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A349299 G.f. A(x) satisfies: A(x) = 1 / ((1 + x) * (1 - x * A(x)^3)).

%I A349299 #20 Jun 06 2024 08:23:55
%S A349299 1,0,1,3,13,57,265,1273,6281,31634,161985,840795,4413937,23395376,
%T A349299 125028541,672958929,3644868105,19850452482,108639736105,597190462341,
%U A349299 3295737021241,18253432801480,101425665527825,565249069709027,3158734991846065
%N A349299 G.f. A(x) satisfies: A(x) = 1 / ((1 + x) * (1 - x * A(x)^3)).
%H A349299 Seiichi Manyama, <a href="/A349299/b349299.txt">Table of n, a(n) for n = 0..1000</a>
%F A349299 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+2*k,3*k) * binomial(4*k,k) / (3*k+1).
%F A349299 a(n) = (-1)^n*F([1/4, 1/2, 3/4, (1+n)/2, (2+n)/2, -n], [1/3, 2/3, 2/3, 1, 4/3], 2^10/3^6), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 14 2021
%F A349299 a(n) ~ sqrt(1 - 2*r) / (4 * 2^(1/6) * sqrt(3*Pi*(1+r)) * n^(3/2) * r^(n + 1/3)), where r = 0.16809738261179529189597734361984743421471587505782537522127218... is the root of the equation 4^4 * r = 3^3 * (1+r)^3. - _Vaclav Kotesovec_, Nov 14 2021
%F A349299 From _Peter Bala_, Jun 02 2024: (Start)
%F A349299 A(x) = 1/(1 + x)*F(x/(1 + x)^3), where F(x) = Sum_{n >= 0} A002293(n)*x^n.
%F A349299 A(x) = 1/(1 + x) + x*A(x)^4. (End)
%t A349299 nmax = 24; A[_] = 0; Do[A[x_] = 1/((1 + x) (1 - x A[x]^3)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t A349299 Table[Sum[(-1)^(n - k) Binomial[n + 2 k, 3 k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 24}]
%o A349299 (PARI) a(n) = sum(k=0, n, (-1)^(n-k) * binomial(n+2*k,3*k) * binomial(4*k,k) / (3*k+1)); \\ _Michel Marcus_, Nov 14 2021
%Y A349299 Cf. A002293, A005043, A346627, A346664, A349289, A349300, A349301, A349302, A349303.
%K A349299 nonn,easy
%O A349299 0,4
%A A349299 _Ilya Gutkovskiy_, Nov 13 2021