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%I #23 Mar 18 2022 13:02:44
%S 1,2,5,14,57,284,1705,11934,95473,859256,8592561,94518170,1134218041,
%T 14744834532,206427683449,3096415251734,49542644027745,
%U 842224948471664,15160049072489953,288040932377309106,5760818647546182121,120977191598469824540,2661498215166336139881
%N a(0) = 1, a(1) = 2; for n>1, a(n) = n*a(n-1) + (-1)^n.
%C A000166 and A001120 have a similar recurrence.
%H Alois P. Heinz, <a href="/A110043/b110043.txt">Table of n, a(n) for n = 0..450</a>
%F a(n) = A001120(n) + n! = A000166(n) + 2*n! for n>0.
%F a(n) = (n-1)*(a(n-1)+a(n-2)), n>2. - _Gary Detlefs_, Apr 11 2010
%F a(n) = 2*n! + floor((n!+1)/e) for n>0. - _Gary Detlefs_, Apr 11 2010
%F E.g.f.: (2*exp(x)*x+1)*exp(-x)/(1-x). - _Alois P. Heinz_, May 07 2020
%p a:= proc(n) option remember;
%p `if`(n<2, n+1, n*a(n-1)+(-1)^n)
%p end:
%p seq(a(n), n=0..23); # _Alois P. Heinz_, May 07 2020
%t a[n_] := Subfactorial[n] + 2 Boole[n > 0] n!;
%t Table[a[n], {n, 0, 23}] (* _Jean-François Alcover_, Mar 18 2022 *)
%Y Cf. A000142, A000166, A001120.
%Y Column k=2 of A334715.
%K nonn,easy
%O 0,2
%A _Philippe Deléham_, Sep 04 2005
%E a(0)=1 prepended and two terms corrected by _Alois P. Heinz_, May 07 2020
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