English

Radical bound for Zaremba's conjecture

Number Theory 2023-10-20 v2

Abstract

Famous Zaremba's conjecture (1971) states that for each positive integer q2q\geq2, there exists positive integer 1a<q1\leq a <q, coprime to qq, such that if you expand a fraction a/qa/q into a continued fraction a/q=[a1,,an]a/q=[a_1,\ldots,a_n], all of the coefficients aia_i's are bounded by some absolute constant k\mathfrak k, independent of qq. Zaremba conjectured that this should hold for k=5\mathfrak k=5. In 1986, Niederreiter proved Zaremba's conjecture for numbers of the form q=2n,3nq=2^n,3^n with k=3\mathfrak k=3 and for q=5nq=5^n with k=4\mathfrak k=4. In this paper we prove that for each number q2n,3nq\neq 2^n,3^n, there exists aa, coprime to qq, such that all of the partial quotients in the continued fraction of a/qa/q are bounded by rad(q)1 \operatorname{rad}(q)-1, where rad(q)\operatorname{rad}(q) is the radical of an integer number, i.e. the product of all distinct prime numbers dividing qq. In particular, this means that Zaremba's conjecture holds for numbers qq of the form q=2n3m,n,mN{0}q=2^n3^m, n,m\in\mathbb N \cup \{0\} with k=5\mathfrak k= 5, generalizing Neiderreiter's result. Our result also improves upon the recent result by Moshchevitin, Murphy and Shkredov on numbers of the form q=pnq=p^n, where pp is an arbitrary prime and nn sufficiently large.

Keywords

Cite

@article{arxiv.2310.09801,
  title  = {Radical bound for Zaremba's conjecture},
  author = {Nikita Shulga},
  journal= {arXiv preprint arXiv:2310.09801},
  year   = {2023}
}

Comments

8 pages, comments appreciated

R2 v1 2026-06-28T12:50:58.998Z