of 10 apart from 1, 2, 5 and 11. (a(n) = 0 only for n = 1,2,5,11 or n = 10k.) A078008 Conjecture: a(n) = the number of fractions in the infinite Farey row of 2^n terms with even denominators. Compare the Salamin & Gosper item in the Beeler et al. link. - _Gary W. Adamson_, Oct 27 2003 A078109 Conjecture : a(n) always exist, a(n)/n^2 is bounded. If initial conditions are u(1)=u(2)=1, u(3)=2n+1, then u(k) reaches a 2-cycle for any k>m large enough (cf. A078098) A078122 Conjecture: the sum of the n-th row equals the number of partitions of 3^n into powers of 3 (A078125). A078128 Conjecture (lower bound): for all k exists b(k) such that a(n)>k for n>b(k); see b(0)=A078129(83)=154 and b(1)=A078130(63)=218. A078130 Conjecture: the sequence is finite; is a(63)=218 the last entry? A078134 Conjecture (lower bound): for all k exists b(k) such that a(n)>k for n>b(k); see b(0)=A078135(12)=23 and b(1)=A078136(15)=39. This is true - see comments by _Hieronymus Fischer_. A078173 Conjecture: Max[t]=n(n+2) and Max[s]=2n, where the maxima are taken over all positive integer solutions r<=s<=t of the above Diophantine equation. A078251 Conjecture: No entry is zero for n > 3. A078254 Conjecture: sequence is finite. A078257 Conjecture: a(n) is not equal to the sequence of denominators presented in A172495 and A172506. [From _Jaroslav Krizek_, Feb 05 2010] A078537 Conjecture: a(n) = sum of the n-th row of lower triangular matrix A078536. A078563 Conjecture: The equation g(N) = n g(N-1), for a fixed positive integer n, is always solvable for N. A078613 Conjecture : a(n) is asymptotic to c*n where c is around 4 - _Benoit Cloitre_, Jan 06 2003 A078671 Conjectures: Is a(n) monotonically increasing for n > 4? Does lim{n->inf} a(n)/a(n+1) = 0.5? - _Ryan Propper_, Jan 04 2008 A079104 Conjectured to equal A079105 (and so have period 3) from a(17) onwards. A079128 Conjecture: gcd(a(n),n)=1. - _Vladeta Jovovic_, Jan 25 2003 A079267 Conjecture: Asymptotically, the n-th row has a Poisson distribution with mean 1. [David Callan, Nov 11 2012] A079402 Conjecture: this is equal to the number of permutations of n^2 distinct integers free of any monotonic increasing or decreasing (n+1)-subsequence. (By the Erdos-Szekeres Theorem, every permutation of n^2+1 distinct integers has such a subsequence.) - _Joseph Myers_, Jan 04 2003 A079665 Conjecture: (b^s+1)/(b^r+1) is an integer if and only if: 1) r<s/2, 2) if s==1 (mod 2) then r is divisor of s, 3) if s=k*2^t with gcd(k,2^t)=1 then r is 2^t*u with u dividing k A080070 Conjecture: only the terms in positions 0,1,2 and 4 are symmetric, i.e. A057164(A080068(n)) = A080068(n) (equivalently: A036044(A080069(n)) = A080069(n)) only when n is one of {0,1,2,4}. If this is true, then the formula given in A079438 is exact. I (AK) have checked this up to n=404631 with no other occurrence of a symmetric (general) tree. A080209 Conjecture: The diagonal of leading successive absolute differences of the Sophie Germain primes consists, except for the initial 2, only of 1's and 3s. A080435 Conjecture: There are infinitely many primes not of the form a(i)+a(j). - _David W. Wilson_, Apr 14 2003 A080439 Conjecture: Only one digit needs to be inserted between each pair of digits of a(n-1) to get a(n); i.e. a(n) contains exactly 2n-1 digits for n > 1. A080440 Conjecture: Only one digit needs to be inserted between each pair of digits of a(n-1) to get a(n); i.e. a(n) contains exactly 2n-1 digits for n > 1. A080441 Conjecture: Only one digit needs to be inserted between each pair of digits of a(n-1) to get a(n); i.e. a(n) contains exactly 2n-1 digits for n > 1. A080442 Conjecture: Only one digit needs to be inserted between each pair of digit of a(n-1) to get a(n); i.e. a(n) contains exactly 2n-1 digits for n > 1. A080572 Conjectured to be less than or equal to lcs(n) (see sequence A063437). The value of a(2^n) is that given in Stinson and van Rees and the value of a(2^n-1) is that given in Fu, Fu and Liao. This function gives an easy way to generate these two constructions. A080612 Conjectured to be finite with last term = 314. Other conjecture log(n)^2*(1/p(2n+1)*sum(k=1,n,p(2k+1)-p(2k)) - 1/p(2*n)*sum(k=1,n,p(2k)-p(2k-1))) -> constant. Weaker : previous formula is bounded. A080674 Conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4 and 5 as a digit . - _Alexandre Wajnberg_, Apr 25 2005 A080735 Conjectures: (Strong) Let x,y be 2 positive integers and define a(n) as a(1)=1, a(n)=x*a(n-1) if a(n-1) is prime, a(n)=a(n-1)+y otherwise; then limit n ->infinity log(a(n))/sqrt(n)=C(x,y) exists. (Weak) log(a(n))/sqrt(n) is bounded. A081071 Conjecture: a(n)= Fibonacci(4n+3) + sum_{k=2..2n} Fibonacci(2k). [From _Alex Ratushnyak_, May 06 2012] A081086 Conjecture: log(a(n+1))/log(a(n)) -> 2. The 9-th term has 69 digits, the 10-th term has 140 digits. The decimal expansion of the continued fraction: [0;2,2,9,91,14201,252238179,...] = 0.404250350307436947086987047594... A081097 Conjecture: if m is an integer and sqrt(m) is irrational, the sequence of n such that n^2 = (1/m)*(n + floor(sqrt(m)*n*floor(sqrt(m)*n))) always satisfies a recurrence of order m. For example: if m=6, the sequence n=b(k) satisfies: b(6k)=4*b(6k-1)+4*b(6k-2)-b(6k-3)-1; b(6k+1)=.... etc. A081199 Conjecture (verified up to a(9)): Number of collinear 4-tuples of points in a 4 X 4 X 4 X... n-dimensional cubic grid [From _R. H. Hardin_, May 24 2010] A081200 Conjecture (verified up to a(9)): Number of collinear 5-tuples of points in a 5 X 5 X 5 X... n-dimensional cubic grid [_R. H. Hardin_, May 23 2010] [_Ron Hardin_, May 24 2010] A081201 Conjecture (verified up to a(8)): Number of collinear 6-tuples of points in a 6 X 6 X 6 X... n-dimensional cubic grid. [_R. H. Hardin_, May 23 2010] A081237 Conjecture: there are no more terms. A081320 Conjecture: for n>12 and n>0 modulo 12: a(n)=a(n-12) and a(12*k)=A065331(k)*144. A081324 Conjecture: for n>1 this is A050804. A081324 Conjecture: sequence consists of numbers of form 2*k^2 such that sigma(2*k^2)==3 (mod 4) and k is not divisible by 5. A081532 Conjecture: T(n,3) = A023402(n-2) for n>2. - _Reinhard Zumkeller_, Jul 18 2003 A081725 Conjecture : a(n)>0 A081847 Conjecture: Every triangular number other than 3, 6 and 10 is a member. (This is a re-arrangement of triangular numbers barring 3,6 and 10). A081884 Conjecture : let b(n,m) denotes the number of steps needed to reach an integer starting with n+1/2^m and iterating the map x-->x*ceiling(x); then sum(k=1,n,b(k,m)) is asymptotic to (m+1)*n. A081885 Conjecture : let b(n,m) denotes the number of steps needed to reach an integer starting with n+1/2^m and iterating the map x-->x*ceiling(x); then sum(k=1,n,b(k,m)) is asymptotic to (m+1)*n. A082108 Conjecture. a(n-1) is the largest integer such that Sum[a(k)/(2k)!, k=1,2,...,n] is less than 1 for n=1,2,3,... . - _John W. Layman_, Jun 26 2008 A082235 Conjecture: Every triangular number other than 3, 6 and 10 is a member. (This is a re-arrangement of triangular numbers barring 3,6 and 10). A082381 Conjecture: The sequence always terminates with 1 or the 4 16 37 58 89 145 42 20 4... loop (cf. A080709). A082382 Conjecture: Each sequence terminates with 1 or the 4 16 37 58 89 145 42 20 4... loop. A082385 Conjecture: The sequence always terminates with one of the following:(tested to n=1000000) 1,55,136,153,160,370,371,407,919 which eventually loop back to themselves. 1,153,370,371,407 loop back in 1 step and are the sum of the cubes of their digits. The others are 55,250,133,55. 136,244,136. 160,217,352,160. 919,1459,919. A046156, A046157 indicate this as a limit of possibilities of numbers that cubed digital roots roll back to the origional number. Proof? - Cino Hilliard, Apr 13 2003 Proof: In A055012 T.D.Noe notes that for n > 1999, A055012(n) < n. This means that by repeatedly applying A055012, we eventually reach a number smaller than 2000. As checked by Cino Hilliard, all numbers below 10^6 end in one of the listed cycles. - Stefan Steinerberger, Sep 05 2007 A082528 Conjecture : define sequence a(n,m) m real >0 as the least k such that x(k)=0 where x(1)=n x(k)=k^m*floor(x(k-1)/k^m) then a(n,m) is asymptotic to (c(m)*n)^(1/(m+1)). where c(m) is a constant depending on m. A082613 Conjecture; The n-th power that divides a(n) is 2^n. A082769 Conjecture: no entry is zero. A082770 Conjecture: no entry is zero. In most cases the number of digits required is 2k+1 where k is the number of digits in A082768(n). What is the first entry that requires more (than 2k+1) digits? A082871 Conjecture: the only solutions for x!+y!=n^2 (x <= y) are x=0,0,0,1,1,1,2,4 and y=4,5,7,4,5,7,2,5 respectively. A082981 Conjectures: (1) the section (a(2n+1)}={1,3,9,19,53,111,...} is A077442, the terms of which are solutions of ax^2+7 = a square, (2) the section {a(4n+1)}={1,9,53,309,1801,...} is A038761, (3) the section {a(4n+2)}={2,14,82,478,2786,...} is A077444, the terms of which are solutions of 2x^2+8 = a square, (4) the sequence {a(4n+2)/2}={1,7,41,239,1393,...} is A002315, the terms of which are solutions of 2x^2+2 = a square, (5) the section {a(4n+4)}={4,24,140,816,4756,...} is A005319, the terms of which are solutions of 2x^2+4=a square, (6) the sequence {a(4n+4)/4}={1,6,35,204,1189,...} is A001109, the terms of which are solutions of 8x^2+1=a square. A082991 Conjecture: despite results for small terms, all even number are reached. (ex. 12 is reached since a(12102)=12). A083125 Conjecture: except for multiples of 10, all natural numbers belong to this sequence. - _Reinhard Zumkeller_, May 02 2003 A083139 Conjecture: sequence is infinite. A083143 Conjecture: no entry is zero. A083144 Conjecture: no entry is zero. A083146 Conjecture: (1). The sequence differs from A083145 only for finitely many values of n. (2) Values of n are 5,6,7,8 and 9. A083146 Conjecture: The sequence differs from A083145 for infinitely many values of n. - _Franklin T. Adams-Watters_, May 16 2006 A083183 Conjecture: no entry is zero. A083184 Conjecture: no entry is zero. A083219 Conjecture: number of roots of P(x) = x^n - x^(n-1) - x^(n-2) - ... - x - 1 in the left half-plane. [_Michel Lagneau_, Apr 09 2013] A083370 Conjecture : start from any initial value f(1)>=2 and define f(n) to be the largest prime factor of f(1)+f(2)+...+f(n-1); then f(n)=n/2+O(log(n)) and there are infinitely primes p such that f(2p)=p. A083371 Conjecture: start from any initial value f(1)>=2 and define f(n) as the largest prime factor of f(1)+f(2)+...+f(n-1) then f(n)=n/2+O(log(n)) and there are infinitely primes p such that f(2p)=p. A083382 Conjectured by Schinzel (Hypothesis H2) to be always positive for n > 1. A083396 Conjecture: for any k >= 1 there will always be a brilliant constellation of the form {n, 2k+n} for some n. (True for all k <= 5000.) A083414 Conjectured to be always positive for n>1. A083434 Conjecture: There are infinitely many zeros in this sequence. A083452 Conjecture: All the terms are all positive for n>1. (n times concatenation of n with itself > n^n), n>1. A083557 Conjecture: if a(1)=m then the sequence becomes cyclic, for any m. A083567 Conjecture: The counting function p(n) satisfies p(n)=c n/log n + o(n/log n). A083754 Conjecture: all odd numbers not of the type 10k+5 are members. A083756 Conjecture: every odd prime is a member. A083758 Conjecture: every prime other than 5 is a member. A083760 Conjecture: every even number is a member. A083762 Conjecture: every natural number belongs to this sequence. A083942 Conjecture: a(n) = Sum [ CatalanNumber[ k ], {k, 1, 2^n-1} ] = Sum[ (2k)!/(k!(k+1)!), {k, 1, 2^n-1} ]. - _Alexander Adamchuk_, Nov 10 2007 A083964 Conjecture: If n is not of the type (5k+1)/2 then a(n) is not zero. A083965 Conjecture: If n is not of the form (5k+1)/2 then a(n) is not zero. A084013 Conjecture: no entry is zero. A084015 Conjecture: no entry is zero. A084023 Conjecture; No entry is zero. A084029 Conjecture: Only one term is zero. A084030 Conjecture: Only one term is zero. A084043 Conjecture: a(n) = 0 iff n = 10k. A084044 Conjecture: a(n) = 0 iff n = 10k. Though many terms match, not the same as A077741. A084189 Conjecture: a(n) is never greater than 3. A084318 Conjecture: fixed point always exists. A084352 Conjecture: a(n) is never greater than 3. A084562 Conjecture: there are no more terms. A084727 Conjecture: All terms exist. A084742 Conjecture: No entry is zero. A084746 Conjecture: no entry is zero. A084829 Conjectured next term: 100 (from results of WenQi Huang and Liang Yu). A084865 Conjecture: A084863(a(n))=1? A084950 Conjecture: also coefficient triangle of the denominators of the (n-th convergents to) the Continued Fraction w/(1+w/(2+w/3+w/.. This C.F. converges to 0.697774657964.. = BesselI[1,2]/BesselI[0,2] for w=1. [_Wouter Meeussen_, Aug 08 2010] A085044 Conjecture: No entry is zero. If n = p^2 where p is an odd prime then a(n) < p^2 or a(n) = p^2 as tau(2p^2) = 6 = tau(p^2) + tau(p^2). The (n,k) pairs are given below. (1,3),(2,10),(3,1),(4,841),(5,3),(6,66),(7,3),(8,37),(9,9),(10,2),(11,3),... Subsidiary sequence:(1) members of this sequence such that a(n) = n. E.g. a(9) = 9. (2)(harder one) Smallest k such that sigma(n) +sigma(k) = sigma(n+k). A085053 Conjecture: no entry is zero; i.e. for every n there exists a prime of the form nk+1, k<=n. A085057 Conjecture: a(n) always equals that lower bound. - Don Reble (djr(AT)nk.ca), Jul 01 2003 A085061 Conjecture 1: a(n+5) > a(n). A085061 Conjecture 2: There are no numbers which occur more than once. A085094 Conjecture: No entry is zero. For every n there exists a k such that n*k-1 is a palindrome. A085096 Conjecture: a(11) = 0, or in other words,if 11k+1 is a palindrome then at least one of the numbers in the sequence k+1,2k+1, 3k+1,...,10k+1 is also a palindrome. E.g. for k = 10 and k = 55, 11k+1 is a palindrome but 10*1 +1 and 55*2 +1 are also palindromes. A085105 Conjecture: Except a(3) no other term is zero. A085106 Conjecture: No entry is zero for n >10. There are only four terms which are zero. A085123 Conjecture: No entry is zero. A085265 Conjecture: a(n) = n + 2 for n > 11. That is, only 1 and 13 are missing. [_Charles R Greathouse IV_, Aug 21 2011] A085267 Conjecture: for n > 50, a(n) = n + 11. [_Charles R Greathouse IV_, Oct 26 2011] A085398 Conjecture: a(n) is defined for all n. A085427 Conjecture: for every n there exists a number k < 3n such that k*2^n - 1 is prime. Comment from T. D. Noe: this fails at n=624, where a(n)=2163. A085574 Conjecture: For all n > 108, a(n) is > 1. A085642 Conjecture: Equals the number of partitions of n with at least one part congruent to 2 mod 4. - _Vladeta Jovovic_, Jul 12 2003. This conjecture was established by _Christine Bessenrodt_ and Jorn B. Olsson (olsson(AT)math.ku.dk), Sep 13 2004. A085710 Conjecture: for every n > 2 there exists a number k < n such that n*k + 1 is a semiprime. A085765 Conjecture: log(a(n))/log(n) grows unboundedly. A085765 Conjecture: a(n) mod 2 repeats the 7-pattern 0,0,1,1,1,0,1. A085929 Conjecture: (1) The sequence is infinite. (2) For every prime signature other than perfect powers, there exists a number k such that k+1 also has the same prime signature. A086004 Conjecture 1: Rotation and addition of primes with even numbers of digits never yields a prime. A086004 Conjecture 2: There are no 5-Rotation Cycle Primes. A086212 Conjecture: there is an infinity of primes in A071998. A086449 Conjecture: all a(n) are even except a(2^k-1) = 1. Also a(2^k-2) = 2^(k-1). [For proof see link.] A086450 Conjecture: a(n) mod 2 repeats the 7-pattern 1,1,0,1,0,0,1 (A011657). A086522 Conjecture: every prime of the type 6k+1 is a member. Comment from _Vim Wenders_, May 27 2008: The conjecture is worng. For example 19 is missing.. A086523 Conjecture: every prime of the form 6k-1 is a member. Comment from _Vim Wenders_, May 27 2008: The conjecture is wrong. For example 11 and 23 are missing. A086528 Conjecture: The sequence is infinite. A086534 Conjecture: sequence is finite. A086537 Conjecture: this is a rearrangement of natural numbers (i.e. every natural number is a member). A086550 Conjecture: No term is zero. A086551 Conjectures: (1) No term is zero. (2) a(n)-1 is a prime. A086552 Conjecture: (1) tau(x)/tau(x-1) = n has solutions for every n. (2) If x is the smallest number for a given n such that tau(x)/tau(x-1) = n >1, then x-1 is a prime. A086553 Conjecture: every number p-1 is a member for all primes p. A086554 Conjecture: sequence is finite. A086558 Conjecture: no term is zero. A086667 Conjecture: T(n,k) -> 10^k - 1. A086711 Conjecture: sequence is infinite. A086766 Conjecture: No term is zero. A086818 Conjecture: Terms contain exactly three 1's digits and the rest 0's digits. A086844 Conjecture: let r(z)= (1/2) *(z+sqrt(z^2+4*z)), for any integral z>=1. Then the sequence a(n)-4n (where a(n) is the sequence of odd numbers m such that the sequence defined by b(1) = m; for k>1, b(k) = floor(r(z)*b(k-1)) contains only odd numbers) becomes ultimately periodic. _Benoit Cloitre_, Aug 10 2003 A086920 Conjecture: no term is zero. A087202 Conjecture: For n > 1, a(n) is prime (compare the conjecture about A037153). A087202 Conjecture holds through 1200 terms. A087304 Conjecture: no term is zero. A087324 Conjecture: (1) Sequence is infinite. (2) For every prime signature there corresponds a term in this sequence. A087325 Conjecture: (1) Sequence is infinite. (2) For every prime signature there corresponds a term in this sequence. A087357 Conjecture: Every number is a member. A087403 Conjecture: No term is zero. A087552 Conjecture: Every prime is a member and this is a rearrangement of the non-composite numbers. A087554 Conjectures: (1) k < 2n. (2) For every r, there exists a number S, nr < S < n(r+1) such that nS+1 is prime. A087582 Conjecture: Sequence is infinite and every even palindrome is a member. A087593 Conjecture: Sequence is infinite. Subsidiary sequence: number of n-digit members. A087594 Conjecture: Sequence is infinite. Subsidiary sequence: number of n-digit members. A087595 Conjecture: Sequence is infinite. A087596 Conjecture: Sequence is infinite. A087597 Conjecture: Sequence is infinite. A087598 Conjecture: Sequence is infinite. A087599 Conjecture: No term is zero. A087600 Conjecture: No term is zero. A087606 Conjecture: a(3n) = 0. No other term is zero. A087714 Conjecture: there are only 4 primes in this sequence. A087716 Conjecture: sequence has only 5 terms. This has been checked for all i <= 150. A087717 Conjecture. The iteration given in the definition above always leads to the 3-cycle {3,5,9,3} or the 6-cycle {19,37,73,145,29,57,19}, thus a(n) takes on only the values 3 or 19 for n=2,3,4,.... This has been verified to n=1000000. A087726 Conjecture: a(n)=n^2 if and only if n is squarefree. [_Ben Branman_, Mar 22 2013] A087942 Conjecture, for m>1: a(m)=0 iff n is an odd semiprime such that m-2 is not prime, i.e. m=A089268(k) for some k. - _Reinhard Zumkeller_, Oct 28 2003 A088023 Conjecture: a(n+1) >= a(n). Comments from Don Reble (djr(AT)nk.ca), Nov 13 2005: The conjecture is plainly true. In fact, a(n+1)-a(n) = 0 or 1. Also a(A091072(n)) = n; a(A091072(n)+1) = n+1. A088054 Conjecture: 3 is the intersection of A002981 and A002982). A088060 Conjecture: sequence is finite. A088072 Conjecture (1): Sequence is infinite. Conjecture (2): There are only finitely many terms which are sandwiched between composite numbers (26 is the only such term below 1000000). A088082 Conjecture: n+1 is a prime. A088090 Conjecture: sequence is finite. A088093 Conjectures: (1) Sequence is infinite. (2) Two successive terms are not equal for n > 5. A088104 Conjecture: No term is zero. Subsidiary sequences: (1) A088754 = number of n-digit primes beginning with prime(n). (2) A088755 = number of n-digit primes beginning with n. A088110 Conjecture: sequence is finite. A088112 Conjectures: (1) Sequence is infinite. (2) For every n there are infinitely many members of the form k^n. A088113 Conjecture: There are finitely many palindromic terms. Question: How many are palindromes? Observation: a(n) is further divisible by 11 for n = 2,3,4,5,6,7,8, A088120 Conjectures: (1) Sequence is infinite. (2) The common difference arising = 1 except for n = 4 and 5 in which case it is 2. A088212 Conjecture: all integers >1 eventually appear. Among first 300000 terms, first absent integers are 113, 119, 122, 124, 127, 130, 134, 136, 137, 139, 140, 142, 143, 145, 146, 148, 149, 151, 152, 154, 155, 157, 158, 160, 161, 163, 164, 166, 167, 169, 170, 172, 173, 175, 176, 178, 179, 181, 182, 184, 185, 186, 187, 188, 190, 191, 193, 194, 196, 197, 199, 200. - _Zak Seidov_, Jun 06 2013 A088226 Conjecture: a(n) = m/2 where m is the smallest even distance from n to a square. - _Ralf Stephan_, Sep 23 2013 A088249 Conjecture: a(n) = 0 iff prime(n) has Most Significant digit one among (2,4,5,6,8). A088250 Conjectures: (1) Sequence is infinite. (2) For every n there are infinitely many arithmetic progressions with n successive primes. A088251 Conjecture: (1) Sequence is infinite. (2) For every n there are infinitely many arithmetic progressions with n successive primes. A088252 Conjectures: (1) Sequence is infinite. (2) For every n there are infinitely many arithmetic progressions with n successive primes. A088254 Conjecture: the sequence is infinite. The smaller member of the twin prime pairs is a term. A088256 Conjecture: sequence is finite. A088258 Conjecture: sequence is infinite. But there are finitely many members which are sandwiched between twin primes. A088260 Conjecture; sequence is finite. A088268 Conjecture: Sequence is finite. A088276 Conjecture: (1) The only zero entries are for values of r == 2 (mod 3). (2) There are infinitely many values of k such that 10^k + r is a prime if r is not == 2 (mod 3). A088280 Conjecture: sequence is finite. A088288 Conjecture: Sequence is finite. A088508 Conjecture: a(n) always exists. A088614 Conjecture: 1 and 5 are the only two odd nonmembers. A088617 Conjecture: The expected number of Us in a Schroeder n-path is asymptotically Sqrt[1/2]*n for large n. - _David Callan_, Jul 25 2008 A088618 Conjecture: Sequence is infinite. And every odd number is a member. A088628 Conjecture: There exists a number k such that for all m > k, m not == 0 or 2 (mod 3), a(m) does not use any extra digit. A088647 Conjecture: sequence is finite. A088677 Conjecture: no number can be expressed such a sum in more than one way. A088719 Conjecture: no number can be expressed as such a sum in more than one way. A088734 Conjecture: 2,3,5,7,11,13,17,19 are the only 1 and 2 digit primes in the sequence. The smallest 8 prime power less 2n primes PPL2NP? A088862 Conjecture: the sequence is defined for all n. A088973 Conjecture: PIPS4(x) -> PIPS4(x+1) always contains 1 or more twin prime pairs. Proof of this would be proof of an infinity of twin primes. A088982 Conjecture: For x > 1 there is at least 1 prime between prime(prime(x) and prime(prime(x+1). A088987 Conjecture: There cannot be an even number total of primes between two consecutive prime-index-primes. A089305 Conjecture: No entry is zero. A089337 Conjecture: Every palindrome is a member. A089339 Conjecture: There are finitely many zeros in this sequence. What is the largest value of n such that a(n) = 0? Perhaps n = 14. A089348 Conjecture: {(a(n)-1)}/{10n} < n, or a(n) < 10*n^2 + 1. A089395 Conjecture: Sequence is infinite. A089705 Conjecture: All the composite positions are occupied by a zero except the most significant digit, which in most cases perhaps is 4. Most of the prime positions for p > 5 are occupied by 2. A089743 Conjecture: every number beginning with 1, 3, 7 or 9 begins infinitely many palindromic primes. A089755 Conjecture: Except for 2 and 5 all primes are members. A089758 Conjecture: (1)Sequence is infinite. A089771 Conjecture: No term is zero. A089781 Conjecture: If a(k+1)-a(k) = n then k < C*n. Can some one find an estimate for the constant C? A089893 Conjecture: a(2^k) = 2^(2^(k+1)-2). [This conjecture is true - _Vladimir Shevelev_, Nov 28 2010] A089893 Conjectures: lim(n->inf, a(2n+1)/a(2n)) = 5, lim(n->inf, a(4n+2)/a(4n+1)) = 17/5, lim(n->inf, a(8n+4)/a(8n+3)) = 257/85 etc. [This follows from the formula, for n>=0, t>=1: ( 4*a(2^t*n+2^(t-1))+1 )/( 4*a(2^t*n+2^(t-1)-1)+1 ) = 3*F_t/(F_t-2), where F_t= A000215(t) - _Vladimir Shevelev_, Nov 28 2010] A090304 Conjecture: sequence is finite. A090335 Conjecture: the sequence is defined for all n. A090461 Conjecture that there is a solution for all n > 24. See A090460 for a count of the number of essentially different solutions. A090469 Conjectures: 1. For n > 1, a(n) is the least composite k not already used such that gcd(n, k) > 1. 2. For any prime p > 5, a(p) = 2p. - _David Wasserman_, Nov 08 2005 A090518 Conjecture: No term is zero. A090519 Conjecture: No term is zero. Subsidiary Sequence: Number of primes in floor[(10^n)/p], p is a prime. a(1) = 3, the primes are 10/2, floor[10/3] and 10/5. A090520 Conjecture: No term is zero. A090521 Conjecture: There exists a number k such that for all n > k, a(n) is prime. Motivation: If p is the least prime >n then all the numbers from n to p-1 divide n!. And most of the numbers from p+1 to q also divide n! where q is the least prime > p, etc. and the dividend is composite in almost all cases. A090523 Conjecture: There are no zeros for n>2. A090524 Conjecture: No term is zero. A090525 Conjecture: No term is zero. A090526 Conjecture: No term is zero. A090527 Conjecture: No term is zero. A090528 Conjecture: No term is zero. A090531 Conjecture: No term is zero. A090534 Conjecture: sequence is infinite. Note:(1); If n==1 (mod 3) then in most cases a(n) is a permutation of (n-1), 1's and a 7, else a(n) is a digit permutation of (n-1) ones and a 3. (2): Obviously all the digits except the most significant one must be odd. A090596 Conjectured to coincide with the sequence of rational knots with n crossings, A018240. A090596 Conjecture derived from: s(n) = k(n) + l(n): definition of sum of rational knots (k) and links (l) s(n) = 6s(n-2) -8s(n-4): see A005418 (Jablan's observation) d(n) = d(n-2) + 2d(n-4): see A001045 (modified Jacobsthal sequence) l(n) = k(n-1) + d(n): conjecture A090597 Conjecture derived from: s(n) = k(n) + l(n): definition of sum of rational knots (k) and links (l) s(n) = 6s(n-2) -8s(n-4): see A005418 (Jablan's observation) d(n) = d(n-2) + 2d(n-4): see A001045 (modified Jacobsthal sequence) l(n) = k(n-1) + d(n): conjecture. A090680 Conjecture: no a(n) is zero. Example: n=15: A033677(15) = 5; 2 primes between 15 and 15+5=20 (17 and 19), so a(15) = 2. A090698 Conjecture: 2^(a(n)-1) - 3 is not prime. - _Vincenzo Librandi_, Feb 04 2013. A090786 Conjectures: By definition a(n)+n!+1 is prime, but is a(n)+n+1=A037153(n) also a prime number for all n>2? Is the growth of b(n) := sum(a(k),k=0..n) quadratic, that is b(n)=O(n^2)? A090825 Conjecture: composite numbers with all prime factors in A053176 are in the sequence. For p prime (3/2)*(1/p)*(2*p+1)*(3^p+1)*B(2*p) == 1 (mod p). There are few terms n with (3/2)*(1/n)*(2*n+1)*(3^n+1)*B(2*n) == 1 (mod n): 91,247,....Is this subsequence finite? A090904 Conjecture: For n > 4 the last term of the n-th group is 2p where p is the largest prime in the (n-1)th group. And these are the Bertrand primes. A090905 Conjecture: For n > 4 the last term of the n-th group is 2p where p is the largest prime in the (n-1)th group. And these are the Bertrand primes. A090906 Conjecture: For n > 4 the last term of the n-th group is 2p where p is the largest prime in the (n-1)th group. And these are the Bertrand primes. A090907 Conjecture: For n > 4 the last term of the n-th group is 2p where p is the largest prime in the (n-1)th group. And these are the Bertrand primes. A090920 Conjecture: If a(n) = n concatenated with k then k < n^2. A090940 Conjectured to include all odd prime numbers. - _David W. Wilson_, Nov 23 2012 A090981 Conjecture: The expected number of ascents in a Schroeder n-path is asymptotically (Sqrt[2]-1)*n for large n. - _David Callan_, Jul 25 2008 A091067 Conjecture: a(n) = A060833(n+1) - 1. A091340 Conjecture: sequence is finite, even though there are quite a lot of known amicable numbers (about 6.0E6 currently). A091457 Conjecture: in the "extremal" expansion x_i = A000058(i) for i=1..n-3. A091590 Conjecture: lim n -> infinity, a(n)/10^n -> C = 12*ln(2)/Pi^2 = 0.842... - _Benoit Cloitre_, May 04 2002 A091680 Conjecture 1: For each k > 0 the trajectory of k eventually leads to a term in the trajectory of some j which belongs to A075421, i.e. whose trajectory (presumably) never leads to a palindrome. Conjecture 2: There is no k > 0 such that the trajectory of k contains more than twelve palindromes, i.e. a(n) = -1 for n > 12. A091849 Conjectured to agree with A091848 except for n = 9 and 13. A092215 Conjecture 1: For each k > 0 the trajectory of k eventually leads to a term in the trajectory of some j which belongs to A075252, i.e. whose trajectory (presumably) never leads to a palindrome. Conjecture 2: There is no k > 0 such that the trajectory of k contains more than eleven base 2 palindromes, i.e. a(n) = -1 for n > 11. A092287 Conjecture: Let p be a prime and let ordp(n,p) denote the exponent of the highest power of p which divides n. For example, ordp(48,2)=4, since 48=3*(2^4). Then we conjecture that the prime factorization of a(n) is given by the formula: ordp(a(n),p) = (floor(n/p))^2 + (floor(n/p^2))^2 + (floor(n/p^3))^2 + . . .. Compare this to the de Polignac-Legendre formula for the prime factorization of n!: ordp(n!,p) = floor(n/p) + floor(n/p^2) + floor(n/p^3) + . . .. This suggests that a(n) can be considered as generalization of n!. See A129453 for the analogue for a(n) of Pascal's triangle. See A129454 for the sequence defined as a triple product of GCD(i,j,k). - _Peter Bala_, Apr 16 2007 A092645 Conjecture: a(n) ~ 10^n / (64 log 10 * n). [_Charles R Greathouse IV_, Oct 24 2011] A092925 Conjecture: No term is zero. A092937 Conjecture: The maximum density occurs at increasing multiples of 6 as the number of primes tested approaches infinity. E.g. the number of nextprime - precprime occurrences for 2n <= 10^10 will be 30 or higher. This appears as a plausable statement since as 2n increases, the probability that the difference between the next and preceding prime will contain larger and larger prime factors. A092940 Conjecture: a(n) = prime(n) only for n = 1 and 2. A092951 Conjecture: No term is zero. A092959 Conjecture: No term is zero. A092960 Conjecture: No term is zero. A092967 Conjecture: a(n)-1 has prime(n) -1 divisors. Subsidiary sequence: Number of primes of the from 2*p*q*r*...+ 1 where p,q,r etc. are distinct odd primes < n. A092969 Conjecture: There are only finitely many zeros in this sequence. In other words the sequence is identical to A092965 barring a finite set of terms which are zero. A092984 Conjecture: There exists a finite k such that a(n) < k for all n. Subsidiary sequence: Indices of the first occurrence of n in this sequence. In case the conjecture is true, this sequence would be finite. A093454 Conjecture: There are finitely many numbers of the form {LCM of next n numbers}/n! which are not integers. A093456 Conjecture: There are finitely many numbers such that a(n)is not == 0 (mod a(n-1). ( Also mentioned in A093455). A093818 Conjecture: every odd prime occurs as a term in the sequence. A093892 Conjecture: there are infinitely many zeros in this sequence. A094043 Conjecture: 2 and 5 are the only two nonmembers. A094044 Conjecture: 5 is the only nonmember. A094045 Conjecture: 2 and 5 are the only two nonmembers. A094076 Conjecture: k > 0 for all n. A094124 Conjecture: This sequence, extended to infinity, contains all odd primes. A094189 Conjecture: for n>11, a(n)>1. A094336 Conjecture: a(n) > 1 for n > 4. A094339 Conjecture: this is a rearrangement of natural numbers. A094340 Conjecture: Every natural number occurs in this sequence. A094358 Conjectured (by Munafo, see link) to be the same as: numbers n such that 2^^n == 1 mod n, where 2^^n is A014221(n). A094383 Conjecture: only 25 primes are not in the sequence, namely 2, 3, 5, 7, 11, 13, 17, 19, 31, 37, 47, 61, 67, 71, 73, 89, 109, 137, 179, 211, 277, 337, 379, 499, 557. - _Alex Ratushnyak_, Sep 08 2012 A094405 Conjecture: For any seed a(1) the sequence "a(n) = (sum of previous terms) mod n" ends with repeating constant. This is true for a(1) = 1,...,941. - _Zak Seidov_, Feb 24 2006 A094540 Conjecture: Ratio of 6's to 8's approaches 1.5 as a limit as this sequence approaches infinity - _J. Lowell_, Jun 26 2007 A094639 Conjecture: For any positive integer n, the polynomial P_n(x) = sum_{k = 0}^n(C_k)^2*x^k (with C_k = binomial(2k, k)/(k+1)) is irreducible over the field of rational numbers. [From _Zhi-Wei Sun_, Mar 23 2013] A094757 Conjecture: For every n there exists a k different from n (possibly k > n) such that n*pi(k) = k*pi(n). (Amarnath Murthy) A094759 Conjecture: There are infinitely many terms such that a(n)<n. A050973 has those n, A050972 has the a(n). A094761 Conjecture (verified up to 727): the numbers not in this sequence are those of A008865. [From _R. J. Mathar_, Jan 23 2009] A094783 Conjecture: If n is in this sequence, then so is the smallest number with n divisors. (This conjecture is definitely false for A002182 (n=840) and A019505 (n=240.) - _J. Lowell_, Jan 24 2008 A094787 Conjecture: every prime is contained in this sequence. A094870 Conjecture: lim_{n->infinity} a(n)/n = 1 (P. Hegarty). A094913 Conjecture: a(n) = A006697(n)-1. - _Vladeta Jovovic_, Sep 19 2005 A094913 Conjecture: a(n) = 2^(m+1)-2 + (n-m)(n-m+1)/2, where m = [ n+1-LambertW( 2^(n+1) * log(2) ) / log(2) ] = integer part of the solution to 2^x = n+1-x. This is based on the reasoning that you can construct the word of length n so that it contains the maximal possible number, max( n-k+1, 2^k ), of different substrings of length k. - M. F. Hasler, Dec 17 2007 A094913 Conjecture: column sums of: A094960 Conjecture: for every integer n>78 there exists a prime p such that the sum of last two base-p digits of n is at least p. This conjecture would imply that this sequence is finite and 60 is the last term. The conjecture is true for n such that n+1 is not a prime or a power of 2. [From _Max Alekseyev_, Jun 04 2012] A095017 Conjecture: a(n) > 0 for all n. This holds for all n <= 100. - _Charles R Greathouse IV_, May 14 2012 A095180 Conjecture: the Benford law limit is 2=Sum[N[Log[10, 1 + 1/dn]], {n, 1, Length[d]}]^2/(( #toralprimes/#totalPrimes)). At 50000 primes total it is 2.05931. - _Roger L. Bagula_ and _Gary W. Adamson_, Jul 02 2008 A095188 Conjecture: a(n) is nonzero if n is not a perfect square. A095189 Conjecture: a(n) is nonzero for all n>1. Generates surprisingly large primes that are easily certified using Elliptic curve techniques (Mathematica's NumberTheory`PrimeQ`). For n=24 no certifiable prime was found with fewer than 1024 digits. - _Wouter Meeussen_, Jun 04 2004 A095223 Conjecture: The sequence is infinite. A095224 Conjecture: The sequence is infinite. A095231 Conjecture: Every natural number occurs infinitely many times. A095237 Conjecture: There are infinitely many primes in this sequence. A095380 Conjecture: ratio of non-sigma numbers tends to one. Increasing majority of numbers is impossible as a sum of divisors. A095689 Conjecture: lim_{n->infinity} a(n)/n = 1 (P. Hegarty). A095720 Conjecture. For even n, the ratio a(n)/a(n-1) is asymptotic to 2 as n becomes large. (At n=3000, the ratio is 2.0004446.) A095849 Conjecture: a number is in this sequence if and only if it is in both A002182 and A095848. - _J. Lowell_, Jun 21 2008 A096097 Conjecture:(1) Every concatenation is squarefree. (2) This is a rearrangement of the noncomposite numbers other than 5. A096097 Conjecture (1) is false. 3^2 divides the concatenation for a(22) and a(30). [From _Sean A. Irvine_, Nov 25 2009] A096098 Conjecture:(1) Every concatenation is squarefree. (2) This is a rearrangement of the squarefree numbers not divisible by 5. A096123 Conjecture: a(n) = n!/(p-1)! for all sufficiently large n, where p is the least prime such that n <= 2*p (Amarnath Murthy). A096974 gives numbers n such that a(n) = n!/(nextprime(n/2)-1)! and indicates that this conjecture is most probably false. A096127 Conjecture: a(n)=n+1 only when n is prime or a power of a prime. A096135 Conjecture: No term is zero. A096136 Conjecture: No term is zero. A096291 Conjecture: a(n) = a(n+1) for infinitely many positive integers. Largest found is n=2577, i.e. sd(pnd(2577^2577)) = sd(pnd(2578^2578)) = 18315. A096293 Conjecture: For any positive integer starting value n, iterations of n -> n + (sum of squares of digits of n) will eventually join A033936 (verified for all n up to 20000). A096301 Conjecture: a(n) = a(n+1) for infinitely many positive integers. Largest found is n=4462, i.e. pnd(sd(4462^4462)) = pnd(sd(4463^4463)) = 126. A096303 Conjecture: For any positive integer starting value n, iterations of n -> n + (number of 1's in binary representation of n) will eventually join A010062. A096306 Conjecture: For any positive integer starting value n, iterations of n -> n + (largest digit of n) will eventually join A045844. A096316 Conjecture: This sequence carried to infinity is non-repeating and non-terminating. If we concatenate the numbers and introduce a decimal point somewhere, we will get an irrational number. A096319 Conjecture: This sequence carried to infinity is non-repeating and non-terminating. If we concatenate the numbers and introduce a decimal point somewhere, we will get an irrational number. A096335 Conjecture: For any positive integer starting value n, iterations of n -> n + tau(n) will eventually join A064491 (verified for all n up to 50000). A096343 Conjecture: a(n) = a(n+1) for infinitely many positive integers. Largest found is n=1091, i.e. n1b(pnd(1091^1091)) = n1b(pnd(1092^1092)) = 1892. A096469 Conjecture: This sequence is infinite. a(n) = 1 if and only if n + 1 is in the sequence A004023, so a(1) = a(18) = a(22) = a(316) = a(1030) = a(49080) = a(86452) = 1 and there are no other n less than 86453 such that a(n) = 1. Every term of this sequence is odd and for each n, 5 doesn't divide a(n). a(50) is greater than 11111. A096913 Conjecture: 0 and 3844 are the only squares in this sequence. A097267 Conjecture: a(15)=a(38)=0. A097387 Conjecture: Terms will always be multiples of ten. Aug 21, 2004: Dean Hickerson proved this. A097406 Conjectures: (1) For every n the highest unique prime factor is of the form kn+1. The values for k are in A097407. (2) For each composite n many factors of the form kn+1 occur intermittently but always singly in any cofactor pair. (3) For each prime n every factor is of the form kn+1. A097587 Conjecture : a(n) is always < 8. A097588 Conjecture : a(n) is always < 8. A097602 Conjecture: a(n) = m^2 iff m mod 3 > 0. A097702 Conjecture: n is a member iff 6*n+3 is squarefree. - _Vladeta Jovovic_, Aug 27 2004 A097703 Conjecture: all numbers of form 3k + 1 are here. Other terms are listed in A097704. A097704 Conjecture: "most" of the terms also belong to [(A067778-1)/2]. Exceptions are {302, 2117, ...}. In other words, most terms satisfy: GCD(2n+1, numerator(B(4n+2))) is not squarefree, with B(n) the Bernoulli numbers. A097913 Conjectured Poincare series for genus 2 Siegel theta series of odd unimodular lattices. A097945 Conjecture: Sum_n=1..inf mu(n)/phi(n) = Sum_n=1..inf a(n)/phi(n)^2 = 0 It is true that Sum_n=1..inf mu(n)/phi(n)^s = 0 at least for s > 1 since: phi(2)=1, phi is multiplicative, so for n's that are squarefree, the phi(n) values can be partitioned in pairs where phi(m)=phi(2m) and mu(m) = -mu(2m). So Sum_i=1..n mu(i)/phi(i)^s < Sum j=[n/2]..n 1/phi(j)^s which approaches 0 as n increases since 1) n^(1-e) < phi(n) < n for any e > 0 and n > N(e) and 2) Sum_i..n 1/n^s converges for s > 1. Conjecture: Sum_n=1..inf mu(n)/phi(n)^z = 0 for Re(z) > 1 A098016 Conjecture: 1/2(prime(x+1) + prime(x)) is not prime for all x. A098021 Conjecture: A098021 is the sequence of positions of 0 in the zero-one sequence [nr+2r]-[nr]-[2r], where r=sqrt(2) and [ ]=floor; see A187967. [From Clark Kimberling, Mar 17 2011 ] A098375 Conjecture: p is an odd prime iff p divides p*(p^(p-1)-1)*B(p-1)-1. Seems to be the equivalent (with integer moduli) to Agoh's conjecture (which involves rational moduli). A098472 Conjecture: Mersenne-prime(12) is a Riesel prime (that is, all numbers k^2*M(12)-1 are composite for all k) and similarly for M(14) and M(15). A098592 Conjecture: if n tends to infinity a(n) may be > 1 and = 2 or 3 or 4 or 5 or 6 or 7. - _Pierre CAMI_, Jun 02 2009 A098833 Conjecture: there are infinitely many consecutive terms in this sequence. For example, sopfr(143335) = 377 and sopfr(143336) = 89 are both Fibonacci numbers. A099728 Conjecture: No term is zero. A099742 Conjecture: No term is zero. A099940 Conjecture: this sequence consists completely of integers. A100056 Conjecture: a(n) provides the maximum sum of difference terms of any permutation of 1 through n. A054065 (or reversals) give the optimal permutations and a(n) gives the sum. A100380 Conjecture: all prime number can be written as + or - p(n) - or + p(k)#. A100481 Conjecture: every prime appears infinitely often in this sequence. A100583 Conjectured to be the number of triangles in an n X n grid of squares with all diagonals drawn. A100682 Conjecture: a(n) = floor((n - 3/2)/24^(1/4)) for n not in {0, 1, 6, 17, 2403, 5318}. - _Charles R Greathouse IV_, May 01 2012 A100759 Conjecture: Every prime is a member. A100800 Conjecture: No term is zero. A100801 Conjecture: No term is zero. A100873 Conjecture 1: If p and p+2 are prime (twin primes), then p+2 divides p^(p+2)+2. Compared to the 1517 twin primes less than 130000, there were 33 pseudoprime occurrences. Conjecture 2: If for a randomly chosen prime p, p+2 divides p^(p+2)+2, then there is a greater than 98% chance that p and p+2 are twin primes. The sequence also contains several Carmichael numbers. In addition, If we relax the condition that p is prime or just odd, we get A001567 341,561,645,1105,1387,1729,1905,2047.. Sarrus numbers. A100952 Conjecture: the sequence is finite and full. A100952 Conjecture: Every positive integer can be represented as p*q-r with distinct primes p, q, r. - _Zak Seidov_, Aug 28 2012 A101183 Conjecture: No term is zero. A101215 Conjecture 1 : for any integer N ending in 1 or 3 or 7 or 9 there is an infinite sequence of numbers n such that the concatenation nN is a prime. Conjecture 2 : for any integer N there is an infinite sequence of numbers n such that the concatenation Nn is a prime A101312 Conjecture: The same basic repeating pattern results if we seek the number of "Sunday the 22nds" or "Wednesday the 8ths" or anything else similar, with the only difference being that the sequence starts at a different point in the repeating pattern. A101462 Conjecture: sequence is defined for all n. First unproved n: 286 Prime(286)=1871, up to date, tested up to k=40959, none 2^k-Prime(286) is prime. A101513 Conjectures from _N. J. A. Sloane_, Apr 22, 2005: A101902 Conjecture that this sequence is complete. Note that, except for n=1 and n=2, n-1 is prime. The case of the 3-term binary form is treated in A025052. For the 4-term case, ab+bc+cd+da, no prime is representable because the 4 terms factor as (a+c)(b+d). A102237 Conjecture: a(n) = 2^n for n>5. (Chandler) A102613 Conjecture: The ratio Pi(x)/(n-Pi(x)) tends to 0 as n tends to infinity. This is evident from the fact that Li(x)/((n-Li(x)) -> 0 as n -> infinity but unfortunately not proof. A102722 Conjecture: a(n) ~ (1-EulerGamma)n A102730 Conjecture: the sequence is bounded. A102751 Conjectured to be infinite. A102818 Conjectures (based on mod values up to n=99): the sequence A001035(m) is (pre)periodic modulo n for all n, the lengths of the ending periods mod n (except n=4) being given by A011773 (which is related to Carmichael's lambda function). A102827 Conjecture: this sequence in various bases never includes a term divisible by the base. A103151 Conjecture: all items for n>=4 are greater than or equal to 1. This is a stronger conjecture than the Goldbach conjecture. A103152 Conjecture: except for the 2nd, 3rd and 4th terms, all other terms are divisible by 3; See also comments in A103151. A103176 Conjecture: In all cases sigma(n)-phi(n)=2, i.e., n is prime. A103267 Conjecture: The greatest prime divisor of x divides y and z. A103515 Conjecture: sequence is defined for all k>=2 A103673 Conjecture: a(n) = 1 for n > 65. - _Charles R Greathouse IV_, Apr 07 2013 A103674 Conjecture: a(n) = 1 for n > 802. - _Charles R Greathouse IV_, Apr 07 2013 A103674 Conjecture checked up to n <= 5*10^5. - _Giovanni Resta_, Apr 07 2013 A103677 Conjecture: the sequence is finite and full; A103683 Conjectured to be not a permutation of the natural numbers. A103794 Conjecture: sequence is defined for all positive indices. A103795 Conjecture: sequence is defined for any n>=2. A103910 Conjecture: all rows except the first add to zero. A103959 Conjecture: a(n) <= prime(n). Tested to n=10000. A103960 Conjecture: All items of this sequence are greater than or equal to 1. Tested to prime(1000000). A103990 Conjecture: Let b(n) be A103991(n), then a(n)/b(n) = (2/3)*(n^3-n^2+2*n+1)/(2*n+1). [From _Gerald McGarvey_, Aug 11 2009] A104102 Conjecture: a(n) is always > p(n+1) A104895 Conjecture: the positions where 0, 1, 2, 3, ... appear are given by A048724; the positions where -1, -2, -3, ... appear are given by A065621. A104896 Conjecture: this is also the number of integers from 0 to 10^n - 1 that lack 0, 1 and 2 as a digit. A105412 Conjecture: There is an infinity of primes p(n) such that p(n)-2 and p(n+k)-2 are both prime for all k > 1. A105413 Conjecture: There are an infinite number of primes p(n) such that p(n)-2 and p(n+k)-2 are both prime for all k > 1. A105414 Conjecture: There are an infinite number of primes p(n) such that p(n)-2 and p(n+k)-2 are both prime for all k > 1. A105595 Conjecture : all terms are odd. A105596 Conjecture : a(2n)=1 mod 4 for all n, a(2n+1)=0 for all n. A105801 Conjecture: for every k>0 there is an index m such that all the a(n) with n>m have the same residue mod 3^k. - _Giovanni Resta_, Nov 17 2010 A105974 Conjectures: a(n) is (pre)periodic modulo m for all m, the lengths of the ending periods being given by 2*A000010(m) if m is a power of 2 and A000010(m) otherwise. A0000100 is Euler's totient function. A106249 Conjecture : number of roots of x^n + 1 in the left half-plane for n > 0. [Michel Lagneau, Oct 31 2012] A107363 Conjectures: { Fib(n) | n in naturals } = { a(n) | n in naturals, a(n) >= 0 } = { a(n) | n in naturals, n not of the form 6*n+2 } (naturals include 0). A107655 Conjecture: A000005(a(n)) <= 12 for all n. [From _Max Alekseyev_, May 07 2010] A108144 Conjecture: There are infinitely many primes and semiprimes in this sequence. A108200 Conjecture: for every n > 4 there exists a number k < n^[n/2] such that k*n + 1 is a golden semiprime, where [] is the floor function. A108393 Conjecture: 23,83,113 and 811 are the only primes with a 0 value in the sequence. There is always a prime of the form p^2+k^2 (1 mod 4) between p^2 and (p+1)^2 for every prime not 23,83,113 or 811. A108399 Conjecture: for every n > 1 there exists a number k < n^3 such that n^2 + k is a golden semiprime. A108847 Conjecture: Inner digits 1 and 2 are the only repeating digits for which s(x) is prime for outer digits 9. I.e. 9444..4449, 9555..5559, 9777..7779 are composite. The cases for inner digits 3,6 and 9 are obvious that s(x) is composite. A108863 Conjecture: this is the Motzkin transform of the sequence of three zeros followed by A001651. [From _R. J. Mathar_, Dec 11 2008] A108957 Conjectures: a. Sequence is infinite. b. There are infinitely many consecutive pairs, such as (5:6), (11:12), (17:18), (53:54), ... (204005:204006). A109132 Conjecture: 77! + 1 is the largest factorial Chen prime. A109224 Conjecture: for n >= 5, a(n) = ((7/2) * (7 - (4 * sqrt(2))) * (1 + sqrt(2))^n) - (8 * n) - 2 + ((7/2) * (7 + (4 * sqrt(2))) * (1 - sqrt(2))^n). Confirmed to n = 13, and known to be an upper bound for all larger n. A109280 Conjecture: Sequence is infinite. A109381 Conjecture: lim_{n->infinity} a(n) = infinity. If true, convergence is very slow, since a(1183893) = 1. Sequence is certainly unbounded, since for n >= 4, there is always a square between n*n! and (n+1)!. A109464 Conjecture: Sequence is infinite. A109474 Conjecture: Except for the first term, a(n)=10*n-a(n-1)-23 (with a(2)=3) [From Vincenzo Librandi, Dec 07 2010] A109730 Conjecture: Sequence is infinite. A109799 Conjecture: 2^127 - 1 is the largest Mersenne Chen prime. A109852 Conjecture: a(2^n) is prime if n is not 0 nor 2. A109853 Conjecture: a(n) is prime if n is not 0 nor 2. A109853 Conjecture: a(n) is the 2n-2th prime for n>1. A109852(2^n-1): 1,3,5,11,17,23,31,41,47,59,67,73. - _Robert G. Wilson v_, Jun 14 2006 A109853 Conjecture: the Union of A109852(2^n-1) & A109852(2^n) is A046022: {1,2,3,4,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79, ...,} and expect for 4, equals A008578: The non-composite numbers. - _Robert G. Wilson v_, Jun 14 2006 A109862 Conjecture : Sequence is infinite. A109864 Conjecture: a(n) = 0 if and only if n > 3 and n begins with 2,4,6,8 or 5 (most significant digit). A109875 Conjecture: sequence is infinite. A109890 Conjectured to be a rearrangement of the natural numbers. A109907 Conjecture: The sequence is infinite. A109908 Conjecture: a(n) > 0 for n > 3. A109909 Conjecture: a(n) > 0 for n > 3. A109912 Conjecture: Sequence is infinite with terms from a(12) onwards alternating between integers with the four digits 4,5,6,9 and integers with the remaining six digits 0,1,2,3,7,8. - _William Rex Marshall_, Jul 19 2005 A109926 Conjecture: a(n)==0 (mod 3) for n > 2. Then n-2^k is not == 0 (mod 3) and a prime is more probable. - _Robert G. Wilson v_, Jul 21 2005 A109926 Conjecture: a(n+15)==0 (mod 30) for n > 4. - Robert G. Wilson v, Jul 21 2005. A109969 Conjecture: a(n) is bounded. A109994 Conjecture: sequence is infinite. A110034 Conjecture: |a(n)| = sum_{k=1,...,n-1} A061646(k). - _J. M. Bergot_, Jun 10 2013 A110073 Conjecture: Each term which is greater than 76 of this sequence is of the form 80p+10 where both p and 8p+1 are primes. A110076 Conjecture: For n>3 a(n) is positive. A110078 Conjecture: For n>2, a(n+1)>a(n). A110194 Conjecture: from a(5) on, are there only substrings <1 1 4> and <1 1 1 1 4 4 >? A110278 Conjecture: sequence is infinite. A110320 Conjecture: A110320(n) = (A051292(n+2)-A051286(n+1))/2. - _Gerald McGarvey_, Jan 14 2007 A110358 Conjecture: The sequence is infinite. A110408 Conjecture: a(n) is 0 iff n is a palindromic prime == 1,3,7 or 9 (mod 10) and/or n ==0 (mod 3). As rev(n) followed by n is == 0 (mod 11), hence every nonzero term contains at least two reverse concatenations of n followed by n. A110434 Conjecture: every number == 1,3,7,9 (mod 10) is a term. A110454 Conjecture a(n) = 0 only for n = 1,2 and 4. A110456 Conjecture a(n) = 0 only for n = 1, or n == 0 (mod 3), n is > 3. Subsidiary sequence (hard): Number of primes generated by concatenation of distinct partitions of n. A110545 Conjecture: a(n)<=n for all positive n's. A110734 Conjecture: for large n, a(n) is nonzero. For a k-digit number there are k-1 gaps and 10^(k-1) candidates, so the chances that one of them is a multiple of n increases with k on the one hand though the probability decreases because n becomes large. A110735 Conjecture: no term is zero. For a k-digit number there are k+1 spaces and 10^(k+1) candidates, so the chances that one of them is a multiple of n increases with k on the one hand although the probability decreases because n becomes large. A110772 Conjecture: every odd number not divisible by 5 is a member. A110875 Conjectures and open problems: 1) It is not known whether the sequence is infinite; 2) It is conjectured that for every n there is corresponding a(n). If Conjecture 2) were proved, Conjecture 1) would follow as a direct consequence. A111015 Conjecture: Starting with 1/1, there is an infinity of primes in the numerators of fractions built according to the rule - add top and bottom to get the new bottom, add top and 2k times bottom to get the new top. k=1,2..infinity. A111108 Conjecture: for odd primes p, p divides a(p). Note that (a(n)) and A001333 have different offsets. A111114 Conjecture: As x-> infinity, there is an infinite number of x's such that a(x) is greater than a(x+1). A111166 Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 2*10^7. A111166 Conjecture: Except for first term, the sequence coincides with A001359. This is true for all primes < 7*10^16. Let n >= 2 be an integer, N +- 1 and M +- 1 two consecutive twin pairs where M>n*N. Finding a counterexample is the same as finding two consecutive primes P1 and P2 with n*N<P1<M and P2-P1 <= n. However, such gaps are unknown even for n=2. A111181 Conjecture: There will always be an x such that a(x+1) - a(x) = k for k=1,2.. However, x becomes large when k > 70. A111183 Conjecture: There will always be an x such that a(x+1) - a(x) = k for k=1,2.. However, x becomes large when k > 70. A111205 Conjecture: The number of terms in this sequence is infinite. A111213 Conjecture: The number of terms in this sequence is infinite. A111252 Conjecture: The number of terms in this sequence is infinite. A111364 Conjecture that the sequence contains no zeros; that is, A033093 is onto the nonnegative integers. Note that most of the values in the sequence are highly divisible, as would expected. a(7) appears to be the last odd value. A111489 Conjecture: The number of primes in this sequence is infinite. A111495 Conjecture: The difference between successive terms is between 2 and 3. A111677 Conjecture: a(n)=0 iff n== 3 (mod 5). [Corrected by _R. J. Mathar_, Aug 20 2007] A112521 Conjecture: Starting with n=1, a(n) is the main diagonal of the array defined as T(1,1) = 1, T(i,j) = 0 if i<1 or j<1, T(n,k) = T(n,k-2) + T(n,k-1) -2*T(n-1,k-1) + T(n-1,k) + T(n-2,k). [From _Gerald McGarvey_, Oct 07 2008] A112970 Conjectures: a(2^n)=a(2^(n+1)+1)=A033638(n); a(2^n-1)=a(3*2^n-1)=1. A112992 Conjecture: the odd values of ceiling(2^mod(2^n,n)/3) are Jacobsthal numbers (and the even values are 1 plus a Jacobsthal number). A113047 Conjecture: a(n) differs from 0 only when n=(3^j-1)/2,j=0,1,... A113052 Conjecture: for p prime, mod(C(p*n,n)/((p-1)*n+1),p) is the indicator function of the sequence (p^n-1)/(p-1). A113249 Conjecture: a(m, 2*n+1) is a perfect square for all m, n. Disregarding signs and/or initial term, we have: m = 0 (A000302), m = 1 (A097948), m = 2 (A056450), m = 3 (a(n)), m = 4 (A113250), m = 5 (A113251), m = 6 (A113252), m = 7 (A113253), m = 8 (A113254), m = 9 (A113255), m = 10 (A113256) A113250 Conjecture: a(m, 2*n+1) is a perfect square for all m (see A113249), Initial terms factored (without regards to sign): 1, 4, (2)^5, (2)^6,(2)^8, (2)^12, (2)^12, (2)^14, (2)^17, (2)^18, (2)^20, (2)^24, (2)^24, (2)^26, (2)^29, (2)^30, (2)^32, (2)^36 A113251 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249), A113252 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249), A113253 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249), A113254 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249), A113255 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249), A113256 Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249), A113459 Conjecture: For n > 1, a(n) = A007918(n). - _David Wasserman_, Jan 08 2006 A113570 Conjecture: that only prime powers are present (A025475). A113571 Conjectured last occurrence of n: 1,3,11,25,137,49,363,761,7129,7381,83711,6617,72072,1117, A113575 Conjecture: No term is zero. A113576 Conjecture: No term is zero. A113598 Conjecture: No term is zero. A113599 Conjecture: No term is zero. The proof should be simple. A113604 Conjecture: a(n) = 0 only if n == 0 (mod 3). A113610 Conjecture: For n > 9 if a(n) > a(n+1) < a(n+2) then prime(n) and prime(n+1) form a twin prime pair. A113676 Conjecture: a(n) for n>1 gives first differences of Lionel Levile's sequence A014644(n) for n>=3. A113676 Conjecture: Final elements of the rows form A014644 except for duplicate 2. A113702 Conjecture: Sequence is not periodic. A113703 Conjecture: Sequence is not periodic. A113917 Conjecture: a(n) is finite for all n A113918 Conjecture: a(n) is finite for all n A114379 Conjecture: The number of terms in this sequence is infinite. A114534 Conjecture: The rank of the matrix M is 2 for every n. A114726 Conjecture: Sequence is infinite and lim n -> infinity a(n+1)/a(n) = e = 2.718281828... A114735 Conjecture that the next two terms are 746443728915 = 3 * 5 * 31 * 929 * 1727939 and 7800513423460801052132265 = 3 * 5 * 31 * 929 * 1727941 * 10450224300389. [From _T. D. Noe_, Apr 13 2010] A114782 Conjecture a(n) = 0 iff prime(n) + 1 == 0 (mod 3). A114888 Conjecture: the sequence is finite. A114948 Conjecture: n^2 + 10 != x^k for all n,x, and k>1. A114962 Conjecture: x^2 + 14 != y^n for all x,y and n > 1. A114963 Conjecture: x^2 + 22 != y^n for all x,y and n > 1. A114964 Conjecture: x^2 + 30 != y^n for all x,y and n > 1. A114965 Conjecture: n^2 + 34 != x^k for all n,x and k > 1. A115038 Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p. In otherwords, there is a one-one mapping of the prime numbers to y^n - x^2 for some x,y,n. Therefore primes of the form y^n - x^2 are infinite in number. A115039 Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p. A115041 Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p. A115257 Conjecture: For any positive integer n, the polynomials sum_{k=0}^n binomial(2k,k)^2*x^k and sum_{k=0}^n binomial(2k,k)^2*x^k/(k+1) are irreducible over the field of rational numbers. [From _Zhi-Wei Sun_, Mar 23 2013] A115509 Conjecture: if 2^(k+1)<=n<3*2^k for k>=1, then a(n)=0. A116473 Conjecture that lim_{n->infinity} a(n) = infinity. Definitely a(n) > 2 for n > 8. A116731 Conjecture: also counts the distinct pairs (flips, iterations) that a bubble sort program generates when sorting all permutations of 1..n. [From _Wouter Meeussen_, Dec 13 2008] A117149 Conjecture: Sequence is not periodic. A117150 Conjecture: Sequence is not periodic. A117351 Conjecture: Sequence is a permutation of the positive integers. A117414 Conjecture: row sums are the Euler numbers A000364. Second column is A024255. Third column is A117415. A117549 Conjecture. Let A be a very(5,1) (respectively very(5,4)) sequence of length n and let Z be a sequence of n-1 0's.. Then AZ(3A)ZA is a very(5,1) (respectively very(5,4)) sequence of length 5n-2. (Here 3A denotes the result of multiplying each term of A by 3, then reducing modulo 5; and juxtaposition of symbols denotes concatenation of sequences.) A118202 Conjecture that a(n) is never -1. A119018 Conjecture: this sequence includes every integer greater than 1. A119498 Conjecture: This sequence never cycles. A119537 Conjecture: a(n>7)=0. - from _Robert G. Wilson v_,Jun 07 2006 A119659 Conjecture: The triples (3,5,11),(5,11,17),(11,17,31) are the only consecutive PIP triples that cannot form a triangle. A120121 Conjecture: 139968 is the largest term. Except for the first term all terms are even. It's interesting that for the number 139968 we have the following relations: 139968=phi((1*3*9*9*6*8)*(1+3+9+9+6+8))=phi(1*3*9*9*6*8) *(1+3+9+9+6+8)=(1*3*9*9*6*8)*phi(1+3+9+9+6+8). A120122 Conjecture: 139968 is the largest term. Except for the first term all terms are even. 139968 is also in the sequences A120121 & A120123, so it has a very pleasant property (see the Comments lines of A120121). A120123 Conjecture: 139968 is the largest term. Except for the first term, all terms are even. 139968 is also in A120121 & A120122, so it has a very pleasant property (see the Comments lines of A120121). A120292 Conjecture: All composite terms are semiprime. A121325 Conjecture: these numbers are all of the form 4m+1. A121627 Conjecture (1): Current nonzero term divides next nonzero term; as an operation, getting a new sequence: 12, 8, 5, 84, 16, 9, 220, 24, 13...; using unsigned terms). Example: a(5) = -480 and a(7)= 40320, the next nonzero term in the sequence. Then 40320/480 = 84. Conjecture (2): If the term in (12, 8, 5...) is generated from a(n)/a(n-1) or a(n)/a(n-2), then n divides each (12, 8, 5...). Example: a(11) = -1277337600 and the previous nonzero term = a(9) = 5806080. Then a(11)/a(9) = 220 and 11 divides 220: 220/11 = 20. A121716 Conjecture: the sequence is finite. A122249 Conjecture - requires proof. A122276 Conjecture: lim {n -> infinity} x_n / y_n = 1, where x_n is the number of j <= n such that A096535(j) = A096535(j-1) + A096535(j-2) and y_n is the number of j <= n such that A096535(j) = A096535(j-1) + A096535(j-2) - j. Computational support: x_n / y_n = 0.9999917 for n = 10^9. A122707 Conjecture: a(n)>1 exists for all n>2. A122869 Conjecture: This sequence is just the primes congruent to 11 or 19 mod 20. - _Charles R Greathouse IV_, May 25, 2011 A122955 Conjecture: Natural density is log 5/log 100. [From _Charles R Greathouse IV_, Nov 15 2010] A122957 Conjecture: except for 1,2,3,4, this is the nonzero values of A066362. A123120 Conjecture: All but (n+6) positive numbers are equal to the sum of n>5 nonzero squares. For all n>5 the only (n+6) positive numbers that are not equal to the sum of n nonzero squares are {1,2,3,...,n-3,n-2,n-1,n+1,n+2,n+4,n+5,n+7,n+10,n+13}. {2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20,21,22,23,24,26,27,28,...} = A000037 Numbers that are not squares (n=1). {1,3,4,6,7,9,11,12,14,15,16,19,21,22,23,24,27,28,30,31,33,35,36,...} = A018825 Numbers that are not the sum of 2 nonzero squares (n=2). {1,2,4,5,7,8,10,13,15,16,20,23,25,28,31,32,37,39,40,47,52,55,58,...} = A004214 Numbers that are not the sum of three nonzero squares (n=3). {1,2,3,5,6,8,9,11,14,17,24,29,32,41,56,96,128,224,384,512,...} = A000534 Numbers that are not the sum of 4 nonzero squares (n=4). {1,2,3,4,6,7,9,10,12,15,18,33} = A047701 Numbers that are not the sum of 5 nonzero squares (n=5). {1,2,3,4,5,7,8,10,11,13,16,19} Numbers that are not the sum of 6 nonzero squares (n=6). {1,2,3,4,5,6,8,9,11,12,14,17,20} Numbers that are not the sum of 7 nonzero squares (n=7). {1,2,3,4,5,6,7,9,10,12,13,15,18,21} Numbers that are not the sum of 8 nonzero squares (n=8). {1,2,3,4,5,6,7,8,10,11,13,14,16,19,22} Numbers that are not the sum of 9 nonzero squares (n=9). A123341 Conjecture: lim_{n->infinity} a(n)/2^n > 0; appears to be about 0.1432645404. A123365 Conjecture. With the exception of the first term a(1)=1, this is exactly the sequence of primes of the form 6k+1 (A002476). This has been verified up to a(n)=2000. A123371 Conjecture: a(n)>0 exists for all n. A123721 Conjecture: For k > 1, the smallest n such that a(n) = k is A123720(k) = 2^k + 2^(k-1) - k. Confirmed for k <= 22. A123722 Conjecture. If n>2 and a(n) is a multiple of 3, then a(n)/3 is a prime; further, all odd primes are given as a(n)/3 for some n. (This has been verified up to n=2000.) A123723 Conjecture. If n>4 and a(n) is a multiple of 4, then a(n)/4 is a prime. (This has been verified up to n = 2000.) A124000 Conjecture: all a(n) except a(1) = 6 and a(2) = 10 are odd. Conjecture: all a(n) except a(5) = 35 are triangular numbers of the form p*(2p +/- 1) that belong to A068443(n) = {6, 10, 15, 21, 55, 91, 253, 703, 1081, 1711, 1891, 2701, ...} Triangular numbers with two distinct prime factors. A124129 Conjecture: there are no other terms. A124431 Conjecture: (n+1)*a(n) -2*(2*n+1)*a(n-1) +2*(n-1)*a(n-2) +2*(5-2*n)*a(n-3) +(n-3)*a(n-4) =0. - _R. J. Mathar_, Aug 09 2012 A124689 Conjecture: -1 only occurs in the first entry of the sequence. 1000 digits precision was used in the calculation. A124690 Conjecture: -1 only occurs in the first entry of the sequence. 2000 digits of precision was used in the calculation. A124900 Conjecture: The sequence a(prime(n)), which begins 2, 6, 20, 42, 110, 156, 272, 342, 506, 812, increases without bound. It appears that a(prime(n)) may equal prime(n)(prime(n)-1), which is A036689. - Artur Jasinski, Feb 26 2011 A125148 Conjecture: for every odd (prime or nonprime) number x>=1 that is not a multiple of 5 there exists a prime p such that p = z + x; where z is a multiple of 10. A125606 Conjectures: a(n) acts like u^n, where u = 2 zeta(3) / zeta(2). In increasing order of strength: (1) lim_{n->infinity} a(n)^(1/n) = u; (2) lim_{n->infinity} a(n+1)/a(n) = u; (3) a(n) = c u^n + o(u^n) for some c. (Conjectures 2 and 3 are very speculative.) A125641 Conjecture: Draw the segments joining every lattice point on axis X with every lattice point on axis Y for 1<=x<=n and 1<=y<=n. The number of regions formed with these segments and axis X and Y is a(n). [_César Eliud Lozada_, Feb 14 2013] A125878 Conjecture: this sequence contains only primes (this would follow if this is indeed the same as A035095). A126142 Conjecture: a(n)^2 is of the form 6*k+1. Examples: 5^2 = 25 = 6*4+1; 7^2 = 49 = 6*8+1; 61^2 = 3721 = 6*620+1. [From _Vincenzo Librandi_, Sep 21 2009; edited by Klaus Brockhaus, Nov 23 2009] A126769 Conjecture: (primality conservation) Every prime q > 3 can in at least one nontrivial way be written as the sum of two or more squares, such that the sum p of the fourth powers of the squared numbers is also prime. q = sum{i}(a_i)^2, p = sum{i}(a_i)^4, p > q. (Tomas Xordan) A126774 Conjectured, as of 2004. Bound cited for this hyperbolic space constant depends on Perelman's proof of Poincare conjecture, which proof is now believed to be true. Can S. R. Finch comment on the conjectured constant as of 2007? A127081 Conjectures: a(8) = 183, a(24) = 144855. A127749 Conjectures: row sums modulo 2 are the Fredholm-Rueppel sequence A036987; row sums of triangle modulo 2 are A111982. Row sums are A127750. A127750 Conjecture: a(n) mod 2 gives Fredholm-Rueppel sequence A036987. Row sums of A127749. A128135 Conjecture: a(n)/a(n-1) tends to sqrt(5); (e.g.: a(10)/a(9) = 2.235294...) A128378 Conjecture: Given the infinite set of sequences generated from using the partial sum operator on A000594, (i.e. A000012^k * A000594, k in succession =1,2,3,...), k=23 and k=24 are the only two sequences in the set with zeros. In A128379, k = 23: (1, -1, -24, 0, 276, 300, -1748, -4300,...). A128379 Conjecture: given A000012^k * A000594, k=23 and 24 are the only k's generating sequences with zeros. k=24 in A128378: (1, 0, -24, -24, 252, 552, -1196, -5496,...). A128380 Conjecture: Given the infinite set of sequences generated from the pairwise operation on A000594 (A097806^k * A000594), k = 24, (A128380) is the only sequence in the set with a zero. The sequence generated from k=23 = (1, -1, -47, 23, 1081, 47, -15939,...). Analogous conjecture with the partial sum operator: (Cf. A128378, A128379); in which zeros are conjectured to occur only with k=23 and k=24. A128380 mod 24 = 1, 0, 0, 0, 0, 0, -4, 0, 0, 16,... A128381 Conjecture: given the bto performed any k times on A000594 (k=1,2,3,...); k=6 and k=24 are the only members of the set with zeros. k=6 generates (1, -18, 0, 688, 4494, 5508,...). A128382 Conjecture: given the inverse Moebius transform operation performed any k times (k=1,2,3,...); k=24 is the only such sequence with zeros. A weaker conjecture: "zero" occurs an infinite number of times in A128382. A128391 Conjecture: Given A054523^k, k = any positive integer, "zero" appears only in the sequence A018391 (k=24). A128392 Conjecture: Given A126988^k, k any positive integer, A128392 is the only sequence in the infinite set with zeros. A128666 Conjecture: a(n)>0 for all n. A128760 Conjecture: there exists c>=23 such that a(n)<2 for n>c. A128913 Conjecture: For each n there is at least one prime p such that 2*a(n) < p < 2*a(n+1). From the conjecture follows that the prime gaps g(n) = p(n+1) - p(n) = O(sqrt(p(n)/log(p(n)))). - Thomas Ordowski, Aug 12 2012 A128930 Conjecture: For each n there is at least one prime p such that a(n) < p < a(n+1). From the conjecture follows that the prime gaps g(n) = p(n+1) - p(n) = O(sqrt(p(n))/log(p(n))). Legendre's hypothesis is that g(n) = O(sqrt(p(n))). - _Thomas Ordowski_, Aug 11 2012 A129150 Conjecture: a strictly increasing sequence. [_J. Lowell_, Sep 10 2008] A129364 Conjecture: a(n) divides A092287(n) for all n - see comments in A129365. A129365 Conjectures: A) a(n) is always an integer. B) If p is a prime then p|a(n) if and only if p <= n/3. Let ordp(n,p) denote the exponent of the largest power of p which divides n. For example, ordp(48,2)=4 since 48=3*(2^4). The precise decomposition of a(n) into primes would follow from the following two conjectures: C) For each positive integer n and prime p, ordp(a(np),p)= ordp(a(np+1),p)= ordp(a(np+2),p)= . . . = ordp(a(np+p-1),p). D) Let b(n)=A004125(n). Then ordp(a(np),p)= b(n)+ b(floor(n/p))+ b(floor(n/p^2))+ b(floor(n/p^3))+ . . .. This is reminiscent of de Polignac's formula (also due to Legendre) for the prime factorization of n! (see the link). A129454 Conjecture: Let p be a prime and let ordp(n,p) denote the largest power of p which divides n. For example, ordp(48,2)=4 since 48 = 3*(2^4). Then we conjecture that the prime factorization of a(n) is given by ordp(a(n),p)=(floor(n/p))^3 + (floor(n/p^2))^3 + (floor(n/p^3))^3 + . . .. Compare with the comments in A092287. A129756 Conjecture: number of roots of P(x) = x^n - x^(n-1) - x^(n-2) - ... - x - 1 in the right half-plane. [_Michel Lagneau_, Apr 09 2013] A129761 Conjecture: All members are of the form ceiling(2^k/3) (A005578). A129809 Conjecture: The number of terms in this sequence is infinite. A129811 Conjecture: The number of terms in this sequence is infinite. A129812 Conjecture: The number of terms in this sequence is infinite. A129813 Conjecture: The number of terms in this sequence is infinite. A129912 Conjecture: every odd prime number must either be adjacent to or a prime distance away from a primorial or primorial product (the distance will be a prime smaller than the candidate). - Bill McEachen, Jun 03 2010 A129924 Conjecture: a(n) = A064384(n+1). A130095 Conjecture: a(n)/a(n-1) tends to phi^2. A130536 Conjecture: For all pairs of relative prime numbers (x, y) there exists at least one number n=2^m and one prime number p such p=x^n+y^n. This sequence show one case of this conjecture where y=x+1. A130654 Conjecture: A092505(n) is always a power of 2. a(n) = Log[ 2, A092505(n) ]. Note that a(n) = 0 iff n is a power of 2; or A002430(2^n) = A046990(2^n) and A092505(2^n) = 1. It appears that a(2k+1) = 1 for k>0. Note that least index k such that a(k) = n is {1, 3, 14, 60, ...} which apparently coincides with A006502(n) = {1, 3, 14, 60, 279, 1251, ...} Related to Fibonacci numbers (see Carlitz reference). A131194 Conjecture: for any initial seed, sequence eventually merges with the first one. A131262 Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n). A131388 Conjecture 1: a( ) is a permutation of the positive integers. Conjecture 2: d( ) is a permutation of the integers. The sequence using "Rule 2" (positive before negative) is A131393. A131393 Conjecture 1: a( ) is a permutation of the positive integers. Conjecture 2: d( ) is a permutation of the integers. The sequence using Rule 1 ("negative before positive") is A131388. A131401 Conjecture: only a(2)=0. I have not found values of a(n)<2*106 less than 100 for n = 43, 47, 74, 82, 83 & 94. A131439 Conjecture: The sequence appears to be (1, 7,...) followed by 4k + 14; k=0,1,2...; thus: (1, 7, 14, 18, 22, 26,...). Inverse binomial transform of A131439 = (1, 6, 1, -4, 7, -10, 13, -16, 19, -22,...). A131626 Conjecture: The number of primes in a row for f'(x,p) = 2*prime(p)*x + prime(p+1) is finite. A131758 Conjecture for the polynomial array: A131883 Conjecture: After omitting multiple occurrences we get A036912. - _Vladeta Jovovic_, Oct 31 2007. This conjecture has been established by Max Alekseyev - see link below. A131978 Conjectured to be always integral. A132089 Conjectures: (1) The first column equals the row sums of Losanitsch's triangle (which equals A005418). (2) Let L^n denote the n-th power of Losanitsch's triangle. Then the first column of L^(n+1) equals the row sums of L^n. A132293 Conjecture from _R. J. Mathar_, Mar 07 2008, Apr 21 2008: The correct sequence is a(n)=A014681(n-1), because a polyomino with a zigzag "stair" shape along a diagonal provides a solution where the number of right angles equals the number of edges: A132827 Conjecture: The array A132827 is the dispersion of the sequence f given by f(n)=floor(n*x+n+1), where x=(golden ratio). Evidence: use f(n_):=Floor[n*x+n+1] in the Mathematica program at A191426. [From Clark Kimberling, Jun 3 2011] A132870 Conjecture: for all integer k, there exists an integer triangle J such that the g.f. of row n of J^(k*n) = (k*n^2 + y)^n for n>=0. A132946 Conjecture: Sequence is not periodic. A132947 Conjecture: Sequence is not periodic. A132990 Conjecture: Sequence is not periodic. A132991 Conjecture: Sequence is not periodic. A133212 Conjecture: a(n)=2*A038503(n+3) if n>0. - _R. J. Mathar_, Oct 23 2007 A133411 Conjecture: subsequence of A019505. A133604 Conjecture: 2, 4 and 8 are the only terms n such that n = 2*A005282(k) for some k. A133745 Conjecture: sequence is infinite. A133832 Conjecture: a(n) exists for all n. These binary trinomials can also be written as f*2^n+1, where f=2^m+1 for some m, which is reminiscent of the Sierpinski problem (see A076336). The conjecture is equivalent to no Sierpinski numbers of the form 2^m+1 existing. The PFGW program was used to find a(32), which produces a 138012-digit probable prime. A133835 Conjecture: for any initial seed, sequence eventually merges with the first one. A133901 Conjecture: Sequence is infinite. A134002 Conjecture. If n a positive integer not a term of this sequence, then n^2+(n+5)^2 is prime. (This has been verified up to n=500.) Examples. For n=1,2,3,4,6,7, n^2+(n+5)^2 is 37, 53,73, 97, 157 and 193, each of which is prime. See A134003 for the complement of this sequence. A134003 Conjecture. If n is in this sequence then n(n+5)=a(a+5)+b(b+5) is not solvable in integers. (This has been verified up to n=500.) See A134002 for the complement of this sequence (in the positive integers). A134460 Conjecture. Except for the first term a(1)=2, this sequence does not contain terms of the form 4k+2. (This has been confirmed for the first several thousand terms.) A134641 Conjecture: This sequence is finite and complete. A134650 Conjectured to be finite, may be complete. A135010 Conjecture: A135060 Conjecture: every number in this sequence is also in A002182. - Lowell Disproved at a(24) = 48188059920 - Chandler. A135096 Conjecture: With any real number c in the interval [L,U] there exists for each natural number n>2 a natural number m and a prime number p so that n=floor(m^c)+p. L is the lower bound of c. U is the upper bound of c. L = 0.5781296526305628137240577579803... U = 1.3652123889685187297293386676777... (A135097). A135303 Conjecture relating to primes with more than one coach: The combined set of integers in the top rows of all coaches of these primes is composed of a permutation of the first q odd integers, where prime p = (4q-1) or (4q+1), (q>0). Example: As shown for 17, this prime has two coaches with the top rows [1} and [3, 7, 5]. 43 has three coaches with q = 11. The top rows are [1, 21, 11], [3, 5, 19], [7, 9, 17, 13, 15]. The comment of Sep 08 2012 in A216371 applies to primes with one coach, in which case "all coaches" is reduced to one and the set of q odd integers is in the top row of the coach. - _Gary W. Adamson_, Sep 10 2012 A135303 Conjecture [Carl Schick]: If 2*n+1 is prime, then these are the number of distinct cycles of f(k) = |(2*n+1) - 2*k| beginning at an odd number 0 < k < 2*n. - _Jonathan Skowera_, Aug 3 2013 A135414 Conjecture: A recursively built tree structure can be obtained from the sequence: A135422 Conjecture: limit n->infinity Phi^(Phi^(n-c))/a(n) = 1 where c is a constant = 3.47348961175710091.... A135499 Conjecture: this is a subset of A008593 (verified for the first 50 thousand terms). - _R. J. Mathar_, Feb 10 2008 A135543 Conjecture: a(n) is not equal A175079(n) - 1 for all n >= 1. [From _Jaroslav Krizek_, Feb 05 2010] A135571 Conjecture. This sequence is just the sequence of positive integers that are either square, twice a square, or triangular, but not both square and triangular (A001110). (This has been verified for n up to 100000.) A135659 Conjecture: All Mersenne Primes A000668 with only one exception number 3 are subset of this sequence A135953 Conjecture: All numbers in this sequence are products of two sums of two squares, e.g. 4181 = 37*113 = (1^2+6^2)(7^2+8^2), 1346269=557*2417 = (14^2+19^2)(4^2+49^2). A135954 Conjecture: All numbers in this sequence are products of three sums of two squares. A135955 Conjecture: All numbers in this sequence are products of four sums of two squares. A135956 Conjecture: all numbers in this sequence are product of 5 or more sum of two squares A136269 Conjecture: For n = 3k, a(n) = (3/4)*10^(4n/3)+5*10^(n/3-1)-1. A136404 Conjecture: square roots of the terms of this sequence are the same terms as A126098 A137264 Conjecture: The only digit that is repeated in the sequence is 0 except for n=2 and n=3 where 2 repeats. So 1 may be followed by 2 or 0; 2 may be followed by 1 or 0; 0 may be followed by 0 or 1 or 2. this has been confirmed for the first million prime gaps. A137328 Conjecture : Each prime number appears in this sequence at least once. A138051 Conjecture: 3^3^n+2 does not meet anti Fermat's conjecture. [From _Vincenzo Librandi_, Aug 01 2010] A138123 Conjecture: The Fibonacci sequence F obeys all the recurrences: A000045(n)=F(n)= L(p)*F(n-p)-(-1)^p*F(n-2p), any p>0, L=A000204. A138123 Conjecture: The Lucas sequence L also obeys all the recurrences: L(n)= L(p)*L(n-p)-(-1)^p*L(n-2p), any p>0, L=A000204. A138217 Conjecture: the sequence contains infinitely many composite numbers. A138227 Conjecture. This sequence is infinite. A138312 Conjectured to be equivalent to 'kappa' = limit_{n -> infty)((sum_(k = 1..n) mu^2(k)/phi(k)) - H_n), where mu(k) is the Mobius function, phi(k) is Euler's Totient and H_n is the n-th harmonic number. A138313 Conjectured to be equivalent to Mertens' constant B_3 less Euler's constant. B_3-gamma is given by sum_(i=1 to infty)log p_i/(p_i(p_i-1)), p_i is the i^th prime. A138572 Conjecture: the sequence is infinite. A138684 Conjecture. Let x(1)=1, x(2)=2), x(n)=2*x(n-1)+x(n-2), and let y(1)=1, y(2)=3, y(n)=y(n-1)+y(n-2). Then if n=x(k), a(n)=y(k); if x(k)<n<2*x(k), a(n)=a(n-x(k)); and if 2*x(k)<=n<x(k+1), a(n)=1. {This has been confirmed for n<500.) [John W. Layman, Nov 26 2011] A138788 Conjecture: All numbers in this sequence are primes. A138808 Conjecture: the row sums of the plane partitions A010766 are upper bounds. [From _R. J. Mathar_, Aug 06 2008] A138904 Conjecture: For binary expansions of length n, there are d(n) distinct values that will show up as symmetries, where d is the divisor function. The symmetry values will be precisely the divisors of n. A138904 Conjecture: For numbers whose binary expansion has length n which has proper divisors which are all coprime: There will be only one number of length n with n symmetries. That number is 2^n - 1. For each proper divisor d (excluding 1), you can generate all numbers of length n that have n/d symmetries like so: (2^0 + 2^d + 2^2d ... 2^(n-d)) * a, where 2^(d-1) <= a < (2^d) - 1. The rest of the expansions of length n will have only the trivial symmetry. A139043 Conjecture: 10^n(10^n+1)/2 - 1 -(10^n)^2/(2*log(10^n)-1) -> a(n) as n -> infinity. Here (10^n)^2/(2*log(10^n)-1) is also conjectured to -> sum of primes < 10^n and is a very good approximation for the sum of primes < 10^n. We know that k^2/(2log(k)-1) diverges as k -> infinity. So if we can prove this limit approaches the sum of the primes <= k, then this implies the sum of primes is infinite and therefore the number of primes is infinite. A139092 Conjecture: all prime divisors in A139089 are distinct A139250 Conjecture: Consider the rectangles in the sieve (including the squares). The area of each rectangle (A = b*c) and the edges (b and c) are powers of 2, but at least one of the edges (b or c) is <= 2. A139250 Conjectures: after 2^k stages, for k >= 2, and for m = 1 to k - 1, there are 4^(m-1) substructures of size s = k - m, where every substructure has 4*s rectangles. The total number of substructures is equal to (4^(k-1)-1)/3 = A002450(k-1). For example: If k = 5 (after 32 stages) we can see that: A139394 Conjecture: a(n) > 0, except n=1,2,3,6,24, when a(n)=0. A139437 Conjecture: the sequence is finite and a(13)=41 is the largest term. A139474 Conjecture of _Kenneth J Ramsey_ from May 16 2006 (see A001109): a(n) is divisible by 2^Prime[n]-1 if and only 2^Prime[n]-1 is Mersenne prime A139476 Conjecture: the squares appear in increasing order. A139490 Conjecture: This sequence is finite and complete (checked for range n<=200 and m<=500) A139544 Conjecture: these numbers do not occur in A139491. A139804 Conjecture: There are no composite numbers in this sequence. A140100 Conjecture: A140100(n) - A140098(n) = only 0 or 1 for n>=1; A140101 Conjecture: A140101(n) - A140099(n) = only 0 or 1 for n>=1. A140254 Conjectures relating to the Mobius sequence A008683: A140320 Conjecture. a(n)=2n*3^(n-1)+1. A140355 Conjecture: this sequence is finite. A140357 Conjecture: starting with any given n and any 1 <= a(n) <= n and applying the rule for the sequence produces a sequence which eventually joins this one. For example, starting with a(9)=5, the sequence continues 10,3,9,11,9, at which point it has joined. A140411 Conjecture 1,9, p. 4, of Goswick et al. "The squarefree numbers in question form a subset of Euler's numeri idonei [A000926], therefore at most one number can be absent from the list above. If such a number does exist, it must exceed 2 * 10^11 and if it is even the Generalized Riemann Hypothesis is false." A140462 Conjecture 1.2, p. 2 of Frohmader. A140608 Conjecture. The sequence is infinite. A140869 Conjecture: If h does not belong to the sequence, then 4*h+5 is prime. - _Vincenzo Librandi_, Nov 18 2012 A140999 Conjecture (false!): includes all members of A094348. A141229 Conjecture. The terms of the sequence have only one prime divisor; moreover, A141419 Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x) (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N]_(i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N]_(k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery Jan 20 2012 A141459 Conjecture: a(n) = denominator of integral_{0..1}(log(1-1/x)^n) dx - _Jean-François Alcover_, Feb 01 2013 A141492 Conjecture: This sequence converges to the number of primes < 10^n or Pi(10^n). A141779 Conjecture: All composite terms of A120292 are semiprime. A141781 Conjecture: All terms are semiprime. A143132 Conjecture: rightmost digit of terms is cyclic: (1, 6, 6, 6,...repeat). A143580 Conjecture (verified for the first 280000 entries): this is the characteristic function of A001969 and therefore a duplicate of A010059. [From _R. J. Mathar_, Sep 05 2008] A143586 Conjecture: if L(n)=A143586(n), then lim(L(n+1)/L(n)) = 3/2. For a more general conjecture see A143589. A143589 Conjecture (following Benoit Cloitre's conjecture at A111090): if L(n) is the number (assumed finite) of terms in row n of K, then L(n)*(2/3)^n approaches a constant. (L= A143590.) A143590 Conjecture (following Benoit Cloitre's conjecture at A111090): A143772 Conjecture: All even numbers are members and the only odd numbers which are members are 1 & 3. [From _Robert G. Wilson v_, Sep 08 2008] A144140 Conjecture: this sequence is finite and complete A144264 Conjecture: the sequence is infinite. A144326 Conjecture: 191, 1217, 1559 and 1901 are not in fact members of this sequence, noting that they are (4, 19) k-figurate numbers; 19 is a member of A138694. Determining whether a Mersenne prime exponent one greater than a (4, 19) k-figurate number exists is sufficient to determine whether these primes are members. A144437 Conjecture: a(n) is the separatix. See A045944. A144621 Conjecture: a(n+1) = determinant of 2^n x 2^n matrix M where M(i,j) = 1 + exponent of 2 in the factorization of i+j-1 = A001511(i+j-1), i,j>0, n>2. - _Ralf Stephan_, Sep 27 2013 A144652 Conjecture: If h does not belong to the sequence, then 4*h+1 is prime. (End) A144686 Conjectured to equal A144685. A145017 Conjecture. For every n>=1 there exist infinitely many primes p of the form 4k+1 for which for a(n)>1 we have sp-(floor(sqrt(sp)))^2 is not a full square for s=1,...,a(n)-1 while a(n)p-(floor(sqrt(a(n)p))^2 is a full square(see A145016(s=1) and A145022, 145023, A145047, A145048, A145149, A145050 correspondingly for s=2, s=5, s=10, s=13, s=17, s=26) [From _Vladimir Shevelev_, Sep 30 2008] A145047 Conjecture. The least positive integer s could take values only from A008784 (see for s=1,2,5,10 sequences A145016, A145022, A145023 and this sequence) A145236 Conjectures: 1) for n>=2 the sequence does not decrease; 2) for n>1 a(n) is odd; 3) a(n) could be equal to a(n+1) only for twins: p_(n+1)-p_n=2 (although there exist also twins for which a(n)<a(n+1)). A145376 Conjecture. For n>=2 the sequence is nondecreasing. A145824 Conjecture: Primes of this form are infinite. A145878 Conjectures: T(3+k,k)=3k+3, T(4+k,k)=(k+1)(k+28)/2, T(5+k,k)=(k+1)(3k+77), T(6+k,k)=(k+1)(k^2+110k+2982)/6, T(7+k,k)=(k+1)(3k^2+235k+7352)/2. A145996 Conjecture: This sequence is finite and complete. A146025 Conjectured to be complete. a(4), if it exists, is greater than 10^15. - _Charles R Greathouse IV_, Apr 06 2012 A151819 Conjecture: the only solutions for x!+y!=n^2 are x=0,0,0,1,1,1,2,4,4,4 and y=4,5,7,4,5,7,2,0,1,5 respectively. A152050 Conjecture: For all primes p <= n there is always a lower twin prime L less A152139 Conjecture that b(m,q) = b(m,4) for all q > 4, is now verified for m to and including 8. Also b(9,q)=b(9,4) for q=5,6,7,8 at least. [From _Paul Leopardi_, Jun 14 2009] A152458 Conjecture: This sequence is finite. A152466 Conjecture: this sequence will never have any numbers k where k+1 is prime. (All similar sequences that start with numbers less than 252 are known to have numbers k where k+1 is prime.) A152519 Conjecture: Number of points on 3 crtitical lines of EKG sequence is finite (and complete?). [From _Artur Jasinski_, Dec 08 2008] A152522 Conjecture (A. Granville, H. te Riele and J. van de Lune, 1989) Let, for even N, p=p(N) be the least prime such that N-p is prime as well. Then p(N)=O((log(N))^2log(log(N))). [From _Vladimir Shevelev_, Dec 08 2008] A152650 Conjecture : All roots of P(n,x) are real, hence negative. - Jean-François Alcover, Oct 10 2012 A152718 Conjecture: this is the inverse Motzkin transform of A054393. A153197 Conjecture: Given any sequence with Hankel transform of [1,1,1,...], performing alternate operations: binomial transform followed by INVERT transform, then binomial transform of the last result (repeat); or INVERT transform starting first, will converge upon A006789 and A153197 as a two sequence limit cycle. The conjecture can be extended to any Hankel transform (and their accompanying sequence set): analogous operations will converge upon a Bessel-type sequence and its binomial transform. A154293 Conjecture: even generalized pentagonal numbers (A193828) divided by 2. - Omar E. Pol, Aug 19 2011 A154800 Conjecture: There are only six 0's in this sequence. A155005 Conjecture: a(5+5n)=1125*10^n for n>0. [From _Robert G. Wilson v_, Jan 23 2009] A155151 Conjecture: Let p = prime number. If 2^p belongs to T(m, n) = 4*m*n + 2*m + 2*n + 2, then 2^p-1 is not a Mersenne prime. - _Vincenzo Librandi_, Dec 12 2012 A155190 Conjectured to be finite. The prime terms are in A065397. A155828 Conjecture: All terms a(n) are of the form 2^m - 1. This has been verified up to n = 1000. A155973 Conjecture: The number of 1's in this sequence is infinite. A156642 Conjecture. For n>=1, a(n)>0. This conjecture does not follow from the validity of the Goldbach binary conjecture because numbers of the form 4n+2, generally speaking, have also decompositions into sums of two primes of the form 4k+1. A156833 Conjecture: for n>1, a(n) = n iff n is prime. A156834 Conjecture: for n>1, a(n) = n iff n is prime. Companion to A156833. A156869 Conjecture: T(2n + m, n + m) = T(2n, n) ( = A156870(n) ) if and only if m >= 0. A156870 Conjecture: A156869(2n + m, n + m) = a(n) if and only if m >= 0. A156907 Conjectured to consist entirely of integers. A156908 Conjectured to consist entirely of integers. A157200 Conjecture: there are infinite numbers of such kind of primes. A157308 Conjecture: a(m) == 1 (mod 2) iff m is a power of 2 or m=0. [From _Paul D. Hanna_, Mar 17 2009] A157656 Conjecture: a(n)=A059100(n-1) holds only for all n<20 as well as n=22 and n=23. (Rustem Aidagulov) A157675 Conjectured to consist entirely of integers. A157751 Conjecture 1. If n>1 is even then F(n,x) has no real roots. A157751 Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r, A157751 Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010] A157830 Conjecture: An higher symmetry elliptical Invariant than E8 exists. The Golay C_24 to C_12 symmetry seems to fit the bill! Using the C_24 Golay enumeration polynomial and the C_12 Golay enumeration polynomial: s[x_]=(1 + 759*x^8 + 2576*x^12 + 759*x^16 + x^24)^3/((24 + 440*x^3 + 264*x^6 + x^12)^3*(1 + 264 x^6 + 440 x^9 + 24 x^12)^3) which is toral inversion symmetric: s[1/x]=s[x] which checks in Mathematica. A157996 Conjecture: for n > 1, a(n) = prime(n+5). [_Charles R Greathouse IV_, Mar 12, 2012] A158110 Conjecture: given q and m are nonnegative integers, then A158233 Conjecture: this contains only the numbers 0,3,4,6 (verified for the first 5000 terms). A158460 Conjecture: If the condition holds, prime(n-1) and prime(n) are twin primes of the form 10k+1 and 10k+3. Ie., the last digits are 1 and 3. A158509 Conjecture: If the condition holds, prime(n-1) and prime(n) are twin primes of the form 10k+1 and 10k+3. Ie., the last digits of the primes are 1 and 3. The 1 ending is evident in the entries here. A158529 Conjecture: If the condition holds, prime(n-1) and prime(n) are twin primes of the form 10k+9 and 10k1+1, i.e. the last digits of the twin prime pairs are 9 and 1. The 9 ending is evident in this sequence. The table of the first 101 terms was computed using Zak Seidov's table. A158979 Conjecture: a(n) exists for all n, i.e.the sequence is well-defined and infinite. A158979 Conjecture: a(n)-n = 1 for infinitely many n. A159629 Conjecture: For every m>2 there exists a minimum index N(m) such that the minimal increasing recursive sequence S_m(n) A159708 Conjecture:if n>4093 there is at least one pair p,q as define such that p+q=2*n A159843 Conjectured asymptotics (based on random matrix theory) on p. 378 of Cohen's book. A159885 Conjecture: a(n) exists for every n>=1. It is easy to see that this conjecture is equivalent to the well-known Collatz 3x+1 conjecture. A159907 Conjecture: with number 1, multiply-anti-perfect numbers m: m divides antisigma(m) = A024816(m). Sequence of fractions antisigma(m) / m: {0, 0, 10, 2157, 2337, 13101, 4455356, …}. - Jaroslav Krizek, Jul 21 2011 A160120 Conjecture: It appears that if n = 2^k, k>0, then, between the other polygons, there appears a new centered hexagon formed by three rhombuses with side length = 2^k/2 = n/2. A160120 Conjecture: Consider the perimeter of the structure. It appears that if n = 2^k, k>0, then the structure is a triangle-shaped polygon with A000225(k)*6 sides and a half toothpick in each vertice of the "triangle". A160120 Conjecture: It appears that if n = 2^k, k>0, then the ratio of areas between the Y-toothpick structure and the unitary triangle is equal to A006516(k)*6. A160217 Conjecture: Let m>3 belong to A003159. Define the sequence b(n) to be the minimal increasing sequence with b(1)=m and the property that b(n) and n are both in or both not in A003159. Then a(n)=b(n) for all n larger than some m-dependent minimum index. A160218 Conjecture 1: Primes occur in A160256 in increasing order. A160218 Conjecture 2: All primes occur in A160256. A160218 Conjecture 3: Except for A160256(4)=4, the least positive integer which does not occur in A160256 up to a given index is always a prime (and thus of the form A160256(a(k)) for some k). A160218 Conjecture 4: A160256(a(n)) is always the least positive integer which did not occur earlier in A160256. A160266 Conjecture. For every n>=1, there exists a finite value of a(n). It is easy to see that this conjecture is equivalent to the well known Collatz 3n+1 conjecture. A160268 Conjecture: For every k in the sequence, the number k^2 is in the sequence as well. A160648 Conjecture: a(n) counts unrestricted partitions of even numbers such that A161579 Conjecture: In every sequence of numbers n such that A010060(n)=A010060(n+k), for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. - _Vladimir Shevelev_, Jul 31 2009 A161580 Conjecture: In every sequence of numbers n such that A010060(n)+A010060(n+k)=1, for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [From _Vladimir Shevelev_, Jul 31 2009] A161622 Conjecture: Lim R(n) as n->oo = 1/2. See also more extensive comment entered with sequence of numerators. This conjecture implies Legendre's conjecture. A161681 Conjecture: The number of primes in x^3-y*2 is infinite. A161681 Conjecture: The number of duplicates for a given prime is finite. Then there is the other side - the primes that are not in the sequence 3, 5, 17, 29, 31, 37, 41, 43, 59, 73, 97, 101, 103, ... Looks like a lot of twin components here. Do these have an analytical form? Is there such a thing as a undecidable sequence? A161682 Conjecture: The sequence is infinite. A161817 Conjecture. In every sequence of numbers n such that A010060(n)=A010060(n+k) for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [From _Vladimir Shevelev_, Jul 31 2009] A161846 Conjecture: this sequence converges to 4. There are several "heuristic demonstrations" but no proofs. A161847 Conjecture: this sequence converges to 4. A161890 Conjecture. In every sequence of numbers n, such that A010060(n)=A010060(n+k), for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [From _Vladimir Shevelev_, Jul 31 2009] A161961 Conjecture: S(n) = Pi(n^2)/(n*Pi(n)) ~ 1/2. The sequence is highly oscillatory but for 51000<n<51000, the ratio is already between 0.46 and 0.48. A162174 Conjecture : primes classified by level are rarefying among prime numbers. A162175 Conjecture: primes classified by level are rarefying among prime numbers. A162180 Conjecture : the sequence is the union of an infinity subsets of the form : {1, 2}, {1, 5, 2, 3}, …{1, p,….,q}, ... where the number 1 is the first element of each subset. A162281 Conjecture: every composite number appears in this sequence. - _Max Alekseyev_ A162311 Conjecture. In every sequence of numbers n, such that A010060(n)=A010060(n+k), for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [From _Vladimir Shevelev_, Jul 31 2009] A162457 Conjecture: The number of terms in this sequence is infinite. While this may be true, there are large gaps with no occurrences of factor triangles. Between 700 and 2000 there are no numbers that form factor triangles. A162657 Conjecture: a(n) is never zero. Checking up to 1000000, the smallest number not found is 210; and a(210) = 26611200. A163138 Conjecture: if q is an integer, then A(x,q) is a power series in x with integer coefficients. A163801 Conjecture: Which features a certain structure (Comparable to A005206 or A135414). If the (below) following two constructs (C and D) are added on top of their ends (either marked with C or D) one will (if starting with one instance of D) receive the above tree (x marks a node): A163846 Conjecture: The graph G constructed above consists of exactly two trees: one containing 7 and all odd primes congruent to 2 modulo 3, and another one containing all primes congruent to 1 modulo 3 except 7. (End) A163873 Conjecture: This recursive structure exists for every sequence of the above mentioned family. The first node of D has always k+1 children nodes where the first one consists of a new copy of D, the second one consists of another node and then D. The remaining children nodes consist of another node and then C. Between the first and the second leaf is always space for k-1 nodes of construct C. Construct C crosses on its way from a(n) to n always exactly k pathes (the right ones from construct D). A163874 Conjecture: This features a certain structure (similar to the G-sequence A005206 or other sequences of this family: A163875 and A163873). If the (below) following two constructs (C and D) are added on top of their ends (either marked with C or D) one will (if starting with one instance of D) receive the above tree (x marks a node, o marks spaces for nodes that are not part of the construct but will be filled by the other construct): A163875 Conjecture: This features a certain structure (similar to the G-sequence A005206 or other sequences of this family: A163874 and A163873). If the (below) following two constructs (C and D) are added on top of their ends (either marked with C or D) one will (if starting with one instance of D) receive the above tree (x marks a node, o marks spaces for nodes that are not part of the construct but will be filled by the other construct): A163921 Conjecture: if F(x) = exp( Sum_{n>=1} L(n)*x^n/n ) is an integer series, A164014 Conjectured finite. Any further terms are greater than 10^8. A164754 Conjecture:Except for the first term, a(n)=4*n^2+13*n+8 [From _Vincenzo Librandi_, May 26 2010] A164848 Conjecture (checked for the first 3000 entries): periodic with a(n+24)=a(n). A164901 Conjecture: No two consecutive primes are absent from this sequence (other than 3 and 5). (Equivalently, if p < q < r are consecutive primes and q > 5 is not a term, then p and r are terms.). See A164980. [From _Rick L. Shepherd_, Sep 03 2009] A164980 Conjecture: Each term is 0 or 1. I have confirmed this for the first 3499 terms while calculating this b-file based on A164901's b-file. A165300 Conjecture. (1) If n>1, and a(n) begins and ends with 1, then a(n+1) is obtained by deleting the initial 1 of a(n); (2) If a(n) begins with 1 and ends with 2 then a(n+1) is obtained by adding a final 1 to a(n); (3) If a(n) begins with 2 and ends with 1 then a(n+1) is obtained by adding a final 2 to a(n); (4) If a(n) begins and ends with 2 then a(n+1) is obtained by adding an initial 1 to a(n). This has been confirmed through a(140), which has 71 digits (and should be fairly easy to prove). [From _John W. Layman_, Sep 22 2009] A165460 Conjecture: a(2n) = 2*A165605(2n) and a(2n+1) = (2/3)*A165605(2n+1). - Antti Karttunen, Oct 5 2009. If true, then implies also the truth of conjecture in A165462. A165462 Conjecture: These are all those terms of A165602 which = 1 modulo 3. If this is true, then A165461 gives also the positions of zeros in A165605. - Antti Karttunen, Oct 5 2009. A165463 Conjecture: a subset of A165603. See conjectures in A165460 and A165462. A165640 Conjecture. a(n) = Floor[4*(n+4)/5] - Floor[2*(n+4)/3]. A165659 Conjecture: with the exception of the second term, 2 <= a(n)/A165660(n) < 3 A165660 Conjecture: with the exception of the second term, 2 <= A165659(n)/a(n) < 3. A165886 Conjecture: a(n)/A141459(n+1) = A006519(n+1). A165940 Conjectured to consist entirely of integers. A165978 Conjecture: The same number will never occur twice in this sequence. - Lowell A166051 Conjecture: There are no more terms after 165. (Checked up to A016813(290511) = 1162045.) If this is true, then also 5, 13 and 37 are only 4k+1 primes in A080114. A166102 Conjecture: a(n)=3*A166103(n). (Checked for terms a(1)-a(92).) A166103 Conjecture: a(n)=A166102(n)/3. (Checked for terms a(1)-a(92).) A166168 Conjectured to consist entirely of integers. A166268 Conjecture: a(n) = A166270(n)/2, and also A165605(A005843(n)). A166269 Conjecture: a(n) = -(1/2) * A166271(n). A166270 Conjecture: a(n) = 2*A166268(n). A166271 Conjecture: a(n) = -2 * A166269(n). A166272 Conjecture: equal to 3*A166100(A166101(n))/A166102(n), i.e. the denominator of the quotient A166100(A166101(n))/A166102(n) in its reduced form is always 3. A166273 Conjecture: a(n) = -3 * A166269(n). A166407 Conjecture: the quotient A166406(i)/A005408(i) has denominator 3 when i is one of the terms of A166101, and it is integral in other cases. If true, then floor in the formula is unnecessary. A166483 Conjecture: sequences for 2-ads, 3-ads (composites of 3 primes), 4-ads, etc., converge to 1/2, 1/4, 1/8,..., respectively. A166678 Conjecture: k <= Pi((Sqrt[p_k!! ]+1)^2) - Pi(p_k!!) A166944 Conjecture. Every record of differences a(n)-a(n-1) more than 5 is greater of twin primes (A006512) A166945 Conjecture. Each term of the sequence is the greater of a pair of twin primes (A006512) A167230 Conjecture: All the non-zero entries in this triangle are Bell numbers (A000110). A167493 Conjectures. 1) For n>=2, every difference a(n)-a(n-1) is 1 or prime; 2)Every record of differences a(n)-a(n-1) more than 3 is greater of twin primes (A006512) A167494 Conjecture. All terms of the sequence are primes. A167495 Conjecture: each term >3 of the sequence is the greater member of a twin prime pair (A006512). A167925 Conjecture: a(6),a(7),a(8),a(9)=coefficient of x^2 in the Maclaurin expansion of 1/(n*x^2+n*x+1) for n=1,2,3,4; a(10),a(11),a(12),a(13),a(14)=-coefficient of x^3 in the Maclaurin expansion of 1/(n*x^2+n*x+1) for n=1,2,3,4,5; a(15),a(16),a(17),a(18),a(19),a(20)=coefficient of x^4 in the Maclaurin expansion of 1/(n*x^2+n*x+1) for n=1,2,3,4,5,6; a(21),a(22),a(23),a(24),a(25),a(26),a(27)=-coefficient of x^5 in the Maclaurin expansion of 1/(n*x^2+n*x+1) for n=1,2,3,4,5,6,7. Etc... [From Francesco Daddi, Aug 04 2011] A168253 Conjecture: a(n)>0 for all n. A168295 Conjecture: "Worpitzky forms" A168296 Conjecture: "Worpitzky forms" A168314 Conjectured convergent of a(n)/a(n-1) = (1+sqrt(2)). a(19)a(18) = A168317 Conjectured convergent of a(n)/a(n-1) = phi^2, 2.6180339... A168670 Conjecture: Numbers n such that 36*n^2+72*n+35 is not equal to p*(p+2), where p, p+2 are twin primes. A168671 Conjecture: Numbers n such that 36*n^2+72*n+35 is not equal to p*(p+2), where p, p+2 are twin primes. A168672 Conjecture: Numbers n such that 36*n^2+72*n+35 is not equal to p*(p+2), where p, p+2 are twin primes. A169597 Conjecture: Numbers n such that 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is not of the form p*(p+2), where p and p+2 are primes. A169598 Conjecture: Numbers n such that 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is not of the form p*(p+2), where p and p+2 are primes. A169599 Conjecture: Numbers n such that 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is not of the form p*(p+2), where p and p+2 are primes. A169600 Conjecture: Numbers n such that 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is not of the form p*(p+2), where p and p+2 are primes. A170941 Conjecture: a(n)/A000085(n) -> 1/e as n -> inf. In other words, the asymptotic proportion of involutions that contain no adjacent transpositions is conjecturally = 1/e. [David Callan, Nov 11 2012] A171141 Conjecture: Numbers n>6 such that 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is not of the form p*(p+2), where p and p+2 are primes. A171783 Conjecture: a(n) = 4 for all prime numbers >= 3 and 3 for all composites A171871 Conjecture: In general, in each column, the last 2^(R-1) values are the same as the first 2^(N-1) values from the corresponding row of A039754. - _Robert Munafo_, Dec 30 2009 A171949 Conjecture. For n>=2, a(n)=A217319(n-1). - _Vladimir Shevelev_, Mar 18, 2013 A172143 Conjecture: sequence consists of an infinite number of subsequences S(m,0) = A172241(n) = (1/18)[8^n-(-1)^n-9], m>0, S(m,n+1) = 4*S(m,n)+1. The first subsequences A172164 Conjecture : The terms are only 18,19,20,21 (From the first thousand turns, there are 2,3% of 18, 36,5% of 19, 46,2% of 20 and 15% of 21). No period found. Probably due to Pi transcendence. A172253 Conjecture (Artur Jasinski): This sequence is infinite. A173238 Conjectures: Given the infinite set of triangles with A000041 in every column shifted down 0, 1, 2,... n times, row sums of n-th triangle (where A173238 = A173263 Conjecture: This sequence is finite and complete (*Artur Jasinski*) A173643 Conjecture: Odd part of a(n) is of form [(1/6)*(8^m-(-1)^m-3)*4^k-1]/3, k,m>0. A173989 Conjecture: always follows the pattern A, A, A+1, A, where A is an odd number A174031 Conjecture: 1 appears infinitely often. A174161 Conjectures: 1)The sequence is a permutation of prime numbers; 2) k=k(n) runs all positive integers. A174162 Conjectures: 1)The sequence is a permutation of prime numbers; 2) k=k(n) runs all positive integers. A174216 Conjecture. a(n+1)/a(n) tends to 2. A174238 Conjecture: this is the inverse Moebius transform of A006519, multiplicative, and as such the Dirichlet g.f. is the Dirichlet g.f. of A006519 multiplied by zeta(s). - R. J. Mathar, Feb 06 2011 A174470 Conjectured to consist entirely of integers. A174531 Conjectures: 1) all terms of the sequence are integers; 2) P_n(0)=(floor((n-1)/2))!*4^{floor((n-1)/2)}; 3)if n is even, then the coefficients of P_n do not exceed the corresponding coefficients of P_(n-1) and the equality holds only for the last ones; 4) all coefficients of P_n, except of the last one, are multiple of n iff n is prime. A174846 Conjecture: limit |a(n)/n!|^(-1/n) = r exists and is finite with r<0.8... A175402 Conjecture: max(a(n)) = 4. A175403 Conjecture: sequence is finite. A175424 Conjecture: max(a(n)) = 4. A175425 Conjecture: sequence a(n) is finite for n = natural number, a(n) = 25 for n = infinity. See A175424. A175498 Conjecture: the lexicographically first permutation of {1,2,...n} for which differences of adjacent numbers are all distinct (cf. A131529) has, for n-->infinity, this sequence as its prefix. [_Joerg Arndt_, May 27 2012] A175522 Conjecture: if there exist infinitely many A-deficient numbers and infinitely many A-abundant numbers, then there exist infinitely many A-perfect numbers. A175931 Conjecture: Sequence is finite and complete with last term 2103. A175938 Conjecture: The permutation has exactly 12 fixed points: 2, 6, 9, 12, 15, 18, 21, 24, 29, 40, 46, 49. A176187 Conjecture: For every n>=0 the sequence {f^(m)(n)} is eventually periodic. Further calculations may show if there exist terms different from 1 and 3. A176494 Conjecture. For every odd prime p, the sequence {|2^n-p|} contains at least one prime. The records of the sequence appear in points 2,10,31,68,341,... and equal to 3, 4, 47, 791,... Note that up to now the value a(341) is not known. _Charles R Greathouse IV_ calculated the following two values: a(815)=16464, a(591)=58091 and noted that a(341)is much larger.-Private communication at 27.05.10. [From _Vladimir Shevelev_, May 29 2010] A176545 Conjecture: If p=2*n^2+14*n+5 then 2^p-1 is not prime. A176547 Conjecture: If p=2*n^2+6*n+1 is prime then 2^p-1 is not prime. A176549 Conjecture: 2^a(n)-1 is not prime; in other words, these primes (except 31) are included in A054723. A176716 Conjectures: 4 divides a(n) iff n == 0 mod 3. 2 divides a(n) iff n == (1, 2) mod 3; as the highest power of 2; for n>1. A176827 Conjecture: {a(n,q), n>=0} are integers when q is a positive integer. A176827 Conjecture: a(n,q) = 1 (mod q) for n>=0 when q is a positive integer. [From _Paul D. Hanna_, Apr 29 2010] A176948 Conjecture: For every n>=4, except of n=6, there exists a n-gonal number N which is not k-gonal for 3<=k<n. A177025 Conjecture. Every positive integer appears in the sequence. A177390 Conjecture: for all integer k, there exists an integer triangle J such that the g.f. of row n of J^(k*n) = (k*n^2 + y)^n for n>=0. A177430 Conjecture: the sequence {a(n)^(m/2), n>=1} forms a logarithmic derivative of an integer sequence only when m is a nonnegative even integer. A177431 Conjecture: the series exp(Sum_{n>=1} a(n)^m*x^n/n) consists entirely of integer coefficients only when m is a nonnegative even integer. A177960 Conjecture: If for every m>=2, to consider triangle of m-nomial coefficients mod 2 converted to decimal, then the sequence lists terms of A001317 which are not in the union of other sequences for m=3 (A038184), 4 (A177882), 5, 6,... A178127 Conjecture: For all n > 570, more than 1/4 of the twin prime pairs < n are both Ramanujan primes. A178253 Conjecture: a(n) exists for all n >= 2. A178375 Conjecture: Records of the sequence are consecutive primes. A178473 Conjecture. If p is prime of the form 4*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 4*k+3, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p. A178482 Conjecture: this is the sequence of numbers k for which f(k) is an integer, where f(x) is the change-of-base function defined at A214969 using b=phi and c=b^2. [Clark Kimberling, Oct 17 2012] A178612 Conjecture: No perfect squares in this sequence (in spite of all numbers being congruent to 0 or 1 mod 4). A178790 Conjecture: the number a(n)=n^{-1}*sum_{k=0}^{n-1}(2k+1)A_k is always an integer. A178790 Conjecture: If p=5,7 (mod 8) is a prime then sum_{k=0}^{p-1}A_k=0 (mod p^2); if p=1,3 (mod 8) is a prime greater than 3 and p=x^2+2y^2 with x,y integers then sum_{k=0}^{p-1}A_k=4x^2-2p (mod p^2). A178916 Conjecture: for every n>1 and 1<=k<=n there is a prime in the interval [k*n!+1,k*n!+3*n*log(n)^2]. [From _Robert Gerbicz_, Dec 28 2010] A178942 Conjecture: a(n) > 0 for all n. A179004 Conjecture: away from boundaries a sliding 101 pattern would fill with a limiting 2/3 1's A179210 Conjecture: a(n) > 0 for all n >= 1. A179238 Conjecture relating palindromic continued fractions to Tan angle A and 2A: A179240 Conjecture: a(n) > 0 for all n. A179256 Conjecture: a(n)>0 for n >= 1. A179328 Conjecture: a(n) > 0 for all n. A179479 Conjecture. a(n)>0 for all n. A179538 Conjecture. All terms are primes except for a finite set of squares of primes. A179771 Conjecture: the last zero (107th) occurs at n=166, after which only ones occur. A179776 Conjecture: the last zero (108th) occurs at n=163, after which only ones occur. A179831 Conjecture: the last zero (119th) occurs at n=164, after which only ones occur. A179873 Conjecture: a(n) = nondecreasing sequence of odd numbers. A179874 Conjecture: a(n) = sequence of odd numbers. A179902 Conjectured mod 2 parity: (1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1,...) A180248 Conjecture: n is a Carmichael number (A002997) if and only if n is a term of A180248 and all r-values of n are even. A180460 Conjecture: Given any even integer E not a power of 2 (see A078426) there exists a positive integer N such that for all n>=N the equation sigma(m)=E^n has at least one solution for m. A180493 Conjecture: This sequence is finite and all its terms are present. (End) A180654 Conjecture 1: this sequence consists entirely of integers. A180654 Conjecture 2: the sequence of coefficients of [x^n/n! ] in the series: A180654 Conjecture 3: the sequence of coefficients of [x^n/n! ] in the series: A180710 Conjecture: sigma(n) | a(n) for n>=1. A181070 Conjecture: this sequence consists entirely of integers. A181074 Conjecture: this sequence consists entirely of integers. A181076 Conjecture: this sequence consists entirely of integers. A181078 Conjecture: this sequence consists entirely of integers. A181080 Conjecture: this sequence consists entirely of integers. A181082 Conjecture: this sequence consists entirely of integers. A181084 Conjecture: this sequence consists entirely of integers. A181316 Conjecture: exp( Sum_{n>=1} (q-1)*((q^n-1)/(q-1))^(n-1)*x^n/n ) is an integer series for all integer q>1. A181546 Conjecture: Given F(n,L) = Sum_{k=0..[n/2]} C(n-k,k)^L, then A181547 Conjecture: Given F(n,L) = Sum_{k=0..[n/2]} C(n-k,k)^L, then A181595 Conjecture 1: For every k>=1, there exist infinitely many entries m for which (2^k)|m and sigma(m)-2^k = 2*m. A181595 Conjecture 2. All entries are even. [Proved to be false, see below. (Ed.)] A181595 Conjecture 3. If the suitable (according to the definition) divisor d of an entry is not a power of 2, then it is not suitable divisor for any other entry. A181595 Conjecture 4. If a suitable divisor for an even entry is odd, then it is a Mersenne prime (A000043). A181630 Conjectured row sums = A023610: (1, 3, 7, 15, 30, 58,...). A181674 Conjecture: Average order is Theta(log log n). -_Charles R Greathouse IV_, May 25, 2011 A182084 Conjectured to be minimal number of nodes in any non-bipartite regular graph of degree n, diameter 2 and girth 4. A182085 Conjecture: N = (30k+7)*(60k+13)*(150k+31) is a Carmichael number if (but not only if) 30k+7, 60k+13 and 150k+31 are all three prime numbers. A182088 Conjecture: The number C = (30n-29)*(60n-59)*(90n-89)*(180n-179) is a Carmichael number if (but not only if) 30n-29, 60n-59, 90n-89 and 180n-179 are all four prime numbers. A182089 Conjecture: C = (330k+7)*(660k+13)*(990k+19)*(1980k+37) is a Carmichael number if 330k+7, 660k+13, 990k+19 and 1980k+37 are all four prime numbers. A182090 Conjecture: Any Carmichael number C divisible by 11 and not divisible by 3 can be written in one of the following four forms: C = 990k+11; C = 990k+11^2; C = 990k+11*41 or C = 990k+11*71, where k is a natural number. A182126 Conjecture: for x>10^9, the most frequent value in a(n), n=0...x, has form 120*k. A182132 Conjecture: The number C = (30n-7)*(90n-23)*(300n-79) is a Carmichael number if (but not only if) 30n-7, 90n-23 and 300n-79 are all three prime numbers. A182133 Conjecture: The number C = (30n+13)*(90n+37)*(150n+61) is a Carmichael number if (but not only if) 30n+13, 90n+37 and 150n+61 are all three prime numbers. A182145 Conjecture: among values less than 60 the most frequent value is 30 (based on frequencies for n < 2126795281, i.e. primes less than 50176000000). A182151 Conjecture: Any Carmichael number C divisible by 31 can be written in one of the following three forms: C = 2790n+31; C = 2790n+31^2 or C = 2790n+31*61, where n is natural. A182167 Conjecture: the most frequent values are 0,1,2,3,5,8,13,21,34,... i.e Fibonacci numbers. A182206 Conjecture: Any Carmichael number C divisible by 37 and 73 can be written as C = 37*73*(18n+91), where n is natural; checked for the first 22 Carmichael numbers divisible by 37 and 73. A182207 Conjecture: Any Carmichael number C divisible by p and 2p-1 (where p and 2p-1 are prime numbers) can be written as C = p*(2p-1)*(n*(2p-2)+p). A182208 Conjecture: Any Carmichael number C divisible by 7 can be written in one of two ways: (1) C=7*(6m+1)*(6n+1), where m and n are natural numbers or (2) C=7*(6m-1)*(6n-1), where m and n are natural numbers. In other words, there aren’t Carmichael numbers divisible by 7 of the form C=7*(6m+1)*(6n-1). Checked for the first 27 Carmichael numbers divisible by 7. Note: a Carmichael number with more than 3 prime divisors can be written (sometimes) in both ways: 41041 = 7*11*13*41 = 7*13*451 (form 1) = 7*11*533 = 7*41*143 (form 2). A182334 Conjecture: a(n) = 35*a(n-3) - 35*a(n-6) + a(n-9). - _Charles R Greathouse IV_, Apr 25 2012 A182336 Conjecture: for n > 96, n + 1 <= a(n) <= 9n/8 + 1. - _Charles R Greathouse IV_, Apr 25 2012 A182372 Conjecture: number of partitions of n into two sorts of parts such that no two successive parts are the same sort (as the order of sorts in a run of identical parts is immaterial, there can be at most two parts of same size), see example. [_Joerg Arndt_, Jun 01 2013] A182381 Conjecture: Any Carmichael number C divisible by 17 and 29 can be written as C = 952*n + 561; checked up to the Carmichael number 105823343809. A182416 Conjecture: N = (60k+13)*(180k+37)*(300k+61) is a Carmichael number if (but not only if) 60k+13, 180k+37 and 300k+61 are all three prime numbers. A182418 Conjecture: As n -> infinity, a(n) -> infinity. A182457 Conjecture: a(n) contains infinitely many zeros. A182458 Conjecture: a(n) contains infinitely many zeros. A182481 Conjecture: a(n)>0; equivalently, for every n, the arithmetic progression {6*k*n-1} contains infinitely many lessers of twin primes (A001359). A182484 Conjecture : a(n) = Product_{k=1..p-1} k^(2k-p-1) is a perfect square if, and only if p = prime(n). A182508 Conjecture: sequence contains infinitely many zeros and ones, but no terms of the form 2^k-1, k>1. A182510 Conjectures: the sequence contains 8 zeros, and more positive terms than negative. A182511 Conjectures: Sequence contains infinitely many terms such that a(n)=n. Infinitely many 1's. A182515 Conjecture: Any Carmichael number C divisible by 23 and 67 can be written as C = 23*67*(66n+23). A182532 Conjecture (1): Any Carmichael number C divisible by 7 and 17 of the form 10k+1 can be written as C = 7*17*(120n+119). We got for the tested numbers the following values for n: 478, 1174, 34348, 38350, 82660, 107110, 158658, 162028, 176746, 179278, 262918, 313570, 377788, 563974, 710950. A182532 Conjecture (2): Any Carmichael number C divisible by 7 and 17 of the form 10k+5 can be written as C = 7*17*(120n+95). We got for the tested numbers the following values for n: 57, 3291, 28965, 357567, 451893. A182532 Conjecture (3): Any Carmichael number C divisible by 7 and 17 of the form 10k+3 can be written as C = 7*17*(120n+47). We got for the tested numbers the following values for n: 111835, 181157, 222733. A182532 Conjecture (4): Any Carmichael number C divisible by 7 and 17 of the form 10k+9 can be written as C = 7*17*(120n+71). We got for the tested number the following value for n: 435446. A182560 Conjecture: sequence contains infinitely many zeros. A182580 Conjecture: for k > 12, a(2k) = a(2k-1) + 1. A182868 Conjecture: the 4 quadrisections of (the family) A064038, A160050, A176126, A178242 (see A181407) come from square spiral. A182963 Conjectured to consist entirely of integers. A183128 Conjecture: this sequence consists entirely of integers. A183129 Conjecture: this sequence consists entirely of integers. A183161 Conjecture: a(n) is never congruent to 3 modulo 4; see A218622. - _Paul D. Hanna_, Nov 03 2012 A183241 Conjectured to consist entirely of integers. A184674 Conjecture: a(n) = A014616(n-1). - R. J. Mathar, Jan 29 2011 A185046 Conjecture: The odd numbers different from a cube are of the form m = p^3 - 2q where p and q are prime numbers. A185061 Conjecture: Only odd numbers occur after 2. - _Antti Karttunen_, Jun 13 2013 A185104 Conjecture: next terms are 1111111111111111111 and 11111111111111111111111. A185150 Conjecture: a(n)>0 for all n>0. A185510 Conjecture: Every row contains infinitely many primes. A185636 Conjecture: a(n)>0 for all n>1. A185645 Conjecture: a(n) > 0 for all n > 0. In general, for any n consecutive primes p_k,...,p_{k+n-1}, there always exists a permutation q_k,...,q_{k+n-1} of p_k,...,p_{k+n-1} with q_{k+n-1} = p_{k+n-1} such that the n-1 numbers |q_k-q_{k+1}|, |q_{k+1}-q_{k+2}|,...,|q_{k+n-2}-q_{k+n-1}| are pairwise distinct. (In the case k = 2, this implies that a(n) > 0.) A186080 Conjecture: If k^4 is a palindrome > 0, then k begins and ends with digit 1, all other digits of k being 0. A186447 Conjecture: All finite binary words appear in the sequence infinitely many times. A187011 Conjecture: (-1)^(n-1)*a(n) > 0 for all n > 1, and |a(n)|^{1/n} tends to the infinity. A187563 Conjectures: there are only 11 primes (5,11,13,31,37,53,61,73,79,97,127) for which k does not exist and there is always at least a pair of twin primes between prime(n) and prime(n)^2. A187757 Conjecture: a(n)>0 for all n>1. A187758 Conjecture: a(n)>0 for all n>4. A187759 Conjecture: If n>200 is not among 211, 226, 541, 701, then a(n)>0. A187762 Conjecture: a(n) exists for all n. (This may follow from the Green-Tao theorem, Dirichlet's theorem and Dickson's conjecture.) A187785 Conjecture: a(n)>0 for all n>48624 not equal to 76106. A187808 Conjecture: a(n)>0 for all n>1. Moreover, if n>5 is different from 9, 191, 329, 641, 711, 979, then 2k-3, 2k+3, n(n-k)-1, n(n+k)-1 are all prime for some 0<k<n. A187809 Conjecture. In the supposition that there are infinitely many twin primes, every term is 2 or in A001359 (lesser of twin primes). A187810 Conjecture. In the supposition that there are infinitely many twin primes, every term beginning the fourth is 2 or in A001359 (lesser of twin primes). A187812 Conjecture. In the supposition that there are infinitely many twin primes, every term beginning with the sixth is 2 or in A001359 (lesser of twin primes). A187866 Conjectures: There are only 11 primes such that k does not exist: (5,11,13,31,37,53,61,73,79,97,127), same as A183563. There are only 20 primes such that k(n)=A187563(n):(2,3,7,17,19,23,41,47,59,89,103,149,167,173,179,191,277,353,433,727) A187941 Conjecture: a(n) = 2^n only if n is prime or if n = 1. A188542 Conjecture: the sequence contains no 0's. Verified up to n = 100000. A188550 Conjecture: if the definition is changed so that k runs through values 2, 3, ..., floor((n-2)/2) then, beginning with n=6, the sequence remains without changes. - Vladimir Shevelev, Apr 10 2011. A188644 Conjecture by C Hohn: Given function f(x, y)=((x^2+y)^(1/2)+x)^2; and constant k=f(x, y); then for all integers x>=1 and y=[+-]1, k may be irrational, but ((k^n)+(k^(-n)))/2 always produces integer sequences; y=-1 results shown here; y=1 results are A188645 A188645 Conjecture by C Hohn: Given function f(x, y)=((x^2+y)^(1/2)+x)^2; and constant k=f(x, y); then for all integers x>=1 and y=[+-]1, k may be irrational, but ((k^n)+(k^(-n)))/2 always produces integer sequences; y=1 results shown here; y=-1 results are A188644 A188646 Conjecture by C Hohn: Given function f(x, y)=((x^2+y)^(1/2)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=-1 results shown here; y=1 results are A188647 A188647 Conjecture by C Hohn: Given function f(x, y)=((x^2+y)^(1/2)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=1 results shown here; y=-1 results are A188646 A188672 Conjecture: a(n) = 0, iff n = 1, 2, 3, 4, 5, 6, 9, 12, 13, 14, 24, 56. A189202 Conjecture: For all n>=2, a(n)>0. A190213 Conjecture: All odd entries are also Mersenne exponents (A000043): primes n such that 2^n-1 is prime. A190638 Conjecture: All terms have the form 12*k+5. A190898 Conjecture: a(n)<(n+1)^2 for all n>0. (See also A185150.) A191004 Conjecture: a(n)>0 for all n>5. A191257 Conjecture: A191257=(A067368)/2. A191303 Conjecture. For every k>=2, there exists a polynomial of degree k taking only k-digit half-palindrome values. A191363 Conjecture: a(1)=3 is the only odd member of the sequence. A191363 Conjecture: All elements of the sequence are of the above form derived from Fermat primes. A191837 Conjecture: For all a(n), a(n)-1 can be found in A014092 (numbers not the sum of two primes), and a(n)+1 can be found in A007921. (numbers not the difference of two primes). - _J. Stauduhar_, Aug 28 2012 A192023 Conjecture: for n>2, A192023(n-2) is the number of 2x2 matrices with all terms in {1,2,...,n} and determinant 2n. [From Clark Kimberling, Mar 31 2012] A192056 Conjecture: a(n)>0 for all n>2. Moreover, if n>2 is not among 12, 18, 105, 522, then there is 0<k<n/2 such that A192062 Conjecture re relation of A192062 to the sequence of primes: T(2*n,j) = A(n,j)*T(n,j) where A(n,j) is from the square array A191971. There, A(3*n,j) = A(n,j)*B(n,j) where B(n,j) are integers. It appears further that B(5*n,j)=B(n,j)*C(n,j); C(7*n,j)= C(n,j)*D(n,j); D(11*n,j) = D(n,j)*E(n,j); E(13*n,j) = E(n,j)*F(n,j) and F(17*n,j) = F(n,j)*G(n,j) where C(n,j), D(n,j) etc. are all integers. My conjecture is that this property continues indefinitely and follows the sequence of primes. A192101 Conjecture: a(n) = (4^n)/3 + O(1). A192189 Conjecture. For every sufficiently large greater q of twin primes, the sequence contains infinite increasing sequence {s_n} of semiprimes beginning with 2*(q-2), such that (s_n)'=s_(n-1). A192205 Conjecture: limit a(n)^(1/n) = 16*sqrt(3)/9 = 3.079201..., which is substantiated by the observation that the sums of the coefficients squared in (1-x-x^2+x^3)^n equals binomial(4n,n) (cf. A005810). A192229 Conjecture: the terms starting with 11 are prime. A192416 Conjecture: a(n) is also a "totally formed" sequence. A192931 Conjecture: The sequence is finite. A192934 Conjecture: total number of digits of a(n) is always 2*n-1. A193262 Conjecture: There exists n_0, such that, for n>n_0, a(n)>0. A193358 Conjecture: Every positive integer is in this sequence. A193358 Conjecture: Odd numbers are not repeated. A193418 Conjecture: log(A005960-a(n)) ~ (log(2)*(2*n-11)). A193437 Conjecture: a(n) is divisible by p^floor(n/p) for prime p == 1 (mod 3). A193438 Conjecture: a(n) is divisible by p^floor(n/p) for prime p == 1 (mod 4). A193447 Conjecture: { k : k >= 7 and f(k) in A000027 } = { p in A000040 : p>=7 }. A193550 Conjecture: given o.g.f. A(x), then sqrt( (A(x) - 1)/x ) is an integer series. A193555 Conjectured next terms are A193555(17)/A193556(17)=19720/121, A193555(18)/A193556(18)=5002/25. A193628 Conjecture: the sequence seems to converge asymptotically to 113. A193918 Conjecture: r(n+1,x) is the product of the following two polynomials whose coefficients are Fibonacci numbers: A194001 Conjecture: r(n+1,x) is the product of the following two polynomials whose coefficients are Fibonacci numbers: A194184 Conjecture: The sequence is unbounded. A194184 Conjecture: The lower asymptotic density of nonzero terms is >0. A194186 Conjecture: The sequence is unbounded. A194217 Conjecture: Asymptotic density of nonzero terms is 3/4. A194659 Conjecture 1. The sequence is unbounded. A194659 Conjecture 2. The asymptotic density of nonzero terms is 2/(e^2+1). A194688 Conjecture. This sequence is self-generated according to the following rule: start with {4} at step 0, then extend by steps, appending {2,2,4} at step n if a(n)=4 or appending {4} if a(n)=2. (This has been verified for several thousand terms.) To illustrate, the first few steps of this process give {4}->{4,2,2,4}, since a(1)=4, ->{4,2,2,4,4}, since a(2)=2, ->{4,2,2,4,4,4}, since a(3)=2,->{{4,2,2,4,4,4,2,2,4}, since a(4)=4, etc. Equivalently, it appears that {a(n)} is the fixed-point of the morphism 2->4, 4->422, starting with 4. A194696 Conjecture: number of toothpicks or D-toothpicks added to the structure of A194440 at stage 2^k+n, if k tends to infinity. It appears that rows of A194441 when written as a triangle converge to this sequence. A194697 Conjecture: number of toothpicks or D-toothpicks added to the structure of A194442 at stage 2^k+n, if k tends to infinity. It appears that rows of A194443 when written as a triangle converge to this sequence. A195095 Conjecture: Mobius transform of A127804. - R. J. Mathar, Sep 14 2011 A195101 Conjecture that for every allowed backstep m > 1 there is some number n such that A005245(n) = A005245(m)+A005245(n-m) and such that for every representation as a product n = u*v with u, v >= 2 or every 1 < = k < m, we have A005245(n) < A005245(u)+A005245(v) and A005245(n) < A005245(k) + A005245(n-k). A195325 Conjecture. For n >= 2, a(n) is a lesser of twin primes (A001359). This implies the twin prime conjecture. [_Vladimir Shevelev_, Sep 15 2011] A195465 Conjecture: a(n)>0 for n>1. This conjecture is equivalent to the conjecture that all terms of A195325 are lessers of twin primes. A195507 Conjecture. {a(n)} satisfies the following recurrence. Write a(n)/n in lowest terms as num/d. Then a(n+1)=d*a(n) - (d-1)*n + 1. llustration: a(4)=10 and a(4)/4=5/2 in lowest terms. Then a(5)=2*10-1*4+1=17. (This has been verified up to a(23)=232792583.) A195538 Conjecture: a(n) = +35*a(n-2) -35*a(n-4) +a(n-6) with bisections A098602 and A076218. - R. J. Mathar, Sep 21 2011 A195539 Conjecture: a(n) = +34*a(n-2) -a(n-4) with a(2n+1) = A000129(4n+4), a(2n)=A046176(n+1). - R. J. Mathar, Sep 21 2011 A195608 Conjectures:1) there are only 3 different first differences 1,4,6; 2) the sequence contains either isolated series of terms, e.g., {1},{7},{25},{31},..., or series of 3 consecutive integers, e.g., {11,12,13}, {19,20,21}, etc.; 3)the first terms m of every series satisfy the condition A(m+1)-A(m-1)=5, where A(n)=A001969(n). A195609 Conjectures: 1) there are only 3 different first differences 1,4,6; 2) the sequence contains either isolated series of terms, e.g., {9},{23},{33},{39},..., or series of 3 consecutive integers, e.g., {3,4,5}, {15,16,17}, etc.; 3)the first terms m of every series satisfy the condition A(m+1)-A(m-1)=5, where A(n)=A000069(n). A195673 Conjecture: The polynomials p(n,x) = sum_{k=0..n} T(n,k)*x^(n-k) based on this simple recurrence for other initial constant values of T(0,0)=p and T(1,0)=q are related to the S-polynomials of A053119: p(n,x,p+1,q+1)-p(n,x,p,q) = S(n,x). A195825 Conjecture: if k is odd then column k contains (k+1)/2 plateaus whose levels are the first (k+1)/2 terms of A210843 and whose lengths are k+1, k-1, k-3, k-5,... 2. Otherwise, if k is even then column k contains k/2 plateaus whose levels are the first k/2 terms of A210843 and whose lengths are k+1, k-1, k-3, k-5,... 3. The sequence A210843 gives the levels of the plateaus of column k, when k -> infinity. For the visualization of the plateaus see the graph of a column, for example see the graph of A210964. - _Omar E. Pol_, Jun 21 2012 A195871 Conjecture: In the supposition that there are infinitely many twin primes, for n>=5 all terms are in A001359 (lesser of twin primes). A195986 Conjecture: a(n) = 1 if A090740 = 1, else a(n) = A090740(n)+1. A196226 Conjectures. (1) If m>=14 is a term of this sequence, then sigma(2,m) is congruent to 5 + m/2 modulo m; (2) If m>=22 is a term of this sequence, then sigma(3,m) is congruent to 9 + m/2 modulo m; If m>=38 is a term of this sequence, then sigma(4,m) is congruent to 17 + m/2 modulo m. (sigma(k,m) denotes the sum of the k-th powers of the divisors of m.) A196660 Conjecture: for every n their exists k < n (apart from a(1)) such that k*n+(n-1) is prime. See A034693. A196697 Conjecture: all elements of this sequence are greater than 0. A196697 Conjecture tested holds up to n=9594. Further test is still running A196698 Conjecture: all elements of this sequence is greater than 0. A196698 Conjecture tested hold up to n=7399. Further test is still running A196778 Conjecture: all elements of this sequence is greater than 0. A196778 Conjecture tested hold up to n=2355. Further test is still running A196779 Conjecture: a(n) has finite value when a>4 A196934 Conjecture: Any prime number greater than 11 (p) can be the center term of arithmetic progressions prime chain p-6k, p, p+6k, while k>0. A196935 Conjecture: a(n) > 0 for all n >= 5. A197082 Conjecture: every proper divisor of a member of this sequence divides infinitely many numbers in the sequence. A197170 Conjecture: For every n such a quadratic field with minimum k exists. A197171 Conjecture: This sequence is infinite. A197521 Conjecture: the constant here, 3.52133782..., is 3 plus the constant in A197383, the latter being the least t>0 satisfying sin(pi*t/6)=(sin pi*t/3)^2. A197702 Conjecture. Let SO(k) be the sum of the first k odd positive integers. Then a(n)=k if n=SO(k). Otherwise, choose k so that SO(k-1)<n<SO(k). Then if SO(k)-n=4, a(n)=k+2, else if SO(k)-n is odd then a(n)=k+1 else a(n)=k. (This has been verified for n up to 200.) A198195 Conjecture. In the supposition that there are infinitely many twin primes, every term beginning with the 20th is 2 or in A001359 (lesser of twin primes). The sequence is unbounded. A198319 Conjecture. In the supposition that there are infinitely many twin primes, every term beginning with the 12th is 2 or in A001359 (lesser of twin primes). A198469 Conjecture. In the supposition that there are infinitely many twin primes, every term beginning with the 15th is 2 or in A001359 (lesser of twin primes). A198472 Conjecture: If b(1)>=4 is an integer and b(k+1)=a(b(k)) for k=1,2,3,..., then b(n)=4 for some n>0. A199331 Conjecture: Only the integers 1, 2, 3, 4, 5, 6, 7, 9, 11, 17, 22, 33 (A072966) cannot be partitioned into a set of two semiprimes. A199800 Conjecture: a(n)>0 for all n>11. A199920 Conjecture: a(n)>0 for all n>11. A200918 Conjecture (*Artur Jasinski*): If another infinite sequences with good Hall's examples occurred, it would have to contain primes from this sequence as constant divisors of the whole sequence, because parts of Danilov's infinite sequence (A200216, A200217, A200218) contain divisors of (3^A014127(1) - 3)/(A014127(1)^2). A200982 Conjecture: The asymptotic density is zero. A201060 Conjecture: a(n) is repunit for n >= 16. A201227 Conjecture: No more infinite families of solutions Mordell curves with extension sqrt(5) than A201227 and A200218. A201250 Conjecture: the sequence is infinite. The graph of A(i)-B(i) provides a visual suggestion but no hint of a proof. In general there is no reason to suppose A_n - B_n = 0. A201267 Conjecture: limsup n-->infty a(n)=infinity. More precisely we claim that log(a(n))/log(n) is bounded and doesn't converge to zero (see related link). Does a(n)=2 infinitely many times or does it exist M>=2 such that a(n)<=M infinitely many times? A201268 Conjecture (*Artur Jasinski*): Extremal points are k-th successive points with maximal coordinates x. A201278 Conjecture (Jasiński) Numbers in this sequence can be multiplicative combinations of: primes congruent to 1 or 2 modulo 4 (A002313) also Pythagorean primes (A002144) + number 2 also norms of Gaussian primes A055025. A201828 Conjecture. For n>=13, every a(n) is the lesser of a pair of cousin primes p and p+4, cf. A023200. Note that it is only conjectured that there are infinitely many pairs of cousin primes. A202137 Conjecture: there are arbitrarily long runs of consecutive integers. A202150 Conjecture: this sequence consists of all odd terms in A202146; the g.f. of A202146 is 1/(1-x) + Sum_{n>=1} x^n/(1-x) * Product_{k=1..n} (1 - x^k) / (1 - x^(2*k+1)), which by the conjecture has an odd coefficient of x^m iff m = 3*n*(n+1) for n>=0. The conjecture holds for at least the initial 30300 terms of A202146. A202211 Conjecture. Problem a) is answered in affirmative, while Problem b) is answered in negative. A202272 Conjecture: sequence contains only A202275 Conjectures: Max a(n) = 15 for n = 195, 403, 434. For n >= 687, a(n) < 0. A202277 Conjecture: sequence is finite with 13 terms. A202319 Conjecture: For every natural number k there are infinitely many semiprime pairs sp and sp' both sandwiched between semiprimes such that sp' - sp = 4k. A202941 Conjecture. If p is an odd prime, then the ((p-1)/2))-th row contains two 1's and (p-3)/2 numbers multiple of p. A202943 Conjecture: the characteristic function of a(n) (mod 2) equals (1+x)*(1+x^2)*(1+x^8) * Sum_{n>0} x^(32*A000695(n)), where A000695 is the sums of distinct powers of 4. A202996 Conjecture: a(n) > 0 for n > 6. A202996 Conjecture: With Sq=sum of q for n=1 to N and Sp=sum of p(n) for n=1 to N, lim sup Sq/Sp = 0. A203182 Conjecture: the sequence is infinite. A203278 Conjecture. If p is odd prime, then a((p-1)/2)==2 (mod p) A203484 Conjecture. If p is prime of the form 3*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 3*k+2, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p. A203498 Conjecture. If prime p==1 (mod 3), then a((p-1)/3)==2 (mod p); if prime p==2 mod 3, then a(2*p-1)/3)==2 (mod p). A203509 Conjecture. If prime p==1 (mod 4), then a((p-1)/4)==2 (mod p); if prime p==3 (mod 4), then a((p-1)/2)==2 (mod p). A204065 Conjecture: For any n > 0 not among 1, 21, 326, 341, 626, we have a(n) < sqrt(n)*log(n). If n > 626 is not equal to 971, then n+k and n+k^2 are both prime for some 0< k < sqrt(n)*log(n). Also, n+k^2 is prime for some 0< k <= sqrt(n) if n > 43181. A204770 Conjecture: density of zero terms slowly approaches 1. A205301 Conjectured lower bound to be increasing for increasing n. A205321 Conjecture : there is at least one k for each n. A205322 Conjecture: there is always at least one k>=0 unless n=3. A205649 Conjecture: The sequence is unbounded. A205901 Conjecture : this sequence is infinite, because it is conjectured that the sequence A002496 is infinite, although this has never been proved. A206328 Conjecture: this sequence is infinite. A206709 Conjecture: The number of primes of the form b^2 + 1 and less than n, is asymptotic to 3*n / 4*log(n). A206911 Conjecture: the difference sequence of A206911 consists of 2s and 3s, and the ratio (number of 3s)/(number of 2s) tends to a number between 3.5 and 3.6. A206912 Conjecture: the difference sequence of A206912 consists of 1s and 2s; see Comments at A206911. A207080 Conjecture: phi(a(n)) divides (a(n)-1)^(n+1). A207480 Conjecture: a(n) > 0 for all n (cf. A062234). A207820 Conjecture. For n>=3, every a(n) is the lesser of a pair of cousin primes p and p+4, cf. A023200. A207969 Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k>=0. A207970 Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k>=0. A208034 Conjectures: For all positive integers k, A208060 Conjecture: limit a(n)^(1/n) = 2*L where L = 2.200161058099... is the geometric mean of Luroth expansions, where log(L) = Sum_{n>=1} log(n)/(n*(n+1)) = 0.7885305659115... (cf. A085361). A208243 Conjecture: a(n)>0 for all n=3,4,... A208244 Conjecture: a(n)>0 for all n>0. A208246 Conjecture: a(n)>0 for all n>8. A208249 Conjecture: a(n)>0 for all n>8. A208460 Conjecture: one of the divisors of T(n,k) is also the k-th divisor of n. In a diagram of the structure of divisors of the natural numbers (see link) the mentioned divisors of the elements of row n are located on a straight line to 45 degrees from the vertical straight line that contains the divisors of n, therefore the divisors of n are predictable. A208548 Conjecture: a(n)>0 for all n (cf. A062234, A207480). A209236 Conjecture: a(n) always exists. In other words, there are infinitely many quintuples (m-2, m-1, m, m+1, m+2) with m-1 and m+1 both prime and m-2, m, m+2 all practical. A209253 Conjecture: a(n)>0 for all n>1. A209254 Conjecture: a(n)>0 for all n>1. A209260 Conjecture 1. 2*n+1 is in the set S_n = {union of first n rows of A209260} if and only if 2*n+1 is composite. (Verified for all n <= 10^6 by Charles R Greathouse IV.) This conjecture has been proved (see [Jeffery]). A209260 Conjecture 2. Let N be an integer, N>0, and let D_N be the set of all positive integral divisors of N. Let B_N be the set of all integral solutions of d + (h-1)/2 for which d in D_N, with h=N/d and N=d*h (some solutions will not be integers and so are not in B_N). Then N appears in row r = min(B_N) of A209260. Also, N appears in row b of A141419, for each b in B_N. - _L. Edson Jeffery_, Feb 13 2013 A209272 Conjecture 1: sequence is unbounded. A209272 Conjecture 2: (a(1)a(2)...a(n))^(1/n) seems to converge to 1.(7)... a limit different from Khintchine's constant (see A002210). A209312 Conjecture: a(n)>0 for all n>2. A209315 Conjecture: a(n)>0 for all n>8. A209320 Conjecture: a(n)>0 for all n>2. A209332 Conjecture: for even n, a(n) > 0. A209554 Conjecture (A209544 and this sequence correspond to cases k=2 and k=3 respectively). A209860 Conjecture: for every a(n), also A054429(a(n)) is in the sequence. Conversely, if i is not in the sequence, then neither is A054429(i). A209861 Conjecture: For all n, A209861(A054429(n)) = A054429(A209861(n)), i.e. A054429 acts as an homomorphism (automorphism) of the cyclic group generated by this permutation. This implies also a weaker conjecture given in A209860. A209862 Conjecture: For all n, a(A054429(n)) = A054429(a(n)), i.e. A054429 acts as a homomorphism (automorphism) of the cyclic group generated by this permutation. This implies also a weaker conjecture given in A209860. A209934 Conjecture: The sequence b_k repeats indefinitely (Example: for n=3, b_k = 9, 8, 8, 8, 8, 8, .... It looks like b_k is 8 for all k > 1) A209998 Conjecture: penultimate term in row n is A199923(n). A210144 Conjecture: all the terms are primes and a(n)<n^2 for all n > 1. A210184 Conjecture: a(n)/p_n > 1/2. A210186 Conjecture: all the terms are primes and a(n) < n^2 for all n > 1. A210444 Conjecture: a(n)>0 for all n>911. A210445 Conjecture: a(n)<n for all n>1, and a(n)<n/2 for all n>47. A210452 Conjecture: a(n)>0 for all n>4. A210456 Conjecture: All entries >1 are divisible by 3. A210456 Conjecture: a(3^n-1)=3^(2*n+1)-3, a(3^n)=3^(2*n+1)+3^(n+1)+3 - Fred W. Helenius (fredh(AT)ix.netcom.com), posting to MathFun, Feb 21 2013 A210465 Conjectures: (1) If q is the nearest prime>a(n), then q-a(n)=4 or 6 and both of these cases occur infinitely many times. (2) If q-a(n)=4 then q is the lesser of twin primes. A210467 Conjectures: (1) If q is the nearest prime > a(n), then q-a(n) = 2 or 6 and both of these cases occur infinitely many times. (2) If q-a(n) = 2, then also q is lesser of a pair of cousin primes q and q+4, see A023200. A210475 Conjecture: for n >= 12, every a(n) is the lesser of a pair of cousin primes p and p+4, (see A023200). A210476 Conjecture: every a(n), except for a(7) = 191, is the lesser of a pair of cousin primes p and p+4, (see A023200). A210479 Conjecture: The sequence a(n)^(1/n) (n=9,10,...) is strictly decreasing to the limit 1. Also, if {b(n)-1,b(n),b(n)+1} is the n-th sandwich of the second kind, then the sequence b(n)^(1/n) (n=1,2,3,...) is strictly decreasing to the limit 1. A210480 Conjecture: a(n)>0 for all n>3. A210505 Conjecture. For every odd prime p there exists infinite sequence of n for which 2*n+p, 4*n+p, ..., 2*(p-1)*n+p are primes. A210528 Conjecture: a(n)>0 for all n>0. Moreover, for any positive integers m and n, if m is greater than one or n is not among 17, 181, 211, 251, 313, 337, then 2n can be written as p+q with p, q and p^{3m}+q^{3m} all practical. A210529 Conjecture: a(n) is defined for all positive integers n. A210531 Conjecture: a(n)>0 for all n>0. Moreover, if n>0 is different from 74, 138, 166, 542, then n+k^3 is practical for some 0<=k<=sqrt(n)*log(n); if n is not equal to 102, then n+k and n+k^3 are both practical for some k=0,...,n-1. A210533 Conjecture: a(n)>0 for all n>2. Moreover, for each m=2,3,4,... any sufficiently large even integer can be written as x+y (x,y>0) with x-1 and x+1 both prime, and x and x^m+y^m both practical. A210537 Conjectures: (1) the sequence contains infinitely many "twins" when such differences equal 6; (2) lim a(n)/prime(n)=1 as n goes to infinity. A210642 Conjecture: a(n) is a prime not exceeding 2n with the only exceptions a(4)=4 and a(6)=9. A210650 Conjecture: for n>1 a(n) is always < n. A210681 Conjecture: a(n)>0 for all n>4. A210721 Conjecture: a(n) < prime(n)^2*(Pi^(-2Pi)+log((prime(n)+1)/prime(n))) A210722 Conjecture: a(n)>0 except for n = 1,...,8, 10, 520, 689, 740. A210826 Conjecture: this is a multiplicative sequence with Dirichlet g.f. zeta(3s)/zeta(s) and inverse Mobius transform in A010057. - _R. J. Mathar_, Mar 31 2012 A211163 Conjecture: sequence of denominators is A141459. A211165 Conjecture: a(n)>0 for all n>5. A211190 Conjecture: a(n)>0 for all n>8. Moreover, for positive integers a<=b<=c, all integers n>=3(a+b+c) with n-a-b-c even can be written as a*p+b*q+c*r with p,q,r terms of A210479, if and only if (a,b,c) is among the following 6 triples: (1,2,3), (1,2,4), (1,2,8), (1,2,9), (1,3,5), (1,3,8). A211204 Conjecture: The sequence contains infinitely many "twins" with a(n)-a(n-1)=6. A211376 Conjecture: a(n) is defined for all integers n > 2. A211376 Conjecture confirmed true up to n = 300000, no exceptions. A211869 Conjecture: A triangular array made from the values of those digits in the palindromes defined by this sequence might be a modification of the Pascal triangle. A211995 Conjectured to be always positive for n > 2. A212223 Conjecture: the sequence never terminates. A212275 Conjecture: a(n)>0. If the conjecture is true, then there exist infinitely many primes of the form m^2+1. A212279 Conjecture: a(n) = A082073(n)^2 + 1 for all n > 159. - _Charles R Greathouse IV_, May 13 2012 A212346 Conjecture stated in Formula holds through a(35). A212392 Conjecture: n divides a(n); see A212391. A212425 Conjecture: (2*n-1) divides a(n); see A212426. A212487 Conjecture: there is always one such k for n > 0. A212488 Conjecture: There is always one such k if n>1. A212652 Conjecture 1. Let D(n) be the set of all positive integral divisors of n. Let B_n be the set of all integral solutions of d + (h-1)/2 for which d in D(n), with h=n/d and n=d*h (some solutions will not be integers and so are not in B_n). Then a(n) = min(M : M in B_n). A212652 Conjecture 2. If Conjecture 1 is true, then n appears in row A212652(n) of A209260. A212843 Conjecture: only Carmichael numbers of the form 10n+1 can have prime divisors of the form 10k+1 (but not all Carmichael numbers of the form 10n+1 have prime divisors of the form 10k+1). A212843 Conjecture: all Carmichael numbers C (not only with three prime divisors) of the form 10n+1 that have only prime divisors of the form 10k+1 can be written as C = (30a+1)*(30b+1)*(30c+1), C = (30a+11)*(30b+11)*(30c+11), or C = (30a+1)*(30b+11)*(30c+11). In other words, there are no numbers of the form C = (30a+1)*(30b+1)*(30c+11). A212844 Conjecture: every integer k >= 0 appears in a(n) at least once. A212882 Conjecture: for any odd number p we have an infinite number of Carmichael numbers of the form n*(2*n - 1)*(p*n - p + 1)*(2*p*n - 2*p + 1). A212916 Conjecture: generally (for tableaux with height <= k), a(n) ~ k^n/Pi^(k/2) * (k/n)^(k*(k-1)/4) * prod(j=1..k,Gamma(j/2)); set k=10 for this sequence. - _Vaclav Kotesovec_, Sep 12 2013 A212920 Conjecture: If m*126 + n = 1729, m*126 > n, then exists a series with infinite many Carmichael terms of the form C mod m*234 = n. A212920 Conjecture: If m*234 + n = 1729, m*234 > n, then exists a series with infinite many Carmichael terms of the form C mod m*234 = n. A212920 Conjecture: If m*342 + n = 1729, m*342 > n, then exists a series with infinite many Carmichael terms of the form C mod m*342 = n. A212920 Conjecture: For any prime factor of a Carmichael number C1 exists a series with infinite many Carmichael terms C2 formed this way: C2 mod m*18*d = n, where m*18*d + n = C1, where d is the prime factor of C1 and m, n are natural numbers, m*18*d < n. A213047 Conjecture. On every interval [1,N] the numbers m for which A212990(m) less than or equal to Prime(m) are in majority. A213187 Conjecture: If b(1)>2 is an integer, and b(k+1)=a(b(k)) for k=1,2,3,..., then b(n)=4 for some n>0. A213202 Conjecture: a(n)>0 for all n>2. A213257 Conjecture. The positive integers that are not in this sequence are given by the positions of 2 in the fixed-point of the morphism 0->01, 1->02, 2->03, 3->01 (see A191255). (This has been confirmed for over 5000 terms of A213257.) To illustrate, the fixed-point of the indicated morphism is {0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,1,0,1,0,2,0,...} and 2 occurs at positions {4,12,20,...}, integers that are missing in A213257. A213258 Conjecture. The terms of this sequence are given by the position s of 2 in the fixed-point of the morphism 0->01, 1->02, 2->03, 3->01 (see A191255). (This has been confirmed for over 5000 terms of A213257.) To illustrate, the fixed-point of the indicated morphism is {0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,1,0,1,0,2,0,...} and 2 occurs at positions {4,12,20,...}, integers in this sequence but missing from A213257. A213325 Conjecture: a(n)>0 for all n>8. A213381 Conjectures: Every k>=0 appears in a(n) infinitely many times. Every integer k>=0 appears in a(n) at least once. Indices of zeros: 2^(x+2)-2, x>=0. A213507 Conjecture: a(n) is not divisible by {2,5,13,29,31,37,41,43,53,71,97}. A213536 Conjecture: Record differences a(n)-a(n-1) (A213537) are a strict subset of the smaller of cousin primes (A023200). (Cousin primes differ by 4.) A213536 Conjecture: Record differences are an infinite sequence. It is widely believed there are infinitely many cousin primes. (Similarly, by Dickson's conjecture and second Hardy-Littlewood conjecture, there are infinitely many pairs of (not necessarily consecutive) primes (p,p+2k) for each natural number k.) A213536 Conjecture: The following pattern makes sequences for every (necessarily even) difference (slight change for 2). For difference d, p is first prime >d that is the smaller of a prime pair (p,p+d). a(1)=2p and a(n)=gcd(n+p-2,a(n-1)) for even n, else gcd(n+p-2-d,a(n-1)). A213648 Conjecture: the fixed points of this sequence are in A000057. A213710 Conjecture: A179016(a(n))= 2^n for all n apart from n=2. This is true if all powers of 2 except 2 itself occur in A179016 as in that case they must occur at positions given by this sequence. A213859 Conjectures: ** A213883 Conjecture: there is always at least one (k,j) solution for each n. A213980 Conjecture: a(n) exists for every n>=2. A214016 Conjecture: There are infinite many Poulet numbers of the form 7200*n^2 + 8820*n + 2701. A214017 Conjecture: There are infinite many Poulet numbers of the form 144*n^2 + 222*n + 85. A214028 Conjecture: A175181(n)/A214027(n)= a(n). This says that the zeros appear somewhat uniformly in a period. The second zero in a period is exactly where n divides the first Lucas number, so this relationship is not really surprising. A214074 Conjecture: a(n) = 3 if and only if n = 330*k-209 for some k>=1. A214305 Conjecture: For any biggest prime factor of a Poulet number p1 with two prime factors exists a series with infinite many Poulet numbers p2 formed this way: p2 mod (p1 - d) = d, where d is the biggest prime factor of p1. Note: it can be seen that the Poulet numbers divisible by 73 bigger than 2701 (7957, 10585, 15841, 31609 etc.) can be written as 1314*n + 73 as well as 2628*m + 73. A214305 Conjecture: Any Poulet number p2 divisible by d can be written as (p1 - d)*n + d, where n is natural, if exists a smaller Poulet number p1 with two prime factors divisible by d. A214305 Conjecture: For any biggest prime factor of a Poulet number p1 exists a series with infinite many Poulet numbers p2 formed this way: p2 mod (p1 - d) = d, where d is the biggest prime factor of p1. A214342 Conjecture: no terms bigger than 35 after a(8)=40. A214415 Conjecture: the sequence is infinite. A214444 Conjecture: for some k and x, a(n)=A214451(n+x) for n>k. A214449 Conjecture: this sequence results from A118168 by deleting the first two 1s. A214495 Conjecture : there is always one such k(n) for each n>0. A214496 Conjecture : there is always one such k for each n>0. A214497 Conjecture : there is always one such k for each n>0. A214498 Conjecture : there is always one such k for each n>0. A214560 Conjecture: for every x>=0 there is i such that a(n)>x for n>i. A214619 Conjecture: there is an X such that among integers bigger than X more than 50% are in the sequence. A214620 Conjecture: there is X such that among integers bigger than X more than 50% are in the sequence. A214621 Conjecture: there is X such that among integers bigger than X more than 50% are in the sequence. A214625 Conjecture: a(n) exists for every n >= 2. A214724 Conjecture: p | a(n) for n>=p when p is a prime of the form m^2+1 (A002496). A214758 Conjecture: A Carmichael number C1 can be written as C1 = (C2 + C3)/2, where C2 and C3 are also Carmichael numbers, only if both C1 and C3 are divisible by C2. A214784 Conjecture: this sequence is the set of n in A214783(n) for which n-4=A214783(n) are all primes. A214834 Conjecture tested true up to n=1000000000. In case the conjecture is not true, zero could be used to represent the missing entries. A214841 Conjecture: a(n)>0 for all n>4. A214852 Conjecture: the sequence is infinite. A214853 Conjecture: the sequence is finite. A214873 Conjecture: for n >= 5, GCD(A000032(a(n)), A000225(a(n))) = 2*a(n) + 1. A214882 Conjecture: For any odd m and for any k, the density of n such that a(n) == k (mod m) is 1/m. A214883 Conjecture: For any m coprime to 5 and for any k, the density of n such that a(n) == k (mod m) is 1/m. A214900 Conjecture: a(n)!=0, that is, all numbers are sums of three squares and one fourth power. A214922 Conjecture: 23 is the only number not in this sequence. A214923 Conjecture: a(n) contains infinitely many positive and infinitely many negative terms. A214970 Conjecture: L(2k-1) and L(2k)+1 are terms of the sequence, for all positive integers k, where L=A000032 (Lucas numbers). A214971 Conjecture: L(2k-1) and L(2k)+1 are terms of this sequence for all positive integers k, where L=A000032 (Lucas numbers) A214971 Conjecture by Peter Moses (Oct. 19, 2012): If D is the difference sequence, then D-3 is the infinite Fibonacci word A096270. If so, then A214971 can be generated as in Program 3 of the Mathematica section. A215010 Conjecture: hexagons are the only simple convex polygons with equal interior angles with such property, due to the fact that cos(pi/3) = 1/2. A215056 Conjecture: If a vector (x1,...,x50) is a zero to a signature S, then it is only also a zero for S' and no other signature T. {S' is the complement signature of S - all the signs are reversed, so if S=(+,-,-,+,-) then S' is (-,+,+,-,+)}. A215174 Conjectures. 1) For every n, a(n)>0; 2) for n>=3, a(n)==0 (mod 10). A215343 Conjecture: there are infinite many Poulet numbers that can be written as (n + 1)*p^2 - n*p, where n is natural, n > 0, and p is another Poulet number. A215343 Conjecture: For any Poulet number p there are infinite many Poulet numbers that can be written as (n + 1)*p^2 - n*p, where n is natural, n > 0. A215461 Conjecture: a(n) is never zero for n > 5, verified to 10^9. A215462 Conjecture: a(n) is never zero for n > 19, verified to 10^9. A215462 Conjecture: a(n) > A215461(n) > A002372(n) for sufficiently large n. A215470 Conjecture: the sequence is infinite. - _Alex Ratushnyak_, Sep 19 2012 A215642 Conjecture: a(243)=34613 is the last term. A215658 Conjecture: Not the same sequence as A192579, which is finite. A215672 Conjecture: Any Poulet number P with three or more prime divisors has at least one prime divisor q for that can be written as P = q*((r + 1)*q - r), where r is a natural number. A215719 Conjecture: The terms of any feasible prime gap triple {a,b,c} to form a quadruple of consequent primes are sums of terms of three consequent sub-sequences of the infinite integer sequence with period (4,2,4,2,4,6,2,6). By this token all possible sequences of quadruples of consequent primes can be generated, including those already in OEIS. A215895 Conjecture: only 130633 primes are not in the sequence: 2, 3, ..., 94532497. A215926 Conjecture: a(n) is 1, 3, or a power of 2. A215926 Conjecture: The first occurence of 2^m happens at A014210(m). A215944 Conjecture: The only Fermat pseudoprimes to base 2 divisible by a smaller Fermat pseudoprime to base 2 that can’t be written as p*((m + 1)*p - m)*((n + 1)*p - n), where m, n, p are natural numbers, are multiples of 5461 and can be written as 5461*(42*k - 13). A215944 Conjecture is checked for the numbers from the sequence above and for the first 15 Poulet numbers with four prime factors. A216094 Conjecture: a(n) ~ prime(n). A216151 Conjecture: a(n) > A057194(n) for all n > 1. A216170 Conjecture: For any m natural, m > 1, there exist a series with infinite many Fermat pseudoprimes to base 2, P, formed this way: P = (n^m + m*n)/(m+1). A216196 Conjecture: This is the maximum for a Xoox-free grid. A216217 Conjecture: there is only one zero term: a(5) = 0. A216231 Conjecture: There exist arbitrary long chains of consecutive prime terms. A216240 Conjecture. There exists x_0 such that for every x>=x_0, the number of a(n)<=x is more than pi(x). A216265 Conjecture: a(n) > 0 for n > 13. A216266 Conjecture: a(n)>0 for n>23. A216275 Conjecture. lim a(n+1)/a(n)=phi as n goes to infinity (phi=golden ratio). A216276 Conjecture: For any Fermat pseudoprime to base 2, p1, there exist infinite many Fermat pseudoprimes to base 2, p2, formed this way: p2 = (p1^n + n*p1)/(n+1), where n natural, n > 1. A216276 Conjecture: For any Carmichael number, c1, there exist infinite many Carmichael numbers, c2, formed this way: c2 = (c1^n + n*c1)/(n + 1), where n natural, n > 1. Note that, in the sequence above, from Fermat pseudoprimes to base 2 that are also Carmichael numbers (1729, 8911, 10585, 41041, 278545, 449065) were obtained too Carmichael numbers. A216368 Conjecture: T(n,n) = A000081(n) for n>=1. It would be nice to have a proof (or a disproof if the conjecture is wrong). A216371 Conjecture relating to odd integers: iff an integer is in the set A216371 and is either of the form 4q - 1 or 4q + 1, (q>0); then the top row of its coach (Cf. A003558) is comprised of a permutation of the first q odd integers. Examples: 11 is of the form 4q - 1, q = 3; with the top row of its coach [1, 5, 3]. 13 is of the form 4q + 1, q = 3; so has a coach of [1, 3, 5]. 37 is of the form 4q + 1, q = 9; so has a coach with the top row comprised of a permutation of the first 9 odd integers: [1, 9, 7, 15, 11, 13, 3, 17, 5]. - _Gary W. Adamson_, Sep 08 2012 A216442 Conjecture: Unlike A004050, which has a limited set of integers expressible in more that one way, this set has no such limit. A216468 Conjecture: only 312722 primes are not in the sequence: 2, 3, ..., 198702899. A216495 Conjecture: only 5 primes are not in the sequence: 2, 3, 5, 13, 37. A216496 Conjecture: a(25)=557 is the last term. A216497 Conjecture: only 653 primes are not in the sequence: 2, 3, ..., 100291. A216498 Conjecture: only 9198 primes are not in the sequence: 2, 3, ..., 2521081. A216590 Conjecture: only 5254157 primes are not in the sequence: 2, 3, ..., 5082095279. A216590 Conjecture: for any k>0 there exists p0 such that for any prime p>p0 there exists a k-term arithmetic progression of primes with p at the end. A216835 Conjecture. lim a(n+1)/a(n)=phi as n goes to infinity (phi=golden ratio). A216944 Conjecture: Any prime >= 11 can be written this way. A216971 Conjecture: every entry in row n is divisible by n. - _Jon Perry_, Sep 21 2012 A216994 Conjecture: Every century has a representation in the sequence. A217016 Conjecture [D. Broadhurst]: a(n) is prime(n)-smooth. A217317 Conjecture: a(n) > 0 for n > 4765516. A217317 Conjecture checked up to n = 5 * 10^10. - _Charles R Greathouse IV_, Mar 21 2013 A217319 Conjecture. For n>=1, a(n)=A171949(n+1). A217396 Conjecture: This sequence is the complete set of k in N such that 2^k - 2 is a) twice/1 a triangular number (see A076046), b) thrice/2 a tetrahedral number (see A217482), or c) both. No other such k in N exists to at least 1.41*10^1505 as per computer check by Charles R Greathouse IV on Physics Forums (Nov 2010). A217448 Conjecture: a(n) exists for all n. A217482 Conjecture: There are no other tetrahedral numbers (Tetra_n = A000292) > 84 such that 3/2*Tetra_n + 2 is a power of 2. This is true to at least 1.41*10^1505 as per computer check by Charles R Greathouse IV on Physics Forums (Nov 2010). A217494 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217495 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217496 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217497 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217498 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217499 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217500 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217501 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217564 Conjecture: this sequence is unbounded, as implied by Dickson's conjecture. - _Charles R Greathouse IV_, Oct 09 2012 A217564 Conjecture: 0 appears infinitely often. - _Jon Perry_, Oct 10 2012 A217606 Conjecture: every prime of the form 12k+1 is a member. A217620 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217621 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A217703 Conjectures: (i) S_n(x) is irreducible over the field of rational numbers for every n=1,2,3,... A217721 Conjecture: a(n) > 0 for n > 1. A217721 Conjecture checked up to n = 2^28 - 1. A217738 Conjectures: the sequence is infinite; all terms are divisible by 12. A217785 Conjecture: For each n=2,3,... there are infinitely many primes of the form 1+2*s+...+n*s^{n-1}, where s is a positive integer; moreover, we have a(n)<12*n^2. A217788 Conjecture: For any integers n >= m > 0, there are infinitely many positive integers s > p_n such that the number sum_{k=m}^n p_k*s^{n-k} (i.e., [p_m,...,p_n] in base s) is prime; morever the smallest such an integer s (denoted by s(m,n)) does not exceed (n+1)*(m+n+1). A217835 Conjecture: For any Fermat pseudoprime p to base 2 there are infinitely many Fermat psudoprimes to base 2 equal to p^2*n - p*n + p, where n is natural. A217835 Conjecture: For any Fermat pseudoprime p to base 2 and any k natural, k > 0, there are infinitely many Fermat psudoprimes to base 2 equal to p^2*n - p*n + p^k, where n is natural. A217853 Conjecture: There are infinitely many Fermat pseudoprimes to base 3 of the form (3^(4*k + 2) - 1)/8, where k is a natural number. A217864 Conjecture: a(n) is unbounded. A217864 Conjecture: a(n)=0 infinitely often. A217866 Conjecture: the last time n appears is always before the first time n+2 appears. A217876 Conjecture: The asymptotic distribution of adjacent transpositions in involutions is Poisson with mean 1. A217976 Conjecture: the last time n appears is always before the first time n+2 appears. A218002 Conjecture: a(n) = number of degree-n permutations of prime order. A218010 Conjecture: There is no absolute Fermat pseudoprime m for which n = (5*m - 1)/24 is a natural number (checked for the first 300 Carmichael numbers; if true, then the formula is a criterion to separate pseudoprimes at least from a subset of primes, because there are 37 primes m from the first 300 primes for which n = (5*m - 1)/24 is a natural number). A218086 Conjecture: This sequence is complete. A218275 Conjectures. 1) a(n) > 0; 2) a(n)/n is between 2 and 3 or between the smaller and larger member of a twin prime pair. A218279 Conjecture: a(n)>0 for all n. A218333 Conjecture: all a(n) >= 2. A218465 Conjecture: Let n be any positive integer. Then a(n) exists, moreover there are infinitely many integers b > 2n+1 such that [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime. Also, the polynomial S_n(x) = sum_{k=0}^n (2k+1)*x^{n-k} is irreducible modulo some prime p < (n+1)(n+2), and the Galois group of S_n(x) over the field of rational numbers is isomorphic to the symmetric group S_n. A218468 Conjectured to be a permutation of the positive integers, which is the case if and only if an infinite number of terms have a digit "1". A218468 Conjecture: For nonnegative integer x, 2*10^x appears at index 2*10^x (see A226179). _Hans Havermann_, May 29 2013 A218483 Conjecture: If a Fermat pseudoprime to base 2 can be written as 8*p*n + p^2, where n is an integer number and p one of it’s prime factors, than can be written this way for any of it’s prime factors. Checked for all pseudoprimes from the sequence above. A218483 Conjecture: If a Fermat pseudoprime to base 2 with two prime factors can be written as 8*p1*n + p1^2, where n is a natural number and p1 one of it’s two prime factors, than can be written too as 8*p2*(-n) + p2^2, where p2 is the other prime factor. Checked for 4033 = 37*109(n = 9), 4369 = 17*257(n = 30), 4681 = 31*151(n = 15), 8321 = 53*157(n = 13), 18721 = 97*193(n = 12), 23377 = 97*241(n = 18), 31417 = 89*353(n = 33), 31609 = 73*433 (n = 45), 65281 = 97*673(n = 72), 85489 = 53*1613 (n = 195). A218483 Conjecture: If a Fermat pseudoprime to base 2 can’t be written as 8*p*n + p^2, where n is an integer number and p one of it’s prime factors, than can’t be written this way for any of it’s prime factors. Checked for the following pseudoprimes: 341, 645, 1387, 2047, 2701, 2821, 3277, 4371, 5461, 7957, 10261, 13741, 13747, 13981, 14491, 15709, 19951, 29341, 31621, 42799, 49141, 49981, 55245, 60701, 60787, 63973, 65077, 68101, 72885, 80581, 83333. A218585 Conjecture: a(n)>0 for all n>1 with the only exception n=8. A218621 Conjecture. For n, p, v, j natural numbers, the conditions on a(n) seem to be the following: A218622 Conjecture: a(n) never equals 3. A218654 Conjecture: a(n)>0 for all n=2,3,4,... A218656 Conjecture: a(n)>0 for all n=1,2,3,... A218754 Conjecture: a(n)>0 for all n=1188,1189,.... A218755 Conjecture: all first differences 36, 288, 180, 180,... of the sequence are multiples of 36. A218797 Conjecture: a(n) > 0 for all n=1715,1716,.... A218825 Conjecture: a(n)>0 for all n>8. A218867 Conjecture: a(n)>0 for all n>50000 with n different from 50627, 61127, 66503. A218976 Conjecture: All terms except a(5) are 1 mod 5. - _R. J. Cano_, Nov 11 2012 A218980 Conjecture: This sequence is infinite. A218981 Conjecture : This sequence is infinite. A219023 Conjecture: a(n)>0 for all n>2732. A219025 Conjecture: a(n)>0 for all n=6,7,... A219026 Conjecture: a(n)>0 except for n=1,2,4,6,10,22,57. A219033 Conjecture: This sequence is infinite. A219033 Conjecture: The sequence only consists of even numbers. A219033 Conjecture: The partitions only consist of even numbers. A219033 Conjecture: None satisfy sigma_3(n) = sigma_3(x) + sigma_3(y). A219033 Conjecture: With the lower partition as 6*A185208(n) and the upper partition 214/3 = 71.3333... of this, then the equalities are satisfied. A219052 Conjecture: a(n) > 0 for all n > 784. A219055 Conjecture: a(n)>0 for all even n>8012 and odd n>15727. A219085 Conjecture: if d is a positive integer and f(x) = x^(1/d), then the solutions of J(k) form a linearly recurrent sequence. A219157 Conjecture: a(n)>0 for all n>30000 with n different from 38451, 46441, 50671, 62371. A219185 Conjecture: a(n)>0 for all odd n>4676 and even n>30986. A219196 Conjecture: a(15) = a(16) = 131070, a(17) through a(32) = 8589934590. A219197 Conjecture: a(n) is never congruent to 3 modulo 4. A219252 Conjecture: except m = 77, all odd number > 9 are of the form m = p + 4*q where p and q are prime numbers. A219303 Conjecture #1: All numbers under the iteration reach 0 or, like the elements of this sequence, reach a finite loop, and none expand indefinitely to infinity. A219303 Conjecture #2: There are an infinite number of such finite loops, though there is often significant distance between them. A219303 Conjecture #3: There are an infinite number of pairs of consecutive integers in this sequence, e.g. 14 and 15, 197 and 198. This argument is strengthened by the fact there are other groupings such as triples - The first of these is 11527, 11528 and 11529 - and also that for randomly chosen numbers of hundreds of digits, N, the nearest pair or grouping appears to be within N +/- 1000. A219431 Conjecture: a(2*n+1) == 2 (mod 4), a(2*n) == 0 (mod 4). A219432 Conjecture: a(n) < prime(n) for n > 21. Conjecture confirmed up to a(122578) = 55296 < prime(122578) = 1620539. A219433 Conjecture: a(n) < prime(n) except for a(522720). A219433 Conjecture tested up to a(1000000). A219543 Conjecture: all the first differences 36, 72, 144, 72,... of the sequence are multiples of 36. A219552 Conjecture: 1 appears infinitely often in this sequence (indicating an infinite number of pairs of consecutive integers in the parent sequence). A219604 Conjecture: except m = 77, all odd number > 9 are of the form m = p + 4*q where p and q are prime numbers. A219782 Conjecture: a(n)>0 if n is not among 1, 8, 10, 18, 20, 41, 46, 58, 78, 116, 440. A219791 Conjecture: a(n)>0 if n is different from 1, 6, 16, 24. A219838 Conjecture: a(n) > 0 for all n > 1. A219842 Conjecture: a(n)>0 for all n>1. Moreover, any integer n>357 can be written as x+y (x>0, y>0) with 2x*y+1 and 2x*y-1 twin primes. A219864 Conjecture: a(n)>0 for all n>7. A219923 Conjecture: a(n)>0 for all n>623. A219930 Conjecture: If n is in the sequence, then the sequence contains an infinite number of multiples of n. A219930 Conjecture: Except for 1 and 3, all members of the sequence are even. If n is odd, it cannot be square-free. A219930 Conjecture: There does not exist N such that for all n > N, a(n) is divisible by 30. A219960 Conjecture #1: All numbers under the iteration reach 0 or, like the elements of this sequence, reach a finite loop, and none expand indefinitely to infinity. A219960 Conjecture #2: There are an infinite number of such finite loops, though there is often significant distance between them. A219960 Conjecture #3: There are an infinite number of pairs of consecutive integers in this sequence despite being less abundant than in A219303. A219961 Conjecture: 1 appears infinitely often in this sequence (indicating an infinite number of pairs of consecutive integers in the parent sequence), despite very few appearances compared with A219552, which is derived from a very similar process. A219966 Conjecture: a(n)>0 for all n>11. A220072 Conjecture: For any n>0, we have a(n) <= n*(n+1), and the Galois group of SF_n(x) = sum_{k=0}^n A005117(k+1)*x^{n-k} over the rationals is isomorphic to the symmetric group S_n. A220091 Conjecture: a(n)>0 for all even n>=8070 and odd n>=18680. A220095 Conjecture: This sequence is complete. A220272 Conjecture: a(n)>0 for all n>2. A220413 Conjecture: a(n)>0 for every n=1,2,3,... Moreover, any integer n>3 not among 7, 22, 31 can be written as p+q (q>0) with p and p^3+2*q^3 both prime. A220419 Conjecture: a(n)>0 for all n>527. A220431 Conjecture: a(n)>0 for all n>3. A220455 Conjecture: a(n)>0 for all n>7. A220554 Conjecture: a(n)>0 for all n>1. A220555 Conjecture 1. Let F_n(x)=sum_{j=0..n} A187660(n,j)*x^{(n-1)*j}. Let f_n in Z[x] be any polynomial in x of degree d such that 0<=d<=(n-1)*(n-2). Then the sequence of coefficients of the series expansion of f_n(x)/F_n(x), when taken over Z/kZ, is periodic with period p <= (n-1)*A220555(n,k), for all n,k > 1. (Cf. [Coleman, et al.] for the case for n=2 (generalized Fibonacci).) A220555 Conjecture 2. If G a cyclic multiplicative group generated by an n X n integer matrix over Z/kZ, then |G|<=T(r,k), for some r<=n. A220555 Conjecture 3. If T(n,k) is optimal, then n is a Queneau number (A054639). A220572 Conjecture: a(n)>0 for every n=1,2,3,.... Moreover, any odd integer greater than 2092 can be written as x+y (x,y>0) with x-3, x+3 and x^18+3*y^18 all prime. A220752 Conjecture: the sequence is infinite. A220846 Conjecture: sequence is injective (all terms of this sequence occur exactly once). A220847 Conjecture: sequence is injective (all terms of this sequence occur only once). A220848 Conjecture: sequence is injective (all terms of this sequence occur only once). A220849 Conjecture: sequence is injective (all terms of this sequence occur only once). A220947 Conjecture: a(n) <= n^2+12 for all n>0. A220949 Conjecture: a(n) <= n^2+22 for all n>0. A220956 Conjecture: a(n) = 0 iff n is an odd prime. A221130 Conjecture: a(n ) = the smallest numbers w such that numbers w, w+1,…, w+k-1 for k=1,2,…n are numbers of form h*2^m + m, where 1<=h <2^m, m = natural number (see A221129). A221218 Conjecture: All a(n)>=570. Conjecture: All sequences B_n are eventually periodic. A221564 Conjecture: a(3k+1) = 2k. A221650 Conjecture 1: P(n,j,k) is the number of partitions of n that contain at least m parts of size k, where m = j/k, if k divides j otherwise P(n,j,k) = 0. A221650 Conjecture 2: P(n,j,k) is the number of parts that are the m-th part of size k in all partitions of n, where m = j/k, if k divides j otherwise P(n,j,k) = 0. A221902 Conjecture: After the first term, 2^a(n)-1 is not prime; in other words, these primes (except 31) are included in A054723. A221903 Conjecture: 2^a(n)-1 is not prime; in other words, these primes are included in A054723. A222066 Conjectured to be center density of densest packing of equal spheres in five dimensions (achieved for example by the D_5 lattice). A222067 Conjectured to be center density of densest packing of equal spheres in six dimensions (achieved for example by the E_6 lattice). A222068 Conjectured to be density of densest packing of equal spheres in four dimensions (achieved for example by the D_4 lattice). A222069 Conjectured to be density of densest packing of equal spheres in five dimensions (achieved for example by the D_5 lattice). A222070 Conjectured to be density of densest packing of equal spheres in six dimensions (achieved for example by the E_6 lattice). A222071 Conjectured to be density of densest packing of equal spheres in 7 dimensions (achieved for example by the E_7 lattice). A222072 Conjectured to be density of densest packing of equal spheres in 8 dimensions (achieved for example by the D_8 lattice). A222073 Conjectured to be density of densest packing of equal spheres in 12 dimensions (achieved for example by the K_12 lattice). A222074 Conjectured to be density of densest packing of equal spheres in 16 dimensions (achieved for example by the Barnes-Wall BW_16 lattice). A222075 Conjectured to be density of densest packing of equal spheres in 24 dimensions (achieved for example by the Leech lattice). A222114 Conjecture: For each n=3,4,..., a(n) is the first prime p>=p_n dividing none of those p_i+p_j-1 (1<=i<j<=n). A222532 Conjecture: For any given prime p, if we define b(1)=p and let b(n+1) be the least prime p_m such that b(n)=p_m-p_{m-1}+...+(-1)^{m-k}p_k for some 0<k<m, then a(n)=b(n') for some positive integers n and n'. In other words, if we take all the primes as vertices of a simple graph T and let two vertices p and q>p adjacent if and only if q is the least prime p_m such that p=p_m-p_{m-1}+...+(-1)^{m-k}p_k for some 0<k<m, then the graph T is a tree! A222566 Conjecture: The graph G constructed above is connected and hence it is a tree! A222579 Conjecture: a(n)<=3n for all n>0. Moreover, a(2n-1)/(2n-1) and a(2n)/(2n) have limits 1 and 2 respectively, as n tends to the infinity. A222580 Conjecture: All the terms are positive. A222603 Conjecture: The graph H constructed above is connected and hence it is a tree. A223174 Conjecture: except m = 25 and 49, all odd numbers > 17 are of the form m = p + 8*q where p and q are prime numbers. A223175 Conjecture: except m = 25 and 49, all odd numbers > 17 are of the form m = p + 8*q where p and q are prime numbers. A223577 Conjecture: A223577(k) = (A223578(k)+k-1)/2, k=1,2,.... A223578 Conjecture: A223568(k) = 2*A223577(k)-k+1, k=1,2,.... A223732 Conjecture: 793 = 6^2 + 9^2 + 26^2 is the largest element of this sequence. - _Alois P. Heinz_, Apr 06 2013 A223733 Conjecture: a(147) = 1885 = 16^2 + 27^2 + 30^2 = 12^2 + 29^2 + 30^2 is the largest element of this sequence. - _Alois P. Heinz_, Apr 06 2013 A223734 Conjecture: a(185) = 4075 = 31^2 + 33^2 + 45^2 = 23^2 + 39^2 + 45^2 = 5^2 + 9^2 + 63^2 is the largest element of this sequence. - _Alois P. Heinz_, Apr 06 2013 A223925 Conjecture: a(2n) is always a divisor of A132049(2n). A223934 Conjecture: a(n) < n*(n+3)/2 for all n>1. A223942 Conjecture: a(n) <= sqrt(7*p_n) for all n>0. A224030 Conjecture: a(n)>0 for all n>4. A224197 Conjecture: (i) For any positive integer k and distinct positive integers a_1< a_2 < ... < a_n with a_n prime, there are infinitely many integers b > a_n^k such that [a_1^k,a_2^k,...,a_n^k] in base b is prime. A224210 Conjecture: a(n) does not exceed the (4n-3)-th prime for each n>0. Moreover, for any integers m>1 and n>0 the polynomial sum_{k=0}^n (k+1)^m*x^{n-k} is irreducible modulo some prime, and its Galois group over the rationals is isomorphic to the symmetric group S_n. Also, for m,n=2,3,... there are infinitely many integers b > n^m such that [n^m,...,2^m,1^m] in base b is prime. A224241 Conjecture: the sequence is infinite. A224274 Conjecture: a(n)== 1 (mod n^3) iff n is an odd prime. A224400 Conjecture: a(n)~7.75n A224416 Conjecture: (i) a(n) does not exceed n^2+n+5 for each n>0, and the Galois group of sum_{k=0}^n C_k*x^{n-k} over the rationals is isomorphic to the symmetric group S_n. A224417 Conjecture: a(n) < 4n^2-1 for all n>0. A224418 Conjecture: a(n) < n^2 for all n > 1. A224480 Conjecture: a(n) <= (n+4)*(n+5)+1 for all n>0. A224487 Conjecture: For any n > 1 we have a(n) < F(n+4); moreover, there are infinitely many integers b > F(n) such that sum_{k=1}^n F(k)*b^{k-1} is prime. A224496 Conjecture: a(n) exist for all n A224511 Conjecture: this is a permutation of odd numbers. A224519 Conjecture: the number of couples such that a(n+1) = a(n) + 1 is infinite. A224523 Conjecture. For n>=1, a(n)>0. A224529 Conjecture 1.1 of Osburn and Sahu is if p is a prime and JacobiSymbol(p , 23) = 1 and n>0 then a(n * p) == a(n) (mod p). - _Michael Somos_, Sep 21 2013 A224612 Conjecture: a(n) exists for all n. A224613 Conjectures: sigma(6n) > sigma(6n - 1) and sigma(6n) > sigma(6n + 1). A224613 Conjectures are false. Try prime 73961483429 for n. One finds sigma(6*73961483429) < sigma(6*73961483429+1). The number n = 105851369791 provides a counterexample for the other case. - _T. D. Noe_, Apr 22 2013 A224794 Conjecture: The odd numbers different from a cube are of the form m = 2p - q^3 where p and q are prime numbers. A224832 Conjecture: if a(n) is a non-twice even perfect numbers, the sum of reciprocals of even divisors equals 2. A224888 Conjecture: a(n) ~ A093343(n). A225001 Conjecture: a(n) = A047203(n), n>1. - _R. J. Mathar_, Apr 28 2013 A225003 Conjecture: a(n) = A047332(n), n>1. - _R. J. Mathar_, Apr 28 2013 A225004 Conjecture: always a divisor of n^2. A225039 Conjecture: the sequence is a permutation of the sequence of all primes. A225040 Conjecture: All terms are positive, or equivalently, the sequence is a permutation of the positive integers. A225091 Conjecture: the sequence contains all primes > 3. A225093 Conjecture: the sequence is a permutation of the sequence of all primes. A225239 Conjecture: all terms are squarefree numbers. A225322 Conjecture: the sequence is infinite. A225389 Conjecture: the sequence is infinite. A225400 Conjecture: a(n) >= 0. A225502 Conjecture: a(n) > 0. A225503 Conjecture: a(n) > 0. A225516 Conjecture: the sequence is a permutation of all primes > 3. A225522 Conjecture: for each integer n > 1, all odd numbers > 2^(n+1) + 1 are of the form p + q*2^n, except for a finite set of integers {n_1, n_2,..., n_r}, where p and q are prime numbers. A225538 Conjecture: the least n's for which a(n) = 0 are 1895, 1985, 2894, 2984, 3893, and 3983. - _Peter Moses_, May 10 2013 A225563 Conjecture: No totient-trajectory can be partitioned into an odd number of sets with the same sum. A225577 Conjecture: a(n) is the least prime p such that 2^p-1 is a Mersenne prime greater than 2n-1. A225611 Conjecture: We have a(n) = 0 if p_n == 3 (mod 4). A225660 Conjecture: if n>9, then a(n) is odd. A225776 Conjecture: a(n)/6^n is always a positive odd integer. Moreover, for any integers r > 1 and n >= 0, the number a(r,n)/2^n is a positive odd integer, where a(r,n) denotes the Hankel determinant |f(r,i+j)|_{i,j=0,...,n} with f(r,k) = sum_{j=0}^k C(k,j)^r. A225819 Conjecture [Sen]: lim inf log_n a(n) >= 5/4. A225835 Conjecture: a(6) = a(7) = 0. _Charles R Greathouse IV_ reports that a(6) must have thousands of digits. - _Michael B. Porter_, May 19 2013 A225919 Conjecture: a(n) is linearly recurrent. See A225918 for details. A225920 Conjecture: a(n) is linearly recurrent. See A225918 for details. A225921 Conjecture: a(n) is linearly recurrent. See A225918 for details. A225922 Conjecture: a(n) is linearly recurrent. See A225918 for details. A226115 Conjecture: sqrt(2*a(n)) > sqrt(p_n)-0.7 for all n > 0, and a(n) is even for any n > 7. A226160 Conjecture: a(n+1)/a(n) converges to 1.8552... A226160 Conjecture confirmed: using series expansion of HarmonicNumber(k) one gets a(n+1)/a(n) -> exp(1/tau) = 1.855276958... [_Jean-François Alcover_, Jun 04 2013] A226161 Conjecture: a(n+1)/a(n) converges to 1.64872... A226163 Conjecture: a(n) = 0 if and only if p_n == 3 (mod 4). A226187 Conjecture: a(n+1)/a(n) converges to 1.39... A226188 Conjecture: a(n+1)/a(n) converges to 1.94... A226188 Conjecture confirmed: series expansion of HarmonicNumber(k) gives a(n+1)/a(n) -> exp(2/3) = 1.947734... [_Jean-François Alcover_, Jun 05 2013] A226318 Conjecture: For any positive even integers d_1,...,d_k, there are infinitely many positive integers n such that p_{n+2j-1}-p_{n+2j-2} = d_j for all j=1,...,k. A226617 Conjecture: a(n)>0 for all n. A226625 Conjecture: Every cycle with the same value of k (k>1) has the same proportion of odd and even elements. Thus if n>1 then A226626(n)/A226625(n) has the same value for each m where A226628(n) <= m < A226628(n+1). A226661 Conjecture: a(n)>0 for all n. A226676 Conjecture: For n>2, a(n)>0. A226677 Conjecture: a(n)>0 for all n. A226678 Conjecture: a(n)>0 for all n. A227012 Conjecture: if f(k) = u/(v*k + w), where u,v,w are integers, and g(n) is a polynomial, then the sequence with nth term m(n) = floor(M(n)) is linearly recurrent. The conjecture extends to these cases, in which a,b,c,d are integers and a > 0: A227036 Conjecture: The perimeter of the n-th iteration of the Harter-Heighway dragon is a(n) segments or a(n)/2^(n/2) base units. A227083 Conjecture: We have a(n) > 0 for all n > 4. A227145 Conjecture: a(F_n) = F_{n-2} for n>1, where F_n is the n-th Fibonacci number. A227156 Conjecture: We have a(n) > 0 except for n = 23. A227202 Conjecture: a(n) < Prime[n*E]. A227202 Conjecture: All primes will join this sequence eventually. A227345 Conjecture: there exists a partition (into distinct parts) of n with boundary size k if and only if 0 < k^2 * 3/4 <= n. [_Patrick Devlin_, Jul 13 2013] A227370 Conjecture 1: This is an involution (self-inverse permutation) of nonnegative integers. (Which would imply that both formulas given in A227368 and A227369 involving A227370 are valid). A227370 Conjecture 2: (which would automatically imply the conjecture 1): the only transpositions (used to compose the permutation) are of adjacent terms 2k-1 and 2k, where A061887 gives the values of k. This is true at least for the first 35 transpositions (up to k=60). A227387 Conjecture: the sequence is infinite. A227419 Conjecture: a(n) > 0. A227456 Conjecture: a(n) > 0 for all n > 0. Similarly, for any positive integer n, there is a circular permutation i_0, i_1, ..., i_n of 0, 1, ..., n such that all the n+1 numbers i_0^2+i_1, i_1^2+i_2, ..., i_{n-1}^2+i_n, i_n^2+i_0 are of the form (p-1)/4 with p a prime congruent to 1 modulo 4. A227470 Conjecture: a(n) > 0. A227500 Conjecture: (b(n)-a(n))/c(n) = 0, 0, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ..., that is two 0 followed by A000045. A227533 Conjecture: a(n) > 0 for all n. A227553 Conjecture: a(2) = 4; if s > 1 then a(2^s) = 2^(2s-1); if p == 1 (mod 4) then a(p^s) = p*(p+1)^(2s-1); if p == 3 (mod 4) then a(p^s) = p*(p-1)^(2s-1). A227569 Conjecture: there are no pair of complements a(n) and b(n) such that F[a(n); b(n)] = 2. A227578 Conjecture: column k is asymptotic to c(k) * (k+1)^(k*n)/n^((k^2-1)/2), where c(k) is a constant dependent only on k. - _Vaclav Kotesovec_, Jul 21 2013 A227609 Conjecture: p_n never divides a(n), and moreover -a(n) is a quadratic residue mod p_n. A227758 Conjecture: a(n) = 0 for Mersenne primes (A000668). [This is easily proved: For Mersenne primes n=2^p-1, sigma(n)=n+1=2^p, sigma(2^p)=2^(p+1)-1, thus a(n)=0. - M. F. Hasler, Jul 30 2013] A227759 Conjecture: a(n) = complement of union A000668 and A227760, where A000668 = Mersenne primes, A227760 = numbers n such that sigma(sigma(n)) - sigma(n) - n > 0. A227760 Conjecture: a(n) = complement of union A000668 and A227759, where A000668 = Mersenne primes, A227759 = numbers n such that sigma(sigma(n)) - sigma(n) - n < 0. A227781 Conjecture: a(n) = 4 if and only if n is divisible by 8 and a(n) = 3 if and only if n is 4 mod 8. Together with A008784 this would completely define the sequence. A227784 Conjecture: a(n) = 15 if n = 9 mod 16 and a(n) = 7 if n = 8 mod 16, otherwise a(n) <= 4. (The associated lower bounds are obvious.) A227786 Conjecture: from n>=2 onward, a(n) gives the positions of 2's in A227761. A227864 Conjecture 1: 0, 1 and 4 are the only values where there is no base in which a digital reversal makes a prime. A227864 Conjecture 2: n = 2 is the only prime p which must be represented in base p+1, i.e. trivially, as a single digit, in order for its reversal to be prime. A227864 Conjecture 3: With the restriction gcd(base,n)==1, a(n) = 0 except for n = 2, 3 and 6k+/-1, for positive integer k, i.e. members of A038179 A227877 Conjecture: a(n) > 0 for all n > 4. A227898 Conjecture: (i) a(n) > 0 for all n > 5. A227899 Conjecture: a(n) > 0 for all n > 3. A227908 Conjecture: a(n) > 0 except for n = 1, 16, 292. A227909 Conjecture: a(n) > 0 for all n > 1. A227920 Conjecture: Any integer n > 2 can be written as x + y + z (x, y, z > 0) such that 6*x-1, 6*y-1, 6x*y-1, 6*z-1 are Sophie Germain primes, and {6*x-1, 6*x+1}, and {6*y-1, 6*y+1} are twin prime pairs. A227923 Conjecture: (i) a(n) > 0 for all n > 1. Moreover, any integer n > 4 not equal to 13 can be written as x + y with x and y distinct and greater than one such that 6*x-1 is a Sophie Germain prime and {6*y-1, 6*y+1} is a twin prime pair. A227938 Conjecture: a(n) < 2*n for all n > 2. A227968 Conjecture: 2*a(n) is always a quadratic residue mod p_n. A227971 Conjecture: If p_n == 1 (mod 4), then a(n) == ((p_n-1)/2)! (mod p_n). If p_n == 3 (mod 4), then a(n) == (2/p_n) (mod p_n). A228005 Conjecture: We have (a(n)/p_n) = ((2(-1)^{(p_n-1)/2}-4)/p_n). A228077 Conjecture: In the case p_n == 1 (mod 4), (2/p_n)*a(n) is a positive odd integer whose prime factors are all congruent to 1 modulo 4, and moreover for some integer b(n) we have b(n) + (2/p_n)*a(n)*sqrt(p_n) = e(p_n)^{(2-(2/p_n))h(p_n)}, where e(p_n) and h(p_n) are the fundamental unit and the class number of the real quadratic field Q(sqrt(p_n)) respectively. A228095 Conjecture: a(n) = 0 if p_n == 5 (mod 6). A228143 Conjecture: a(n)/24^n is always a positive integer. Similarly, if b(n) denotes the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to A005258(i+j) for all i,j = 0,...,n, then b(n)/10^n is always a positive integer; also, if p is a prime with floor[p/10] odd and p is not congruent to 31 or 39 modulo 40, then p divides b((p-1)/2). A228144 Conjecture: there is always at least one k for each n. A228289 Conjecture: If p_n == 1 (mod 3) and p_n = x^2 + 3*y^2 with x and y integers, then we have a(n) == (-1)^{(p_n-1)/2}*(4*x^2-2*p_n) (mod p_n^2). In the case p_n == 2 (mod 3), we have a(n) == 0 (mod p_n^2). A228304 Conjecture: Let p be any odd prime, and let A(p) be the p X p determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,p-1. Then A(p) == (-1)^{(p-1)/2} (mod p). Similarly, if c(n) = sum_{k=0}^n (-1)^k*C(n,k)^2*C(2k,k)*C(2(n-k),n-k) and C(p) is the p X p determinant with (i,j)-entry equal to c(i+j) for all i,j = 0,...,p-1, then we have C(p) == 1 (mod p). A228379 Conjecture: If n > 2, then (-1)^{n(n-1)/2}*a(n) > 0 and 2*product_{k=1}^n(k!*(2k-1)!) divides a(n). A228456 Conjecture: (i) a(n) is always positive and odd, and not congruent to 7 modulo 8. A228469 Conjecture: a(n) is the least positive integer c for which there is a positive integer b for which trace(b,c) consists of the first n letters of 01010101010101... A228471 Conjecture: a(n) is the least positive integer c for which there is a positive integer b for which trace(b,c) consists of the first n letters of 101010101010101..., where "trace" is as defined at A228469. A228487 Conjectures: (1) trace(pi) is not periodic; (2) if k is a positive integer, then trace(sqrt(k)) is purely periodic; (3) if r is a quadratic irrational, or if r = s*e where s is a positive rational number, or if r = s*e^2 where s is a positive rational number, then trace(r) is eventually periodic. A228511 Conjecture: Let p be any odd prime. A228514 Conjecture: Let p > 3 be a prime. A228518 Conjecture: a(n) = 0 iff n = 1, 2, 3, 4, 5, 6, 9, 12, 14, 24, 56. A228548 Conjecture: a(n) ia always nonzero. Moreover, |a(n)|^(1/n) tends to infinity. A228549 Conjecture: a(n) is always nonzero. Moreover, |a(n)|^(1/n) tends to infinity. A228558 Conjecture: All terms are prime. A228559 Conjecture: a(n) is nonzero if n is greater than 55 and not congruent to 5 modulo 6. A228561 Conjecture: a(n) is nonzero for each n > 28. A228574 Conjecture: a(2*n-1) = 0 for all n > 0, and a(2*n) is nonzero when n > 9. A228591 Conjecture: a(n) = 0 for no n > 15. A228615 Conjecture: a(n) is nonzero for any n > 35. A228616 Conjecture: a(n) is nonzero for any n > 7. A228623 Conjecture: a(n) is nonzero if n is odd and greater than 120. A228624 Conjecture: a(n) is nonzero for any n > 21. A228626 Conjecture: a(n) > 0 for all n > 4. In other words, for each n = 5,6,... there is a permutation i_1,...,i_n of 1,...,n such that |i_1-i_2|, |i_2-i_3|, ..., |i_{n-1}-i_n| and |i_n-i_1| are all prime. A228638 Conjecture: a(n) is nonzero for any n > 3. A228728 Conjecture: For any n distinct real numbers a_1 < a_2 < ... < a_n, if there is a permutation b_1,b_2,...,b_n of a_1,...,a_n with |b_1-b_2|, |b_2-b_3|, ..., |b_{n-1}-b_n|, |b_n-b_1| pairwise distinct, then there exists a permutation c_1,c_2,...,c_n of a_1,...,a_n with c_1 = a_1 and c_n = a_n such that the n numbers |c_1-c_2|, |c_2-c_3|, ..., |c_{n-1}-c_n|, |c_n-c_1| are pairwise distinct. A228762 Conjecture: a(n) > 0 for any odd number n > 1. In general, if G is an additive abelian group with |G| = n odd and greater than one, then there is a permutation a_1,...,a_{n-1} of all the nonzero elements of G such that a_1-a_2, a_2-a_3, ..., a_{n-2}-a_{n-1}, a_{n-1}-a_1 are pairwise distinct. A228766 Conjecture: a(n) > 0 for all n > 3. In general, if a_1,...,a_n are n > 2 distinct elements of a finite additive abelian group G with n odd or |G| not divisible by n, then there exists a circular permutation b_1,...,b_n of a_1,...,a_n such that b_1+b_2, b_2+b_3, ..., b_{n-1}+b_n, b_n+b_1 are pairwise distinct. A228772 Conjecture: For any subset A of an additive abelian group G with |A| = n > 3, there is a numbering a_1,...,a_n of the elements of A such that the n sums a_1+a+2+a_3, a_2+a_3+a_4, ..., a_{n-2}+a_{n-1}+a_n, a_{n-1}+a_n+a_1, a_n+a_1+a_2 are pairwise distinct. A228828 Conjecture: the sequence is infinite. A228858 Conjecture: there exists an integer corresponding to the common area of exactly n distinct triangles (x_i, y_i, z_i) for i = 1,2,...,n such that A(x_1, y_1, z_1)= A(x_2, y_2, z_2)= ... = A(x_n, y_n, z_n) for n = 1, 2, ... A228860 Conjecture: a(n) > 0 except for n = 3. A228865 Conjecture: the sequence is infinite. A228884 Conjecture: (i) a(n) is always positive and divisible by Phi(n)^{Phi(n)}*sum_{d|n}Phi(d)*n/d, where Phi(n) is Euler's totient function. A228885 Conjecture: If n is squarefree, then (-1)^{n(n-1)/2}*a(n) > 0. A228886 Conjecture: a(n) > 0 except for n = 2, 4. A228916 Conjecture: for n>1 a(n) is a multiple of 3. A228917 Conjecture: a(n) > 0 for all n > 0. A228931 Conjecture: The terms from a(3) are all positive and can be generated by the recurrence relation a(k+1) = a(k)^2 - 2. A228933 Conjecture: The golden ratio (phi) expansion exhibits from the fourth term the recurrence relation a(n) = a(n-1)^2 - 2 described in A228931. A228934 Conjecture: The terms of the expansion of sqrt(x) are all negative starting from a(4) and satisfy these recurrence relations for n>=3: a(2n) = 4*a(2n-1) - 4 and a(2n+1) = -2*a(2n-1)^2 + 1. A228942 Conjecture: this sequence is always positive, analogous to sequence A045917 for the strong Goldbach conjecture. - Jaycob Coleman, Sep 8 2013 A228943 Conjecture: (a) This sequence is strictly increasing beginning with n=5. (b) For all n>2, if p is the greatest prime with p<A002182(n)-1, then A002182(n)-p is prime. This is a strengthening of a conjecture regarding A117825. - Jaycob Coleman, Sep 8 2013 A228944 Conjecture: this sequence is always positive, analogous to sequence A202472 for strong Goldbach conjecture. - Jaycob Coleman, Sep 8 2013 A228945 Conjecture: (a) This sequence is strictly increasing beginning with n=7. (b) If p is the smallest prime with p>A002182(n)+1, then p-A002182(n) is prime. This is a strengthening of a conjecture regarding A117825. A228956 Conjecture: a(n) > 0 for all n > 0. A228993 Conjecture: if x is an irrational number between 3 and 4, then L(x) = L(Pi). A229005 Conjecture: a(n) > 0 except for n = 2, 4. A229019 Conjecture: all terms are positive. A229038 Conjecture: a(n) > 0 if 2*n+1 is a prime greater than 13. A229068 Conjecture: generally (for tableaux with height <= k), a(n) ~ k^n/Pi^(k/2) * (k/n)^(k*(k-1)/4) * prod(j=1..k, Gamma(j/2)); set k=12 for this sequence. A229075 Conjecture: the expression p^2+q^2+c with p and q consecutive primes and c=21 generates more primes than any other value of c in the range 1..150. Hence, c=21 is considered for this sequence. A229082 Conjecture: a(n) > 0 except for n = 4. A229086 Conjecture: a(n) for n > 1 is not equal to A161387, sequence also contains composites. A229130 Conjecture: a(n) > 0 except for n = 2, 5, 11. Similarly, for any positive integer n not equal to 4, there is a permutation i_0, i_1, ..., i_n of 0, 1, ..., n with i_0 = 0 and i_n = n such that the n+1 numbers i_0^2-i_1, i_1^2-i_2, ..., i_{n-1}^2-i_n, i_n^2-i_0 are all coprime to both n-1 and n+1. A229136 Conjecture: a(n) = A131885(n)*2^n. A229141 Conjecture: a(n) > 0 if 2*n+1 is a prime greater than 11. A229159 Conjecture: for all prime p > 2 there exists an integer-sided triangles where at least one side is of p length. A229166 Conjecture: a(n) > 0 for all n > 0. Moreover, if n > 0 is not among 1, 3, 60, then there are positive integers x and y with x*(x+1)/2 + y = n such that y*(y+1)/2 + 1 is prime. A229232 Conjecture: a(n) > 0 for all n > 5 with n not equal to 13. A229461 Conjecture: These 59 numbers are all such exceptions. A229570 Conjecture: The expression p*q + c with p and q consecutive primes and c = 30 generates more primes than any other value of c in the range 1 < c < 100 and p = 48611 which is 5000-th prime. Hence, c = 30 is considered for this sequence. A229597 Conjecture: if x is an irrational number between 2 and 3, then L(x) = L(e). A229613 Conjecture: The expression p*q - c with p and q consecutive primes and c = 30 generates more primes than any other value of c in the range 1 < c < 100 and p = 48611 that is 5000-th prime. Hence, c = 30 is considered for this sequence. A229827 Conjecture: (i) a(p) > 0 for any prime p > 3. Moreover, for any finite field F(q) with q elements and a polynomial P(x) over F(q) of degree smaller than q-1, if P(x) is not of the form c-x, then there is a circular permutation a_1, ..., a_q of all the elements of F(q) with A229857 Conjecture: a(n) < f(n) = number of primes of the form k*2^(n+2) + 1 with k odd that exist between a = 2^(n+2) + 1 and b = floor((2^(2^n) + 1)/(3*2^(n+2) + 1)). A229878 Conjecture: a(n) > 0 for all n > 6. In other words, for any prime p = 2*n+1 > 13, there is a circular permutation a_1, ..., a_n of the n quadratic residues modulo p such that a_1+a_2, a_2+a_3, ..., a_{n-1}+a_n, a_n+a_1 give all the n quadratic residues modulo p. A229899 Conjecture: a(n) > 0 except for n = 1, 2, 3, 4, 6, 10, 18. In other words, for any prime p > 7 not equal to 13 or 29 or 61, there are three consecutive integers which are primitive roots modulo p. A229910 Conjecture: a(n) > 0 for all n > 6. In other words, for any prime p > 13, there is a primitive root g modulo p such that g + g^{-1} is also a primitive root modulo p, where g^{-1} is the inverse of g modulo p. A229910 Conjecture: Let F be a finite field with |F| = q > 13. Then there is a primitive root g of F (i.e., a generator of the cyclic group F\{0}) such that g + g^{-1} is also a primitive root of F. If q > 61, then there exists a primitive root g of F such that g - g^{-1} is also a primitive root of F. A229927 Conjecture: there is no "Riesel" number of the form 4^n-1; that is, a(n) exists for all n. A229969 Conjecture: a(n) > 0 for all n > 5. Moreover, any integer n > 6 can be written as x + y + z with x among 3, 4, 6, 10, 15 such that 2*y-1, 2*z-1, 2*x*y-1, 2*x*z-1, 2*y*z-1 are prime. A229974 Conjecture: (i) a(n) > 0 for all n > 3. Moreover, any integer n > 3 can be written as x + y + z with x among 2, 3, 6 such that {2*x*y-1, 2*x*y+1} and {2*x*y*z-1, 2*x*y*z+1} are twin prime pairs. A229989 Conjectures: (1) a(n+1) - a(n) = 1 for infinitely many n; (2) a(n+1) - a(n) = -1 for infinitely many n; (3) a(n+1) - a(n) = -1 if and only if n = 2*prime(m+1) - 1. A230037 Conjecture: a(n) > 0 for all n > 2. Moreover, any integer n > 2 can be written as x + y + z with x = 1 or 5 such that {6*y-1, 6*y+1}, {6*z-1, 6*z+1} and {6*x*y-1, 6*x*y+1} are twin prime pairs. A230040 Conjecture: a(n) > 0 for all n > 2. A230107 Conjecture: the b sequence, for any starting value n, will eventually merge with one of A000004 (the zero sequence), A004207, A016052 or A016096. A230121 Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 5, 7, 12. Moreover, for each n = 20, 21, ... there are three distinct positive integers x, y and z with x + y + z = n such that x*(x+1)/2 + y*(y+1)/2 + z*(z+1)/2 is a triangular number. A230140 Conjecture: (i) a(n) > 0 for all n > 2, i.e., 6*n-3 with n > 2 can be expressed as a sum of three terms from A230138. Moreover, for any integer n > 12, there are three distinct positive integers x, y, z with x + y + z = n such that 6*x-1, 6*y-1, 6*z-1 are primes in A230138. A230141 Conjecture: a(n) > 0 for all n > 2. Also, any integer n > 2 can be written as x + y + z (x, y, z > 0) such that 6*x-1, 6*y-1, 6*z-1 are terms of A230138 and 6*y*z-1 is prime. A230168 Conjecture: each term in the sequence ends with digit 9. A230194 Conjecture: a(n) > 0 for all n > 5. A230219 Conjecture: a(n) > 0 for all n > 6. A230223 Conjecture: Any even number greater than 35 can be written as a sum of four terms of this sequence. A230224 Conjecture: a(n) > 0 for all n > 17. A230227 Conjecture: For any integer n > 4 not equal to 76, we have 2*n = p + q for some terms p and q from the sequence. A230230 Conjecture: a(n) > 0 for all n > 3. A230241 Conjecture: a(n) > 0 for all n > 5. A230243 Conjecture: a(n) > 0 for all n > 4. A230252 Conjecture: (i) a(n) > 0 for all n > 1. A230254 Conjecture: a(n) > 0 for all n > 3. A230261 Conjecture: (i) a(n) > 0 for all n > 3. Also, any odd number greater than 6 can be written as p + q (q > 0) with p, p + 6 and q^2 + 1 all prime. A230351 Conjecture: a(n) > 0 for all n > 3. A230362 Conjecture: 0 < a(n) < sqrt(2n)*(log n) except for n = 1, 2, 3, 232, 1478, 6457. A230451 Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7. A230493 Conjecture: a(n) > 0 for all n > 6. A230494 Conjecture: (i) a(n) > 0 for all n > 1. Moreover, if n > 1 is not among 2, 69, 76, then there are positive integers x and y such that x^2 + y is equal to n and 2*y^2 - 1 is prime. A230496 Conjecture: for all n, a(n) > 0. A230497 Conjecture: For all n, a(n) > 0. A230498 Conjecture: For all n, a(n) > 0. A230502 Conjecture: a(n) > 0 for all n > 6. A230507 Conjecture: (i) a(n) > 0 for all n > 2. A230514 Conjecture: a(n) > 0 for all n > 5. A230516 Conjecture: a(n) > 0 for all n > 5. A230578 Conjecture: all a(n) are prime numbers. A230596 Conjecture: (i) a(n) > 0 except for n = 1, 2, 4, 7. A230665 Conjecture: The expression k^n with k= 38 generates more primes than any other value of k in the range 1 < k < 100 and n= 100. Hence, 38 is considered for this sequence such that digit sum of 38^n is prime. A230747 Conjecture: a(n) > 0 for all n > 17. A230852 Conjecture: if n>8 there is always a solution. A231168 Conjecture: a(n) > 0 for all n > 5.

## Contents

## A051451 conjecture

I am not sure this holds true for n=135...I am conversing with the author--Bill McEachen 12:41, 6 May 2014 (UTC)

- the conjecture has subsequently been reworded...--Bill McEachen 15:21, 11 May 2014 (UTC)

The generalized form of the conjecture (solution set) includes numbers dictated by A053319 as per the comment there.--Bill McEachen 22:32, 16 May 2014 (UTC)

## A000006

A subsequent comment to that shown above (see A000006) says it is equivalent to Legendre's conjecture.--Bill McEachen 00:15, 7 May 2014 (UTC)

## A073641

A subsequent comment (see A073641) refutes the conjecture.

## A000392

The conjecture is true and all the pieces needed to prove it are already in OEIS, scattered between A036561, A001047, and A000392 itself. What's the appropriate way to edit it? Are there general guidelines on updating conjectures which are theorems? Peter J. Taylor 07:35, 22 September 2014 (UTC)