A081623 - OEIS

%I #20 Nov 05 2024 10:45:08

%S 1,1,6,126,12870,5200300,9075135300,63205303218876,

%T 1832624140942590534,212392290424395860814420,

%U 100891344545564193334812497256,191645966716130525165099506263706416,1480212998448786189993816895482588794876100

%N 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.

%H Alois P. Heinz, <a href="/A081623/b081623.txt">Table of n, a(n) for n = 0..57</a>

%H Brian Hayes, <a href="http://bit-player.org/wp-content/extras/bph-publications/AmSci-2000-09-Hayes-Ising.pdf">The World in a Spin</a>, American Scientist 88:5 (September-October 2000), pp. 384-388. [<a href="http://www.americanscientist.org/issues/pub/the-world-in-a-spin">alternate link</a>]

%H James Grime, <a href="http://www.youtube.com/watch?v=wdDF7_vfLcE">Maths Problem: Complete Noughts and Crosses (Burnside's Lemma)</a>

%H Noah Lordi, Maedee Trank-Greene, Akira Kyle, and Joshua Combes, <a href="https://arxiv.org/abs/2410.22287">Quantum permutation puzzles with indistinguishable particles</a>, arXiv:2410.22287 [quant-ph], 2024. See p. 8.

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

%F a(n) = binomial(n^2,floor(n^2/2)). - _Alois P. Heinz_, Jul 21 2017

%e a(2) = C(4,2) = 6.

%e a(3) = C(9,5) = 126.

%p a:= n-> (s-> binomial(s, floor(s/2)))(n^2):

%p seq(a(n), n=0..15);  # _Alois P. Heinz_, Jul 21 2017

%o (Mathcad or Microsoft Excel): f(n)=combin(n^2,trunc((n^2+1)/2))

%o (PARI) a(n)=binomial(n^2,n^2\2) \\ _Charles R Greathouse IV_, May 09 2013

%Y A082963 is the equivalent sequence up to reflection and rotation.

%K easy,nonn

%O 0,3

%A _A. Timothy Royappa_, Apr 22 2003

%E a(0)=1 prepended by _Alois P. Heinz_, Jul 21 2017