Number of solutions to x + y + z = 0 mod (2n+1) such that x,y,z are units modulo 2n+1, i.e., gcd(x, 2n+1) = gcd(y, 2n+1) = gcd(z, 2n+1) = 1.

A061780

Number of solutions to x + y + z = 0 mod (2n+1) such that x,y,z are units modulo 2n+1, i.e., gcd(x, 2n+1) = gcd(y, 2n+1) = gcd(z, 2n+1) = 1.