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A110149 a(0) = 1, a(1) = 3; for n>1, a(n) = n*a(n-1) + (-1)^n.

%I A110149 #19 May 08 2020 18:16:09
%S A110149 1,3,7,20,81,404,2425,16974,135793,1222136,12221361,134434970,
%T A110149 1613219641,20971855332,293605974649,4404089619734,70465433915745,
%U A110149 1197912376567664,21562422778217953,409686032786141106,8193720655722822121,172068133770179264540
%N A110149 a(0) = 1, a(1) = 3; for n>1, a(n) = n*a(n-1) + (-1)^n.
%C A110149 A000166, A001120 and A110043 have a similar recurrence.
%C A110149 a(n) = (n-1)*(a(n-1)+a(n-2)), n>2. - _Gary Detlefs_, Apr 11 2010
%H A110149 Alois P. Heinz, <a href="/A110149/b110149.txt">Table of n, a(n) for n = 0..450</a>
%F A110149 a(n) = A110043(n) + n! = A001120(n) + 2*n! = A000166(n) + 3*n! for n>0.
%F A110149 a(n) = 3*n! + floor((n!+1)/e) for n>0. - _Gary Detlefs_, Apr 11 2010
%F A110149 E.g.f.: (3*exp(x)*x+1)*exp(-x)/(1-x). - _Alois P. Heinz_, May 07 2020
%p A110149 a:= proc(n) option remember;
%p A110149       `if`(n<2, 2*n+1, n*a(n-1)+(-1)^n)
%p A110149     end:
%p A110149 seq(a(n), n=0..23);  # _Alois P. Heinz_, May 07 2020
%t A110149 RecurrenceTable[{a[1]==3,a[n]==n a[n-1]+(-1)^n},a,{n,20}]  (* _Harvey P. Dale_, Nov 21 2011 *)
%Y A110149 Column k=3 of A334715.
%K A110149 nonn,easy
%O A110149 0,2
%A A110149 _Philippe Deléham_, Sep 04 2005
%E A110149 a(0)=1 prepended by _Alois P. Heinz_, May 07 2020