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a(n) = n*a(n-1) + n^(1+n mod 2), a(0) = 1.
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%I #28 May 19 2021 03:25:14

%S 1,2,6,19,80,401,2412,16885,135088,1215793,12157940,133737341,

%T 1604848104,20863025353,292082354956,4381235324341,70099765189472,

%U 1191696008221025,21450528147978468,407560034811590893,8151200696231817880,171175214620868175481

%N a(n) = n*a(n-1) + n^(1+n mod 2), a(0) = 1.

%H Alois P. Heinz, <a href="/A344317/b344317.txt">Table of n, a(n) for n = 0..450</a>

%F E.g.f.: (1+(x+1)*sinh(x))/(1-x).

%F a(n) = A155521(n-1) + A344262(n) for n > 0.

%F Lim_{n->infinity} a(n)/n! = 1+2*sinh(1) = 1+e-1/e = 1+A174548. - _Amrit Awasthi_, May 19 2021

%p a:= proc(n) a(n):= n*a(n-1) + n^(1+n mod 2) end: a(0):= 1:

%p seq(a(n), n=0..23);

%Y Cf. A155521, A174548, A344229, A344262.

%K nonn

%O 0,2

%A _Alois P. Heinz_, May 14 2021