cp's OEIS Frontend

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.

%I A381984 #22 Mar 14 2025 09:04:27
%S A381984 1,2,9,94,1649,40146,1246057,47004014,2087644449,106709890114,
%T A381984 6170322084041,398219508589662,28376096583546769,2212797385807852754,
%U A381984 187441592012756668329,17139223549605292448686,1682551982313514625386817,176505773149909540258262274,19704960849698723062181296009
%N 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.
%C A381984 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
%F A381984 a(n) = n! * Sum_{k=0..n} A001764(k)/(n-k)!.
%F A381984 From _Peter Bala_, Mar 13 2025: (Start)
%F A381984 a(n) = hypergeom([-n, 1/3, 2/3], [3/2], -27/4).
%F A381984 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)
%F A381984 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
%p A381984 seq(simplify(hypergeom([-n, 1/3, 2/3], [3/2], -27/4)), n = 0..18); # _Peter Bala_, Mar 13 2025
%t A381984 Table[HypergeometricPFQ[{-n, 1/3, 2/3}, {3/2}, -27/4], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 14 2025 *)
%o A381984 (PARI) a(n) = n!*sum(k=0, n, binomial(3*k+1, k)/((3*k+1)*(n-k)!));
%Y A381984 Cf. A381985, A381986, A381987.
%Y A381984 Cf. A001763, A001764.
%K A381984 nonn,easy
%O A381984 0,2
%A A381984 _Seiichi Manyama_, Mar 11 2025