A232845 a(0) = 0, a(1) = 1, and for n>=2, a(n) = (n-2)*a(n-1) - (n-1)*a(n-2).

0, 1, 0, -2, -4, -4, 4, 44, 236, 1300, 8276, 61484, 523804, 5024036, 53478980, 624890236, 7946278604, 109195935284, 1612048228276, 25439293045580, 427278358483196, 7609502950269124, 143217213477235364, 2840152418116021916, 59189357288576068780

Offset

0,4

Links

Robert Israel, Table of n, a(n) for n = 0..452

Formula

a(n) ~ (n-3)!. - Vaclav Kotesovec, Jan 20 2014

E.g.f.: (1-x)^2*exp(x-1)*(Ei(1)-Ei(1-x))/2 -(1-x)^2*exp(x) - x/2 + 1. - Robert Israel, Jan 08 2018

a(n) = (-1)^(n+1)*C(n-1, 1) where C(n, x) are the Charlier polynomials (with parameter a=1) as given in A137338. (Evaluation at x = -1 gives the left factorials A003422.) - Peter Luschny, Nov 28 2018

Example

a(8) = 6*44 - 7*4 = 236.

Maple

f:= gfun:-rectoproc({a(n)=(n-2)*a(n-1)-(n-1)*a(n-2), a(0)=0, a(1)=1}, a(n), remember):
map(f, [$0..30]); # Robert Israel, Jan 08 2018
# Alternative:
C := proc(n, x) option remember; if n > 0 then (x-n)*C(n-1, x)-n*C(n-2, x)
elif n = 0 then 1 else 0 fi end: A232845 := n -> (-1)^(n+1)*C(n-1, 1):
seq(A232845(n), n=0..24); # Peter Luschny, Nov 28 2018

Mathematica

Flatten[{0, RecurrenceTable[{(-1+n) a[-2+n]+(2-n) a[-1+n]+a[n]==0, a[1]==1, a[2]==0}, a, {n, 20}]}] (* Vaclav Kotesovec, Jan 20 2014 *)

Crossrefs

Cf. A153229, A137338, A003422.

Sequence in context: A009145 A009292 A009622 * A269300 A182215 A036443

Adjacent sequences: A232842 A232843 A232844 * A232846 A232847 A232848

Keyword

sign,easy

Author

Philippe Deléham, Nov 30 2013

Status

approved