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A071304 a(n) = (1/2) * (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).

%I A071304 #46 Nov 07 2022 02:25:49
%S A071304 1,4,120,14400,9360000,29016000000,457002000000000,
%T A071304 35646156000000000000,13946558535000000000000000,
%U A071304 27230655539587500000000000000000,266009466302345390625000000000000000000,12987912192212013697265625000000000000000000000
%N A071304 a(n) = (1/2) * (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 A071304 Also, number of n X n orthogonal matrices over GF(5) with determinant 1. - _Max Alekseyev_, Nov 06 2022
%H A071304 Jessie MacWilliams, <a href="https://doi.org/10.2307/2317262">Orthogonal Matrices Over Finite Fields</a>, The American Mathematical Monthly 76:2 (1969), 152-164.
%H A071304 Jianing Song, <a href="/A060968/a060968.txt">Structure of the group SO(2,Z_n)</a>.
%F A071304 a(2k+1) = 5^k * Product_{i=0..k-1} (5^(2k) - 5^(2i)); a(2k) = (5^k - 1) * Product_{i=1..k-1} (5^(2k) - 5^(2i)) (see MacWilliams, 1969). - _Max Alekseyev_, Nov 06 2022
%F A071304 From _Petros Hadjicostas_, Dec 20 2019: (Start)
%F A071304 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(n-1) + 5*b(n-2) - 25*b(n-3) for n >= 3 with b(0) = 0, b(1) = 2, and b(2) = 4.
%F A071304 We have b(n) = A330607(n, k=1) for n >= 0.
%F A071304 We conjecture that a(n+1) = a(n)*b(n+1) for n >= 1. (End)
%e A071304 From _Petros Hadjicostas_, Dec 17 2019: (Start)
%e A071304 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 A071304 (a) those with 1 = det(M) mod 5:
%e A071304 [[1,0],[0,1]]; [[0,4],[1,0]]; [[0,1],[4,0]]; [[4,0],[0,4]].
%e A071304 These form the abelian group SO(2, Z_5). See the comments for sequence A060968.
%e A071304 (b) those with 4 = det(M) mod 5:
%e A071304 [[0,1],[1,0]]; [[0,4],[4,0]]; [[1,0],[0,4]]; [[4,0],[0,1]].
%e A071304 Note that, for n = 3, we have 2*a(3) = 2*120 = 240 = A264083(5). (End)
%o A071304 (PARI) { a071304(n) = my(t=n\2); prod(i=0, t-1, 5^(2*t)-5^(2*i)) * if(n%2, 5^t, 1/(5^t+1)); } \\ _Max Alekseyev_, Nov 06 2022
%Y A071304 Cf. A060968, A071302, A071303, A071305, A071306, A071307, A071308, A071309, A071310, A071900, A087784, A208895, A264083, A318609, A330607.
%K A071304 nonn
%O A071304 1,2
%A A071304 _R. H. Hardin_, Jun 11 2002
%E A071304 Terms a(7) onward from _Max Alekseyev_, Nov 06 2022