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1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 -2n +2 (Wielandt).
Let Phi_k(x) be the k-th cyclotomic polynomial and form the sequence Phi_k(0), Phi_k(1), Phi_k(2), ... This gives A000027 (k=2), A002061 (k=3), A002522 (k=4), A053699 (k=5), A002061 (k=6), A053716 (k=7), A002523 (k=8), A060883 (k=9), A060884 (k=10), A060885 (k=11), A060886 (k=12), A060887 (k=13), A060888 (k=14), A060889 (k=15), A060890 (k=16), A060891 (k=18), A060892 (k=20), A060893 (k=24), A060894 (k=30), A060895 (k=32), A060896 (k=36).
As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre (benoit7848c(AT)orange.fr), Dec 07 2001
a(n) is one less than the arithmetic mean of its neighbors: a(n) = {a(n-1) + a(n+1)}/2 -1. e.g. 2 = (1+5)/2, 5 = (2+10)/2 -1. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jul 29 2003. Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,....). - Franz Vrabec (franz.vrabec(AT)aon.at), Jan 23, 2006.
Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.
The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sebastien Dumortier (sdumortier(AT)ac-limoges.fr), Jun 16 2005
Comment from Cino Hilliard (hillcino368(AT)gmail.com), Feb 21 2006: "Also, except for the first term, numbers that cannot be expressed as a perfect power, i.e. x^2 + 1 != y^n for all x,y,n > 1. Proof. We assume the truth of the following theorem. Proofs can be found in elementary texts on number theory and online. Theorem I: A number N is a sum of two squares if and only if all prime factors of N of the form 4m+3 have even exponents.
"We are now ready to prove x^2 + 1 != y^n for all x,y,n > 1. We assume equality and seek a contradiction for n even and n odd. If n is even = 2k, x^2 + 1 = y^2k = (y^k)^2 and (y^k - x)(y^k + x) = 1. This implies y^k-x = y^k+x = 1 or 2x = 0 contrary to x > 1. So n must be odd for equality to hold.
"Then x^2+1 = y^(2k+1) implies all prime factors of y, including those of the form 4m+3 are raised to an odd exponent contrary to Theorem I. So we have shown x^2+1 = y^n is false for n even or n odd. Therefore x^2 + 1 != y^n as was desired."
Also, numbers m such m^3-m^2 is a perfect square, (n*(1 + n^2))^2. - Zak Seidov
1 + 2/2 + 2/5 + 2/10 +...= Pi*coth Pi [Jolley] - Gary W. Adamson, Dec 21 2006
For n>=1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd (rshepherd2(AT)hotmail.com), Nov 18 2007
Sequence allows us to find X values of the equation: X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 29 2007
For n>0: a(n-1)=A143053(A000290(n))-1, - Reinhard Zumkeller, Jul 20 2008
A143053(a(n))=A000290(n+1). - Reinhard Zumkeller, Jul 20 2008
a(n) = A156798(n)/A087475(n). [From Reinhard Zumkeller, Feb 16 2009]
{a(k): 0 <= k < 4} = divisors of 10. [From Reinhard Zumkeller, Jun 17 2009]
Number of units of a(n) belongs to a periodic sequence: 1, 2, 5, 0, 7, 6, 7, 0, 5, 2. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009]
a(n)=A170949(A002061(n+1)); A170949(a(n))=A132411(n+1); A170950(a(n))=A002061(n+1). [From Reinhard Zumkeller, Mar 08 2010]
Contribution from Johannes W. Meijer, Jun 12 2010: Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... .
For n>0, continued fraction [n,n] = n/a(n); e.g. [5,5] = 5/26 [From Gary W. Adamson, Jul 15 2010]
The only real solution of the form f(x)= A*x^p with negative p which satisfies f^(m)(x) = f^[ -1](x), x>=0, m>=1, with f^(m) the m-th derivative and f^[ -1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=product(x-j,j=0..k-1)(falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). [From Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Oct 21 2010]
n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period length 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.
a(n-1) counts configurations of non-attacking bishops on a 2Xn strip [Chaiken et al, Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011
Also numbers n such that 4*n-4 is a square. Hence this sequence is the union of A053755 and A069894. [From Arkadiusz Wesolowski, Aug 02 2011]
a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011
Left edge of the triangle in A195437: a(n+1) = A195437(n,0). [Reinhard Zumkeller, Nov 23 2011]
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REFERENCES
| S. J. Cyvin and I. Gutman, Kekule structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.
Wielandt, H. 1950. Unzerlegbare nicht negativen Matrizen, Math. Z. 52, 642-648.
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, New York, 2001. [From Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Oct 21 2010]
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..1000Guo-Niu Han, Enumeration of Standard Puzzles
S. J. Leon, Linear Algebra with Applications: THE PERRON-FROBENIUS THEOREM
T. Mansour and J. West, Avoiding 2-letter signed patterns.
Eric Weisstein's World of Mathematics, Number Picking
Eric Weisstein's World of Mathematics, Near-Square Prime
R. Zumkeller, Enumerations of Divisors [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 17 2009]
Index entries for sequences related to linear recurrences with constant coefficients, signature (3,-3,1).
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FORMULA
| O.g.f.: (1-x+2*x^2)/((1-x)^3). [From Eric Werley, Jun 27 2011]
Sequences of the form a(n)=n^2+K with offset 0 have o.g.f. (K-2*K*x+K*x^2+x+x^2)/(1-x)^3 and recurrence a(n)=3*a(n-1)-3*a(n-2)+a*(n-3). - R. J. Mathar, Apr 28 2008
a(n)*a(n-2) = (n-1)^4 + 4. [From Reinhard Zumkeller, Feb 12 2009]
For n>1, a(n)^2 +(a(n)+1)^2 +... +(a(n)+n-2)^2 +(a(n)+n-1+a(n)+n)^2 = (n+1) *(6*n^4+18*n^3+26*n^2+19*n+6) /6 = (a(n)+n)^2 +... +(a(n)+2*n)^2. - Charlie Marion, Jan 10 2011
a(n) = 2*a(n-1)-a(n-2)+2. a(n)=a(n-1)+2*n-1.[From Eric Werley, Jun 27 2011]
a(n) = (n-1)^2 + 2(n-1) + 2 = 122 read in base n-1 (for n>3). - Jason Kimberley, Oct 20 2011
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MATHEMATICA
| a[n_]:=n^2+1; [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 15 2008]
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PROG
| (MAGMA) [n^2 + 1: n in [0..50]]; // Vincenzo Librandi, May 01 2011
(PARI) a(n)=n^2+1 \\ Charles R Greathouse IV, Jun 10 2011
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CROSSREFS
| Left edge of A055096.
Cf. A059100, A117950, A087475, A117951, A114949, A117619 (sequences of form n^2 + K).
a(n+1) = A101220(n, n+1, 3).
Cf. A059592, A124808.
Cf. A117950, A132411, A132414, A028872.
Cf. A005408, A000124, A016813, A086514, A000125, A058331, A161701, A161702, A161703, A000127, A161704, A161706, A161707, A161708, A161710, A080856, A161711, A161712, A161713, A161715, A006261. [From Reinhard Zumkeller, Jun 17 2009]
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), this sequence (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Sequence in context: A059591 A082607 A159547 * A069987 A119114 A062493
Adjacent sequences: A002519 A002520 A002521 * A002523 A002524 A002525
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| More terms from Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 28 2006
Partially edited by Joerg Arndt (arndt(AT)jjj.de), Mar 11 2010
In the last comment line an end bracket was missing in my recent comment:A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1), I corrected it. WL Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Oct 28 2010
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