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%I #20 Apr 20 2024 03:53:35
%S 1,1,3,25,369,7881,220845,7677363,319307665,15487290535,859400072837,
%T 53749578759526,3743585586509849,287496351622105328,
%U 24143937833744911767,2201703647718624364913,216700738558116024114289,22900073562659910815354339,2586409916780162599516986945,310947096149155992699450689912,39650252031533561961437812566315
%N G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n/A(x)^n.
%H Paul D. Hanna, <a href="/A295765/b295765.txt">Table of n, a(n) for n = 0..300</a>
%F G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n)/(n+1) for n>=0.
%F a(n) ~ c * exp(n) * n^(n - 5/2), where c = exp(3/2 - exp(-2)) / sqrt(2*Pi) = 1.56162380971247949723297... - _Vaclav Kotesovec_, Oct 17 2020, updated Apr 20 2024
%e G.f.: A(x) = 1 + x + 3*x^2 + 25*x^3 + 369*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + 859400072837*x^10 + ...
%e such that
%e A(x) = 1 + x/A(x) + 4*(x/A(x))^2 + 35*(x/A(x))^3 + 506*(x/A(x))^4 + 10472*(x/A(x))^5 + 285384*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)^2*(x/A(x))^n + ...
%e RELATED SERIES.
%e Define B(x) = A(x*B(x)) and A(x) = B(x/A(x)) then B(x) begins
%e B(x) = 1 + x + 4*x^2 + 35*x^3 + 506*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + ... + binomial((n+1)^2,n)/(n+1)^2*x^n + ...
%e ILLUSTRATION OF DEFINITION.
%e The table of coefficients of x^k in A(x)^(n+1) begins:
%e [1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, ...];
%e [1, 2, 7, 56, 797, 16650, 460291, 15862152, 655825337, ...];
%e [1, 3, 12, 94, 1293, 26409, 719922, 24587202, 1010428347, ...];
%e [1, 4, 18, 140, 1867, 37272, 1001476, 33887832, 1384043365, ...];
%e [1, 5, 25, 195, 2530, 49366, 1306860, 43802060, 1777652015, ...];
%e [1, 6, 33, 260, 3294, 62832, 1638166, 54370836, 2192294775, ...];
%e [1, 7, 42, 336, 4172, 77826, 1997688, 65638294, 2629075183, ...];
%e [1, 8, 52, 424, 5178, 94520, 2387940, 77652024, 3089164371, ...];
%e [1, 9, 63, 525, 6327, 113103, 2811675, 90463365, 3573805950, ...]; ...
%e in which the main diagonal begins:
%e [1, 2, 12, 140, 2530, 62832, 1997688, ..., binomial((n+1)^2,n)/(n+1), ...].
%t terms = 21; A[_] = 1; Do[A[x_] = Sum[Binomial[(n+1)^2, n]/(n+1)^2*x^n/ A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
%t CoefficientList[A[x], x] (* _Jean-François Alcover_, Jan 14 2018 *)
%o (PARI) {a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2,m)/(m+1) - V[m+1])/(m+1);); A[n+1]}
%o for(n=0,20,print1(a(n),", "))
%Y cf. A295764, A295763, A143669.
%K nonn,changed
%O 0,3
%A _Paul D. Hanna_, Jan 06 2018
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