
FORMULA

From Petros Hadjicostas, Dec 20 2019: (Start)
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.
We have b(n) = A330607(n, k=1) for n >= 0.
We conjecture that a(n+1) = a(n)*b(n+1) for n >= 1. (End)


EXAMPLE

From Petros Hadjicostas, Dec 17 2019: (Start)
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:
(a) those with 1 = det(M) mod 5:
[[1,0],[0,1]]; [[0,4],[1,0]]; [[0,1],[4,0]]; [[4,0],[0,4]].
These form the abelian group SO(2, Z_5). See the comments for sequence A060968.
(b) those with 4 = det(M) mod 5:
[[0,1],[1,0]]; [[0,4],[4,0]]; [[1,0],[0,4]]; [[4,0],[0,1]].
Note that, for n = 3, we have 2*a(3) = 2*120 = 240 = A264083(5). (End)
