%I #14 Jun 06 2024 08:24:05
%S 1,0,1,5,31,200,1351,9430,67531,493505,3665981,27602081,210179437,
%T 1615820402,12524590873,97775503808,768083233899,6067097140799,
%U 48159634951855,383965003803985,3073379977522321,24688458872260007,198968304164411309
%N G.f. A(x) satisfies: A(x) = 1 / ((1 + x) * (1 - x * A(x)^5)).
%H Seiichi Manyama, <a href="/A349301/b349301.txt">Table of n, a(n) for n = 0..500</a>
%F a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+4*k,5*k) * binomial(6*k,k) / (5*k+1).
%F a(n) ~ sqrt(1 - 4*r) / (2^(6/5) * 3^(7/10) * sqrt(5*Pi*(1+r)) * n^(3/2) * r^(n + 1/5)), where r = 0.11589193448796656683553561932282471511616945026903125324... is the real root of the equation 6^6 * r = 5^5 * (1+r)^5. - _Vaclav Kotesovec_, Nov 14 2021
%F From _Peter Bala_, Jun 02 2024: (Start)
%F A(x) = 1/(1 + x)*F(x/(1 + x)^5), where F(x) = Sum_{n >= 0} A002295(n)*x^n.
%F A(x) = 1/(1 + x) + x*A(x)^6. (End)
%t nmax = 22; A[_] = 0; Do[A[x_] = 1/((1 + x) (1 - x A[x]^5)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t Table[Sum[(-1)^(n - k) Binomial[n + 4 k, 5 k] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 22}]
%Y Cf. A002295, A005043, A346627, A346666, A349291, A349299, A349300, A349302, A349303.
%K nonn,easy
%O 0,4
%A _Ilya Gutkovskiy_, Nov 13 2021