A052124 Expansion of e.g.f. exp(-2*x)/(1-x)^3.

1, 1, 4, 16, 88, 568, 4288, 36832, 354688, 3781504, 44199424, 561823744, 7714272256, 113769309184, 1793341407232, 30085661765632, 535170830467072, 10060645294440448, 199287423535808512, 4148644277780217856, 90545807649965080576, 2067407731760475406336, 49285894020028992323584

Offset: 0

Author

N. J. A. Sloane, Jan 23 2000

Keywords

• nonn

References

• R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.64(b).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A052127, A082031.

Cf. A000023, A087981.

Cf. A000153, A137775.

Programs

Maple

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

Mathematica

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 *)
With[{nn=20},CoefficientList[Series[Exp[(-2x)]/(1-x)^3,{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Oct 23 2017 *)

PARI

my(x='x+O('x^25)); Vec(serlaplace( exp(-2*x)/(1-x)^3)) \\ Michel Marcus, Oct 25 2021

Python

from math import factorial
from fractions import Fraction
def A052124(n): return int((n+5)*(n+2)*factorial(n)*sum(Fraction((-1 if k&1 else 1)*(k+3)<Chai Wah Wu, Apr 20 2023

Formula

a(n) = n*a(n-1) + 2*(n-1)*a(n-2). - Detlef Pauly (dettodet(AT)yahoo.de), Sep 22 2003

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

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

a(n) ~ n!*(n+5)*(n+2)/(2*exp(2)). - Vaclav Kotesovec, Jun 15 2013

From Peter Bala, Sep 20 2013: (Start)

a(n) ~ (1/2)*n^2*n!/e^2 for large n.

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)!.

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)

a(n) = A087981(n+2)/(2*(n+1)). - Seiichi Manyama, Apr 25 2025