A107595 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^(n^2).
1, 1, 2, 7, 31, 158, 884, 5292, 33385, 219797, 1500449, 10573815, 76688602, 571232869, 4363912280, 34161879247, 273906591562, 2248935278231, 18909284838057, 162842178607893, 1436660527685476, 12988076148036405, 120345643023918566, 1143054910071718088, 11129160383826078389
Offset: 0
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 + 5292*x^7 +... Let A = g.f. A(x) then A = 1 + x*A^1 + x^2*A^4 + x^3*A^9 + x^4*A^16 + x^5*A^25 ... = 1 + x*(1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 +...) + x^2*(1 + 4*x + 14*x^2 + 56*x^3 + 257*x^4 + 1312*x^5 +...) + x^3*(1 + 9*x + 54*x^2 + 291*x^3 + 1557*x^4 + 8568*x^5 +..) + x^4*(1 + 16*x + 152*x^2 + 1152*x^3 + 7836*x^4 +...) + x^5*(1 + 25*x + 350*x^2 + 3675*x^3 + 32625*x^4 +...) + x^6*(1 + 36*x + 702*x^2 + 9912*x^3 + 114201*x^4 +...) +... = 1 + x + 2*x^2 + 7*x^3 + 31*x^4 + 158*x^5 + 884*x^6 +...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..200
Programs
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Mathematica
m = 25; A[_] = 0; Do[A[x_] = 1 + Sum[x^k A[x]^(k^2) + O[x]^j, {k, 1, j}], {j, m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 05 2019 *) -
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^2)+x*O(x^n)));polcoeff(A,n)} for(n=0,30,print1(a(n),", "))
Formula
G.f. A(x) = (1/x)*Series_Reversion(x/F(x)) and thus A(x) = F(x*A(x)) where F(x) is the g.f. of A107594.
G.f. A(x) = x/Series_Reversion(x*G(x)) and thus A(x) = G(x/A(x)) where G(x) is the g.f. of A107596.
From Paul D. Hanna, Apr 23 2010: (Start)
Let A = g.f. A(x), then A satisfies the continued fraction:
A = 1/(1 - A*x/(1 - (A^3-A)*x/(1 - A^5*x/(1 - (A^7-A^3)*x/(1 - A^9*x/(1- (A^11-A^5)*x/(1 - A^13*x/(1 - (A^15-A^7)*x/(1 - ...)))))))))
due to an identity of a partial elliptic theta function. (End)
From Paul D. Hanna, May 04 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^(4k-3)) / (1 - x*A^(4k-1))
due to a q-series identity. (End)
Comments