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Erdős problem 884

For a natural number n, let 1=d1<<dτ(n)=n1 = d_1 < \dotsc < d_{\tau(n)} = n denote the divisors of n in increasing order. Does it hold that 1i<jτ(n)1djdi1+1i<τ(n)1di+1di\sum_{1 \le i < j \le \tau(n)} \frac{1}{d_j - d_i} \ll 1 + \sum_{1 \le i < \tau(n)} \frac{1}{d_{i + 1} - d_i} for nn \to \infty`, i.e. \sum_{1 \le i < j \le \tau(n)} \frac{1}{d_j - d_i} \in O \left( 1 + \sum_{1 \le i < \tau(n)} \frac{1}{d_{i + 1} - d_i}) \right)?

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FormalConjectures/ErdosProblems/

884.lean

Retained formal statement1 of 2

For a natural number n, let 1=d1<<dτ(n)=n1 = d_1 < \dotsc < d_{\tau(n)} = n denote the divisors of n in increasing order. Does it hold that 1i<jτ(n)1djdi1+1i<τ(n)1di+1di\sum_{1 \le i < j \le \tau(n)} \frac{1}{d_j - d_i} \ll 1 + \sum_{1 \le i < \tau(n)} \frac{1}{d_{i + 1} - d_i} for nn \to \infty`, i.e. \sum_{1 \le i < j \le \tau(n)} \frac{1}{d_j - d_i} \in O \left( 1 + \sum_{1 \le i < \tau(n)} \frac{1}{d_{i + 1} - d_i}) \right)?

This conjecture has been disproved: - In September 2025, Terence Tao gave a conditional _negative_ answer assuming the prime tuples conjecture, see erdos_884_false_of_hardy_littlewood for this implication. - Daniel Larsen subsequently gave an [unconditional disproof](https://github.com/Larsen-Daniel/Erdos-884/blob/main/884.pdf).

*Reference:* [erdosproblems.com/884](https://www.erdosproblems.com/884)

FormalConjectures/ErdosProblems/884.leanErdos884.erdos_8841 lineExact file
FalseErdos884.Erdos884Prop
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.

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