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%I
%S 1,1,2,5,19,86,436,2378,13731,83077,523275,3416329,23051600,160440679,
%T 1150435934,8492238919,64508971958,504172573079,4053925852485,
%U 33535370139607,285391912938870,2498255748837089,22489737035242848,208124346717364948,1978949027666465869,19321957528006663637,193581292284734286398,1988536950750112238165
%N G.f. A(x) = Sum_{n>=0} x^n*(A(x)^n - 1)^n/(1 - x*A(x)^n)^(n+1).
%H Paul D. Hanna, <a href="/A324618/b324618.txt">Table of n, a(n) for n = 0..200</a>
%F G.f. A(x) satisfies:
%F (1) A(x) = Sum_{n>=0} x^n*(A(x)^n - 1)^n / (1 - x*A(x)^n)^(n+1).
%F (2) A(x) = Sum_{n>=0} x^n*(A(x)^n + 1)^n / (1 + x*A(x)^n)^(n+1).
%F (3) A(x) = Sum_{n>=0} x^n*Sum_{k=0..n} binomial(n,k) * (A(x)^n - A(x)^k)^(n-k).
%F (4) A(x) = Sum_{n>=0} x^n*Sum_{k=0..n} (-1)^k * binomial(n,k) * (A(x)^n + A(x)^k)^(n-k).
%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 19*x^4 + 86*x^5 + 436*x^6 + 2378*x^7 + 13731*x^8 + 83077*x^9 + 523275*x^10 + 3416329*x^11 + 23051600*x^12 + ...
%e such that
%e A(x) = 1/(1 - x) + x*(A(x) - 1)/(1 - x*A(x))^2 + x^2*(A(x)^2 - 1)^2/(1 - x*A(x)^2)^3 + x^3*(A(x)^3 - 1)^3/(1 - x*A(x)^3)^4 + x^4*(A(x)^4 - 1)^4/(1 - x*A(x)^4)^5 + x^5*(A(x)^5 - 1)^5/(1 - x*A(x)^5)^6 + ...
%e also
%e A(x) = 1/(1 + x) + x*(A(x) + 1)/(1 + x*A(x))^2 + x^2*(A(x)^2 + 1)^2/(1 + x*A(x)^2)^3 + x^3*(A(x)^3 + 1)^3/(1 + x*A(x)^3)^4 + x^4*(A(x)^4 + 1)^4/(1 + x*A(x)^4)^5 + x^5*(A(x)^5 + 1)^5/(1 + x*A(x)^5)^6 + ...
%t m = 35; A[_] = 0; Unprotect[Power]; 0^0 = 1; Protect[Power];
%t Do[A[x_] = Sum[ x^n (A[x]^n - 1)^n/(1 - x A[x]^n)^(n + 1), {n, 0, k}] + O[x]^k, {k, m}];
%t CoefficientList[A[x], x] (* _Jean-François Alcover_, Oct 21 2019 *)
%o (PARI) {a(n) = my(A=[1,1]); for(i=0,n, A = concat(A,0);
%o A[#A] = polcoeff( sum(n=0,#A+1, x^n*(Ser(A)^n - 1)^n/(1 - x*Ser(A)^n)^(n+1) ),#A-1));
%o polcoeff(Ser(A),n)}
%o for(n=0,40,print1(a(n),", "))
%o (PARI) {a(n) = my(A=[1,1]); for(i=0,n, A = concat(A,0);
%o A[#A] = polcoeff( sum(n=0,#A+1, x^n*(Ser(A)^n + 1)^n/(1 + x*Ser(A)^n)^(n+1) ),#A-1));
%o polcoeff(Ser(A),n)}
%o for(n=0,40,print1(a(n),", "))
%Y Cf. A324619.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Mar 11 2019
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