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A159594 G.f.: A(x) = x*exp( Sum_{n>=1} [ D^n A(x) ]^n/n ), where differential operator D = x*d/dx. 0

%I #2 Mar 30 2012 18:37:17

%S 1,1,3,16,125,1301,17070,272976,5218727,118508219,3224104875,

%T 108226321884,4740041705554,291705715765328,26728599026539162,

%U 3688459631229579912,751246585455211054713,208348432365596381718906

%N G.f.: A(x) = x*exp( Sum_{n>=1} [ D^n A(x) ]^n/n ), where differential operator D = x*d/dx.

%F G.f.: A(x) = x*exp( Sum_{n>=1} [ Sum_{k>=1} k^n*a(k)*x^k ]^n/n ) where A(x) = Sum_{k>=1} a(k)*x^k.

%e G.f.: A(x) = x + x^2 + 3*x^3 + 16*x^4 + 125*x^5 + 1301*x^6 +...

%e A(x) = x*exp( Sum_{n>=1} [x + 2^n*a(2)*x^2 + 3^n*a(3)*x^3 +...]^n/n ).

%e D^n A(x) = x + 2^n*x^2 + 3^n*3*x^3 + 4^n*16*x^4 + 5^n*125*x^5 +...

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

%K nonn

%O 1,3

%A _Paul D. Hanna_, May 03 2009

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