发表机构
Central University of Finance and Economics（中央财经大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文确定了1-平面图中无$tC_5$子图且谱半径最大的唯一极值图，通过结构分解与打包-缺陷不等式解决了Li-Wang-Zhao问题1。
AI 中文摘要
设$\mathcal P_1$表示1-平面图的类，设$tC_5$是$t$个$C_5$副本的不交并。对于每个固定的$t\ge3$和所有足够大的$n$，我们确定了$\mathcal P_1$中具有最大邻接谱半径的唯一的$n$顶点$tC_5$-自由图，回答了Li、Wang和Zhao的问题1。证明首先给出每个极值图的结构描述。在移除两个支配顶点后，剩余部分由七顶点图$B=K_1\vee2K_3$的副本以及至多一个连通的有界核心组成。这来自1-平面$K_2$-连接的二族覆盖定理和尖锐的打包-缺陷不等式\\[ 12v(Q)-7e(Q)\ge1-10\nu_5(Q). \\] 同一不等式对有界五边形打包的剩余部分产生了精确的边极值结果。然后，归一化预解式抵消了重复的$B$-分量，有限矩比较将所有打包和所有非零缺陷强制到一个核心中，并在每个模7剩余类中唯一地识别该核心。$t=3$的情况是无五边形边界情况，并通过一个精确的有限分量引理完成。
英文摘要
Let $\mathcal P_1$ denote the class of 1-planar graphs and let $tC_5$ be the disjoint union of $t$ copies of $C_5$. For every fixed $t\ge3$ and all sufficiently large $n$, we determine the unique $n$-vertex $tC_5$-free graph in $\mathcal P_1$ with maximum adjacency spectral radius, answering Problem 1 of Li, Wang and Zhao. The proof first gives a structural description of every extremizer. After two dominating vertices are removed, the remainder consists of copies of the seven-vertex graph $B=K_1\vee2K_3$ together with at most one bounded connected core. This follows from a two-family covering theorem for 1-planar $K_2$-joins and the sharp packing--defect inequality \[ 12v(Q)-7e(Q)\ge1-10ν_5(Q). \] The same inequality yields an exact edge-extremal result for remainders with bounded pentagon packing. A normalized resolvent then cancels the repeated $B$-components, and finite moment comparisons force all packing and all nonzero defect into one core and identify that core uniquely in each residue class modulo $7$. The case $t=3$ is the pentagon-free boundary case and is completed by one exact finite component lemma.
Comments20 pages, 1 figure
