A081623 Number of ways in which the points on an n X n square lattice can be equally occupied with spin "up" and spin "down" particles. If n is odd, we arbitrarily take the lattice to contain one more spin "up" particle than the number of spin "down" particles.

1, 1, 6, 126, 12870, 5200300, 9075135300, 63205303218876, 1832624140942590534, 212392290424395860814420, 100891344545564193334812497256, 191645966716130525165099506263706416, 1480212998448786189993816895482588794876100

Offset

0,3

Links

Alois P. Heinz, Table of n, a(n) for n = 0..57

Brian Hayes, The World in a Spin, American Scientist 88:5 (September-October 2000), pp. 384-388. [alternate link]

James Grime, Maths Problem: Complete Noughts and Crosses (Burnside's Lemma)

Noah Lordi, Maedee Trank-Greene, Akira Kyle, and Joshua Combes, Quantum permutation puzzles with indistinguishable particles, arXiv:2410.22287 [quant-ph], 2024-2025. See p. 8.

Formula

a(n) = binomial(n^2, (n^2+1)/2) if n is odd and binomial(n^2, n^2/2) if n is even.

a(n) = binomial(n^2,floor(n^2/2)). - Alois P. Heinz, Jul 21 2017

Largest coefficient of (1 + x)^(n^2). - Ilya Gutkovskiy, Apr 24 2025

a(n) ~ 2^(n^2+1/2) / (n * sqrt(Pi)). - Amiram Eldar, Nov 18 2025

Example

a(2) = C(4,2) = 6.
a(3) = C(9,5) = 126.

Maple

a:= n-> (s-> binomial(s, floor(s/2)))(n^2):
seq(a(n), n=0..15);  # Alois P. Heinz, Jul 21 2017

Mathematica

a[n_] := Binomial[n^2, Floor[n^2/2]]; Array[a, 13, 0] (* Amiram Eldar, Nov 18 2025 *)

Prog

(PARI) a(n)=binomial(n^2, n^2\2) \\ Charles R Greathouse IV, May 09 2013

Crossrefs

A082963 is the equivalent sequence up to reflection and rotation.

Sequence in context: A390891 A255900 A133792 * A223210 A324093 A177756

Adjacent sequences: A081620 A081621 A081622 * A081624 A081625 A081626

Keyword

easy,nonn

Author

A. Timothy Royappa, Apr 22 2003

Extensions

a(0)=1 prepended by Alois P. Heinz, Jul 21 2017

Status

approved