A126042 Expansion of f(x^3)/(1-x*f(x^3)), where f(x) is the g.f. of A001764, whose n-th term is binomial(3n,n)/(2n+1).

1, 1, 1, 2, 3, 4, 8, 13, 19, 38, 64, 98, 196, 337, 531, 1062, 1851, 2974, 5948, 10468, 17060, 34120, 60488, 99658, 199316, 355369, 590563, 1181126, 2115577, 3540464, 7080928, 12731141, 21430267, 42860534, 77306428, 130771376, 261542752, 473018396, 803538100

Offset: 0

Author

Paul Barry, Dec 16 2006

Keywords

• easy
• nonn

Comments

Row sums of number triangle A111373.

Interleaves T(3n,2n), T(3n+1,2n+1) and T(3n+2,2n+2) for T(n,k) = A047089(n,k).

One step forward and two steps back: number of nonnegative walks of n steps where the steps are size 1 forwards and size 2 backwards. - David Scambler, Mar 15 2011

Brown's criterion ensures that the sequence is complete (see formulae). - Vladimir M. Zarubin, Aug 05 2019

Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 3. - Emanuele Munarini, Jun 20 2024

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Paul Barry, The Triple Riordan Group, arXiv:2412.05461 [math.CO], 2024. See pp. 7, 10.
• Eric Weisstein's World of Mathematics, Brown's Criterion.

Crossrefs

Cf. A047089, A111373.

Programs

Magma

[n lt 3 select 1 else Binomial(n, Floor(n/3)) - (&+[Binomial(n,j): j in [0..Floor(n/3)-1]]): n in [0..40]]; // G. C. Greubel, Jul 30 2022

Maple

a:= proc(n) option remember; `if`(n<4, [1$3, 2][n+1], (a(n-1)*
       2*(20*n^4-14*n^3-31*n^2-n+8)-6*(3*n-1)*(5*n-6)*a(n-2)
      +9*(n-2)*(15*n^3-48*n^2+15*n+14)*a(n-3)-54*(n-2)*(n-3)*
      (5*n^2-n-2)*a(n-4))/(2*(2*n+1)*(n+1)*(5*n^2-11*n+4)))
    end:
seq(a(n), n=0..45);  # Alois P. Heinz, Sep 07 2022

Mathematica

Table[Binomial[n, Floor[n/3]] -Sum[Binomial[n,i], {i,0,Floor[n/3] -1}], {n,0,40}] (* David Callan, Oct 26 2017 *)
a[n_] := Binomial[n, Floor[n/3]] (1 + Hypergeometric2F1[1, -n + Floor[n/3], 1 + Floor[n/3], -1]) - 2^n; Table[a[n], {n, 0, 38}] (* Peter Luschny, Jun 20 2024 *)

PARI

{a(n)=polcoeff((1/x)*serreverse(x*(1+x)^2/((1+x)^3+x^3+x*O(x^n))),n)}

PARI

n=30;
{a0=1;a1=1;a2=1;for(k=1, n/3,print1(a0,", ",a1,", ",a2,", ");
a0=2*a2;a1=2*a0-binomial(3*k,k)/(2*k+1);a2=2*a1-binomial(3*k+1,k)/(k+1))
} \\ Vladimir M. Zarubin, Aug 05 2019

SageMath

[binomial(n, (n//3)) - sum(binomial(n,j) for j in (0..(n//3)-1)) for n in (0..40)] # G. C. Greubel, Jul 30 2022

Formula

a(n) = Sum_{k=0..n} binomial(3*floor((n+2k)/3) - 2k, floor((n+2k)/3)-k)*(k+1)/(2*floor((n+2k)/3) - k + 1)(2*cos(2*Pi*(n-k)/3) + 1)/3.

G.f.: (1/x)*Series_Reversion( x*(1+x)^2/((1+x)^3+x^3) ). - Paul D. Hanna, Mar 15 2011

From Vladimir M. Zarubin, Aug 05 2019: (Start)

a(0) = 1, a(1) = 1, a(2) = 1 and for k>0

a(3*k) = 2*a(3*k-1),

a(3*k+1) = 2*a(3*k) - binomial(3*k,k)/(2*k+1),

a(3*k+2) = 2*a(3*k+1) - binomial(3*k+1,k)/(k+1),

where binomial(3*k,k)/(2*k+1) = A001764(k)

and binomial(3*k+1,k)/(k+1) = A006013(k). (End)

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} (n-3*k+1) * binomial(n+1,k). - Seiichi Manyama, Jan 27 2024