%I #2 Mar 30 2012 18:37:22
%S 1,1,2,7,53,1305,177559,247880897,9252468642944,39211218810149955871,
%T 155393322265527728403643204013,
%U 12102963126133848174768954972642433817636369
%N G.f: A(x) = exp(Sum_{n>=1} A(L(n)*x)^n*x^n/n) = exp(Sum_{n>=1} L(n)*x^n/n) where L(n) = A180721(n).
%C Terms seem to be doubly-exponential: log(a(n+1))/log(a(n)) ~ c where c = 1.44...
%e G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 53*x^4 + 1305*x^5 + 177559*x^6 +...
%e The logarithm of g.f. A(x) begins:
%e log(A(x)) = x + 3*x^2/2 + 16*x^3/3 + 183*x^4/4 + 6236*x^5/5 + 1057176*x^6/6 + 1733913028*x^7/7 + 74017764628367*x^8/8 +...+ A180721(n)*x^n/n +...
%e which also equals the series:
%e log(A(x)) = A(x)*x + A(3*x)^2*x^2/2 + A(16*x)^3*x^3/3 + A(183*x)^4*x^4/4 + A(6236*x)^5*x^5/5 + A(1057176*x)^6*x^6/6 + A(1733913028*x)^7*x^7/7 +...+ A(A180721(n)*x)^n*x^n/n + ...
%o (PARI) {a(n)=local(A=1+x+x*O(x^n),L);for(i=1,n,L=Vec(deriv(log(A))); A=exp(sum(m=1,#L,subst(A,x,L[m]*x)^m*x^m/m)+x*O(x^n)));polcoeff(A,n)}
%Y Cf. A180721.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 19 2010