A303132 Expansion of Product_{n>=1} (1 + (25*x)^n)^(-1/5).

1, -5, -50, -3875, 2500, -2046250, -12409375, -1087687500, 13232343750, -907225000000, 1545669140625, -362705679687500, 6007095839843750, -224713698632812500, 2118331116210937500, -226812683210205078125, 4765872641563720703125

Offset: 0

Author

Seiichi Manyama, Apr 19 2018

Keywords

• sign

Comments

This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = 1/5, g(n) = -25^n.

In general, for h>=1, if g.f. = Product_{k>=1} (1 + (h^2*x)^k)^(-1/h), then a(n) ~ (-1)^n * exp(Pi*sqrt(n/(6*h))) * h^(2*n) / (2^(7/4) * 3^(1/4) * h^(1/4) * n^(3/4)). - Vaclav Kotesovec, Apr 20 2018

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..500

Crossrefs

Expansion of Product_{n>=1} (1 + ((b^2)*x)^n)^(-1/b): A081362 (b=1), A298993 (b=2), A303130 (b=3), A303131 (b=4), this sequence (b=5).

Cf. A303125, A303136.

Programs

Mathematica

CoefficientList[Series[(2/QPochhammer[-1, 25*x])^(1/5), {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 20 2018 *)

Formula

a(n) ~ (-1)^n * exp(Pi*sqrt(n/30)) * 5^(2*n - 1/4) / (2^(7/4) * 3^(1/4) * n^(3/4)). - Vaclav Kotesovec, Apr 20 2018