A052842 E.g.f. A(x) = series reversion of (1-x)*(1-exp(-x)).

0, 1, 3, 23, 290, 5104, 115374, 3185972, 103946688, 3912527016, 166884627360, 7955159511672, 419106982360560, 24182042474691984, 1516563901865906880, 102717031449780063360, 7472238163167018081024

Offset: 0

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

• easy
• nonn

Comments

A simple grammar.

Examples

			E.g.f.: A(x) = x + 3*x^2/2! + 23*x^3/3! + 290*x^4/4! + 5104*x^5/5! +... which satisfies: A(x) = -log(1 - x/(1-A(x))).
		

Links

• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 809
• Vladimir Kruchinin, The method for obtaining expressions for coefficients of reverse generating functions, arXiv:1211.3244 [math.CO], 2012.

Crossrefs

Cf. A226571.

Programs

Maple

spec := [S,{C=Prod(Z,B),S=Cycle(C),B=Sequence(S)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
# second Maple program:
A052842 := proc (n) option remember; `if`(n = 0, 0, add(pochhammer(n, k)*abs(Stirling1(n, k+1)), k = 0..n-1)) end:
seq(A052842(n), n = 0..16); # Mélika Tebni, Jun 02 2023

Mathematica

CoefficientList[InverseSeries[Series[(-1 + E^(-x))*(x-1),{x,0,20}],x],x] * Range[0,20]! (* Vaclav Kotesovec, Jan 08 2014 *)

Maxima

a(n):=sum((sum((sum((stirling2(i+n-1,j)*binomial(j,j-i))/(i+n-1)!,i,0,j))*(-1)^(n+j-1)/(k-j)!,j,0,k))*(n+k-1)!,k,0,n-1); /* Vladimir Kruchinin, Feb 06 2012 */

PARI

{a(n)=n!*polcoeff(serreverse((1-exp(-x+O(x^(n+2))))*(1-x)),n)} /* Paul D. Hanna, Jun 22 2011 */

Formula

E.g.f. satisfies: A(x) = -log(1 - x/(1-A(x))). [From Encyclopedia of Combinatorial Structures]

a(n) = sum(k=0..n-1, (sum(j=0..k, (sum(i=0..j, (stirling2(i+n-1,j)*C(j,j-i))/ (i+n-1)!))*(-1)^(n+j-1)/(k-j)!))*(n+k-1)!), n>0. - Vladimir Kruchinin, Feb 06 2012

a(n) ~ n^(n-1) * c^n / (sqrt(1+c) * exp(n) * (c-1)^(2*n-1)), where c = LambertW(exp(2)) = 1.5571455989976114... (see A226571). - Vaclav Kotesovec, Jan 08 2014

For n >= 1, a(n) = Sum_{k=0..n-1} Pochhammer(n, k)*|Stirling1(n, k+1)|. - Mélika Tebni, Jun 02 2023

Extensions

Name from a comment by Paul D. Hanna, Jun 22 2011