A090355 Satisfies A^4 = BINOMIAL(A)^3.

1, 3, 15, 109, 1086, 14178, 232906, 4647006, 109376595, 2967406345, 91130074437, 3123199831983, 118106517900868, 4883161763750820, 219076867059030300, 10597531747143624820, 549768536732090716371, 30443800514118532762329

Offset

0,2

Comments

See comments in A090353.

Links

Table of n, a(n) for n=0..17.

Formula

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

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)*3^k = A032033(n) = 3*A050352(n).

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)*4^k = A201354(n) = 4*A050352(n) for n >= 1. A(x) = B(x)^3 and BINOMIAL(A(x)) = B(x)^4 where B(x) = 1 + x + 4*x^2 + 28*x^3 + 286*x^4 + ... is the o.g.f. for A090353. See also A019538. (End)

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)^3 + 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^4+B^3); polcoeff(A, n, x))}

Crossrefs

Cf. A090353, A090354; A019538, A032033, A050352, A201354.

Sequence in context: A110328 A217061 A054201 * A083483 A089468 A109498

Adjacent sequences: A090352 A090353 A090354 * A090356 A090357 A090358

Keyword

nonn,easy

Author

Paul D. Hanna, Nov 26 2003

Status

approved