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%I #18 Apr 22 2021 08:30:08
%S 1,0,1,0,0,2,0,2,3,0,4,4,1,10,5,6,16,6,14,28,10,32,40,18,63,60,42,112,
%T 83,84,187,124,172,300,186,320,456,302,581,684,507,982,1004,874,1624,
%U 1476,1508,2566,2174,2582,3981,3262,4338,6002,4945,7138,8947,7660
%N Expansion of Product_{k>=1} (1+x^(3*k-1))^k.
%C In general, if s>0, t>0, GCD(s,t)=1 and g.f. = Product_{k>=1} (1 + x^(s*k-t))^k then a(n) ~ 2^(t^2/(2*s^2) - 3/4) * s^(2/3) * Zeta(3)^(1/6) * exp(-Pi^4 * t^2 / (1296 * s^2 * Zeta(3)) + Pi^2 * t * 2^(1/3) * 3^(2/3) * s^(2/3) * n^(1/3) / (36 * s^2 * Zeta(3)^(1/3)) + 3^(4/3) * Zeta(3)^(1/3) * n^(2/3) / (2^(4/3) * s^(2/3)) ) / (3^(1/3) * s * sqrt(Pi) * n^(2/3)). - _Vaclav Kotesovec_, Oct 12 2015
%H Vaclav Kotesovec, <a href="/A262878/b262878.txt">Table of n, a(n) for n = 0..10000</a>
%H Vaclav Kotesovec, <a href="http://arxiv.org/abs/1509.08708">A method of finding the asymptotics of q-series based on the convolution of generating functions</a>, arXiv:1509.08708 [math.CO], Sep 30 2015
%F a(n) ~ exp(2^(-4/3) * 3^(2/3) * Zeta(3)^(1/3) * n^(2/3) + Pi^2 * n^(1/3) / (2^(5/3)*3^(8/3) * Zeta(3)^(1/3)) - Pi^4/(11664*Zeta(3))) * Zeta(3)^(1/6) / (2^(25/36) * 3^(2/3) * sqrt(Pi) * n^(2/3)).
%p with(numtheory):
%p b:= n-> `if`(n<3, n-1, (p-> [0, -r, 2*r, 0, 0, 2*r+1][p]
%p )(1+irem(n+3, 6, 'r'))):
%p a:= proc(n) option remember; `if`(n=0, 1, add(add(
%p d*b(d), d=divisors(j))*a(n-j), j=1..n)/n)
%p end:
%p seq(a(n), n=0..60); # _Alois P. Heinz_, Oct 05 2015
%t nmax=100; CoefficientList[Series[Product[(1+x^(3k-1))^k, {k, 1, nmax}], {x, 0, nmax}], x]
%t nmax=100; CoefficientList[Series[E^Sum[(-1)^(j+1)/j*x^(2*j)/(1-x^(3j))^2,{j,1,nmax}],{x,0,nmax}],x]
%Y Cf. A026007, A027346, A035528, A262876, A262877, A262879, A262884, A262948, A263138, A263145.
%K nonn
%O 0,6
%A _Vaclav Kotesovec_, Oct 04 2015
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