The board / number theory / #964
Erdős Problem #964
PROVED LEANdatabase updated 2026-02-14
The statement as posed is true, and a proof exists. A solution has also been checked by machine, in Lean.
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\(\tau(n+1)/\tau(n)\) as a real number, where \(\tau = \sigma_0\) counts divisors. For \(n = 0\) this is the junk value \(0\), so the statements below use \(n \geq 1\).
Let \(\tau(n)\) count the number of divisors of \(n\). Is the sequence \[\frac{\tau(n+1)}{\tau(n)}\] everywhere dense in \((0,\infty)\)? The answer is yes, proved by Eberhard [Eb25].
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 1 further variant / known partial result in the Lean file. The authoritative wording is on erdosproblems.com/964.Prizenone
Topicsnumber theory, divisors
Statement in LeanFormal Conjectures
Solution in Leanyes
Original pageerdosproblems.com/964
Attempts
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