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%I #42 Apr 21 2023 02:47:30
%S 1,1,4,16,88,568,4288,36832,354688,3781504,44199424,561823744,
%T 7714272256,113769309184,1793341407232,30085661765632,535170830467072,
%U 10060645294440448,199287423535808512,4148644277780217856
%N Expansion of e.g.f. exp(-2*x)/(1-x)^3.
%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.64(b).
%H Vincenzo Librandi, <a href="/A052124/b052124.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = n*a(n-1) + 2*(n-1)*a(n-2). - Detlef Pauly (dettodet(AT)yahoo.de), Sep 22 2003
%F a(n) = (n+5)*(n+2)*n! * Sum_{k=0..n} (-1)^k*2^(k+2)*(k+3)/(k+5)!. - _Vaclav Kotesovec_, Oct 28 2012
%F G.f.: 1/Q(0), where Q(k) = 1 + 2*x - x*(k+3)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, Apr 22 2013
%F a(n) ~ n!*(n+5)*(n+2)/(2*exp(2)). - _Vaclav Kotesovec_, Jun 15 2013
%F From _Peter Bala_, Sep 20 2013: (Start)
%F a(n) ~ (1/2)*n^2*n!/e^2 for large n.
%F Letting n -> infinity in the above series for a(n) given by Kotesovec gives the series expansion 1/e^2 = Sum_{k >= 0} (-1)^k*(k+3)*2^(k+3)/(k+5)!.
%F The sequence b(n) := (1/2)*n!*(n+2)*(n+5) satisfies the recurrence for a(n) given above by Pauly but with the starting values b(0) = 5 and b(1) = 9. This leads to the finite continued fraction expansion a(n) = (1/2)*n!*(n+2)*(n+5)( 1/(5 + 4/(1 + 2/(2 + 4/(3 + ... + 2*(n-1)/n)))) ), valid for n >= 2. Letting n -> infinity in the previous result gives the infinite continued fraction expansion 1/e^2 = 1/(5 + 4/(1 + 2/(2 + 4/(3 + ... + 2*(n-1)/(n + ...))))). Cf. A082031. (End)
%p A052124 := proc(n) option remember; if n <=1 then 1 else n*A052124(n-1)+2*(n-1)*A052124(n-2); fi; end; # Detlef Pauly
%t Table[(n+5)*(n+2)*n!*Sum[(-1)^k*2^(k+2)*(k+3)/(k+5)!,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 28 2012 *)
%t With[{nn=20},CoefficientList[Series[Exp[(-2x)]/(1-x)^3,{x,0,nn}],x] Range[ 0,nn]!] (* _Harvey P. Dale_, Oct 23 2017 *)
%o (PARI) my(x='x+O('x^25)); Vec(serlaplace( exp(-2*x)/(1-x)^3)) \\ _Michel Marcus_, Oct 25 2021
%o (Python)
%o from math import factorial
%o from fractions import Fraction
%o def A052124(n): return int((n+5)*(n+2)*factorial(n)*sum(Fraction((-1 if k&1 else 1)*(k+3)<<k+2,factorial(k+5)) for k in range(n+1))) # _Chai Wah Wu_, Apr 20 2023
%Y Cf. A052127, A082031.
%K nonn
%O 0,3
%A _N. J. A. Sloane_, Jan 23 2000
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