发表机构
School of Mathematics, Shandong University; Institute for Basic Science (IBS)（山东大学数学学院; 基础科学研究院）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
该研究将Erdős–Gallai界扩展到不含t个连续偶环长度的图，解决了Verstraëte的猜想，还确定了强制特定模长环的边阈值，证明采用了增强稳定性的次线性扩张器方法。
AI 中文摘要
Erdős和Gallai在1959年证明了一个具有里程碑意义的结果：每个不含长度至少为2t+2的环的n顶点图，其边数最多为(2t+1)/2 · (n-1)。我们证明了该结论的一个扩展：对于每个足够大的t，相同的数值也是不含t个连续偶环长度的图的严格极值界，从而解决了Verstraëte的一个猜想。因此，在Erdős–Gallai阈值下，强制整个偶环长度区间的代价不超过强制该区间中最长偶环的代价。更确切地说，每个边数e(G) ≥ (2t+1)(n-1)/2的n顶点图G，要么包含t个连续偶环长度，要么达到等式且G是连通的，其每个块都同构于K_{2t+1}。作为推论，对于每个足够大的偶数k，我们确定了强制出现长度为0 mod k或2 mod k的环的严格边阈值，分别回答了Bai、Grzesik、Li和Prorok以及Gao、Li、Ma和Xie提出的问题。该证明发展了一种增强稳定性的次线性扩张器方法，其主要新要素是一种稠密情形分解，通过将灵活的稠密核心与核心外顶点的有根环族相结合，来恢复扩张器提取过程中损失的长度。
英文摘要
Erdős and Gallai in 1959 proved the seminal result that every $n$-vertex graph with no cycle of length at least $2t+2$ has at most $\tfrac{2t+1}{2}(n-1)$ edges. We prove the extension that, for every sufficiently large $t$, the same quantity is also the sharp extremal bound for graphs with no $t$ consecutive even cycle lengths, resolving a conjecture of Verstraëte. Thus, at the Erdős-Gallai threshold, forcing an entire interval of even cycle lengths costs no more than forcing its longest member. More precisely, every $n$-vertex graph $G$ with $e(G)\ge \tfrac{(2t+1)(n-1)}2$ $\bullet$ either contains $t$ consecutive even cycle lengths, $\bullet$ or equality holds and $G$ is connected with every block isomorphic to $K_{2t+1}$. As consequences, for every sufficiently large even $k$ we determine the sharp edge thresholds forcing a cycle of length $0\pmod k$ or $2\pmod k$, answering questions of Bai, Grzesik, Li, and Prorok and of Gao, Li, Ma and Xie, respectively, for sufficiently large even $k$. The proof develops a stability-enhanced sublinear expander method. Its main new ingredient is a dense-case decomposition that recovers the lengths lost in the expander extraction by combining a flexible dense core with rooted cycle families in the vertices outside the core.
Comments39 pages, 5 pages appendix, 3 figures (The proof for sparse expanders is simplified)
