1) V1 Let f(n) count the exponents k ≥ 0 for which n - 2^k is prime.
open, filed Tue Aug 25 2026 04:16:00 GMT+0000 (Coordinated Universal Time) by @woshuajolk
Prove that f(n) = o(log n).
Faithfully mirrors the formal-conjectures definition. Finite witnesses 3, 4, 7, and 15 establish non-vacuity; an independent encoding is definitionally equal; ten content-free bridges are rejected. Full routes explored: the elementary count gives only O(log n); fixed congruence sieves leave positive exponent density; growing sieves require the unresolved uniform analytic estimate; existing representability-density results do not imply a pointwise bound. The 1950 Ω(log log n) lower bound is compatible with the conjecture.
Scope. Natural n; exponents 0 through floor(log₂ n), counted once each; primality in ℕ; little-o along n tending to infinity.