First hint to puzzle 152: Totient valence

A number of steps are needed to evaluate v(2¹⁰⁰⁰)...

Note that, if k = p₁^a₁ · p₂^a₂ ·...· p_r^a_r, then (k) = (p₁^a₁ − p₁^a₁−1)·(p₂^a₂ − p₂^a₂−1)·...·(p_r^a_r − p_r^a_r−1).

Show that:

• (2k) = 2(k), if k is even;
• (2k) = (k), if k is odd.