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Expansion of g.f. A(x) satisfying A(x)^2 = (1+x) * A(x*A(x)) with A(0) = 1.
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%I #18 Jan 09 2024 08:50:34

%S 1,1,1,3,14,91,737,7096,78766,987842,13797282,212285511,3567147508,

%T 65001800879,1276876433466,26901212485406,605144725934525,

%U 14477882030623891,367106501841287465,9834718739970491625

%N Expansion of g.f. A(x) satisfying A(x)^2 = (1+x) * A(x*A(x)) with A(0) = 1.

%C Note that if F(x)^2 = (1+x) * F(x*F(x)) with F(0) = 0, then F(x) is the g.f. of A367386. - _Paul D. Hanna_, Jan 08 2024

%H Vaclav Kotesovec, <a href="/A120056/b120056.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) ~ c * n^n / (exp(n) * (log(2))^n), where c = 0.8876659274678... . - _Vaclav Kotesovec_, Aug 08 2014

%F From _Paul D. Hanna_, Jan 08 2024: (Start)

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n and B(x) = x*A(x) and B(x) = x*A(x) satisfy the following formulas.

%F (1) A(x)^2 = (1+x) * A(x*A(x)).

%F (2) A(x) = (1+x)^(1/2) * (1 + B(x))^(1/4) * (1 + B(B(x)))^(1/8) * (1 + B(B(B(x))))^(1/16) * (1 + B(B(B(B(x)))))^(1/32) * ..., an infinite product involving iterations of B(x) = x*A(x).

%F (3.a) B(B(x)) = x*A(x)^3 / (1+x).

%F (3.b) B(B(B(x))) = x*A(x)^7 / ((1+x)^3 * (1 + x*A(x))).

%F (3.c) B(B(B(B(x)))) = x*A(x)^15 / ((1+x)^7 * (1 + x*A(x))^3 * (1 + x*A(x)^3/(1+x))).

%F (3.d) B(B(B(B(B(x))))) = x*A(x)^31 / ((1+x)^15 * (1+x*A(x))^7 * (1 + x*A(x)^3/(1+x))^3 * (1 + x*A(x)^7/((1+x)^3*(1+x*A(x))))).

%F The compositions of g.f. A(x) with the iterations of B(x) = x*A(x) begin

%F (4.a) A(B(x)) = A(x)^2 / (1+x).

%F (4.b) A(B(B(x))) = A(x)^4 / ((1+x)^2 * (1 + x*A(x))).

%F (4.c) A(B(B(B(x)))) = A(x)^8 / ((1+x)^4 * (1 + x*A(x))^2 * (1 + x*A(x)^3/(1+x))).

%F (4.d) A(B(B(B(B(x))))) = A(x)^16 / ((1+x)^8 * (1+x*A(x))^4 * (1 + x*A(x)^3/(1+x))^2 * (1 + x*A(x)^7/((1+x)^3*(1+x*A(x))))).

%F (End)

%e A(x) = 1 + x + x^2 + 3*x^3 + 14*x^4 + 91*x^5 + 737*x^6 +...

%e A(x)^2 = 1 + 2*x + 3*x^2 + 8*x^3 + 35*x^4 + 216*x^5 + 1693*x^6 +...

%e A(x*A(x)) = 1 + x + 2*x^2 + 6*x^3 + 29*x^4 + 187*x^5 + 1506*x^6 +...

%o (PARI) {a(n)=local(A); if(n<1, n==0, A=1+x; for(i=1,n, A=subst(A,x,x*A+x*O(x^n))*(1+x)/A); polcoeff(A, n))}

%Y Cf. A088792, A367386.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Jun 05 2006