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%I #41 Mar 24 2022 04:21:00
%S 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,
%T 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,
%U 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
%N Number of ways to represent n as a sum of an even number of consecutive integers.
%C Equals number of odd divisors of n greater than sqrt(2*n). [Hirschhorn and Hirschhorn]
%C a(n) + A082647(n) = A001227. This follows immediately from the definitions. - _N. J. A. Sloane_, Dec 07 2020
%C 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]
%C Indices of nonzero terms give A281005. - _Omar E. Pol_, Mar 04 2018
%C Indices of zero terms give A082662. - _Omar E. Pol_, Mar 20 2022
%D 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]
%H Antti Karttunen, <a href="/A131576/b131576.txt">Table of n, a(n) for n = 1..65537</a>
%H M. D. Hirschhorn and P. M. Hirschhorn, <a href="http://www.maa.org/sites/default/files/3004420056860.pdf.bannered.pdf">Partitions into Consecutive Parts</a>, Mathematics Magazine: 2005, Volume 78, Number 5, Pages: 396-397.
%F G.f.: Sum_{k>=1} x^(k*(2*k+1))/(1-x^(2*k)). [Corrected by _N. J. A. Sloane_, Dec 18 2020]
%F 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
%F Conjectures: a(n) = (A001227(n) - A067742(n))/2 = A082647(n) - A067742(n). - _Omar E. Pol_, Feb 22 2017
%e 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.
%p 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
%p 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
%t 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 *)
%o (PARI) a(n) = my(s=sqrt(2*n)); sumdiv(n, d, (d % 2) && (d > s)); \\ _Michel Marcus_, Jan 15 2020
%Y Cf. A082647, A082662, A001227, A237593, A279387, A281005.
%K easy,nonn
%O 1,21
%A _Vladeta Jovovic_, Aug 28 2007, Sep 16 2007
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