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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 3, stated on p. 3, with Conjectures 1 and 2 on p. 3; proved under Conjecture 1 on p. 6 (Subsection 3.1) and under Conjecture 2 on pp. 7--8 (Subsection 3.3).

Dependencies. Conjecture 1 or Conjecture 2 below, both unproven; the bound $\tau(2^k-1)\geq2^{\tau(k)-2}$ (the paper's (5), p. 4); the size of $\tau$ at highly composite numbers (the paper's (8), p. 5); Bang's primitive-divisor theorem.

Bears on. #893: the theorem gives $f(2n)/f(n)\to\infty$ only conditionally, under either of two conjectures the paper does not prove.

Definitions and conjectures

Let $f(n)=\sum_{1\leq k\leq n}\tau(2^k-1)$. Call $N$ an index of a highly composite Mersenne number when $\tau(2^N-1)>\tau(2^m-1)$ for all $1\leq m<N$; this does not require $2^N-1$ itself to be highly composite. Let $\Phi_d$ be the $d$th cyclotomic polynomial and $\omega(m)$ the number of distinct prime factors of $m$.

• Conjecture 1 (p. 3). As $N\to\infty$ through indices of highly composite Mersenne numbers, $\tau(2^N+1)/N\to\infty$.
• Conjecture 2 (p. 3). The inequality $\omega(\Phi_d(2))\leq10\log d$ holds for all positive integers $d\geq2$, with at most finitely many exceptions.

Statement

If either Conjecture 1 or Conjecture 2 holds, then

$$\lim_{n\to\infty}\frac{f(2n)}{f(n)}=\infty.$$

Proof pointer

Under Conjecture 1: take $N$ the largest index of a highly composite Mersenne number with $N\leq n$. Since $2^{2m}-1=(2^m-1)(2^m+1)$ has more divisors than $2^m-1$, $N>n/2$ and $2N\in(n,2n]$. The ratio $\sum_{n<k\leq2n}\tau(2^k-1)\big/f(n)$ is then at least $\tau(2^N+1)/(2N)$, which tends to infinity by Conjecture 1.

Under Conjecture 2 the paper derives Conjecture 1. Factoring $2^N-1=\prod_{d\mid N}\Phi_d(2)$ and applying the conjectured bound gives $\tau(2^N-1)\leq2^{O(\tau(N)(\log N)^2)}$, while the index property, the bound $\tau(2^k-1)\geq2^{\tau(k)-2}$ and (8) give $\log_2\tau(2^N-1)\geq 2^{(1+o(1))\log N/\log\log N}$. Hence $\tau(N)\gg2^{(1+o(1))\log N/\log\log N}$, and since $\tau(N)\ll\tau(M)\log N$ for the odd part $M$ of $N$, also $\tau(M)\geq2^{(1+o(1))\log N/\log\log N}$. Each divisor $d$ of $N$ with $N/d$ odd gives $2^d+1\mid2^N+1$, so Bang's theorem yields $\tau(2^N+1)\geq2^{\tau(M)-2}$, and $\tau(2^N+1)/N\to\infty$.

Evidence reported

Section 4 (pp. 8--12) reports computations that the authors read as supporting the divergence and both conjectures: the 30 indices $N\leq1206$ of highly composite Mersenne numbers with $\tau(2^N+1)/N$ (Table 1, p. 10), and $\omega(\Phi_d(2))\leq1.51\log d$ for every $1\leq d\leq1206$ (p. 11). These are evidence, not proof.