erdos_897.variants.parts.ii

Let $f(n)$ be an additive function (so that $f(ab)=f(a)+f(b)$ if $(a,b)=1$) such that $\limsup_{p,k} f(p^k) \log(p^k) = ∞$ and $f(p^k) = f(p)$ or $f(p^k) = kf(p)$. Is it true that $\limsup_n f(n+1)/f(n) = ∞$?

theorem erdos_897.variants.parts.ii : answer(sorry) ↔ ∀ (f : ℕ → ℝ),
    (∀ᵉ (a > 0) (b > 0), a.Coprime b → f (a * b) = f a + f b) →
    ((Filter.atTop ⊓ Filter.principal {(p, k) : ℕ × ℕ | p.Prime}).limsup
      (fun (p, k) => (f (p^k) / (p^k : ℝ).log : EReal)) = ⊤) →
    (∀ k p, p.Prime → f (p^k) = f p) ∨ (∀ (k p : ℕ), p.Prime → f (p^k) = k*f p) →
    Filter.atTop.limsup (fun (n : ℕ) => (f (n+1) / f n : EReal)) = ⊤

Recent Submissions

No submissions yet. Be the first to attempt this problem.