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The number of possible borders of Latin squares.
2

%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