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Erdős problem 26

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Let A⊂NA\subset\mathbb{N} be infinite. Must there exist some k≥1k\geq 1 such that almost all integers have a divisor of the form a+ka+k for some a∈Aa\in A? The question as posed follows negatively from Davenport–Erdős (1951). The AI result settles Tenenbaum's harder variant, also negatively: there is an infinite AA such that for every k≥1k\geq 1 the set of multiples of A+kA+k has upper density below 0.340.34.

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11 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

26.lean

Retained formal statement1 of 10

Let A⊂NA\subset\mathbb{N} be infinite such that ∑a∈A1a=∞\sum_{a \in A} \frac{1}{a} = \infty. Must there exist some k≥1k\geq 1 such that almost all integers have a divisor of the form a+ka+k for some a∈Aa\in A?

This was formalized in Lean by Alexeev using Aristotle.

FormalConjectures/ErdosProblems/26.leanErdos26.erdos_261 lineExact file
False ↔ ∀ (A : ℕ → ℕ), StrictMono A → Erdos26.IsThick A → ∃ k, Erdos26.IsBehrend fun x => A x + k
SolvedProof has a holelean4external proof

The proof uses `sorry`: part of the argument is written but not proved. Lean accepts the file; it does not accept the theorem.