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%I #8 Nov 03 2019 04:31:11
%S 1,3,20,329,22584,7938470,12605643936,84977963809781,
%T 2379247465188706528,273419351336298753589802,
%U 128009562526607810326874017088,242979581192696030760182903464959706
%N G.f. satisfies: A(x) = exp( Sum_{n>=1} (2^n + A(x))^n * x^n/n ).
%C More generally, we have the following identity:
%C If A(x,q) = exp( Sum_{n>=1} (q^n + A(x,q))^n * x^n/n ), then
%C A(x,q) = 1/(1-x*A(x,q))*exp( Sum_{n>=1} q^(n^2)/(1-q^n*x*A(x,q))^n*x^n/n ).
%C Conjecture: if q is an integer, then A(x,q) is a power series in x with integer coefficients.
%C Setting q=1 defines the g.f. of the large Schroeder numbers (A006318).
%F G.f.: A(x) = 1/(1-x*A(x))*exp( Sum_{n>=1} 2^(n^2)/(1 - 2^n*x*A(x))^n * x^n/n ).
%e G.f.: A(x) = 1 + 3*x + 20*x^2 + 329*x^3 + 22584*x^4 + 7938470*x^5 +...
%e log(A(x)) = [2 + A(x)]*x + [2^2 + A(x)]^2*x^2/2 + [2^3 + A(x)]^3*x^3/3 +...
%e log(A(x)*(1-xA(x))) = 2/(1-2xA(x))*x + 2^4/(1-4xA(x))^2*x^2/2 + 2^9/(1-8xA(x))^3*x^3/3 +...
%e log(A(x)) = 3*x + 31*x^2/2 + 834*x^3/3 + 86227*x^4/4 + 39339038*x^5/5 +...
%t m = 12; A[_] = 1; Do[A[x_] = Exp[Sum[(2^n + A[x])^n x^n/n, {n, 1, m}]] + O[x]^m, {m}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Nov 03 2019 *)
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (2^m+A+x*O(x^n))^m*x^m/m))); polcoeff(A, n)}
%Y Cf. A202518, A155200.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Aug 07 2009
%E Comment corrected by _Paul D. Hanna_, Aug 08 2009