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--- title: 'Two kinds of hook length formulas for complete $m$-ary trees ' --- Yidong Sun$^\dag$ and Huajun Zhang$^\ddag$ $^\dag$Department of Mathematics, Dalian Maritime University, 116026 Dalian, P.R. China\ $^\ddag$Department of Mathematics, Zhejiang Normal University, 321004 Jinhua, P.R. China [*$^\<PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>, $^\<PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>* ]{} 0.5cm Abstract {#abstract .unnumbered} -------- In this paper, we define two kinds of hook length for internal vertices of complete $m$-ary trees, and deduce their corresponding hook length formulas, which generalize the main results obtained by Du and Liu. [**Keywords**]{}: Hook length formulas, $m$-ary trees [2000 Mathematics Subject Classification]{}: 05A15, 05A19 Introduction ============ Postnikov’s hook length formula [@post] states that $$\begin{aligned} \frac{n!}{2^n}\sum_{T}\prod_{v}\Big(1+\frac{1}{h_v}\Big)=(n+1)^{n-1},\end{aligned}$$ where the sum is over all unlabeled complete binary trees $T$ with $n$ internal vertices, the product is over all internal vertices $v$ of $T$, and $h_v$ is the “hook length" of $v$ in $T$, namely, the number of internal vertices in the subtree of $T$ rooted at $v$. Postnikov derived the formula indirectly and asked for a combinatorial proof which was provided by Seo [@seo], Chen and Yang [@chenyang]. Later, Lascoux conjectured that $$\begin{aligned} \sum_{T}\prod_{v}\Big(x+\frac{1}{h_v}\Big)=\frac{1}{(n+1)!}\prod_{i=0}^{n-1}\Big((n+1+i)x+n+1-i\Big).\end{aligned}$$ This is equivalent to the more suggestive form $$\begin{aligned} \label{eqn formua1.1} \sum_{T}\prod_{v}\frac{(h_v+1)x-h_v+1}{2h_v}=\frac{1}{n+1}\binom{(n+1)x}{n},\end{aligned}$$ which was proved by Du and Liu [@duliu]. Moreover, they generalized (\[eqn formua1.1\]) from counting complete binary trees to counting complete $(m+1)$-ary trees and obtained the following formula for $(m+1)$-ary trees: $$\begin{aligned} \label{eqn formua1.2a} \sum_{T\in \mathcal {T}_{n,m+1}}\prod_{v}\frac{(mh_v+1)x-h_v+1}{(m+1)h_v}=\frac{1}{mn+1}\binom{(mn+1)x}{n},\end{aligned}$$ or equivalently $$\begin{aligned} \label{eqn formua1.2b} \sum_{T\in \mathcal {T}_{n,m+1}}\prod_{v}\Big(x+\frac{1}{h_v}\Big)=\frac{x+1}{n!}\prod_{i=1}^{n-1}\Big((mn+i+1)(x+1)-(m+1)i\Big).\end{aligned}$$ where $\mathcal {T}_{n,m+1}$ denotes the set of complete $(m+1)$-ary trees with $n$ internal vertices, the product is over all internal vertices $v$ of $T$. Recall that a [*plane forest*]{} is a forest of plane trees that are linearly ordered. Let $\mathcal{F}(n)$ denote the set of plane forests with $n$ vertices. For any vertex $v$ of $F\in \mathcal{F}(n)$, the hook length $H_v$ of $v$ is defined as the number of vertices in the subtree rooted at $v$. Note that this definition is slightly different to that of hook length defined above for $(m+1)$-ary trees. Du and Liu [@duliu] investigated the hook length polynomials for plane forests and obtained that $$\begin{aligned} \label{eqn formua1.3a} \sum_{F\in \mathcal {F}(n)}\prod_{v\in V(F)}\Big(x+\frac{1}{H_v}\Big)=\frac{(x+1)}{n!}\prod_{i=1}^{n-1}\Big((2n+1-i)(x+1)-i\Big),\end{aligned}$$ or equivalently, $$\begin{aligned} \label{eqn formua1.3b} F_n(x)=\sum_{F \in \mathcal {F}(n) }\prod_{v\in V(F)}\frac{(2h_v-1)x-H_v+1}{H_v}=\frac{1}{2n+1}\binom{(2n+1)x}{n},\end{aligned}$$ where $V(F)$ is the set of vertices of $F$. It is well known that there exists a simple bijection between plane forests and complete binary trees. For the sake of completeness, we present it here. Given any plane forest $F\in \mathcal {F}(n)$, we pick the first plane tree $T$ of $F$ with root $u$. Let $T'$ denote the plane forest deduced from $T$ by removing the root $u$. Then the bijection can be defined recursively as follows: $\psi(F)$ is the complete binary tree with root $u$ such that it has the left subtree $\psi(T')$ and the right subtree $\psi(F\backslash T)$. It is clear that the bijection maps the hook length of $v$ in $V(F)$ to the number of internal vertices of the left component of $v$ of $\psi(F)$. This motivates us to define the first kind of