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A090723 a(1) = 1; for n > 1, number of partitions of n into distinct odd parts such that the two largest parts differ by 2.

%I A090723 #15 Jun 15 2025 04:25:48
%S A090723 1,0,0,1,0,0,0,1,1,0,0,1,1,0,1,2,1,0,1,2,2,1,1,3,3,1,2,4,3,2,3,5,5,3,
%T A090723 4,7,6,4,6,9,8,6,8,11,11,9,10,15,15,11,14,19,18,16,19,24,24,21,24,31,
%U A090723 31,27,32,40,39,36,41,49,50,47,52,63,64,59,67,79,79,77,85,98,101,97,106
%N A090723 a(1) = 1; for n > 1, number of partitions of n into distinct odd parts such that the two largest parts differ by 2.
%C A090723 Also number of partitions of n such that if k is the largest part, then 1 occurs twice, each of 2,3,...,k-1 occurs a positive even number of times and k occurs an odd number of times. Example: a(25)=3 because we have [5,4,4,3,3,2,2,1,1], [3,3,3,3,3,2,2,2,2,1,1] and [3,2,2,2,2,2,2,2,2,2,2,1,1]. - _Emeric Deutsch_, Mar 29 2006
%F A090723 From _Emeric Deutsch_, Mar 29 2006: (Start)
%F A090723 G.f.: x+sum(x^(4k)*product(1+x^(2j-1), j=1..k-1), k=1..infinity).
%F A090723 G.f.: sum(x^(k^2)/product(1-x^(2j), j=2..k), k=1..infinity). (End)
%F A090723 a(n) ~ Pi * exp(Pi*sqrt(n/6)) / (2^(9/4) * 3^(3/4) * n^(5/4)). - _Vaclav Kotesovec_, Jun 15 2025
%e A090723 a(25)=3 as 25=1+3+5+7+9=5+9+11=1+11+13.
%p A090723 g:=sum(x^(k^2)/product(1-x^(2*j),j=2..k),k=1..10): gser:=series(g,x=0,95): seq(coeff(gser,x,n),n=1..92); # _Emeric Deutsch_, Mar 29 2006
%t A090723 nmax = 100; Rest[CoefficientList[Series[x + Sum[x^(4*k)*Product[(1 + x^(2*j - 1)), {j, 1, k-1}], {k, 1, nmax}], {x, 0, nmax}], x]] (* _Vaclav Kotesovec_, Jun 15 2025 *)
%o A090723 (PARI) { v=[1,0,0,1,0,0,0,1,1,0,0,1,1,0,1,2,1,0,1,2,2,1,1,3,3,1,2,4,3,2]; for (i=2,30,c=0; forstep (j=i,1,-2,c+=v[j]); print1(c",")) }
%Y A090723 a(n) = A000700(n) - A000700(n-2), n > 2.
%K A090723 nonn
%O A090723 1,16
%A A090723 _Jon Perry_, Feb 06 2004
%E A090723 More terms from _Vladeta Jovovic_, Feb 10 2004