FormalConjectures/ErdosProblems/698.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

4 declarations, 4 contain sorry, 0 declared as axioms

Erdős Problem 698

References:

Imports (1)
  1. theorem Erdos698.erdos_698 sorry line 51 source

    depends on sorry what this means

    Statement uses: Iff, True, Exists, Nat, And, Filter.Tendsto, Filter.atTop, LE.le, OfNat.ofNat, LT.lt, HDiv.hDiv, Nat.gcd, Nat.choose

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

    12 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

  2. theorem Erdos698.erdos_698.variants.erdos_szekeres Documented declaration: erdos_szekeres sorry line 64 source

    Source documentation

    A problem of Erdős and Szekeres, who observed that $$\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \geq \frac{\binom{n}{i}}{\binom{j}{i}} \geq 2^i$$ (in particular the greatest common divisor is always $>1$).

    depends on sorry what this means

    Statement uses: Nat, LE.le, OfNat.ofNat, LT.lt, HDiv.hDiv, And, Real, Nat.choose, Nat.gcd, HPow.hPow

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

    34 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_698.variants.erdos_szekeres (n i j : ℕ) (hi : 1 ≤ i) (hij : i < j)
        (hj : j ≤ n / 2) :
        (n.choose i : ℝ) / (j.choose i : ℝ) ≤ (Nat.gcd (n.choose i) (n.choose j) : ℝ) ∧
          (2 : ℝ) ^ i ≤ (n.choose i : ℝ) / (j.choose i : ℝ) := by
    …
  3. theorem Erdos698.erdos_698.variants.erdos_szekeres_sharp This inequality is sharp for $i=1$, $j=p$, and $n=2p$ sorry line 74 source

    Source documentation

    This inequality is sharp for $i=1$, $j=p$, and $n=2p$.

    depends on sorry what this means

    Statement uses: Nat, Nat.Prime, LT.lt, OfNat.ofNat, And, Eq, Real, Nat.gcd, Nat.choose, HMul.hMul, HDiv.hDiv, HPow.hPow

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

    32 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_698.variants.erdos_szekeres_sharp (p : ℕ) (hp : p.Prime) (hp2 : 2 < p) :
        (Nat.gcd ((2 * p).choose 1) ((2 * p).choose p) : ℝ) =
            ((2 * p).choose 1 : ℝ) / (p.choose 1 : ℝ) ∧
          ((2 * p).choose 1 : ℝ) / (p.choose 1 : ℝ) = (2 : ℝ) ^ 1 := by
    …
  4. theorem Erdos698.erdos_698.variants.bergman Documented declaration: bergman sorry line 86 source

    Source documentation

    This was resolved by Bergman [Be11], who proved that for any $2\leq i<j\leq n/2$ $$\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \gg n^{1/2}\frac{2^i}{i^{3/2}},$$ where the implied constant is absolute.

    depends on sorry what this means

    Statement uses: Exists, Real, And, LT.lt, OfNat.ofNat, Nat, LE.le, HDiv.hDiv, HMul.hMul, Real.sqrt, HPow.hPow, Nat.gcd, Nat.choose

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

    44 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_698.variants.bergman :
        ∃ c : ℝ, 0 < c ∧ ∀ n i j : ℕ, 2 ≤ i → i < j → j ≤ n / 2 →
          c * (Real.sqrt (n : ℝ) * (2 : ℝ) ^ i / ((i : ℝ) * Real.sqrt (i : ℝ))) ≤
            (Nat.gcd (n.choose i) (n.choose j) : ℝ) := by
    …

    Mentions: c