Kovac 2025 number divisors mersenne numbers

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proposition_2: Proves that the summatory function of 2 to the power tau(k) has doubling ratios tending to infinity, the engine of the unboundedness theorem.

theorem_1: Proves that the ratios f(2n)/f(n) of the summed divisor counts of the Mersenne numbers have limit superior infinity, so no finite limit exists.

theorem_3: Proves that f(2n)/f(n) tends to infinity assuming either a conjecture on highly composite Mersenne indices or a logarithmic bound on the prime factors of cyclotomic values at 2.

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Vjekoslav Kovač, Florian Luca, On the number of divisors of Mersenne numbers. arXiv preprint (2025). arXiv:2506.04883. The arXiv record names arXiv's non-exclusive distribution license (arXiv:2506.04883), every other right reserved. The copy read for this card is arXiv v4 (3 February 2026).

Kovač and Luca study f(n) = sum_{k<=n} tau(2^k - 1), the divisor counts of the Mersenne numbers 2^k - 1 with k <= n added up, and take up Erdős's question of whether f(2n)/f(n) tends to a limit. Theorem 1 proves limsup f(2n)/f(n) = infinity, so the ratio is unbounded and any limit cannot be a real number; a naive heuristic based on tau(m) behaving like log m had suggested the limit 4. The proof goes through Proposition 2, which shows the modified summatory function f'(n) = sum of 2^{tau(k)} satisfies f'(2n)/f'(n) -> infinity, using properties of highly-composite numbers studied by Ramanujan, Erdős and Nicolas; Theorem 1 follows from Proposition 2 and the bound f(n) >= f'(n)/4, inequality (6). Theorem 3 gives the conditional divergence f(2n)/f(n) -> infinity assuming either Conjecture 1 (tau(2^N + 1)/N -> infinity along indices N of highly-composite Mersenne numbers) or Conjecture 2 (omega(Phi_d(2)) <= 10 log d for all d >= 2 with at most finitely many exceptions), and Section 4 supplies extensive computation using the OEIS tables of Eldar and Alekseyev and Gillies' approximation. For #893 the paper answers unconditionally that no finite limit exists, and shows divergence to infinity only conditionally; the problem page lists it among its references.

Source: https://arxiv.org/abs/2506.04883.

Bears on. #893: Theorem 1 rules out every finite limit of f(2n)/f(n); Theorem 3 gives f(2n)/f(n) -> infinity only under Conjecture 1 or Conjecture 2, neither proved.

Results to transcribe.

• Theorem 1: limsup_{n->infinity} f(2n)/f(n) = infinity, where f(n) = sum_{k<=n} tau(2^k - 1); the doubling ratios are unbounded.
• Proposition 2: for f'(n) = sum_{k<=n} 2^{tau(k)}, the doubling ratios satisfy f'(2n)/f'(n) -> infinity, proved via highly-composite numbers.
• Theorem 3: assuming either Conjecture 1 (highly-composite Mersenne numbers) or Conjecture 2 (prime divisors of Phi_d(2)), f(2n)/f(n) -> infinity.
• Section 4 (experiments): Tables of tau(2^n - 1) and omega(Phi_n(2)) for n <= 1206 and of tau(2^n + 1) for n <= 1128, then partial factorization with Gillies' approximation extending the doubling-ratio plots from n <= 603 to n <= 1000, support both the divergence claim and Conjectures 1 and 2.

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