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Problem 893

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claims/: The 2 claim pages of Problem 893, one per claimant's result; the problem's standing derives from them.


Statement. If τ(n)\tau(n) counts the divisors of nn then let

f(n)=∑1≤k≤nτ(2k−1).f(n)=\sum_{1\leq k\leq n}\tau(2^k-1).

Does f(2n)/f(n)f(2n)/f(n) tend to a limit?

Formulation. The question whether f(2n)/f(n)f(2n)/f(n) tends to a limit is read, as the site and Kovač and Luca read it, as allowing the limit +∞+\infty. After Kovač and Luca proved the ratios unbounded, the site kept the problem OPEN, and they describe what remains as deciding whether f(2n)/f(n)→∞f(2n)/f(n)\to\infty or the ratio has no limit. The formal-conjectures statement, as last changed on 18 September 2026, likewise asks whether the ratio tends to infinity. Read as asking for a finite limit, the question is answered no by their Theorem 1.

Status. The site labels the problem OPEN. Every finite limit is ruled out by Kovač and Luca's unboundedness theorem; divergence to infinity is proved only under unproven conjectures, on their conditional page.

Source. erdosproblems.com/893, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #893, https://www.erdosproblems.com/893.

References.

  • [Er98] Erdős, Paul, Some of my new and almost new problems and results in combinatorial number theory. Number theory (Eger, 1996) (1998), 169-180.
  • [KoLu25] V. Kovač and F. Luca, On the number of divisors of Mersenne numbers. arXiv:2506.04883 (2025); published in Experimental Mathematics (online 20 May 2026), doi:10.1080/10586458.2026.2636060.

Formalization. Statement in formal-conjectures.

Current assessment

The site labels the problem OPEN. Under the formulation above, the open question is whether f(2n)/f(n)→∞f(2n)/f(n)\to\infty. Kovač and Luca (card) prove the ratios unbounded, which rules out every finite limit (claim page). They also derive f(2n)/f(n)→∞f(2n)/f(n)\to\infty from either of two unproven conjectures (conditional claim page), and their computations support divergence. Erdős [Er98] expected no simple asymptotic formula for f(n)f(n), since it grows too fast.

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