---

Source. Theorem 4, Section 1, p. 2 of Peter Shiu, The denominators of harmonic numbers (Revised), arXiv:1607.02863v2 (30 July 2024; the paper is dated 29 July 2024); proof in Section 5, pp. 5--6, through Lemma 1 (p. 5). Preprint, not published in a journal (arXiv listing checked). Notation: $H_n=c_n/d_n$ in lowest terms and $D_n=\mathrm{lcm}(1,\ldots,n)=d_nq_n$.

Statement

Theorem 4 (p. 2). Let $2<p_1<p_2<\cdots<p_k$ be primes, and suppose the numbers

$$\theta_i=\frac{\log p_1}{\log p_i},\qquad i=1,2,\ldots,k,$$

are linearly independent. Then there is an $n$ with $p_1p_2\cdots p_k\mid q_n$, that is, $p_1p_2\cdots p_k\,d_n\mid D_n$ (the form in the abstract).

As printed, the hypothesis does not name the field of scalars, and the list includes $\theta_1=1$. The paper remarks (p. 2) that the hypothesis is probably unnecessary and is a consequence of Schanuel's conjecture, citing Lang's Introduction to Transcendental Numbers, pp. 30--31.

Proof pointer and sketch (Section 5)

Take $m_i=p_i-1\in E_{p_i}$. By Theorem 2 every $n$ in $[m_ip_i^{a_i},(m_i+1)p_i^{a_i})$ has $p_i\mid q_n$, so it suffices to choose exponents $a_1>a_2>\cdots>a_k$ that make these $k$ intervals nested. Lemma 1 (p. 5) supplies, for $0<\delta<1$, exponents with $(1-\delta)p_1^{a_1}<p_{i+1}^{a_{i+1}}<p_i^{a_i}\le p_1^{a_1}$ for $1\le i\le k-1$, from Kronecker's theorem on simultaneous approximation (Hardy and Wright, Theorem 443) applied to the $\theta_i$; a suitable choice of $\delta$ in terms of the $p_i$ then nests the intervals. The proof is about a page, read for structure here and not verified.

Dependencies and read depth

Theorem 2, Lemma 1, and Kronecker's theorem. Read depth: claims checked; the proof read for structure only.

Relation to Problem 291

In the notation of Problem 291, $q_n=(a_n,L_n)$. Under its hypothesis the theorem gives $n$ with $(a_n,L_n)$ divisible by a prescribed product of distinct odd primes. The second question ($(a_n,L_n)>1$ infinitely often) is already answered unconditionally by Theorem 2, so this adds no new case of it, and the theorem says nothing about the first question. Wu and Yan's conditional theorem, recorded on the problem page, also applies Kronecker's theorem under a linear-independence hypothesis.

Bears on. #291 (a conditional strengthening on the side of the second question; nothing on the first).