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A090357 G.f. satisfies A^5 = BINOMIAL(A)^4; also equals A090356^4.

%I A090357 #19 May 28 2025 07:33:06
%S A090357 1,4,26,244,3131,52600,1111940,28559320,865622825,30250881420,
%T A090357 1196941704454,52860066623036,2576115583371739,137274420821505776,
%U A090357 7937914900025008984,494941882189888642832,33096552232229291234923
%N A090357 G.f. satisfies A^5 = BINOMIAL(A)^4; also equals A090356^4.
%C A090357 See comments in A090356.
%F A090357 G.f.: A(x)^5 = A(x/(1-x))^4/(1-x)^4.
%F A090357 From _Peter Bala_, May 26 2015: (Start)
%F A090357 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)*4^k = A094417(n) = 4*A050353(n) for n >= 1.
%F A090357 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)*(-5)^k = A201365(n) = 5*A050353(n) for n >= 1.
%F A090357 A(x) = B(x)^4 and BINOMIAL(A(x)) = B(x)^5 where B(x) = 1 + x + 5*x^2 + 45*x^3 + 495*x^4 + ... is the o.g.f. for A090356. See also A019538. (End)
%F A090357 G.f.: Product_{k>=1} 1/(1 - k*x)^((1/5) * (4/5)^k). - _Seiichi Manyama_, May 26 2025
%F A090357 a(n) ~ (n-1)! / (5 * log(5/4)^(n+1)). - _Vaclav Kotesovec_, May 28 2025
%t A090357 nmax = 16; sol = {a[0] -> 1};
%t A090357 Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^5 - A[x/(1 - x)]^4/(1 - x)^4 + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
%t A090357 sol /. Rule -> Set;
%t A090357 a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 02 2019 *)
%o A090357 (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^5+B^4);polcoeff(A,n,x))}
%Y A090357 Cf. A090356, A019538, A050353, A201365.
%Y A090357 Cf. A090352, A090355, A090362.
%K A090357 nonn,easy
%O A090357 0,2
%A A090357 _Paul D. Hanna_, Nov 26 2003