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G.f. A(x) satisfies: A(x) = 1 / ((1 + x) * (1 - x * A(x)^4)).
6

%I #18 Jun 06 2024 08:24:00

%S 1,0,1,4,21,114,651,3844,23301,144169,906866,5782350,37289431,

%T 242793439,1593918916,10538988984,70121101825,469133993094,

%U 3154115695476,21299373321344,144402246424591,982506791975780,6706724412165956,45917245477282994

%N G.f. A(x) satisfies: A(x) = 1 / ((1 + x) * (1 - x * A(x)^4)).

%H Seiichi Manyama, <a href="/A349300/b349300.txt">Table of n, a(n) for n = 0..500</a>

%F a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+3*k,4*k) * binomial(5*k,k) / (4*k+1).

%F a(n) = (-1)^5*F([1/5, 2/5, 3/5, 4/5, (1+n)/3, (2+n)/3, (3+n)/3, -n], [1/4, 1/2, 1/2, 3/4, 3/4, 1, 5/4], 3^3*5^5/2^16), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 14 2021

%F a(n) ~ sqrt(1 - 3*r) / (2 * 5^(3/4) * sqrt(2*Pi*(1+r)) * n^(3/2) * r^(n + 1/4)), where r = 0.136824361675510443450981569282313811786270109272790613523286... is the root of the equation 5^5 * r = 4^4 * (1+r)^4. - _Vaclav Kotesovec_, Nov 14 2021

%F From _Peter Bala_, Jun 02 2024: (Start)

%F A(x) = 1/(1 + x)*F(x/(1 + x)^4), where F(x) = Sum_{n >= 0} A002294(n)*x^n.

%F A(x) = 1/(1 + x) + x*A(x)^5. (End)

%t nmax = 23; A[_] = 0; Do[A[x_] = 1/((1 + x) (1 - x A[x]^4)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t Table[Sum[(-1)^(n - k) Binomial[n + 3 k, 4 k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 23}]

%o (PARI) a(n) = sum(k=0, n, (-1)^(n-k) * binomial(n+3*k,4*k) * binomial(5*k,k) / (4*k+1)); \\ _Michel Marcus_, Nov 14 2021

%Y Cf. A002294, A005043, A346627, A346665, A349290, A349299, A349301, A349302, A349303.

%K nonn,easy

%O 0,4

%A _Ilya Gutkovskiy_, Nov 13 2021