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A189981 E.g.f. satisfies: A(x) = Sum_{n>=0} log(1 + x*A(x)^n)^n/n!. 8

%I #19 Jul 14 2014 02:11:17

%S 1,1,2,12,120,1600,28500,621138,16017792,480474720,16390969920,

%T 626786792280,26584872779520,1238524175509608,62873918454756864,

%U 3455537675553482400,204449393824639488000,12958008875933613962880

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

%C The definition of the e.g.f. A(x) is an application of the identity:

%C * Sum_{n>=0} log(1 + q^n*x)^n/n! = Sum_{n>=0} binomial(q^n, n)*x^n at q = A(x).

%C Consider the function G(x) such that G(x) = 1 + x*G(x)^p, then

%C * G(x) = Sum_{n>=0} log(1 + x*G(x)^p)^n/n! (trivially), and

%C * G(x) = Sum_{n>=0} binomial(p*n+1,n)*x^n/(p*n+1) for fixed p;

%C does an analogous expression exist for the e.g.f. of this sequence?

%C Note that terms a(70)-a(83) are negative. - _Vaclav Kotesovec_, Jul 13 2014

%H Vaclav Kotesovec, <a href="/A189981/b189981.txt">Table of n, a(n) for n = 0..106</a>

%F E.g.f. also satisfies:

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

%F (2) A(x) = Sum_{n>=0} x^n * Sum_{k=0..n} Stirling1(n,k) * A(x)^(n*k)/n!.

%e E.g.f.: A(x) = 1 + x + 2*x^2/2! + 12*x^3/3! + 120*x^4/4! + 1600*x^5/5! +...

%e where A(x) satisfies:

%e A(x) = 1 + log(1 + x*A(x)) + log(1 + x*A(x)^2)^2/2! + log(1 + x*A(x)^3)^3/3! +...

%e The e.g.f. also satisfies:

%e A(x) = 1 + A(x)*x + A(x)^2*(A(x)^2-1)*x^2/2! + A(x)^3*(A(x)^3-1)*(A(x)^3-2)*x^3/3! + A(x)^4*(A(x)^4-1)*(A(x)^4-2)*(A(x)^4-3)*x^4/4! +...+ binomial(A(x)^n, n)*x^n +...

%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=sum(m=0,n,log(1+x*(A+x*O(x^n))^m)^m/m!));n!*polcoeff(A,n)}

%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=sum(m=0,n,binomial((A+x*O(x^n))^m,m)*x^m));n!*polcoeff(A,n)}

%o (PARI) {Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}

%o {a(n)=local(A=1+x);for(i=1,n,A=sum(m=0,n,sum(k=0,m,Stirling1(m,k)*(A+x*O(x^n))^(m*k))*x^m/m!));n!*polcoeff(A,n)}

%Y Cf. A014070, A221101, A224797.

%K sign

%O 0,3

%A _Paul D. Hanna_, May 03 2011

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