A162314 Row sums of A162313.

1, 4, 24, 208, 2400, 34624, 599424, 12107008, 279467520, 7257355264, 209403009024, 6646303019008, 230126121738240, 8632047179874304, 348695526455476224, 15091839203924574208, 696733490476660162560

Offset: 0

Author

Peter Bala, Jul 01 2009

Keywords

• easy
• nonn

Comments

Conjecture: for fixed k = 1,2,..., the sequence obtained by reducing a(n) modulo k is eventually periodic with the exact period dividing phi(k), where phi(k) is the Euler totient function A000010. For example, modulo 24 the sequence becomes [1, 4, 0, 16, 0 16, 0, 16, ...] with an apparent period of 2 beginning at a(2). - Peter Bala, Jul 08 2022

Crossrefs

Cf. A000629, A162313.

Programs

Maple

#A162314
with(combinat):
a:= n -> 2^n*add(k!*Stirling2(n+1,k+1), k = 0..n):
seq(a(n), n = 0..16);

Formula

a(n) = 2^n*A000629(n) = 2^n*Sum_{k = 0..n} k!*Stirling2(n+1,k+1).

E.g.f.: exp(2*x)/(2-exp(2*x)) = 1 + 4*x + 24*x^2/2! + 208*x^3/3! + ....

G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 8*x*(k+1)/(8*x*(k+1) - 1 + 4*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013

From Peter Bala, Jul 08 2022: (Start)

a(n) = Sum_{k = 0..n} (-2)^(n+k)*k!*Stirling2(n,k).

Conjectural o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 4*x/(1 - 2*x/(1 - 8*x/(1 - 4*x/(1 - ... - 3*n*x/(1 - 2*n*x/(1 - ...))))))). (End)