A060542 a(n) = (1/6)*multinomial(3*n;n,n,n).

1, 15, 280, 5775, 126126, 2858856, 66512160, 1577585295, 37978905250, 925166131890, 22754499243840, 564121960420200, 14079683012144400, 353428777651788000, 8915829964229105280, 225890910734335847055, 5744976449471863238250, 146603287914300510042750

Offset: 1

Author

Henry Bottomley, Apr 02 2001

Keywords

• nonn
• easy

Comments

Number of ways of dividing 3n labeled items into 3 unlabeled boxes with n items in each box.

From Antonio Campello (campello(AT)ime.unicamp.br), Nov 11 2009: (Start)

A060542(t) is the number of optimal [n,2,d] binary codes that correct at most t errors, i.e., having Hamming distance 2*t + 1 (achieved on length n = 3*t + 2). These codes are all isometric.

It is also the number of optimal [n,2,d] binary codes that detect 2*t + 1 errors, i.e., having Hamming distance 2t+2 (obtained by adding an overall parity check to the n = 3*t + 2 optimal codes). These codes are also all isometric.

For t = 0, we have the famous MDS, cyclic, simplex code {(000), (101), (110), (011)}. (End)

Also the number of distinct adjacency matrices of the complete tripartite graph K_{n,n,n}. - Eric W. Weisstein, Apr 21 2017

Links

• Harry J. Smith, Table of n, a(n) for n = 1..100
• Eric Weisstein's World of Mathematics, Adjacency Matrix
• Eric Weisstein's World of Mathematics, Complete Tripartite Graph

Crossrefs

Row 3 of A060540.

Cf. A006480, A025035, A361717.

Programs

Maple

a:= n-> combinat[multinomial](3*n,n$3)/3!:
seq(a(n), n=1..18);  # Alois P. Heinz, Jul 29 2023

Mathematica

Table[(3*n)!/(n!^3*6),{n,1,20}] (* Vaclav Kotesovec, Sep 23 2013 *)
Table[Multinomial[n, n, n], {n, 20}]/6 (* Eric W. Weisstein, Apr 21 2017 *)

PARI

{ a=1/6; for (n=1, 100, write("b060542.txt", n, " ", a=a*3*(3*n - 1)*(3*n - 2)/n^2); ) } \\ Harry J. Smith, Jul 06 2009

Formula

a(n) = (3*n)!/((n!)^3*6) = a(n-1)*3*(3*n - 1)*(3*n - 2)/n^2 = A060540(3,n) = A006480(n)/6. - corrected by Vaclav Kotesovec, Sep 23 2013

a(n) ~ 3^(3*n-1/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 23 2013

a(n) = 1/(8*n^3) * Sum_{k = 0..2*n} (-1)^(n+k) * k*(2*n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Oct 11 2024

a(n) = 1/(n^2) * Sum_{k = 0..n} (-1)^(n+k+1) * (n-k)^2 * binomial(2*n, k)^3. - Peter Bala, Nov 03 2024

Extensions

Definition revised by N. J. A. Sloane, Feb 02 2009