A188193 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x))^n/n!.
1, 2, 10, 100, 2500, 224728, 77611032, 95603336016, 411188458873152, 6215509773143124736, 334390128406134844422816, 64839530694681966290325813952, 45813418110052719651124682371286592, 119029898667160345925612801976119712375168, 1145669207542037312420485502021473483147684627584
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 2*x + 10*x^2 + 100*x^3 + 2500*x^4 + 224728*x^5 +... which equals the series: A(x) = 1 + log(1+2*x*A(x)) + log(1+4*x*A(x))^2/2! + log(1+8*x*A(x))^3/3! +... Let B(x) equal the g.f. of A014070, which begins: B(x) = 1 + 2*x + 6*x^2 + 56*x^3 + 1820*x^4 +...+ C(2^n,n)*x^n +... then B(x) = A(x/B(x)) and A(x) = B(x*A(x)), so that: A(x) = 1 + 2*x*A(x) + 6*x^2*A(x)^2 + 56*x^3*A(x)^3 + 1820*x^4*A(x)^4 +...+ C(2^n,n)*x^n*A(x)^n +...
Programs
-
PARI
{a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, log(1+2^m*x*A+x*O(x^n))^m/m!)); polcoeff(A, n)}
Formula
Extensions
More terms from Michel Marcus, Apr 21 2025