A263148 Expansion of Product_{k>=1} (1+x^(5*k-4))^k.

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, 7, 40, 63, 33, 3, 8, 60, 112, 74, 14, 9, 80, 187, 161, 46, 11, 110, 300, 308, 120, 23, 140, 455, 568, 283, 53, 182, 672, 968, 594, 145, 228, 963, 1609, 1172

Offset: 0

Author

Vaclav Kotesovec, Oct 10 2015

Keywords

• nonn

Comments

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)).

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..10000
• Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015

Crossrefs

Cf. A263145, A263146, A263147, A263144.

Programs

Mathematica

nmax = 100; CoefficientList[Series[Product[(1+x^(5k-4))^k, {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 100; CoefficientList[Series[E^Sum[(-1)^(j+1)/j*x^j/(1 - x^(5*j))^2, {j, 1, nmax}], {x, 0, nmax}], x]

Formula

G.f.: exp(Sum_{j>=1} (-1)^(j+1)/j*x^j/(1 - x^(5*j))^2).

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)).