A107591 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n*(n+1)/2).
1, 1, 2, 6, 22, 91, 408, 1939, 9635, 49614, 263140, 1431301, 7959568, 45152340, 260847526, 1532825675, 9154581802, 55537885743, 342147577227, 2140251570508, 13594688301758, 87702596534110, 574815620158265, 3829029514213952
Offset: 0
Examples
A = 1 + x*A^1 + x^2*A^3 + x^3*A^6 + x^4*A^10 + x^5*A^15 ... = 1 + (x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 91*x^6 +...) + (x^2 + 3*x^3 + 9*x^4 + 31*x^5 + 120*x^6 +...) + (x^3 + 6*x^4 + 27*x^5 + 116*x^6 +...) + (x^4 + 10*x^5 + 65*x^6 +...) +... = 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 91*x^5 + 408*x^6 +...
Programs
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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)}
Formula
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.
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.
Contribution from Paul D. Hanna, Apr 24 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
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- ...)))))))))
due to an identity of a partial elliptic theta function.
(End)
Contribution from Paul D. Hanna, May 05 2010: (Start)
Let A = g.f. A(x), then A satisfies:
A = Sum_{n>=0} x^n*A^n*Product_{k=1..n} (1-x*A^(2k-1))/(1-x*A^(2k))
due to a q-series identity.
(End)