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A090353 G.f. satisfies A^4 = BINOMIAL(A^3).

%I A090353 #28 May 26 2025 09:56:28
%S A090353 1,1,4,28,286,3886,66260,1361972,32784353,904412593,28124223808,
%T A090353 973106096392,37073604836768,1541948625066176,69513081435903392,
%U A090353 3376138396206853792,175739519606046355540,9760024269508314079444
%N A090353 G.f. satisfies A^4 = BINOMIAL(A^3).
%C A090353 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 A090353 Vaclav Kotesovec, <a href="/A090353/b090353.txt">Table of n, a(n) for n = 0..320</a>
%F A090353 G.f. satisfies: A(x)^4 = A(x/(1-x))^3/(1-x).
%F A090353 a(n) ~ (n-1)! / (12 * (log(4/3))^(n+1)). - _Vaclav Kotesovec_, Nov 19 2014
%F A090353 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-1) = A050352(n) = 1/3*A032033(n) for n >= 1. - _Peter Bala_, May 26 2015
%F A090353 G.f. satisfies [x^n] 1/A(x)^(3*n-3) = [x^n] 1/A(x)^(4*n-4) for n >= 0. - _Paul D. Hanna_, Apr 28 2025
%F A090353 G.f.: Product_{k>=1} 1/(1 - k*x)^((1/12) * (3/4)^k). - _Seiichi Manyama_, May 26 2025
%e A090353 A^4 = BINOMIAL(A090355), since A090355=A^3. Also, BINOMIAL(A) = A090354^2.
%t A090353 nmax = 17; sol = {a[0] -> 1};
%t A090353 Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^4 - A[x/(1 - x)]^3/(1 - x) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
%t A090353 sol /. Rule -> Set;
%t A090353 a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 02 2019 *)
%t A090353 With[{m=40}, CoefficientList[Series[Exp[Sum[Sum[3^(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 A090353 (PARI) {a(n) = my(A); if(n<0,0,A=1+x +x*O(x^n); for(k=1,n, B = subst(A^3,x,x/(1-x))/(1-x)+x*O(x^n); A = A - A^4 + B); polcoef(A,n,x))}
%o A090353 for(n=0,20,print1(a(n),", "))
%o A090353 (Magma)
%o A090353 m:=40;
%o A090353 f:= func< n,x | Exp((&+[(&+[3^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
%o A090353 R<x>:=PowerSeriesRing(Rationals(), m+1); // A090353
%o A090353 Coefficients(R!( f(m,x) )); // _G. C. Greubel_, Jun 09 2023
%o A090353 (SageMath)
%o A090353 m=50
%o A090353 def f(n, x): return exp(sum(sum(3^(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 A090353 def A090353_list(prec):
%o A090353     P.<x> = PowerSeriesRing(QQ, prec)
%o A090353     return P( f(m,x) ).list()
%o A090353 A090353_list(m-9) # _G. C. Greubel_, Jun 09 2023
%Y A090353 Cf. A032033, A050352, A084784, A090351, A090354, A090355, A090356, A090358.
%K A090353 nonn,easy
%O A090353 0,3
%A A090353 _Paul D. Hanna_, Nov 26 2003