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

1, 1, 4, 28, 286, 3886, 66260, 1361972, 32784353, 904412593, 28124223808, 973106096392, 37073604836768, 1541948625066176, 69513081435903392, 3376138396206853792, 175739519606046355540, 9760024269508314079444

Offset: 0

Author

Paul D. Hanna, Nov 26 2003

Keywords

• nonn
• easy

Comments

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.

Examples

			A^4 = BINOMIAL(A090355), since A090355=A^3. Also, BINOMIAL(A) = A090354^2.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..320

Crossrefs

Cf. A032033, A050352, A084784, A090351, A090354, A090355, A090356, A090358.

Programs

Magma

m:=40;
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]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A090353
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 09 2023

Mathematica

nmax = 17; sol = {a[0] -> 1};
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}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)
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 *)

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))}
for(n=0,20,print1(a(n),", "))

SageMath

m=50
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)))
def A090353_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090353_list(m-9) # G. C. Greubel, Jun 09 2023

Formula

G.f. satisfies: A(x)^4 = A(x/(1-x))^3/(1-x).

a(n) ~ (n-1)! / (12 * (log(4/3))^(n+1)). - Vaclav Kotesovec, Nov 19 2014

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

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

G.f.: Product_{k>=1} 1/(1 - k*x)^((1/12) * (3/4)^k). - Seiichi Manyama, May 26 2025