Erdős problem 1000
Let be an infinite sequence of integers, and let count the number of such that the fraction does not have denominator for when written in lowest form; equivalently, for all .
Sources
FormalConjectures/ErdosProblems/
1000.lean
Retained formal statement
In fact he proved that if then .
∀ (n : ℕ → ℕ), StrictMono n → 0 < n 0 → Filter.liminf (fun k => ↑(Erdos1000.phiSeq n k) / ↑(n k)) Filter.atTop = 0 → Filter.limsup (fun k => ↑(Erdos1000.phiSeq n k) / ↑(n k)) Filter.atTop = 1SolvedStatement only, no proof