AI 中文总结
本文改进了关于无等差数列子集计数的问题，证明了对于k≥5，无限多个n值下计数结果为2^{r_k(n)(1+o(1))}，并为k≥3的所有n提供了2^{O(r_k(n))}的计数结果。
AI 中文摘要
Cameron和Erdős问的是，不含长度为k的等差数列的集合数量是否为2^{r_k(n)(1+o(1))}，其中r_k(n)是{1,…,n}中最大大小的k-AP-free子集。Balogh、Liu和Sharifzadeh在这一问题上取得了重要进展，证明对于无限序列的n，其数量为2^{O(r_k(n))}。我们以两种方式改进了他们的结果。一方面，我们证明对于k≥5，[n]中k-AP-free集合的数量为2^{r_k(n)(1+o(1))}，对于无限序列的n，解决了Cameron和Erdős的问题，对于无限多个值。另一方面，我们还证明对于k≥3和所有n，[n]中k-AP-free集合的数量为2^{O(r_k(n))}。这些结果实际上是我们在计数排除特定算术模式的集合族的通用框架中的特殊案例，只要相应的极值阈值满足某种Behrend型下界。作为进一步的例子，我们得到了解决几乎所有线性方程组解集的类似结果，以及多维Szemerédi定理的计数版本。
英文摘要
Cameron and Erdős asked if the number of sets free of arithmetic progressions of length $k$ is $2^{r_k(n)(1+o(1))}$, where $r_k(n)$ is the maximum cardinality of a $k$-AP-free subset of $\{1, \dots, n\}$. Balogh, Liu and Sharifzadeh made significant progress on this question showing that it is $2^{O(r_k(n))}$ for an infinite sequence of $n$. We improve their result in two ways. On the one hand, we prove that, for $k\geq 5$, the number of $k$-AP-free sets in $[n]$ is $2^{r_k(n)(1+o(1))}$ for an infinite sequence of $n$, solving the question of Cameron and Erdős for infinitely many values. On the other hand, we also prove that for $k \geq 3$ and all $n$ the number of $k$-AP-free sets in $[n]$ is $2^{O(r_k(n))}$. These results are in fact special cases of a general framework that we develop to count families of sets excluding certain arithmetic patterns, which applies as long as the corresponding extremal threshold satisfies certain Behrend-type lower bounds. As further examples, we get analogous results for solution sets to almost all systems of linear equations as well as counting versions of the multidimensional Szemerédi theorem.
Comments30 pages
