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
Examples
a(2) = C(4,2) = 6. a(3) = C(9,5) = 126.
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. See p. 8.
Crossrefs
A082963 is the equivalent sequence up to reflection and rotation.
Programs
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Maple
a:= n-> (s-> binomial(s, floor(s/2)))(n^2): seq(a(n), n=0..15); # Alois P. Heinz, Jul 21 2017
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PARI
a(n)=binomial(n^2,n^2\2) \\ Charles R Greathouse IV, May 09 2013
Formula
a(n) = C(n^2, (n^2+1)/2) if n is odd and C(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
Extensions
a(0)=1 prepended by Alois P. Heinz, Jul 21 2017