The board / number theory / #673
Erdős Problem #673
PROVEDdatabase updated 2025-08-31
The statement as posed is true, and a proof exists.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
If \(1=d_1<\cdots <d_{\tau(n)}=n\) are the divisors of \(n\) (here `Nat.nth (· ∣ n) i` is \(d_{i+1}\)), then \[G(n) = \sum_{1\leq i<\tau(n)}\frac{d_i}{d_{i+1}}.\]
Let \(1=d_1<\cdots <d_{\tau(n)}=n\) be the divisors of \(n\) and \[G(n) = \sum_{1\leq i<\tau(n)}\frac{d_i}{d_{i+1}}.\] Is it true that \(G(n)\to \infty\) for almost all \(n\)? The answer is yes: Tao observed that \(\tau(n/m)/m\leq G(n)\leq \tau(n)\) for any \(m\mid n\) with \(m>1\), so that \(G(n)\) behaves very similarly to \(\tau(n)\). In [Er82e] Erdős recalls this conjecture and observes that it is indeed trivial that \(G(n)\to \infty\) for almost all \(n\).
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 2 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/673.Prizenone
Topicsnumber theory, divisors
Statement in LeanFormal Conjectures
Solution in Leanno
Original pageerdosproblems.com/673
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