A132695 G.f. A(x) satisfies: a(n+1) = [x^n] A(x)^(2^n) for n>=0, with a(0)=1.

1, 1, 2, 14, 280, 13500, 1494432, 397108896, 295442310016, 790590934285280, 8917538639501922816, 411296902906980232373760, 73215772219125676749036230656, 49200595120135859231689635315995648

Offset: 0

Author

Paul D. Hanna, Aug 26 2007

Keywords

• nonn

Examples

			In the following table of initial powers 2^n of A(x),
we see that the coefficients of [x^n] along the diagonal
form this sequence shift left:
A^(2^0) =(1)+ x + 2x^2 + 14x^3 + 280x^4 + 13500x^5 +(1494432)x^6 +...;
A^(2^1) = 1 +(2)x + 5x^2 + 32x^3 + 592x^4 + 27616x^5 +...;
A^(2^2) = 1 + 4x +(14)x^2 + 84x^3 + 1337x^4 + 57920x^5 +...;
A^(2^3) = 1 + 8x + 44x^2 +(280)x^3 + 3542x^4 + 128888x^5 +...;
A^(2^4) = 1 + 16x + 152x^2 + 1264x^3 +(13500)x^4 + 339088x^5 +...;
A^(2^5) = 1 + 32x + 560x^2 + 7392x^3 + 90552x^4 +(1494432)x^5 +...;
A^(2^6) = 1 + 64x + 2144x^2 + 50624x^3 + 967792x^4 + 17063232x^5 +...;
		

Programs

PARI

{a(n)=local(A=[1,1]);for(i=1,n,A=concat(A,Vec(Ser(A)^(2^(#A-1)))[ #A]));A[n+1]}

Formula

G.f. A(x) satisfies: A(x) = 1 + x*Sum_{n>=0} log( A(2^n*x) )^n / n! = 1 + x*[1 + log(A(2x)) + log(A(4x))^2/2! + log(A(8x))^3/3! +...]. - Paul D. Hanna, Jan 05 2008