子树计数与最大平均子树阶
Subtree Counts and the Maximum Mean Subtree Order
- Xiamen University（厦门大学）
- Qinghai Minzu University（青海民族大学）
- Xinjiang Normal University（新疆师范大学）
机构由 AI 辅助整理，请以论文原文为准。
AI总结：
本文证明固定阶图中子树计数与完全图之比随子树大小单调递减，从而完全图最大化平均子树阶，并解决相关猜想。
AI中文摘要：
对于正阶有限简单图$G$，设$s_k(G)$为其$k$顶点树子图的数量。我们证明，在固定阶$n$的图中，比值$s_k(G)/s_k(K_n)$随$k$形成非递增序列。由此得出，完全图使平均子树阶最大化：$\mu(G)\leq\mu(K_n)$，当且仅当$G$为完全图时取等号。这一单调性定理还证实了近期关于极值平均子树阶工作中提出的$s_{n-1}$与$s_n$之比的猜想。证明将已知的连续阶有根Cayley树形状的耦合提升到$K_n$的均匀标记树子图，这些子图在每个样本点处嵌套。事件包含关系给出计数不等式，双重求和恒等式给出均值界。
英文摘要：
For a finite simple graph $G$ of positive order, let $s_k(G)$ be the number of its $k$-vertex tree subgraphs. We prove that, among graphs of a fixed order $n$, the ratios $s_k(G)/s_k(K_n)$ form a nonincreasing sequence in $k$. It follows that the complete graph maximizes the mean subtree order: $μ(G)\leqμ(K_n)$, with equality if and only if $G$ is complete. This monotonicity theorem also establishes the $s_{n-1}$-to-$s_n$ ratio conjecture posed in recent work on extremal mean subtree order. The proof lifts a known coupling of rooted Cayley-tree shapes of consecutive orders to uniform labeled tree subgraphs of $K_n$ that are nested at every sample point. Event inclusion gives the count inequalities, and a double-sum identity gives the mean bound.
