A034841 a(n) = (n^2)! / (n!)^n.

1, 1, 6, 1680, 63063000, 623360743125120, 2670177736637149247308800, 7363615666157189603982585462030336000, 18165723931630806756964027928179555634194028454000000, 53130688706387569792052442448845648519471103327391407016237760000000000

Offset: 0

Author

Erich Friedman

Keywords

• nonn

Comments

The number of arrangements of 1,2,...,n^2 in an n X n matrix such that each row is increasing. - Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 12 2001

a(n) == 0 (mod n!). In fact (n^2)! == 0 (mod (n!)^n) by elementary combinatorics, a better result is (n^2)! == 0 (mod (n!)^(n+1)). - Amarnath Murthy, Jul 13 2005

a(n) is also the number of lattice paths from {n}^n to {0}^n using steps that decrement one component by 1. a(2) = 6: [(2,2), (1,2), (0,2), (0,1), (0,0)], [(2,2), (1,2), (1,1), (0,1), (0,0)], [(2,2), (1,2), (1,1), (1,0), (0,0)], [(2,2), (2,1), (1,1), (0,1), (0,0)], [(2,2), (2,1), (1,1), (1,0), (0,0)], [(2,2), (2,1), (2,0), (1,0), (0,0)]. - Alois P. Heinz, May 06 2013

Given n^2 distinguishable balls and n distinguishable urns, a(n) = the number of ways to place n balls in the i-th urn for all 1 <= i <= n, where n = n_1 + n_2 + ... + n_n. - Ross La Haye, Dec 28 2013

Links

• Alois P. Heinz and Tilman Piesk, Table of n, a(n) for n = 0..26 (first 20 terms from Alois P. Heinz)
• Noah Lordi, Maedee Trank-Greene, Akira Kyle, and Joshua Combes, Quantum permutation puzzles with indistinguishable particles, arXiv:2410.22287 [quant-ph], 2024. See p. 8.

Crossrefs

Cf. A000142, A039622, A229050, A229050, A340590.

Cf. A000984, A006480, A008977, A008978, A022915.

Main diagonal of A089759, A187783.

Programs

Magma

[Factorial(n^2) / Factorial(n)^n: n in [0..10]]; // Vincenzo Librandi, Oct 29 2014

Maple

a:= n-> (n^2)! / (n!)^n:
seq(a(n), n=0..10);  # Alois P. Heinz, Jul 24 2012

Mathematica

Prepend[Table[nn = n^2;nn! Coefficient[Series[(x^n/n!)^n, {x, 0, nn}], x^nn], {n, 1, 15}], 1] (* Geoffrey Critzer, Mar 08 2015 *)

PARI

a(n) = (n^2)! / (n!)^n; \\ Michel Marcus, Oct 28 2014

Formula

Using a higher order version of Stirling's formula (the "standard" formula appears in A000142) we have the asymptotic expression: a(n) ~ sqrt(2*Pi) * e^(-1/12) * n^(n^2 - n/2 + 1) / (2*Pi)^(n/2). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001

a(n) = Product_{k=1..n} binomial(k*n, n). - Vaclav Kotesovec, Mar 10 2019

Extensions

a(0)=1 prepended by Tilman Piesk, Oct 28 2014