AI 中文总结
本文研究d维格点立方体$[n]^d$的无和子集数量，证明固定$d\geq3$时其数量为$2^{M([n]^d)+O_d(n^{d-1})}$，验证了Elsholtz与Rackham的猜想，所用方法结合了多种数学工具且避开了容器引理等。
AI 中文摘要
d维格点立方体$[n]^d$的一个子集是无和的，若它不包含方程$x+y=z$的解。我们研究这类子集的总数。对于$d=1$，Cameron和Erdős猜想$[n]$的无和子集数量为$O(2^{n/2})$，该结论被Green与Sapozhenko独立证明；Ghosal近期的工作解决了$d=2$的情形。本文中，我们考虑所有剩余维度，证明对每个固定整数$d\geq3$，$[n]^d$的无和子集数量为$2^{M([n]^d)+O_d(n^{d-1})}$，其中$M([n]^d)$是$[n]^d$的无和子集的最大可能规模，这验证了Elsholtz和Rackham的一个猜想。我们的证明结合了Keevash与Lim在其关于$M([n]^d)$的工作中构造的对偶权重、Ghosal给出的一维计数估计、Zhao的二分交换引理以及Madiman和Tetali的强分数熵不等式，且避免使用容器引理或先推导稳定性定理。
英文摘要
A subset of the $d$-dimensional lattice cube $[n]^d$ is sum-free if it contains no solution to the equation $x+y=z$. We study the total number of such subsets. For $d=1$, Cameron and Erdős conjectured that the number of sum-free subsets of $[n]$ is $O(2^{n/2})$, and this was proved independently by Green and Sapozhenko. A recent work by Ghosal solved the case $d = 2$. In this paper, we consider all remaining dimensions and prove that for every fixed integer $d \geqslant 3$, the number of sum-free subsets of $[n]^d$ is $2^{M([n]^d) + O_d(n^{d-1})}$, where $M([n]^d)$ is the maximum possible size of a sum-free subset of $[n]^d$. This verifies a conjecture of Elsholtz and Rackham. Our proof combines the dual weights constructed by Keevash and Lim in their work for $M([n]^d)$, a one-dimensional counting estimate due to Ghosal, a bipartite swapping lemma of Zhao, and a strong fractional entropy inequality of Madiman and Tetali, and it avoids the use of the container lemma or deriving a stability theorem first.
Comments26 + 10 pages. 3 figures. Comments are welcome
