%I
%S 1,1,3,13,73,621,5491,60313,743793,10115353,158914531,2815311621,
%T 55094081593,1142894689093,25142695616403,594557634923281,
%U 15084112106943841,407999468524242993,11669035487641120963
%N Eigenvector of triangle A118588; E.g.f. satisfies: A(x) = exp(x)*A(x^2+x^3).
%C E.g.f. of triangle A118588 is exp(x + y*(x^2+x^3)); note the similarity to the e.g.f. of this sequence. More generally, the e.g.f. of an eigenvectors can be determined from the e.g.f. of a triangle as follows. [ Given a triangle with e.g.f.: exp(x + y*x*F(x)) such that F(0) = 0, then the eigenvector has e.g.f.: exp(G(x)) where o.g.f. G(x) satisfies: G(x) = x + G(x*F(x)). ]
%F Log(A(x)) = o.g.f. of A014535 (Btrees of order 3 with n leaves).
%e A(x) = 1 + x + 3*x^2/2! + 13*x^3/3! + 73*x^4/4! + 621*x^5/5! +...
%e log(A(x)) = o.g.f. of A014535 =
%e x + x^2+ x^3+ x^4+ 2*x^5+ 2*x^6+ 3*x^7+ 4*x^8+ 5*x^9+ 8*x^10 +...
%o (PARI) {a(n)=if(n==0,1,sum(k=0,n\2,a(k)*n!*polcoeff(polcoeff(exp(x+y*(x^2+x^3)+x*O(x^n)+y*O(y^k)),n,x),k,y)))}
%Y Cf. A118588 (triangle), A118589 (row sums), A014535 (log(A(x)).
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 08 2006
