A349311 - OEIS

%I #13 Nov 15 2021 08:56:34

%S 1,2,12,112,1232,14832,189184,2512064,34358784,480745984,6848734464,

%T 99003237376,1448575666176,21411827808256,319255531155456,

%U 4796005997940736,72520546008219648,1102912584949792768,16859182461720526848,258886644574700699648

%N G.f. A(x) satisfies: A(x) = (1 + x * A(x)^5) / (1 - x).

%H Seiichi Manyama, <a href="/A349311/b349311.txt">Table of n, a(n) for n = 0..500</a>

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

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

%F a(n) ~ sqrt(1 + 4*r) / (2 * 5^(3/4) * sqrt(2*Pi) * (1-r)^(1/4) * n^(3/2) * r^(n + 1/4)), where r = 0.0600920016324256496641829206872407657377702010870270617... is the real root of the equation 4^4 * (1-r)^5 = 5^5 * r. - _Vaclav Kotesovec_, Nov 15 2021

%t nmax = 19; A[_] = 0; Do[A[x_] = (1 + x A[x]^5)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t Table[Sum[Binomial[n + 4 k, 5 k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 19}]

%Y Cf. A002294, A006318, A346626, A346647, A349310, A349312, A349313, A349314.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Nov 14 2021