Define sopfr(n) = sum of prime factors of n with repetition.
Because sopfr is completely additive:
sopfr(a × b) = sopfr(a) + sopfr(b)
Now compute Δsopfr for the k-times table:
Δsopfr(k·n) = sopfr(k·n) − sopfr(k·(n−1))
= [sopfr(k) + sopfr(n)] − [sopfr(k) + sopfr(n−1)]
= sopfr(n) − sopfr(n−1)
The multiplier sopfr(k) cancels exactly. Therefore Δ is invariant — it depends only on n, not on k. ∎