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%I #13 Jul 19 2016 11:33:32
%S 1,1,2,5,18,80,430,2590,17140,121992,923641,7375525,61742602,
%T 539333870,4898377418,46123950289,449236837762,4517200710720,
%U 46817951543477,499470969336575,5478324090579153,61712245317522148,713306845141673578,8452729850796180097,102612120337242520914
%N G.f. A(x) satisfies: A(x) = exp( Sum_{n>=1} B(x^n)/n ) where B(x) = Series_Reversion(x/A(x)) = g.f. of A179325.
%F G.f.: A(x) = 1/Product_{n>=1} (1 - x^n)^A179325(n); i.e., this sequence equals the Euler transform of A179325.
%F Given g.f. A(x), let B(x) be the g.f. of A179325, then A(x) and B(x) satisfy:
%F _ B(x) = x*A(B(x)) = Series_Reversion(x/A(x)),
%F _ B(x) = x*exp( Sum_{n>=1} B(B(x)^n)/n ).
%F G.f. satisfies: A(x) = exp( Sum_{n>=1} Sum_{k>=1} x^(n*k)/(n*k) * {[y^(k-1)] A(y)^k} ), where [y^(k-1)] A(y)^k denotes the coefficient of y^(k-1) in A(y)^k. - _Paul D. Hanna_, Feb 13 2013
%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 18*x^4 + 80*x^5 + 430*x^6 +...
%e which satisfies:
%e log(A(x)) = B(x) + B(x^2)/2 + B(x^3)/3 + B(x^4)/4 +...
%e where B(x) = x*A(B(x)) = g.f. of A179325 begins:
%e B(x) = x + x^2 + 3*x^3 + 12*x^4 + 59*x^5 + 331*x^6 + 2062*x^7 +...
%o (PARI) a(n)=local(A=1+x,B=x);for(i=1,n,B=serreverse(x/(A+x*O(x^n)));A=exp(sum(m=1,n,subst(B,x,x^m+x*O(x^n))/m)));polcoeff(A,n)
%o (PARI) /* As the Euler transform of A179325: */
%o {a(n)=local(A=1+x,B=x);for(i=1,n,B=serreverse(x/(A+x*O(x^n)));A=1/prod(k=1,n,(1-x^k+x*O(x^n))^polcoeff(B,k)));polcoeff(A,n)}
%o (PARI) a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=1, n, x^(m*k)/(m*k)*polcoeff(A^k+O(x^k), k-1)) +x*O(x^n)))); polcoeff(A, n)
%o for(n=0,25,print1(a(n),", ")) \\ _Paul D. Hanna_, Feb 13 2013
%Y Cf. A179325.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 06 2011
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