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A090356 G.f. A(x) satisfies A(x)^5 = BINOMIAL(A(x)^4); that is, the binomial transform of the coefficients in A(x)^4 yields the coefficients in A(x)^5.

%I A090356 #25 May 26 2025 09:56:17
%S A090356 1,1,5,45,595,10475,231255,6148495,191276600,6815243040,273601200136,
%T A090356 12217471594856,600580173151560,32224787998758280,1873909224391774760,
%U A090356 117388347849375956328,7880739469498103077588,564440024187816634143380
%N A090356 G.f. A(x) satisfies A(x)^5 = BINOMIAL(A(x)^4); that is, the binomial transform of the coefficients in A(x)^4 yields the coefficients in A(x)^5.
%C A090356 In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.
%H A090356 Vaclav Kotesovec, <a href="/A090356/b090356.txt">Table of n, a(n) for n = 0..310</a>
%F A090356 G.f. satisfies: A(x)^5 = A(x/(1-x))^4/(1-x).
%F A090356 a(n) ~ (n-1)! / (20 * (log(5/4))^(n+1)). - _Vaclav Kotesovec_, Nov 19 2014
%F A090356 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-1) = A050353(n) = 1/4*A094417(n) for n >= 1. - _Peter Bala_, May 26 2015
%F A090356 G.f.: Product_{k>=1} 1/(1 - k*x)^((1/20) * (4/5)^k). - _Seiichi Manyama_, May 26 2025
%e A090356 G.f.: A(x) = 1 + x + 5*x^2 + 45*x^3 + 595*x^4 + 10475*x^5 + 231255*x^6 + ...
%e A090356 The coefficients in A(x)^4 are given by A090357 and begin
%e A090356 A(x)^4: [1, 4, 26, 244, 3131, 52600, 1111940, ..., A090357(n), ...].
%e A090356 The binomial transform of A090357 yields the coefficients of A(x)^5:
%e A090356 A(x)^5: [1, 5, 35, 335, 4280, 70976, 1479800, ...]
%e A090356 as shown by
%e A090356 1 = 1*1,
%e A090356 5 = 1*1 + 1*4,
%e A090356 35 = 1*1 + 2*4 + 1*26,
%e A090356 335 = 1*1 + 3*4 + 3*26 + 1*244,
%e A090356 4280 = 1*1 + 4*4 + 6*26 + 4*244 + 1*3131, ...
%t A090356 nmax = 17; sol = {a[0] -> 1};
%t A090356 Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^5 - A[x/(1 - x)]^4/(1 - x) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
%t A090356 sol /. Rule -> Set;
%t A090356 a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 02 2019 *)
%t A090356 With[{m = 40}, CoefficientList[Series[Exp[Sum[Sum[4^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* _G. C. Greubel_, Jun 09 2023 *)
%o A090356 (PARI) {a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^4,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^5+B);polcoeff(A,n,x))}
%o A090356 (Magma)
%o A090356 m:=40;
%o A090356 f:= func< n,x | Exp((&+[(&+[4^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
%o A090356 R<x>:=PowerSeriesRing(Rationals(), m+1); // A090356
%o A090356 Coefficients(R!( f(m,x) )); // _G. C. Greubel_, Jun 09 2023
%o A090356 (SageMath)
%o A090356 m=40
%o A090356 def f(n, x): return exp(sum(sum(4^(j-1)*factorial(j)* stirling_number2(k,j)*x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
%o A090356 def A090356_list(prec):
%o A090356     P.<x> = PowerSeriesRing(QQ, prec)
%o A090356     return P( f(m,x) ).list()
%o A090356 A090356_list(m) # _G. C. Greubel_, Jun 09 2023
%Y A090356 Cf. A050353, A084784, A090353, A090357, A090358, A094417.
%K A090356 nonn,easy
%O A090356 0,3
%A A090356 _Paul D. Hanna_, Nov 26 2003