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A007310 Numbers congruent to 1 or 5 mod 6. 141
1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003

Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).

Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004

Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007

Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016

A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008

Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008

For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008

A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009

Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010

If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010

A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010

Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010

Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011

A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013

(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013

Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014

Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014

a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015

a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015

Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016

Conjecture: a(n) are the only values of m such that both p^m+6^k and p^m-6^k are positive primes, for all k>0, and for multiple values of prime p.  This does NOT hold true if a constant other than 6 is used, nor if, instead of using primes, the p values are selected from a(n). - Richard R. Forberg, Jun 30 2016

This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016

REFERENCES

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961.

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

R. J. Cano, b-file generator written in C.

B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers

William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N)

Eric Weisstein's World of Mathematics, Rough Number

Eric Weisstein, Pi Formulas [From Jaume Oliver Lafont, Oct 23 2009]

Index entries for sequences related to smooth numbers [From Michael B. Porter, Oct 09 2009]

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

FORMULA

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito (antonio.b.esposito(AT)italtel.it), Jan 18 2002

a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula

a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006

a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006

1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006

For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007

Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ... O.g.f.: x(1+4x+x^2)/((1+x)(1-x)^2). - R. J. Mathar, May 23 2008

a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008

1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009

A169611(a(n)) = 0. - Juri-Stepan Gerasimov, Dec 03 2009

a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010

a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010

a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010

Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011

a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011

a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012

a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012

1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013

1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013

GCD(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013

a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014

a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016

From Mikk Heidemaa, Feb 11 2016: (Start)

a(n) = 2*floor(3*n/2) - 1.

a(n) = A047238(n+1) - 1. (Suggested by Michel Marcus.) (End)

E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016

From Bruno Berselli, Apr 27 2017: (Start)

a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:

k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;

k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;

k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;

k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)

From Antti Karttunen, May 20 2017: (Start)

a(A273669(n)) = 5*a(n) = A084967(n).

a((5*n)-3) = A255413(n).

A126760(a(n)) = n. (End)

EXAMPLE

G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...

MAPLE

P:=(j, n)-> sum(k^j, k=1..n): for n from 1 to 149 do if (P(14, n) mod n = 0) then print(n) fi od; # Gary Detlefs, Dec 27 2011

# Alternative:

seq(seq(6*i+j, j=[1, 5]), i=0..100); # Robert Israel, Sep 08 2014

MATHEMATICA

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* or *) LinearRecurrence[{1, 1, -1}, {1, 5, 7}, 50] (* Harvey P. Dale, Aug 27 2013 *)

rem156[n_] := (6 n + (-1)^n - 3)/2; Array[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)

Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)

Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)

PROG

(PARI) j=[1]; for(n=0, 1000, if(sqrtint(4!*(n+1) + 1) == ceil(sqrt(4!*(n+1) + 1)), j=concat(j, sqrtint(4!*(n+1) + 1)))); j \\ Alexander R. Povolotsky, Sep 09 2008

(PARI) isA007310(n) = gcd(n, 6)==1 \\ Michael B. Porter, Oct 09 2009

(PARI) A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014

(PARI) \\ given an element from the sequence, find the next term in the sequence.

nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016

(Sage) [i for i in range(150) if gcd(6, i) == 1] # Zerinvary Lajos, Apr 21 2009

(Haskell)

a007310 n = a007310_list !! (n-1)

a007310_list = 1 : 5 : map (+ 6) a007310_list

-- Reinhard Zumkeller, Jan 07 2012

(C) See Cano Link.

(MAGMA) [n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016

CROSSREFS

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.

Cf. A000330, A001580, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.

For k-rough numbers with other values of k, see A000027, A005408, A007310, A007775, A008364-A008366, A166061, A166063.

Cf. A126760 (a left inverse).

Row 3 of A260717 (without the initial 1).

Sequence in context: A106571 A067291 A286265 * A069040 A070191 A231810

Adjacent sequences:  A007307 A007308 A007309 * A007311 A007312 A007313

KEYWORD

nonn,easy,uned,changed

AUTHOR

C. Christofferson (Magpie56(AT)aol.com)

STATUS

approved

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Last modified May 27 15:15 EDT 2017. Contains 287207 sequences.