A249934 G.f. A(x) satisfies: x = Sum_{n>=1} 1/A(x)^(3*n) * Product_{k=1..n} (1 - 1/A(x)^(2*k-1)).

1, 1, 1, 4, 19, 107, 671, 4600, 34218, 276415, 2439426, 23724674, 256361107, 3091554768, 41560590331, 618957882104, 10119509431084, 179887355572358, 3446915545155744, 70686674091569072, 1542478858735415921, 35650141769790146478, 869385516566240903091, 22299067147713040916568

Offset: 0

Author

Paul D. Hanna, Nov 27 2014

Keywords

• nonn

Comments

Compare the g.f. to the identity:

G(x) = Sum_{n>=0} 1/G(x)^(2*n) * Product_{k=1..n} (1 - 1/G(x)^(2*k-1))

which holds for all power series G(x) such that G(0)=1.

Examples

			A(x) = 1 + x + x^2 + 4*x^3 + 19*x^4 + 107*x^5 + 671*x^6 + 4600*x^7 + 34218*x^8 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^4 + (A(x)-1)*(A(x)^3-1)/A(x)^10 + (A(x)-1)*(A(x)^3-1)*(A(x)^5-1)/A(x)^18 + (A(x)-1)*(A(x)^3-1)*(A(x)^5-1)*(A(x)^7-1)/A(x)^28 +
(A(x)-1)*(A(x)^3-1)*(A(x)^5-1)*(A(x)^7-1)*(A(x)^9-1)/A(x)^40 +...
		

Links

• Paul D. Hanna and Vaclav Kotesovec, Table of n, a(n) for n = 0..240 (first 100 terms from Paul D. Hanna)

Crossrefs

Cf. A214692.

Programs

Mathematica

nmax = 20; aa = ConstantArray[0,nmax]; aa[[1]] = 1; Do[AGF = 1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[SeriesCoefficient[Sum[Product[(1-1/AGF^(2m-1))/AGF^3,{m,1,k}],{k,1,j}],{x,0,j}]==0,koef][[1]]; aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* More efficient than PARI program, Vaclav Kotesovec, Nov 30 2014 *)

PARI

{a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0);
A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(3*m)*prod(k=1, m, 1-1/Ser(A)^(2*k-1))), #A-1)); A[n+1]}
for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) satisfies: x = Sum_{n>=1} 1/A(x)^(n*(n+3)) * Product_{k=1..n} (A(x)^(2*k-1) - 1).

a(n) ~ exp(Pi^2/24) * 12^n * n^(n-1) / (sqrt(6) * exp(n) * Pi^(2*n-1)). - Vaclav Kotesovec, Dec 01 2014