OPEN
This is open, and cannot be resolved with a finite computation.
If $\tau(n)$ counts the divisors of $n$ then let\[f(n)=\sum_{1\leq k\leq n}\tau(2^k-1).\]Does $f(2n)/f(n)$ tend to a limit?
Erdős
[Er98] says that 'probably there is no simple asymptotic formula for $f(n)$ since $f(n)$ increases too fast'.
Kovač and Luca
[KoLu25] (building on a heuristic independently found by Cambie (personal communication)) have shown that there is no finite limit, in that\[\limsup_{n\to \infty}\frac{f(2n)}{f(n)}=\infty,\]and provide both theoretical and numerical evidence that suggests $\lim \frac{f(2n)}{f(n)}=\infty$.
View the LaTeX source
Additional thanks to: Stijn Cambie
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #893, https://www.erdosproblems.com/893, accessed 2025-11-16
