A381987 E.g.f. A(x) satisfies A(x) = exp(x) * B(x), where B(x) = 1 + x*B(x)^4 is the g.f. of A002293.

1, 2, 11, 160, 3941, 134486, 5851327, 309520436, 19283504585, 1382980764106, 112223497464371, 10165461405056552, 1016801830348902061, 111312715288354681310, 13237965546409421546471, 1699516550894276788156156, 234263144339070269872076177, 34507561203827621878485498386

Offset

0,2

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 5 the sequence becomes [1, 2, 1, 0, 1, 1, 2, 1, 0, 1, ...] with period 5. In particular, a(5*n+3) == 0 (mod 5). Cf. A047974. - Peter Bala, Mar 13 2025

Links

Table of n, a(n) for n=0..17.

Formula

a(n) = n! * Sum_{k=0..n} A002293(k)/(n-k)!.

From Peter Bala, Mar 13 2025: (Start)

a(n) = hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27).

3*(3*n - 1)*(3*n + 1)*a(n) = n*(256*n^2 - 303*n + 95)*a(n-1) - 3*(n - 1)*(256*n^2 - 485*n + 245)*a(n-2) + 3*(256*n - 375)*(n - 1)*(n - 2)*a(n-3) - 256*(n - 1)*(n - 2)*(n - 3)*a(n-4) with a(0) = 1, a(1) = 2, a(2) = 11 and a(3) = 160. (End)

a(n) ~ 2^(8*n+1) * n^(n-1) / (3^(3*n + 3/2) * exp(n - 27/256)). - Vaclav Kotesovec, Mar 14 2025

Maple

seq(simplify(hypergeom([-n, 1/2, 1/4, 3/4], [2/3, 4/3], -256/27)), n = 0..17); # Peter Bala, Mar 13 2025

Mathematica

Table[HypergeometricPFQ[{-n, 1/2, 1/4, 3/4}, {2/3, 4/3}, -256/27], {n, 0, 20}] (* Vaclav Kotesovec, Mar 14 2025 *)

Prog

(PARI) a(n) = n!*sum(k=0, n, binomial(4*k+1, k)/((4*k+1)*(n-k)!));

Crossrefs

Cf. A381984, A381988, A381989.

Cf. A002293, A365340.

Sequence in context: A275923 A288560 A318173 * A349639 A067968 A295269

Adjacent sequences: A381984 A381985 A381986 * A381988 A381989 A381990

Keyword

nonn,easy

Author

Seiichi Manyama, Mar 12 2025

Status

approved