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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)!).

%I A180609 #29 Jul 26 2018 11:31:37
%S A180609 1,-1,3,-16,110,-540,-9240,292320,14908320,-1639612800,-33013854720,
%T A180609 21046667685120,-549927873855360,-637881314775344640,
%U A180609 76198391578224115200,41404329870413936025600,-12499862617277304901632000,-5212560012919105291193548800,3436632117109253032257698611200,1146156616720354265092896141312000,-1615552168543480516126725021634560000,-379914190499326491647463301427478528000,1268235921756889621556352102589895172096000
%N 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)!).
%H A180609 Paul D. Hanna, <a href="/A180609/b180609.txt">Table of n, a(n) for n = 1..150</a>
%H A180609 M Manetti, G Ricciardi, <a href="http://arxiv.org/abs/1509.09032">Universal Lie formulas for higher antibrackets</a>, arXiv preprint arXiv:1509.09032 [math.QA], 2015-2016.
%F A180609 G.f. satisfies: L(x) = (1+x)*log(1+x) * L( log(1+x) ) /x.
%F A180609 Let E_n(x) = E_{n-1}(exp(x)-1) denote the n-th iteration of exp(x)-1, then
%F A180609 . L(E_n(x)) = L(x) * x * E_n'(x) / E_n(x) for all n.
%F A180609 G.f. L(x) forms column 0 in the matrix log of the Riordan array ((exp(x)-1)/x, exp(x)-1).
%F A180609 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
%e A180609 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!) -+...
%e A180609 The Riordan array ((exp(x)-1)/x, exp(x)-1) begins:
%e A180609 1;
%e A180609 1/(1!2!), 1;
%e A180609 2/(2!3!), 2/(1!2!), 1;
%e A180609 6/(3!4!), 7/(2!3!), 3/(1!2!), 1;
%e A180609 24/(4!5!), 36/(3!4!), 15/(2!3!), 4/(1!2!), 1;
%e A180609 120/(5!6!), 248/(4!5!), 108/(3!4!), 26/(2!3!), 5/(1!2!), 1;
%e A180609 720/(6!7!), 2160/(5!6!), 1032/(4!5!), 240/(3!4!), 40/(2!3!), 6/(1!2!), 1; ...
%e A180609 where the g.f. of column k = ((exp(x)-1)/x)^(k+1) for k>=0.
%e A180609 ...
%e A180609 The matrix log of the above array begins:
%e A180609 0;
%e A180609 1/(1!2!), 0;
%e A180609 -1/(2!3!), 2/(1!2!), 0;
%e A180609 3/(3!4!), -2/(2!3!), 3/(1!2!), 0;
%e A180609 -16/(4!5!), 6/(3!4!), -3/(2!3!), 4/(1!2!), 0;
%e A180609 110/(5!6!), -32/(4!5!), 9/(3!4!), -4/(2!3!), 5/(1!2!), 0;
%e A180609 -540/(6!7!), 220/(5!6!), -48/(4!5!), 12/(3!4!), -5/(2!3!), 6/(1!2!), 0;
%e A180609 -9240/(7!8!), -1080/(6!7!), 330/(5!6!), -64/(4!5!), 15/(3!4!), -6/(2!3!), 7/(1!2!), 0; ...
%e A180609 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.
%t A180609 K[1] = 1;
%t A180609 K[n_] := K[n] = -2/((n+2)(n-1)) Sum[StirlingS2[n+1, i] K[i], {i, 1, n-1}];
%t A180609 a[n_] := n! K[n];
%t A180609 Array[a, 23] (* _Jean-François Alcover_, Jul 26 2018, from the Manetti-Ricciardi recurrence *)
%o A180609 (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]}
%o A180609 for(n=1,30,print1(a(n),", "))
%Y A180609 Cf. A180610.
%Y A180609 The fractions a(n)/n! are A134242(n)/A134243(n).
%K A180609 sign,easy
%O A180609 1,3
%A A180609 _Paul D. Hanna_, Sep 12 2010