A107590 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n*(n-1)/2).
1, 1, 1, 2, 5, 15, 50, 181, 698, 2837, 12062, 53374, 244923, 1162536, 5697119, 28786266, 149814059, 802436166, 4420515689, 25031466730, 145616087486, 869760092469, 5330945435272, 33508699787635, 215863606818041
Offset: 0
Examples
A = 1 + x + x^2*A^1 + x^3*A^3 + x^4*A^6 + x^5*A^10 +... = 1 + x + (x^2 + x^3 + x^4 + 2*x^5 + 5*x^6 + 15*x^7 +...) + (x^3 + 3*x^4 + 6*x^5 + 13*x^6 + 33*x^7 +...) + (x^4 + 6*x^5 + 21*x^6 + 62*x^7 +...) + (x^5 + 10*x^6 + 55*x^7 +...) +... = 1 + x + x^2 + 2*x^3 + 5*x^4 + 15*x^5 + 50*x^6 + 181*x^7 +...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..100
Programs
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PARI
{a(n) = my(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)} for(n=0,30,print1(a(n),", "))
Formula
G.f. A(x) = x/series-reversion(x*F(x)) and thus A(x) = F(x/A(x)) where F(x) = A(x*F(x)) is the g.f. of A107591.
G.f. A(x)^2 = x/series-reversion(x*G(x)^2) and thus A(x) = G(x/A(x)^2) where G(x) = A(x*G(x)^2) is the g.f. of A107592.
Contribution from Paul D. Hanna, Apr 25 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
A = 1/(1- x/(1- (A-1)*x/(1- A^2*x/(1- A*(A^2-1)*x/(1- A^4*x/(1- A^2*(A^3-1)*x/(1- A^6*x/(1- A^3*(A^4-1)*x/(1- ...)))))))))
due to an identity of a partial elliptic theta function.
(End)