%I
%S 1,4,120,14400,9360000,29016000000
%N 1/2 times the number of n X n 0..4 matrices M with MM' mod 5 = I, where M' is the transpose of M and I is the n X n identity matrix.
%C Even though only 6 terms are known for this sequence (as of the time of this note), our conjecture below is based on the work of _Jianing Song_ for sequence A318609 with the mod 3 case. (Like 3, the number 5 is also a prime number.)  _Petros Hadjicostas_, Dec 18 2019
%H Jianing Song, <a href="/A060968/a060968.txt">Structure of the group SO(2,Z_n)</a>.
%F From _Petros Hadjicostas_, Dec 20 2019: (Start)
%F Let b(n) be the number of solutions to the equation Sum_{i = 1..n} x_i^2 = 1 (mod 5) with x_i in 0..4. We have that b(n) = 5*b(n1) + 5*b(n2)  25*b(n3) for n >= 3 with b(0) = 0, b(1) = 2, and b(2) = 4.
%F We have b(n) = A330607(n, k=1) for n >= 0.
%F We conjecture that a(n+1) = a(n)*b(n+1) for n >= 1. (End)
%e From _Petros Hadjicostas_, Dec 17 2019: (Start)
%e For n = 2, the 2*a(2) = 8 n x n matrices M with elements in {0,1,2,3,4} that satisfy MM' mod 5 = I are the following:
%e (a) those with 1 = det(M) mod 5:
%e [[1,0],[0,1]]; [[0,4],[1,0]]; [[0,1],[4,0]]; [[4,0],[0,4]].
%e These form the abelian group SO(2, Z_5). See the comments for sequence A060968.
%e (b) those with 4 = det(M) mod 5:
%e [[0,1],[1,0]]; [[0,4],[4,0]]; [[1,0],[0,4]]; [[4,0],[0,1]].
%e Note that, for n = 3, we have 2*a(3) = 2*120 = 240 = A264083(5). (End)
%Y Cf. A060968, A071302, A071303, A071305, A071306, A071307, A071308, A071309, A071310, A071900, A087784, A208895, A264083, A318609, A330607.
%K nonn,more
%O 1,2
%A _R. H. Hardin_, Jun 11 2002
