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On a Conjecture of Cusick on a sum of Cantor sets

Number Theory 2025-06-09 v2

Abstract

In 1971 Cusick proved that every real number x[0,1]x\in[0,1] can be expressed as a sum of two continued fractions with no partial quotients equal to 11. In other words, if we define a set S(k):={x[0,1]:an(x)k for all nN}, S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{ for all } n\in\mathbb{N} \}, then S(2)+S(2)=[0,1]. S(2)+S(2) = [0,1]. He also conjectured that this result is unique in the sense that if you exclude partial quotients from 11 to k1k-1 with k3k\geq3, then the Lebesgue measure λ\lambda of the set of numbers which can be expressed as a sum of two continued fractions with no partial quotients from {1,,k1}\{1,\ldots,k-1\} is equal to 00, that is λ(S(k)+S(k))=0 for k3.\lambda\Bigl( S(k)+S(k) \Bigl)= 0 \text{ for }k\geq 3. In this paper, we disprove the conjecture of Cusick by showing that S(k)+S(k)[0,1k1]. S(k)+S(k) \supseteq \left[0,\frac{1}{k-1}\right]. The proof is constructive and does not rely on ideas from previous works on the topic. We also show the existence of countably many 'gaps' in S(k)+S(k)S(k)+S(k), that is intervals, for which the endpoints lie in S(k)+S(k)S(k)+S(k), while none of the elements in the interior do so. Finally, we prove several results on the sums S(m)+S(n) S(m)+S(n) for mnm\neq n.

Keywords

Cite

@article{arxiv.2411.17379,
  title  = {On a Conjecture of Cusick on a sum of Cantor sets},
  author = {Nikita Shulga},
  journal= {arXiv preprint arXiv:2411.17379},
  year   = {2025}
}

Comments

22 pages, 1 figure, comments are appreciated