A131576 Number of ways to represent n as a sum of an even number of consecutive integers.

0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 2, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 2, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 1, 1, 0, 1, 1, 3, 1, 1, 2, 1, 0, 2, 1, 1, 1, 2, 1, 2, 0, 1, 2, 1, 1, 2, 1, 2, 0, 1, 1, 2, 1, 1, 2, 1, 0, 4

Offset

1,21

Comments

Equals number of odd divisors of n greater than sqrt(2*n). [Hirschhorn and Hirschhorn]

a(n) + A082647(n) = A001227. This follows immediately from the definitions. - N. J. A. Sloane, Dec 07 2020

Conjecture: a(n) is also the number of pairs of subparts in the symmetric representation of sigma(n) which are mirror images of each other in the main diagonal. (Cf. A279387). - Omar E. Pol, Feb 22 2017 [Conjecture clarified by N. J. A. Sloane, Dec 16 2020]

Indices of nonzero terms give A281005. - Omar E. Pol, Mar 04 2018

Indices of zero terms give A082662. - Omar E. Pol, Mar 20 2022

References

M. D. Hirschhorn and P. M. Hirschhorn, Partitions into Consecutive Parts, Mathematics Magazine, 78:5 (2005), 396-398. [Please do not delete this reference. - N. J. A. Sloane, Dec 16 2020]

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

M. D. Hirschhorn and P. M. Hirschhorn, Partitions into Consecutive Parts, Mathematics Magazine: 2005, Volume 78, Number 5, Pages: 396-397.

Formula

G.f.: Sum_{k>=1} x^(k*(2*k+1))/(1-x^(2*k)). [Corrected by N. J. A. Sloane, Dec 18 2020]

a(A000040(i))=1 for i=1,2,3,... a(A000079(j))=0 for j=0,1,2,3,... - R. J. Mathar, Sep 13 2007

Conjectures: a(n) = (A001227(n) - A067742(n))/2 = A082647(n) - A067742(n). - Omar E. Pol, Feb 22 2017

Example

a(11)=1 because we have 11=5+6; a(21)=2 because we have 21=10+11=1+2+3+4+5+6; a(75)=3 because we have 75=37+38=10+11+12+13+14+15=3+4+5+6+7+8+9+10+11+12.

Maple

G:=sum(x^(k*(2*k+1))/(1-x^(2*k)), k=1..10): Gser:=series(G, x=0, 85): seq(coeff(Gser, x, n), n=1..80); # Emeric Deutsch, Sep 08 2007
A131576 := proc(n) local dvs, a, k, r; dvs := numtheory[divisors](n) ; a := 0 ; for k in dvs do r := n/k+1 ; if r mod 2 = 0 then if r/2-k >= 1 then a := a+1 ; fi ; fi ; od: RETURN(a) ; end: seq(A131576(n), n=1..120) ; # R. J. Mathar, Sep 13 2007

Mathematica

With[{m = 105}, Rest@ CoefficientList[Series[Sum[x^(k (2 k + 1))/(1 - x^(2 k)), {k, m}], {x, 0, m}], x]] (* Michael De Vlieger, Mar 04 2018 *)

Prog

(PARI) a(n) = my(s=sqrt(2*n)); sumdiv(n, d, (d % 2) && (d > s)); \\ Michel Marcus, Jan 15 2020

Crossrefs

Cf. A082647, A082662, A001227, A237593, A279387, A281005.

Sequence in context: A345007 A026920 A060763 * A341675 A358218 A100073

Adjacent sequences: A131573 A131574 A131575 * A131577 A131578 A131579

Keyword

easy,nonn

Author

Vladeta Jovovic, Aug 28 2007, Sep 16 2007

Status

approved