A156910 G.f.: A(x) = exp( Sum_{n>=1} 2^(n^2)/(1 - 2^n*x)^n * x^n/n ).

1, 2, 14, 268, 21462, 7872396, 12585797612, 84949155244024, 2379063526056509734, 273414369715003663482380, 128009001272184822673783879332, 242979321424122460096958142064785384

Offset: 0

Author

Paul D. Hanna, Feb 17 2009

Keywords

• nonn

Comments

An example of this logarithmic identity at q=2:

Sum_{n>=1} [q^(n^2)/(1 - q^n*x)^n]*x^n/n = Sum_{n>=1} [(1 + q^n)^n - 1]*x^n/n.

Examples

			G.f.: A(x) = 1 + 2*x + 14*x^2 + 268*x^3 + 21462*x^4 +...
log(A(x)) = 2/(1-2*x)*x + 2^4/(1-2^2*x)^2*x^2/2 + 2^9/(1-2^3*x)^3*x^3/3 +...
log(A(x)) = (3-1)*x + (5^2-1)*x^2/2 + (9^3-1)*x^3/3 + (17^4-1)*x^4/4 +...
		

Crossrefs

Cf. A155201, A155200.

Programs

PARI

{a(n)=polcoeff(exp(sum(m=1, n+1, 2^(m^2)/(1-2^m*x)^m*x^m/m)+x*O(x^n)), n)}

PARI

/* As First Differences of A155201: */
{a(n)=polcoeff((1-x)*exp(sum(m=1, n+1, (2^m+1)^m*x^m/m)+x*O(x^n)), n)}

Formula

G.f.: A(x) = (1-x)*exp( Sum_{n>=1} (1 + 2^n)^n * x^n/n );

Equals the first differences of A155201.