FormalConjectures.ErdosProblems.«1093»

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module

public import FormalConjecturesUtil

Erdős Problem 1093

Reference: erdosproblems.com/1093

@[expose] public sectionnamespace Erdos1093open Finset Nat

If defined, the deficiency is the count of $0 \le i < k$ such that $n - i$ is $k$-smooth, that is, divisible only by primes $\le k$.

Nat.smoothNumbers m is the set of positive naturals all of whose prime factors are $< m$, so "$k$-smooth" in the sense above is Nat.smoothNumbers (k + 1).

noncomputable def deficiency (n k : ℕ) : ℕ :=
  #{i ∈ range k | n - i ∈ smoothNumbers (k + 1)}

Are there infinitely many binomial coefficients with deficiency 1?

@[category research open, AMS 5]
theorem erdos_1093.parts.i :
    answer(sorry) ↔ {x : ℕ × ℕ | let k := x.1; let n := x.2; 2 * k ≤ n ∧ deficiency n k = 1 ∧
      ∀ p, p.Prime → (p ∣ choose n k) → k < p}.Infinite := by⊢ True ↔
  {x |
      let k := x.1;
      let n := x.2;
      2 * k ≤ n ∧ deficiency n k = 1 ∧ ∀ (p : ℕ), Nat.Prime p → p ∣ n.choose k → k < p}.Infinite
  sorryAll goals completed! 🐙

Are there only finitely many binomial coefficients with deficiency > 1?

@[category research open, AMS 5]
theorem erdos_1093.parts.ii :
    {x : ℕ × ℕ | let k := x.1; let n := x.2; 2 * k ≤ n ∧ deficiency n k > 1 ∧
      ∀ p, p.Prime → (p ∣ choose n k) → k < p}.Finite := by⊢ {x |
    let k := x.1;
    let n := x.2;
    2 * k ≤ n ∧ deficiency n k > 1 ∧ ∀ (p : ℕ), Nat.Prime p → p ∣ n.choose k → k < p}.Finite
  sorryAll goals completed! 🐙

end Erdos1093