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Large Subsets of \(\mathbb{Z}_m^n\) without Arithmetic Progressions
For integers \(m\) and \(n\), we study the problem of finding good lower bounds for the size of progression-free sets in \((\mathbb{Z}_{m}^{n},+)\). Let \(r_{k}(\mathbb{Z}_{m}^{n})\) denote the maximal size of a subset of \(\mathbb{Z}_{m}^{n}\) without arithmetic progressions of length \(k\) and let...
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Published in: | arXiv.org 2022-11 |
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Main Authors: | , , |
Format: | Article |
Language: | English |
Subjects: | |
Online Access: | Get full text |
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Summary: | For integers \(m\) and \(n\), we study the problem of finding good lower bounds for the size of progression-free sets in \((\mathbb{Z}_{m}^{n},+)\). Let \(r_{k}(\mathbb{Z}_{m}^{n})\) denote the maximal size of a subset of \(\mathbb{Z}_{m}^{n}\) without arithmetic progressions of length \(k\) and let \(P^{-}(m)\) denote the least prime factor of \(m\). We construct explicit progression-free sets and obtain the following improved lower bounds for \(r_{k}(\mathbb{Z}_{m}^{n})\): If \(k\geq 5\) is odd and \(P^{-}(m)\geq (k+2)/2\), then \[r_k(\mathbb{Z}_m^n) \gg_{m,k} \frac{\bigl\lfloor \frac{k-1}{k+1}m +1\bigr\rfloor^{n}}{n^{\lfloor \frac{k-1}{k+1}m \rfloor/2}}. \] If \(k\geq 4\) is even, \(P^{-}(m) \geq k\) and \(m \equiv -1 \bmod k\), then \[r_{k}(\mathbb{Z}_{m}^{n}) \gg_{m,k} \frac{\bigl\lfloor \frac{k-2}{k}m + 2\bigr\rfloor^{n}}{n^{\lfloor \frac{k-2}{k}m + 1\rfloor/2}}.\] Moreover, we give some further improved lower bounds on \(r_k(\mathbb{Z}_p^n)\) for primes \(p \leq 31\) and progression lengths \(4 \leq k \leq 8\). |
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ISSN: | 2331-8422 |
DOI: | 10.48550/arxiv.2211.02588 |