login
Expansion of g.f. A(x) satisfying A(x)^2 = A(x*A(x)) / (1-x) with A(0) = 1.
3

%I #20 Jan 09 2024 08:50:46

%S 1,1,2,7,38,279,2535,27106,330998,4526827,68402684,1130591607,

%T 20282026273,392427087780,8146462590683,180629845193090,

%U 4261044215912232,106571080784076813,2817142524095598771

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

%C Note that if F(x)^2 = F(x*F(x)) / (1-x) with F(0) = 0, then F(x) is the g.f. of A367387. - _Paul D. Hanna_, Jan 08 2024

%H Vaclav Kotesovec, <a href="/A088792/b088792.txt">Table of n, a(n) for n = 0..220</a>

%F a(n) ~ c * n^(n + log(2)) / (exp(n) * (log(2))^n), where c = 0.9415987277996... . - _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) satisfies the following formulas.

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

%F (2) A(x) = 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 The iterations of B(x) = x*A(x) begin

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

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

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

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

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

%F (4.a) A(B(x)) = (1-x) * A(x)^2.

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

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

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

%F (End)

%e A(x) = 1 + x + 2*x^2 + 7*x^3 + 38*x^4 + 279*x^5 + 2535*x^6 +...

%e A(x)^2 = 1 + 2*x + 5*x^2 + 18*x^3 + 94*x^4 + 662*x^5 + 5829*x^6 +...

%e A(x*A(x)) = 1 + x + 3*x^2 + 13*x^3 + 76*x^4 + 568*x^5 + 5167*x^6 +...

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

%o for(n=0,20, print1(a(n),", "))

%Y Cf. A367387, A120056.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 16 2003; corrected Jun 05 2006