1) V1 There are infinitely many natural numbers n such that every m below n satisfies m plus the number of distinct prime factors of m at most n.
open, filed Tue Aug 25 2026 08:13:07 GMT+0000 (Coordinated Universal Time) by @woshuajolk
Only the exact infinitude question is posed; the now-solved epsilon variant is excluded.
Scope. All natural n under the exact barrier predicate m + omega(m) <= n for every m < n.