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%I #2 Mar 30 2012 18:37:22
%S 1,-2,15,-240,6420,-249120,12729360,-799249920,59539354560,
%T -5472506188800,708462047462400,-129547455170918400,
%U 24363744478955481600,-2357904494544779980800,-506327043021030975744000
%N G.f. satisfies: L(x) = L(x*exp(x))/(1+x) = Sum_{n>=1} a(n)*x^n/n!^2.
%F G.f. L(x) satisfies: L(x) = (1 + W(x))*L(W(x)) where W(x) = Sum_{n>=1} (-1)^(n-1)*n^n*x^n/n! is the LambertW function.
%F Let E_n(x) = E_{n-1}(x*exp(x)) denote the n-th iteration of x*exp(x), then
%F . L(E_2(x)) = L(x)*(1+x)*(1+x*exp(x));
%F . L(E_3(x)) = L(x)*(1+x)*(1+x*exp(x))*(1+x*exp(x*exp(x)));
%F . L(E_n(x)) = L(x)*x*E_n'(x)/E_n(x) = L(x)*Product_{k=0..n-1}(1+E_k(x)).
%F G.f. L(x) forms column 0 in the matrix log of the Riordan array (exp(x), x*exp(x)).
%e G.f: L(x) = x - 2*x^2/2!^2 + 15*x^3/3!^2 - 240*x^4/4!^2 + 6420*x^5/5!^2 - 249120*x^6/6!^2 + 12729360*x^7/7!^2 -+...
%e The Riordan array (exp(x), x*exp(x)) begins:
%e 1;
%e 1, 1;
%e 1/2!, 2, 1;
%e 1/3!, 4/2!, 3, 1;
%e 1/4!, 8/3!, 9/2!, 4, 1;
%e 1/5!, 16/4!, 27/3!, 16/2!, 5, 1;
%e 1/6!, 32/5!, 81/4!, 64/3!, 25/2!, 6, 1; ...
%e The matrix log of the above array begins:
%e 0;
%e 1, 0;
%e -2/2!^2, 2, 0;
%e 15/3!^2, -4/2!^2, 3, 0;
%e -240/4!^2, 30/3!^2, -6/2!^2, 4, 0;
%e 6420/5!^2, -480/4!^2, 45/3!^2, -8/2!^2, 5, 0;
%e -249120/6!^2, 12840/5!^2, -720/4!^2, 60/3!^2, -10/2!^2, 6, 0; ...
%e in which the g.f. of column k equals (k+1)*L(x) for k>=0 where L(x) is the g.f. of this sequence.
%o (PARI) {a(n)=local(M=matrix(n+1,n+1,r,c,if(r>=c,polcoeff(exp(c*x+x*O(x^n)),r-c))),L=sum(n=1,#M,-(M^0-M)^n/n));n!^2*L[n+1,1]}
%K eigen,sign
%O 1,2
%A _Paul D. Hanna_, Sep 11 2010
%E Formula corrected by _Paul D. Hanna_, Sep 19 2010
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