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A080227 a(n) = n*a(n-1) + (1/2)*(1+(-1)^n), a(0)=0.

%I A080227 #24 Mar 27 2023 07:51:06
%S A080227 0,0,1,3,13,65,391,2737,21897,197073,1970731,21678041,260136493,
%T A080227 3381774409,47344841727,710172625905,11362762014481,193166954246177,
%U A080227 3477005176431187,66063098352192553,1321261967043851061,27746501307920872281,610423028774259190183
%N A080227 a(n) = n*a(n-1) + (1/2)*(1+(-1)^n), a(0)=0.
%H A080227 Alois P. Heinz, <a href="/A080227/b080227.txt">Table of n, a(n) for n = 0..450</a>
%F A080227 E.g.f.: (exp(x) + exp(-x) - 2)/(2*(1 - x)).
%F A080227 a(n) = floor((cosh(1)-1)*n!). - _Benoit Cloitre_, Feb 14 2003
%F A080227 a(n) = (n-1)*(a(n-1) + a(n-2)) + 1 for n > 1. - _Gary Detlefs_, Jun 22 2010
%F A080227 a(n) = (1/2)*(exp(-1)*Gamma(n+1,-1) + exp(1)*Gamma(n+1,1)) - Gamma(n+1,0). - _Martin Clever_, Mar 26 2023
%t A080227 c=CoefficientList[Series[(e^x+e^(-x)-2)/(2(1-x)), {x, 0, 25}], x]; For[n = 0, n < 26, n++; Print[c[[n]]*(n - 1)! ]]
%t A080227 Join[{0},RecurrenceTable[{a[1]==0,a[2]==1,a[n]==(n-1)(a[n-1]+a[n-2])+ 1}, a[n],{n,30}]] (* _Harvey P. Dale_, Jul 21 2011 *)
%Y A080227 Cf. A009179.
%K A080227 easy,nonn
%O A080227 0,4
%A A080227 Mario Catalani (mario.catalani(AT)unito.it), Feb 07 2003