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arXiv 2403.05072math.CO
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钩长平方和与容平方和

Sum of squares of hook lengths and contents

Krishna Menon

AI总结:

针对整数分划杨图的钩长平方和与容平方和关系,给出双射证明解决Stanley提出的问题,得到杨图矩形数量公式,并提及钩长和容其他次幂和的结果。

摘要:

已知对于整数$n$的任意分划的杨图,其所有格子钩长的平方和恰好比容的平方和多$n^2$。即对于整数$n$的任意分划$λ$,有$\begin{equation*}\n \forall u \in λ, \sum_{u \in λ} h(u)^2 = n^2 + \sum_{u \in λ} c(u)^2. \end{equation*}$我们给出了这一事实的双射证明,从而解决了Stanley提出的问题。在此过程中,我们得到了分划杨图中矩形数量的公式。我们还提及了钩长和容的其他次幂和的相关结果。

英文摘要:

It is known that for the Young diagram of any partition of an integer $n$, the sum of squares of the hook lengths of its cells is exactly $n^2$ more than that of the contents of its cells. That is, for any partition $λ$ of an integer $n$, \begin{equation*} \sum_{u \in λ} h(u)^2 = n^2 + \sum_{u \in λ} c(u)^2. \end{equation*} We provide a bijective proof of this fact, thus solving a problem posed by Stanley. Along the way, we obtain a formula for the number of rectangles in the Young diagram of a partition. We also mention a result for sums of other powers of hook lengths and contents.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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