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A188194 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x)^2)^n/n!.

%I A188194 #5 Mar 30 2012 18:37:26
%S A188194 1,2,14,168,3756,261560,80733232,96730287424,412733638204832,
%T A188194 6222933783425122080,334514554099356252794912,
%U A188194 64846889330532757107162199040,45814974387230048629026769270192768
%N A188194 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x)^2)^n/n!.
%F A188194 G.f. A(x) satisfies:
%F A188194 (1) A(x) = Sum_{n>=0} C(2^n,n)*x^n*A(x)^(2n),
%F A188194 (2) A(x) = sqrt((1/x)*Series_Reversion(x/B(x)^2)),
%F A188194 (3) A(x) = B(x*A(x)^2) and B(x) = A(x/B(x)^2),
%F A188194 where B(x) = Sum_{n>=0} C(2^n,n)*x^n is the g.f. of A014070.
%F A188194 (4) A(x) = F(x*A(x)) and F(x) = A(x/F(x)), where F(x) is the g.f. of A188193.
%e A188194 G.f.: A(x) = 1 + 2*x + 14*x^2 + 168*x^3 + 3756*x^4 + 261560*x^5 +...
%e A188194 which equals the series:
%e A188194 A(x) = 1 + log(1+2*x*A(x)^2) + log(1+4*x*A(x)^2)^2/2! + log(1+8*x*A(x)^2)^3/3! +...
%e A188194 Let B(x) equal the g.f. of A014070, which begins:
%e A188194 B(x) = 1 + 2*x + 6*x^2 + 56*x^3 + 1820*x^4 +...+ C(2^n,n)*x^n +...
%e A188194 then B(x) = A(x/B(x)^2) and A(x) = B(x*A(x)^2), so that:
%e A188194 A(x) = 1 + 2*x*A(x)^2 + 6*x^2*A(x)^4 + 56*x^3*A(x)^6 + 1820*x^4*A(x)^8 +...+ C(2^n,n)*x^n*A(x)^(2n) +...
%o A188194 (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, log(1+2^m*x*A^2+x*O(x^n))^m/m!)); polcoeff(A, n)}
%Y A188194 Cf. A014070, A188193.
%K A188194 nonn
%O A188194 0,2
%A A188194 _Paul D. Hanna_, Mar 23 2011