A255673 Coefficients of A(x), which satisfies: A(x) = 1 + x*A(x)^3 + x^2*A(x)^6.

1, 1, 4, 21, 127, 833, 5763, 41401, 305877, 2309385, 17739561, 138197876, 1089276972, 8670856834, 69606939717, 562879492551, 4580890678781, 37490975387565, 308369889858450, 2547741413147700, 21133987935358776, 175947462569886786, 1469656053534121804

Offset: 0

Author

Werner Schulte, Jul 10 2015

Keywords

• nonn

Comments

This sequence is the next after A001006 and A006605.

Examples

			A(x) = 1 + x + 4*x^2 + 21*x^3 + 127*x^4 + 833*x^5 + 5763*x^6 ...
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 13.

Crossrefs

Cf. A001006, A006605.

Cf. A365183, A365189.

Programs

Maple

a:= n-> coeff(series(RootOf(1-A+x*A^3+x^2*A^6, A), x, n+1), x, n):
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 15 2015
# second Maple program:
a:= proc(n) option remember; `if`(n<2, 1, 9*(((3*n-1))*
     (2*n-1)*(3*n-2)*(9063*n^4-18126*n^3+8403*n^2+660*n-280)*a(n-1)
     +(27*(n-1))*(3*n-1)*(3*n-4)*(3*n-2)*(3*n-5)*(57*n^2-2)*a(n-2))
      /((5*(5*n+2))*(5*n-1)*(5*n+1)*(5*n-2)*n*(57*n^2-114*n+55)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 16 2015

Mathematica

m = 30; A[_] = 0;
Do[A[x_] = 1 + x A[x]^3 + x^2 A[x]^6 + O[x]^m, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 04 2019 *)

PARI

a(n) = sum(k=0, n\2, binomial(n-k, k)*binomial(3*n+1, n-k))/(3*n+1); \\ Seiichi Manyama, Sep 02 2023

Formula

a(n+1) = Sum_{j=0..3*n+4} binomial(j,2*j-5*n-7) * binomial(3*n+4,j) / (3*n+4). (conjectured). [Vladimir Kruchinin, Mar 09 2013]

a(n) = 1/(3*n+1) * Sum_{k=0..n} (-1)^k * binomial(3*n+1, k) * binomial(6*n+2-2*k, n-k). (conjectured)

G.f. A(x) satisfies A(x) = G(x*A(x)), where G is g.f. of A006605.

G.f. A(x) satisfies A(x) = H(x*A(x)^2), where H is g.f. of A001006.

From Peter Bala, Jul 27 2023: (Start)

Define b(n) = [x^n] (1 + x + x^2)^(3*n). Then A(x)^3 = exp(Sum_{n >= 1} b(n)*x^n/n).

A(x^3) = (1/x) * series reversion of x/(1 + x^3 + x^6) = 1 + x^3 + 4*x^6 + 21*x^9 + 127*x^12 + .... (End)

a(n) = (1/(3*n+1)) * Sum_{k=0..floor(n/2)} binomial(n-k,k) * binomial(3*n+1,n-k). - Seiichi Manyama, Sep 02 2023

Extensions

More terms from Alois P. Heinz, Jul 15 2015