Euler Theorem. (3)

     If p is a prime number, then j(p^k) = p^k - p^k-1, in particular, j(p) = p - 1.

     Proof. Among numbers: 1, 2, 3, ... , p^k only numbers p, 2p, 3p, ... , p^k-1p have common divisor with p^k, p^k-1 numbers in all. The other p^k - p^k-1 numbers are co-prime with p, what was to be proved.

     Using these theorems it's easy to calculate j(n) for any n.