A129123 Number of 4-tuples of standard tableau with height less than or equal to 2.

1, 1, 2, 17, 98, 882, 7812, 78129, 815474, 8955650, 101869508, 1194964498, 14374530436, 176681194276, 2212121332488, 28145258688369, 363177582488274, 4745064935840178, 62687665026816228, 836447728509168930, 11261240896657686660, 152847558411986548260

Offset: 0

Author

Mike Zabrocki, Mar 29 2007

Keywords

• nonn

Comments

Number of pairs of Dyck paths of semilength n with equal midpoint. - Alois P. Heinz, Oct 07 2022

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..839
• F. Bergeron, L. Favreau and D. Krob, Conjectures on the enumeration of tableaux of bounded height, Discrete Math, vol. 139, no. 1-3 (1995), 463-468.

Crossrefs

Cf. A000108, A001405, A003161, A120730, A355481, A357652.

Column k=4 of A357824.

Programs

Magma

[(&+[((n-2*j+1)/(n-j+1))^4*Binomial(n,j)^4: j in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Nov 08 2022

Maple

b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
     `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
    end:
a:= n-> add(b(n, n-2*j)^4, j=0..n/2):
seq(a(n), n=0..21);  # Alois P. Heinz, Mar 25 2025

Mathematica

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^4, {k,0, Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Dec 16 2017 *)

PARI

a(n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^4);

Python

from math import comb
def A129123(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**4 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

SageMath

def A129123(n): return sum(((n-2*j+1)/(n-j+1))^4*binomial(n,j)^4 for j in range((n//2)+1))
[A129123(n) for n in range(31)] # G. C. Greubel, Nov 08 2022

Formula

a(n) = Sum_{k=0..n} A120730(n,k)^4. - Philippe Deléham, Oct 18 2008

From Vaclav Kotesovec, Dec 16 2017: (Start)

Recurrence: n*(n+1)^3*(15*n^2 - 34*n + 7)*a(n) = 2*n*(90*n^5 - 309*n^4 + 147*n^3 + 124*n^2 - 135*n + 35)*a(n-1) + 4*(n-1)^2*(4*n - 5)*(4*n - 3)*(15*n^2 - 4*n - 12)*a(n-2).

a(n) ~ 3* 2^(4*n - 1/2) / (Pi^(3/2) * n^(7/2)). (End)

a(n) = A357652(n) - A355481(n). - Alois P. Heinz, Oct 13 2022

a(n) = Sum_{j=0..floor(n/2)} ((n-2*j+1)/(n-j+1))^4 * binomial(n,j)^4. - G. C. Greubel, Nov 08 2022

a(n) = Sum_{k=0..n} binomial(n,k) * ( binomial(n,k) - binomial(n,k-1) )^3. - Seiichi Manyama, Mar 25 2025