A386366 Expansion of e.g.f. (Sum_{k>=0} binomial(4*k,k) * x^k)^(1/4).

1, 1, 11, 225, 6729, 264885, 12933675, 753953445, 51089936625, 3945857018985, 342128949720075, 32905744117871625, 3476617058554464825, 400259518407091468125, 49874289081145099245675, 6687208401827555535058125, 960003161392360306947350625, 146914452707464363053984476625

Offset: 0

Author

Seiichi Manyama, Jul 19 2025

Keywords

• nonn

Comments

In general, if m > 1 and e.g.f. = (Sum_{k>=0} binomial(m*k,k) * x^k)^(1/m), then a(n) ~ n! * m^(m*n + 1/(2*m)) / (Gamma(1/(2*m)) * 2^(1/(2*m)) * n^(1 - 1/(2*m)) * (m-1)^((m-1)*n + 1/(2*m))). - Vaclav Kotesovec, Jul 19 2025

Crossrefs

Cf. A000142, A007696, A386365.

Cf. A378484.

Programs

Mathematica

nmax = 20; CoefficientList[Series[Sum[Binomial[4*k,k] * x^k, {k, 0, nmax}]^(1/4), {x, 0, nmax}], x] * Range[0,nmax]! (* Vaclav Kotesovec, Jul 19 2025 *)

PARI

my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, binomial(4*k, k)*x^k)^(1/4)))

Formula

a(n) ~ sqrt(Pi) * 2^(8*n + 5/8) * n^(n - 3/8) / (Gamma(1/8) * exp(n) * 3^(3*n + 1/8)). - Vaclav Kotesovec, Jul 19 2025