A381984 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^3 is the g.f. of A001764.

1, 2, 9, 94, 1649, 40146, 1246057, 47004014, 2087644449, 106709890114, 6170322084041, 398219508589662, 28376096583546769, 2212797385807852754, 187441592012756668329, 17139223549605292448686, 1682551982313514625386817, 176505773149909540258262274, 19704960849698723062181296009

Offset: 0

Author

Seiichi Manyama, Mar 11 2025

Keywords

• nonn
• easy

Comments

For each positive integer k, the sequence obtained by reducing a(n) modulo k is a periodic sequence with period dividing k. For example, modulo 6 the sequence becomes [1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, ...] with period 6. Cf. A047974. - Peter Bala, Mar 13 2025

Crossrefs

Cf. A381985, A381986, A381987.

Cf. A001763, A001764.

Programs

Maple

seq(simplify(hypergeom([-n, 1/3, 2/3], [3/2], -27/4)), n = 0..18); # Peter Bala, Mar 13 2025

Mathematica

Table[HypergeometricPFQ[{-n, 1/3, 2/3}, {3/2}, -27/4], {n, 0, 20}] (* Vaclav Kotesovec, Mar 14 2025 *)

PARI

a(n) = n!*sum(k=0, n, binomial(3*k+1, k)/((3*k+1)*(n-k)!));

Formula

a(n) = n! * Sum_{k=0..n} A001764(k)/(n-k)!.

From Peter Bala, Mar 13 2025: (Start)

a(n) = hypergeom([-n, 1/3, 2/3], [3/2], -27/4).

2*(2*n + 1)*a(n) = (27*n^2 - 19*n + 4)*a(n-1) - 2*(n - 1)*(27*n - 25)*a(n-2) + 27*(n - 1)*(n - 2)*a(n-3) with a(0) = 0, a(1) = 2 and a(2) = 9. (End)

a(n) ~ 3^(3*n + 1/2) * n^(n + 1/2) / (2^(2*n + 3/2) * exp(n - 4/27) * n^(3/2)). - Vaclav Kotesovec, Mar 14 2025