A090352 G.f. satisfies A^3 = BINOMIAL(A)^2, where A = A090351^2.

1, 2, 7, 36, 255, 2370, 27713, 393352, 6582068, 126888632, 2767912036, 67362737168, 1808596304964, 53083358012760, 1690443996202428, 58039582729688320, 2136931230333535178, 83981145793974066484

Offset: 0

Author

Paul D. Hanna, Nov 26 2003

Keywords

• nonn
• easy

Comments

See comments in A090351.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..390

Crossrefs

Cf. A004123, A019538, A050351, A084785, A090351, A201339.

Cf. A090355, A090357, A090362.

Programs

Magma

m:=40;
f:= func< n, x | Exp((&+[(&+[2^j*Factorial(j)*StirlingSecond(k, j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1);  // A090352
Coefficients(R!( f(m, x) )); // G. C. Greubel, Jul 07 2023

Mathematica

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - A[x/(1 - x)]^2/(1 - x)^2 + 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 *)

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^3+B^2); polcoeff(A,n,x))}

SageMath

m=50
def f(n, x): return exp(sum(sum(2^j*factorial(j)*stirling_number2(k, j)*x^k/k for j in range(1, k+1)) for k in range(1, n+2)))
def A090352_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m, x) ).list()
A090352_list(m-9) # G. C. Greubel, Jul 07 2023

Formula

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

From Peter Bala, May 26 2015: (Start)

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 = A004123(n+1) = 2*A050351(n) for n >= 1. Cf. A084785.

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)*(-3)^k = A201339(n) = 3*A050351(n) for n >= 1.

A(x) = B(x)^2 and BINOMIAL(A(x)) = B(x)^3 where B(x) = 1 + x + 3*x^2 + 15*x^3 + 108*x^4 + ... is the o.g.f. for A090351. See also A019538. (End)

G.f.: Product_{k>=1} 1/(1 - k*x)^((1/3) * (2/3)^k). - Seiichi Manyama, May 26 2025

a(n) ~ (n-1)! / (3 * log(3/2)^(n+1)). - Vaclav Kotesovec, May 28 2025