OFFSET
0,3
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
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)
MAPLE
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 *)
PROG
(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)<<k+2, factorial(k+5)) for k in range(n+1))) # Chai Wah Wu, Apr 20 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jan 23 2000
STATUS
approved