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A107591 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n*(n+1)/2).

%I A107591 #13 Feb 12 2014 22:22:14
%S A107591 1,1,2,6,22,91,408,1939,9635,49614,263140,1431301,7959568,45152340,
%T A107591 260847526,1532825675,9154581802,55537885743,342147577227,
%U A107591 2140251570508,13594688301758,87702596534110,574815620158265,3829029514213952
%N A107591 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n*(n+1)/2).
%F A107591 G.f. A(x) = (1/x)*series-reversion(x/F(x)) and A(x) = F(x*A(x)) where F(x) = A(x/F(x)) is the g.f. of A107590.
%F A107591 G.f. A(x) = x/series-reversion(x*G(x)) and A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A107592.
%F A107591 Contribution from _Paul D. Hanna_, Apr 24 2010: (Start)
%F A107591 Let A = g.f. A(x), then A satisfies the continued fraction:
%F A107591 A = 1/(1- A*x/(1- A*(A-1)*x/(1- A^3*x/(1- A^2*(A^2-1)*x/(1- A^5*x/(1- A^3*(A^3-1)*x/(1- A^7*x/(1- A^4*(A^4-1)*x/(1- ...)))))))))
%F A107591 due to an identity of a partial elliptic theta function.
%F A107591 (End)
%F A107591 Contribution from _Paul D. Hanna_, May 05 2010: (Start)
%F A107591 Let A = g.f. A(x), then A satisfies:
%F A107591 A = Sum_{n>=0} x^n*A^n*Product_{k=1..n} (1-x*A^(2k-1))/(1-x*A^(2k))
%F A107591 due to a q-series identity.
%F A107591 (End)
%e A107591 A = 1 + x*A^1 + x^2*A^3 + x^3*A^6 + x^4*A^10 + x^5*A^15 ...
%e A107591 = 1 + (x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 91*x^6 +...)
%e A107591 + (x^2 + 3*x^3 + 9*x^4 + 31*x^5 + 120*x^6 +...)
%e A107591 + (x^3 + 6*x^4 + 27*x^5 + 116*x^6 +...)
%e A107591 + (x^4 + 10*x^5 + 65*x^6 +...) +...
%e A107591 = 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 91*x^5 + 408*x^6 +...
%o A107591 (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(k=1,n,A=1+sum(j=1,n,x^j*A^(j*(j+1)/2)+x*O(x^n)));polcoeff(A,n)}
%Y A107591 Cf. A107590, A107592, A155805, A219359.
%K A107591 eigen,nonn
%O A107591 0,3
%A A107591 _Paul D. Hanna_, May 17 2005, May 05 2010