Erdős Problem #884
The statement as posed is false. A solution has also been checked by machine, in Lean.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
The sum \(\sum_{1 \le i < j \le \tau(n)} \frac{1}{d_j - d_i}\) over all pairs of divisors \(d_i < d_j\) of \(n\).
The sum \(\sum_{1 \le i < \tau(n)} \frac{1}{d_{i + 1} - d_i}\) over consecutive divisors of \(n\).
For a natural number n, let \(1 = d_1 < \dotsc < d_{\tau(n)} = n\) denote the divisors of \(n\) in increasing order. Does it hold that \(\sum_{1 \le i < j \le \tau(n)} \frac{1}{d_j - d_i} \ll 1 + \sum_{1 \le i < \tau(n)} \frac{1}{d_{i + 1} - d_i}\) for \(n \to \infty\), 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_
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