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The number of possible borders of Latin squares with the top row fixed.
2

%I #25 Nov 10 2020 15:44:31

%S 1,1,2,26,924,81624,13433520,3706068240,1582042381920,987057348842880,

%T 861632512758823680,1016677874552767660800,1576819957670934809817600,

%U 3140963381712726319842892800,7880571655922780897709237811200,24492587962448960350527019884595200

%N The number of possible borders of Latin squares with the top row fixed.

%C The definition is not quite right, and should be corrected.

%H Alois P. Heinz, <a href="/A173103/b173103.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-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 The only two configurations for n=3, given the top row is 123:

%e 123 123

%e 2 1 3 2

%e 312 231

%e Two arbitrary configurations for n=4, given the top row is 1234:

%e 1234 1234

%e 2 1 4 3

%e 3 2 3 2

%e 4123 2341

%p d:= proc(n) d(n):= `if`(n<=1, 1-n, (n-1)*(d(n-1)+d(n-2))) end:

%p a:= proc(n) a(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 seq(a(n), n=1..20); # _Alois P. Heinz_, Aug 18 2013

%t d = Subfactorial;

%t a[n_] := If[n <= 3, {1, 1, 2}[[n]], (n-2)! (((2n-5) d[n-2]^2)/(n-3) + 2d[n-1] d[n-2] + ((n-1) d[n-1]^2)/(n-2))];

%t Array[a, 20] (* _Jean-François Alcover_, Nov 10 2020 *)

%Y Related to A000166. Equals A173104 divided by n!.

%K nonn

%O 1,3

%A _Johan de Ruiter_, Feb 09 2010