%I #22 Nov 10 2020 17:08:43
%S 1,2,12,624,110880,58769280,67704940800,149428671436800,
%T 574091539551129600,3581833707481042944000,34393612685291413069824000,
%U 486990328595374993951457280000,9818890674272030616178239406080000,273823820339488809857168046768783360000
%N The number of possible borders of Latin squares.
%C The definition is not quite right, and should be corrected.
%H Alois P. Heinz, <a href="/A173104/b173104.txt">Table of n, a(n) for n = 1..100</a>
%H J. de Ruiter, <a href="http://liacs.leidenuniv.nl/assets/Bachelorscripties/10-04-JohandeRuiter.pdf">On Jigsaw Sudoku Puzzles and Related Topics</a>, Bachelor Thesis, Leiden Institute of Advanced Computer Science, 2010.
%F For n>3, a(n)=n!(n-2)!((n-1)/(n-2)d[n-1]^2+2d[n-1]d[n-2]+(2n-5)/(n-3)d[n-2]^2), where d[k] is the number of derangements of k elements (A000166).
%e Two arbitrary configurations for n=3:
%e 123 312
%e 2 1 1 3
%e 312 231
%e Two arbitrary configurations for n=4:
%e 1234 1432
%e 2 1 3 4
%e 3 2 4 1
%e 4123 2143
%p d:= proc(n) d(n):= `if`(n<=1, 1-n, (n-1)*(d(n-1)+d(n-2))) end:
%p b:= proc(n) b(n):= `if`(n<4, [1, 1, 2][n], (n-2)!*((n-1)/
%p (n-2)*d(n-1)^2+2*d(n-1)*d(n-2)+(2*n-5)/(n-3)*d(n-2)^2))
%p end:
%p a:= n-> n!*b(n):
%p seq(a(n), n=1..20); # _Alois P. Heinz_, Aug 18 2013
%t d = Subfactorial;
%t a[n_] := If[n <= 3, {1, 2, 12}[[n]], n! (n-2)! ((n-1)/(n-2) d[n-1]^2 + 2d[n-1] d[n-2] + (2n-5)/(n-3) d[n-2]^2)];
%t Array[a, 20] (* _Jean-François Alcover_, Nov 10 2020 *)
%Y Related to A000166. Equals A173103 multiplied by n!.
%K nonn
%O 1,2
%A _Johan de Ruiter_, Feb 09 2010