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A090355 G.f. satisfies A^4 = BINOMIAL(A)^3.

%I A090355 #20 May 28 2025 07:31:10
%S A090355 1,3,15,109,1086,14178,232906,4647006,109376595,2967406345,
%T A090355 91130074437,3123199831983,118106517900868,4883161763750820,
%U A090355 219076867059030300,10597531747143624820,549768536732090716371,30443800514118532762329
%N A090355 G.f. satisfies A^4 = BINOMIAL(A)^3.
%C A090355 See comments in A090353.
%F A090355 G.f.: A(x)^4 = A(x/(1-x))^3/(1-x)^3.
%F A090355 From _Peter Bala_, May 26 2015: (Start)
%F A090355 O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*3^k = A032033(n) = 3*A050352(n).
%F A090355 BINOMIAL(A(x)) = exp( Sum_{n >= 1} c(n)*x^n/n ) where c(n) = (-1)^n*Sum_{k = 1..n} k!*Stirling2(n,k)*4^k = A201354(n) = 4*A050352(n) for n >= 1. A(x) = B(x)^3 and BINOMIAL(A(x)) = B(x)^4 where B(x) = 1 + x + 4*x^2 + 28*x^3 + 286*x^4 + ... is the o.g.f. for A090353. See also A019538. (End)
%F A090355 G.f.: Product_{k>=1} 1/(1 - k*x)^((1/4) * (3/4)^k). - _Seiichi Manyama_, May 26 2025
%F A090355 a(n) ~ (n-1)! / (4 * log(4/3)^(n+1)). - _Vaclav Kotesovec_, May 28 2025
%t A090355 nmax = 17; sol = {a[0] -> 1};
%t A090355 Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^4 - A[x/(1 - x)]^3/(1 - x)^3 + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
%t A090355 sol /. Rule -> Set;
%t A090355 a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 02 2019 *)
%o A090355 (PARI) {a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^4+B^3);polcoeff(A,n,x))}
%Y A090355 Cf. A090353, A090354; A019538, A032033, A050352, A201354.
%Y A090355 Cf. A090352, A090357, A090362.
%K A090355 nonn,easy
%O A090355 0,2
%A A090355 _Paul D. Hanna_, Nov 26 2003