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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

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OFFSET

0,2

COMMENTS

See comments in A090351.

LINKS

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

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

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 *)

PROG

(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))}

(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<x>:=PowerSeriesRing(Rationals(), m+1); // A090352

Coefficients(R!( f(m, x) )); // G. C. Greubel, Jul 07 2023

(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)))