A303353 - OEIS

%I #19 Apr 25 2018 03:29:38

%S 1,-3,15,-120,915,-7086,56661,-462405,3819165,-31843110,267610443,

%T -2263491255,19246265025,-164379723735,1409306287470,-12122528944620,

%U 104575462390842,-904411297029585,7839310835762475,-68086561401745275,592417977205534017

%N Expansion of Product_{n>=1} 1/(1 + 9*x^n)^(1/3).

%C This sequence is obtained from the generalized Euler transform in A266964 by taking f(n) = 1/3, g(n) = -9.

%H Vaclav Kotesovec, <a href="/A303353/b303353.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) ~ c * (-9)^n / n^(2/3), where c = 1 / (Gamma(1/3) * QPochhammer[-1/9]^(1/3)) = 0.361746646328749408912877789757526727... - _Vaclav Kotesovec_, Apr 25 2018

%p seq(coeff(series(mul(1/(1+9*x^k)^(1/3), k = 1..n), x, n+1), x, n), n = 0..25); # _Muniru A Asiru_, Apr 22 2018

%t nmax = 20; CoefficientList[Series[Product[1/(1 + 9*x^k)^(1/3), {k, 1, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Apr 25 2018 *)

%Y Expansion of Product_{n>=1} 1/(1 + b^2*x^n)^(1/b): A081362 (b=1), A303352 (b=2), this sequence (b=3).

%Y Cf. A303349.

%K sign

%O 0,2

%A _Seiichi Manyama_, Apr 22 2018