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Source. Vjekoslav Kovač and Florian Luca, On the number of divisors of Mersenne numbers, arXiv:2506.04883v4 (3 February 2026), Theorem 1, stated on p. 2 and proved on p. 5.

Dependencies. Proposition 2 and the inequality $f(n)\geq\tfrac14 f'(n)$, the paper's (6) on p. 4.

Bears on. #893: the theorem rules out every finite value of $\lim f(2n)/f(n)$; it does not decide whether the ratio tends to $+\infty$ or has no limit.

Statement

Let $\tau$ count divisors and put

$$f(n)=\sum_{1\leq k\leq n}\tau(2^k-1).$$

Then

$$\limsup_{n\to\infty}\frac{f(2n)}{f(n)}=\infty,$$

that is, the sequence $(f(2n)/f(n))_{n\geq1}$ is unbounded.

Proof pointer

Each divisor $d$ of $k$ other than $1$ and $6$ gives $2^k-1$ a primitive prime factor of $2^d-1$ (Bang, Zsigmondy), so $\omega(2^k-1)\geq\tau(k)-2$ and $\tau(2^k-1)\geq 2^{\tau(k)}/4$. Summing gives $f(n)\geq f'(n)/4$ with $f'(n)=\sum_{k\leq n}2^{\tau(k)}$. If $f(2n)/f(n)\leq C$ for all $n$, then $f(2^m)\leq C^m$; but $f'(2^m)$ is, up to the factor $f'(1)=2$, the product of the ratios $f'(2^{\ell+1})/f'(2^\ell)$ for $\ell<m$, and these tend to infinity by Proposition 2, so $f(2^m)$ outgrows $C^m$ for every finite $C$.