|
%I #23 Apr 25 2018 03:50:09
%S 1,-3,-12,-39,-246,-1578,-11487,-84054,-635781,-4893357,-38292969,
%T -303553209,-2432865630,-19678331838,-160427322399,-1316796234933,
%U -10872602692581,-90242886252945,-752488383572787,-6300541703215803,-52949782408528290
%N Expansion of Product_{n>=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 Seiichi Manyama, <a href="/A303348/b303348.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) ~ -c * 3^(2*n-1) / (Gamma(2/3) * n^(4/3)), where c = QPochhammer[1/9]^(1/3) = 0.95703379660353017269195329... - _Vaclav Kotesovec_, Apr 25 2018
%p seq(coeff(series(mul((1-9*x^k)^(1/3), k=1..n), x, n+1), x, n), n=0..25); # _Muniru A Asiru_, Apr 22 2018
%o (PARI) N=66; x='x+O('x^N); Vec(prod(k=1, N, (1-9*x^k)^(1/3)))
%Y Expansion of Product_{n>=1} (1 - b^2*x^n)^(1/b): A010815 (b=1), A303347 (b=2), this sequence (b=3).
%Y Cf. A303349, A303351.
%K sign
%O 0,2
%A _Seiichi Manyama_, Apr 22 2018
|
