Problem Statement
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?
Categories:
Number Theory Divisors
Progress
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 $\lim \frac{f(2n)}{f(n)}=\infty$.
Source: erdosproblems.com/893 | Last verified: January 19, 2026