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%I #12 Jul 18 2022 19:46:42
%S 1,4,24,208,2400,34624,599424,12107008,279467520,7257355264,
%T 209403009024,6646303019008,230126121738240,8632047179874304,
%U 348695526455476224,15091839203924574208,696733490476660162560
%N Row sums of A162313.
%C 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
%F a(n) = 2^n*A000629(n) = 2^n*Sum_{k = 0..n} k!*Stirling2(n+1,k+1).
%F E.g.f.: exp(2*x)/(2-exp(2*x)) = 1 + 4*x + 24*x^2/2! + 208*x^3/3! + ....
%F 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
%F From _Peter Bala_, Jul 08 2022: (Start)
%F a(n) = Sum_{k = 0..n} (-2)^(n+k)*k!*Stirling2(n,k).
%F 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)
%p #A162314
%p with(combinat):
%p a:= n -> 2^n*add(k!*Stirling2(n+1,k+1), k = 0..n):
%p seq(a(n), n = 0..16);
%Y Cf. A000629, A162313.
%K easy,nonn
%O 0,2
%A _Peter Bala_, Jul 01 2009
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