%I
%S 1,3,6,57,294,1884,13011,95178,712293,5448495,42444375,335392941,
%T 2681006280,21639853488,176113016241,1443450932445,11903668996713,
%U 98695838478585,822212761531101,6878755556938029,57767592614370576,486792969548157129
%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.
%C In general, if h > 1 and g.f. = Product_{k>=1} (1 + h^2*x^k)^(1/h), then a(n) ~ (1)^n * c^(1/h) * h^(2*n1) / (Gamma(1  1/h) * n^(1 + 1/h)), where c = Product_{k>=2} (1 + (1)^k / h^(2*k2)).  _Vaclav Kotesovec_, Apr 22 2018
%F a(n) ~ (1)^n * c^(1/3) * 3^(2*n1) / (Gamma(2/3) * n^(4/3)), where c = Product_{k>=2} (1 + 9*(1/9)^k) = 1.09874828793226302381837574278380702...  _Vaclav Kotesovec_, Apr 22 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
%t nmax = 30; CoefficientList[Series[Product[(1 + 9*x^k)^(1/3), {k, 1, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, 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): A000009 (b=1), A303350 (b=2), this sequence (b=3).
%K sign
%O 0,2
%A _Seiichi Manyama_, Apr 22 2018
