cp's OEIS Frontend

A156911 G.f.: A(x) = exp( Sum_{n>=1} 3^(n^2)/(1 - 3^n*x)^n * x^n/n ).

%I A156911 #2 Mar 30 2012 18:37:16
%S A156911 1,3,54,7470,11326446,173007630594,25222890606413004,
%T A156911 34295263336258106333292,429734207324188407742780371030,
%U A156911 49292144072318945019920850119049478578
%N A156911 G.f.: A(x) = exp( Sum_{n>=1} 3^(n^2)/(1 - 3^n*x)^n * x^n/n ).
%C A156911 An example of this logarithmic identity at q=3:
%C A156911 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.
%F A156911 G.f.: A(x) = (1-x)*exp( Sum_{n>=1} (1 + 3^n)^n * x^n/n );
%F A156911 Equals the first differences of A155204.
%e A156911 G.f.: A(x) = 1 + 3*x + 54*x^2 + 7470*x^3 + 11326446*x^4 +...
%e A156911 Log(A(x)) = 3/(1-3*x)*x + 3^4/(1-3^2*x)^2*x^2/2 + 3^9/(1-3^3*x)^3*x^3/3 +...
%e A156911 Log(A(x)) = (4-1)*x + (10^2-1)*x^2/2 + (28^3-1)*x^3/3 + (82^4-1)*x^4/4 +...
%o A156911 (PARI) {a(n)=polcoeff(exp(sum(m=1, n+1, 3^(m^2)/(1-3^m*x)^m*x^m/m)+x*O(x^n)), n)}
%o A156911 (PARI) /* As First Differences of A155204: */
%o A156911 {a(n)=polcoeff((1-x)*exp(sum(m=1, n+1, (3^m+1)^m*x^m/m)+x*O(x^n)), n)}
%Y A156911 Cf. A155204, A155203, A155205, A156910.
%K A156911 nonn
%O A156911 0,2
%A A156911 _Paul D. Hanna_, Feb 17 2009