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Source. Vjekoslav Kovač and Florian Luca, On the number of divisors of Mersenne numbers, arXiv:2506.04883v4 (3 February 2026), Proposition 2, stated on p. 2 and proved on pp. 4--5.

Dependencies. The size of the divisor function at highly composite numbers, $\tau(N)=2^{(1+o(1))\log N/\log\log N}$ as $N\to\infty$ along highly composite $N$ (the paper's (8), p. 5: the upper bound is Wigert's, the lower bound follows from Ramanujan's work on highly composite numbers).

Bears on. #893: through the inequality $f(n)\geq\tfrac14 f'(n)$ it yields Theorem 1; it is a statement about $f'$, not about the Mersenne sum $f$ itself.

Statement

Put

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

Then

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

Proof pointer

It suffices that $\sum_{n<k\leq2n}2^{\tau(k)}\big/\sum_{k\leq n}2^{\tau(k)}$ tends to infinity. A positive integer $N$ is highly composite when $\tau(N)>\tau(m)$ for all $1\leq m<N$. Take $N$ the largest highly composite number not exceeding $n$; then $N>n/2$, so $2N$ lies in $(n,2n]$, and the ratio is at least $2^{\tau(2N)-\tau(N)+O(\log N)}$. Writing $N=2^{e_1}3^{e_2}\cdots p_t^{e_t}$ with non-increasing exponents, $\tau(2N)-\tau(N)=\tau(N)/(e_1+1)\gg\tau(N)/\log N$, and the size of $\tau(N)$ at highly composite $N$ makes this exponent tend to infinity.