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%I #16 Apr 20 2018 08:40:13
%S 1,-5,-50,-3875,2500,-2046250,-12409375,-1087687500,13232343750,
%T -907225000000,1545669140625,-362705679687500,6007095839843750,
%U -224713698632812500,2118331116210937500,-226812683210205078125,4765872641563720703125
%N Expansion of Product_{n>=1} (1 + (25*x)^n)^(-1/5).
%C This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = 1/5, g(n) = -25^n.
%C 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
%H Seiichi Manyama, <a href="/A303132/b303132.txt">Table of n, a(n) for n = 0..500</a>
%F 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
%t CoefficientList[Series[(2/QPochhammer[-1, 25*x])^(1/5), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Apr 20 2018 *)
%Y 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).
%Y Cf. A303125, A303136.
%K sign
%O 0,2
%A _Seiichi Manyama_, Apr 19 2018
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