FormalConjectures/ErdosProblems/897.lean

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Explanation legend Lean4 Domain Explanation Language (L4DEL)

L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL
Lean4 Domain Explanation Language

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

5 declarations, 5 contain sorry, 0 declared as axioms

Erdős Problem 897

References:

Imports (1)
  1. theorem Erdos897.erdos_897.parts.i sorry line 48 source

    depends on sorry what this means

    Statement uses: Iff, False, Nat, Real, GT.gt, OfNat.ofNat, Nat.Coprime, Eq, HMul.hMul, HAdd.hAdd, EReal, Filter.limsup, Prod, Erdos897.erdos_897.parts.i.match_1, HDiv.hDiv, HPow.hPow, Real.log, Min.min, Filter, Filter.atTop and 5 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    60 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  2. theorem Erdos897.erdos_897.parts.ii sorry line 64 source

    depends on sorry what this means

    Statement uses: Iff, False, Nat, Real, GT.gt, OfNat.ofNat, Nat.Coprime, Eq, HMul.hMul, HAdd.hAdd, EReal, Filter.limsup, Prod, Erdos897.erdos_897.parts.i.match_1, HDiv.hDiv, HPow.hPow, Real.log, Min.min, Filter, Filter.atTop and 4 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    52 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  3. theorem Erdos897.erdos_897.variants.log_growth sorry line 77 source

    depends on sorry what this means

    Statement uses: Nat, Real, GT.gt, OfNat.ofNat, Nat.Coprime, Eq, HMul.hMul, HAdd.hAdd, LE.le, abs, HSub.hSub, Exists, And, Asymptotics.IsBigO, Filter.atTop, Real.log

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    40 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  4. theorem Erdos897.erdos_897.variants.parts.i 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 $f(p^k) = kf(p)$ sorry line 96 source

    Source documentation

    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 $\limsup_n (f(n+1)−f(n))/ \log n = ∞$?

    The known counterexample does not satisfy either of these extra hypotheses, so this variant remains open.

    depends on sorry what this means

    Statement uses: Iff, True, Nat, Real, GT.gt, OfNat.ofNat, Nat.Coprime, Eq, HMul.hMul, HAdd.hAdd, EReal, Filter.limsup, Prod, Erdos897.erdos_897.parts.i.match_1, HDiv.hDiv, HPow.hPow, Real.log, Min.min, Filter, Filter.atTop and 6 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    62 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_897.variants.parts.i : 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) / (n : ℝ).log : EReal)) = ⊤ := by
    …
  5. theorem Erdos897.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 $f(p^k) = kf(p)$ sorry line 114 source

    Source documentation

    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 $\limsup_n f(n+1)/f(n) = ∞$?

    The known counterexample does not satisfy either of these extra hypotheses, so this variant remains open.

    depends on sorry what this means

    Statement uses: Iff, True, Nat, Real, GT.gt, OfNat.ofNat, Nat.Coprime, Eq, HMul.hMul, HAdd.hAdd, EReal, Filter.limsup, Prod, Erdos897.erdos_897.parts.i.match_1, HDiv.hDiv, HPow.hPow, Real.log, Min.min, Filter, Filter.atTop and 5 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    54 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    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)) = ⊤ := by
    …