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Expansion of Product_{k>=1} (1+x^(5*k-4))^k.
5

%I #9 Oct 12 2015 10:33:36

%S 1,1,0,0,0,0,2,2,0,0,0,3,4,1,0,0,4,10,6,0,0,5,16,14,3,0,6,28,32,10,0,

%T 7,40,63,33,3,8,60,112,74,14,9,80,187,161,46,11,110,300,308,120,23,

%U 140,455,568,283,53,182,672,968,594,145,228,963,1609,1172

%N Expansion of Product_{k>=1} (1+x^(5*k-4))^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)).

%H Vaclav Kotesovec, <a href="/A263148/b263148.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 G.f.: exp(Sum_{j>=1} (-1)^(j+1)/j*x^j/(1 - x^(5*j))^2).

%F a(n) ~ 2^(57/100) * 3^(2/3) * 5^(2/3) * Zeta(3)^(1/6) * exp(-Pi^4/(2025*Zeta(3)) + Pi^2 * 3^(2/3) * 2^(1/3) * 5^(2/3) * n^(1/3) / (225*Zeta(3)^(1/3)) + Zeta(3)^(1/3) * 3^(4/3) * 2^(2/3) * 5^(1/3) * n^(2/3) / 20) / (30 * sqrt(Pi) * n^(2/3)).

%t nmax = 100; CoefficientList[Series[Product[(1+x^(5k-4))^k, {k, 1, nmax}], {x, 0, nmax}], x]

%t nmax = 100; CoefficientList[Series[E^Sum[(-1)^(j+1)/j*x^j/(1 - x^(5*j))^2, {j, 1, nmax}], {x, 0, nmax}], x]

%Y Cf. A263145, A263146, A263147, A263144.

%K nonn

%O 0,7

%A _Vaclav Kotesovec_, Oct 10 2015