A180609 G.f. L(x) satisfies: L(x) = L(exp(x)-1)*(1-exp(-x))/x = Sum_{n>=1} a(n)*x^n/(n!*(n+1)!).

1, -1, 3, -16, 110, -540, -9240, 292320, 14908320, -1639612800, -33013854720, 21046667685120, -549927873855360, -637881314775344640, 76198391578224115200, 41404329870413936025600, -12499862617277304901632000, -5212560012919105291193548800, 3436632117109253032257698611200, 1146156616720354265092896141312000, -1615552168543480516126725021634560000, -379914190499326491647463301427478528000, 1268235921756889621556352102589895172096000

Offset: 1

Author

Paul D. Hanna, Sep 12 2010

Keywords

• sign
• easy

Examples

			G.f.: L(x) = x/(1!*2!) - x^2/(2!*3!) + 3*x^3/(3!*4!) - 16*x^4/(4!*5!) + 110*x^5/(5!*6!) - 540*x^6/(6!*7!) - 9240*x^7/(7!*8!) + 292320*x^8/(8!*9!) -+...
The Riordan array ((exp(x)-1)/x, exp(x)-1) begins:
1;
1/(1!2!), 1;
2/(2!3!), 2/(1!2!), 1;
6/(3!4!), 7/(2!3!), 3/(1!2!), 1;
24/(4!5!), 36/(3!4!), 15/(2!3!), 4/(1!2!), 1;
120/(5!6!), 248/(4!5!), 108/(3!4!), 26/(2!3!), 5/(1!2!), 1;
720/(6!7!), 2160/(5!6!), 1032/(4!5!), 240/(3!4!), 40/(2!3!), 6/(1!2!), 1; ...
where the g.f. of column k = ((exp(x)-1)/x)^(k+1) for k>=0.
...
The matrix log of the above array begins:
0;
1/(1!2!), 0;
-1/(2!3!), 2/(1!2!), 0;
3/(3!4!), -2/(2!3!), 3/(1!2!), 0;
-16/(4!5!), 6/(3!4!), -3/(2!3!), 4/(1!2!), 0;
110/(5!6!), -32/(4!5!), 9/(3!4!), -4/(2!3!), 5/(1!2!), 0;
-540/(6!7!), 220/(5!6!), -48/(4!5!), 12/(3!4!), -5/(2!3!), 6/(1!2!), 0;
-9240/(7!8!), -1080/(6!7!), 330/(5!6!), -64/(4!5!), 15/(3!4!), -6/(2!3!), 7/(1!2!), 0; ...
in which the g.f. of column k equals (k+1)*L(x) for k>=0 and L(x) is the g.f. of this sequence.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..150
• M Manetti, G Ricciardi, Universal Lie formulas for higher antibrackets, arXiv preprint arXiv:1509.09032 [math.QA], 2015-2016.

Crossrefs

Cf. A180610.

The fractions a(n)/n! are A134242(n)/A134243(n).

Programs

Mathematica

K[1] = 1;
K[n_] := K[n] = -2/((n+2)(n-1)) Sum[StirlingS2[n+1, i] K[i], {i, 1, n-1}];
a[n_] := n! K[n];
Array[a, 23] (* Jean-François Alcover, Jul 26 2018, from the Manetti-Ricciardi recurrence *)

PARI

{a(n)=local(M=matrix(n+1,n+1,r,c,if(r>=c,polcoeff(((exp(x+x^2*O(x^n))-1)/x)^c,r-c))),L=sum(n=1,#M,-(M^0-M)^n/n));n!*(n+1)!*L[n+1,1]}
for(n=1,30,print1(a(n),", "))

Formula

G.f. satisfies: L(x) = (1+x)*log(1+x) * L( log(1+x) ) /x.

Let E_n(x) = E_{n-1}(exp(x)-1) denote the n-th iteration of exp(x)-1, then

. L(E_n(x)) = L(x) * x * E_n'(x) / E_n(x) for all n.

G.f. L(x) forms column 0 in the matrix log of the Riordan array ((exp(x)-1)/x, exp(x)-1).

Manetti-Ricciardi Theorem 4.4 give a recurrence for K_n := a(n)/n! in terms of Stirling numbers. - N. J. A. Sloane, May 25 2016