A090351 - OEIS

%I #18 Jul 07 2023 05:44:19

%S 1,1,3,15,108,1032,12388,179572,3052986,59555338,1310677726,

%T 32114051862,866766965308,25547102523604,816335926158372,

%U 28107705687291892,1037367351120788551,40852168787823027351,1709792654612819858341

%N G.f. satisfies A^3 = BINOMIAL(A^2).

%C 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 Vaclav Kotesovec, <a href="/A090351/b090351.txt">Table of n, a(n) for n = 0..390</a>

%F G.f. satisfies: A(x)^3 = A(x/(1-x))^2/(1-x).

%F a(n) ~ (n-1)! / (6 * (log(3/2))^(n+1)). - _Vaclav Kotesovec_, Nov 18 2014

%F 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)*2^(k-1) = A050351(n) = 1/2*A004123(n+1) for n >= 1. - _Peter Bala_, May 26 2015

%e A^3 = BINOMIAL(A090352), since A090352=A^2.

%t nmax = 18; sol = {a[0] -> 1};

%t Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - A[x/(1 - x)]^2/(1 - x) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];

%t sol /. Rule -> Set;

%t a /@ Range[0, nmax] (* _Jean-François Alcover_, Nov 02 2019 *)

%t With[{m=40}, CoefficientList[Series[Exp[Sum[Sum[2^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* _G. C. Greubel_, Jun 08 2023 *)

%o (PARI) {a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^2,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^3+B);polcoeff(A,n,x))}

%o (Magma)

%o m:=40;

%o f:= func< n,x | Exp((&+[(&+[2^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;

%o R<x>:=PowerSeriesRing(Rationals(), m+1);  // A090351

%o Coefficients(R!( f(m,x) )); // _G. C. Greubel_, Jun 08 2023

%o (SageMath)

%o m=50

%o def f(n, x): return exp(sum(sum(2^(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 def A090351_list(prec):

%o     P.<x> = PowerSeriesRing(QQ, prec)

%o     return P( f(m,x) ).list()

%o A090351_list(m-9) # _G. C. Greubel_, Jun 08 2023

%Y Cf. A004123, A050351, A084784, A090352, A090353, A090356, A090358.

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, Nov 26 2003