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A188193 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x))^n/n!.

%I A188193 #10 Apr 21 2025 13:22:45
%S A188193 1,2,10,100,2500,224728,77611032,95603336016,411188458873152,
%T A188193 6215509773143124736,334390128406134844422816,
%U A188193 64839530694681966290325813952,45813418110052719651124682371286592,119029898667160345925612801976119712375168,1145669207542037312420485502021473483147684627584
%N A188193 G.f. satisfies: A(x) = Sum_{n>=0} log(1 + 2^n*x*A(x))^n/n!.
%F A188193 G.f. A(x) satisfies:
%F A188193 (1) A(x) = Sum_{n>=0} C(2^n,n)*x^n*A(x)^n,
%F A188193 (2) A(x) = (1/x)*Series_Reversion(x/B(x)),
%F A188193 (3) A(x) = B(x*A(x)) and B(x) = A(x/B(x)),
%F A188193 where B(x) = Sum_{n>=0} C(2^n,n)*x^n is the g.f. of A014070.
%F A188193 (4) A(x) = G(x/A(x)) and G(x) = A(x*G(x)), where G(x) is the g.f. of A188194.
%e A188193 G.f.: A(x) = 1 + 2*x + 10*x^2 + 100*x^3 + 2500*x^4 + 224728*x^5 +...
%e A188193 which equals the series:
%e A188193 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 A188193 Let B(x) equal the g.f. of A014070, which begins:
%e A188193 B(x) = 1 + 2*x + 6*x^2 + 56*x^3 + 1820*x^4 +...+ C(2^n,n)*x^n +...
%e A188193 then B(x) = A(x/B(x)) and A(x) = B(x*A(x)), so that:
%e A188193 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 A188193 (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 A188193 Cf. A014070, A188194.
%K A188193 nonn
%O A188193 0,2
%A A188193 _Paul D. Hanna_, Mar 23 2011
%E A188193 More terms from _Michel Marcus_, Apr 21 2025