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G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x))^n/n!.
1

%I #5 Mar 30 2012 18:37:26

%S 1,2,10,100,2500,224728,77611032,95603336016,411188458873152,

%T 6215509773143124736,334390128406134844422816,

%U 64839530694681966290325813952,45813418110052719651124682371286592

%N G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x))^n/n!.

%F G.f. A(x) satisfies:

%F (1) A(x) = Sum_{n>=0} C(2^n,n)*x^n*A(x)^n,

%F (2) A(x) = (1/x)*Series_Reversion(x/B(x)),

%F (3) A(x) = B(x*A(x)) and B(x) = A(x/B(x)),

%F where B(x) = Sum_{n>=0} C(2^n,n)*x^n is the g.f. of A014070.

%F (4) A(x) = G(x/A(x)) and G(x) = A(x*G(x)), where G(x) is the g.f. of A188194.

%e G.f.: A(x) = 1 + 2*x + 10*x^2 + 100*x^3 + 2500*x^4 + 224728*x^5 +...

%e which equals the series:

%e 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! +...

%e Let B(x) equal the g.f. of A014070, which begins:

%e B(x) = 1 + 2*x + 6*x^2 + 56*x^3 + 1820*x^4 +...+ C(2^n,n)*x^n +...

%e then B(x) = A(x/B(x)) and A(x) = B(x*A(x)), so that:

%e 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 +...

%o (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)}

%Y Cf. A014070, A188194.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Mar 23 2011