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Claim. There is an absolute constant $C>0$ such that infinitely many positive integers $n$ satisfy $\omega(n-k)\le\Omega(n-k)\le C\log k$ for every integer $1<k<n$. This is Theorem 1.3 of Cheuk Fung (Joshua) Lau, On the number of prime factors of consecutive integers, arXiv:2604.15042 (v1 2026-04-16, v2 2026-06-24), stated for $\Omega$, the number of prime factors counted with multiplicity, and hence for $\omega\le\Omega$; it is the descending companion of the paper's Theorem 1.1 and is proved by the same argument, a quantitative refinement of Tao and Teräväinen's probabilistic sieve. The source card is Lau 2026.

Covers. The second question of Problem 413: there is an $\epsilon>0$ such that infinitely many $n$ satisfy $m+\epsilon\,\omega(m)\le n$ for all $m<n$. It follows from Theorem 1.3 with $\epsilon=1/(C\log2)$. For an $n$ given by the theorem put $N=n-1$; every $m<N$ is $m=n-k$ with $k=N-m+1\ge2$, so

$$\epsilon\,\omega(m)\le\frac{C\log(N-m+1)}{C\log 2}=\log_2(N-m+1)\le N-m,$$

since $\log_2(j+1)\le j$ for every integer $j\ge1$. Hence $m+\epsilon\,\omega(m)\le N$ for all $m<N$, and infinitely many $n$ give infinitely many such $N$. The first question, whether $\omega$ has infinitely many barriers ($m+\omega(m)\le n$ for all $m<n$), is not covered: the paper's Corollary 1.4 gives infinitely many $n$ with $\omega(n-k)\le k$ for all sufficiently large $k<n$, which the site's commentary restates as $m+\omega(m)\le n$ for all $1\le m\le n-C$, and that weaker form settles no part of the problem.

Depends on. Nothing in this wiki; the argument is self-contained in the preprint.

Acceptance. None. The preprint has no journal record known here, and the site labels the problem OPEN while its commentary credits Lau [La26] with a positive answer to the second question and a weaker version of the first (page last edited 2026-04-17); commentary on a problem the site labels open is not acceptance. No Lean development of the result is known.