
COMMENTS

It seems that a(n) = n! * 2^(binomial(n+1,2)  1) for n = 1, 2, 3, 4, 5, while for n = 6, a(n) is twice this number. The number n! * 2^(binomial(n+1,2)  1) appears in Proposition 6.1 in Eriksson and Linusson (2000) as an upper bound to the number of threedimensional permutation arrays of size n (see column k = 3 of A330490).  Petros Hadjicostas, Dec 16 2019


EXAMPLE

From Petros Hadjicostas, Dec 16 2019: (Start)
For n = 2, here are the 2*a(2) = 16 2 x 2 matrices M with elements in {0,1,2,3} that satisfy MM' mod 4 = I:
(a) With 1 = det(M) mod 4:
[[1,0],[0,1]]; [[0,1],[3,0]]; [[0,3],[1,0]]; [[1,2],[2,1]];
[[2,1],[3,2]]; [[2,3],[1,2]]; [[3,0],[0,3]]; [[3,2],[2,3]].
These form the abelian group SO(2, Z_n). See the comments for sequence A060968.
(b) With 3 = det(M) mod 4:
[[0,1],[1,0]]; [[0,3],[3,0]]; [[1,0],[0,3]]; [[1,2],[2,3]];
[[2,1],[1,2]]; [[2,3],[3,2]]; [[3,0],[0,1]]; [[3,2],[2,1]].
Note that, for n = 3, we have 2*a(3) = 2*192 = 384 = A264083(4). (End)
