arXiv math/99120531999-12-07math.CO

Enumeration，菱形镶嵌六边形，中央有三角形孔

Enumeration of lozenge tilings of hexagons with a central triangular hole

Mihai Ciucu, Theresia Eisenkölbl, C. Krattenthaler, D. Zare

备注：: 57 pages, AmS-LaTeX, uses TeXDraw
发表信息：: J. Combin. Theory Ser. A 95 (2001), 251-334

摘要：

本文研究边长为$a,b+m,c,a+m,b,c+m$的六边形菱形镶嵌的未加权和加权枚举问题，其中边长为$m$的等边三角形已从中心移去.我们给出了封闭的公式为平原枚举和一定的$(-1)$ -枚举这些菱形平铺。在这种情况下，$a=b=c$，我们还提供了封闭的公式，某些加权枚举的菱形镶嵌是循环对称的。对于$m=0$，后面的公式专用于关于循环对称平面划分的加权枚举的语句。一个这样的专业化提供了一个证明的猜想Stembridge对一定的加权计数的循环对称平面分区。在我们的证明中采用的工具是不相交格路径和行列式评价理论的非标准应用。特别是，我们评估的决定因素$\det_{0\le i,j\le n-1}\big(\om δ_{ij}+\binom {m+i+j}j\big)$，其中$\om$是任何6次根的单位。这些行列式的评价是一个著名的结果，由于安德鲁斯（发明。53（1979），193- 225），其对应于$\om=1$。

英文摘要：

We deal with unweighted and weighted enumerations of lozenge tilings of a hexagon with side lengths $a,b+m,c,a+m,b,c+m$, where an equilateral triangle of side length $m$ has been removed from the center. We give closed formulas for the plain enumeration and for a certain $(-1)$-enumeration of these lozenge tilings. In the case that $a=b=c$, we also provide closed formulas for certain weighted enumerations of those lozenge tilings that are cyclically symmetric. For $m=0$, the latter formulas specialize to statements about weighted enumerations of cyclically symmetric plane partitions. One such specialization gives a proof of a conjecture of Stembridge on a certain weighted count of cyclically symmetric plane partitions. The tools employed in our proofs are nonstandard applications of the theory of nonintersecting lattice paths and determinant evaluations. In particular, we evaluate the determinants $\det_{0\le i,j\le n-1}\big(\om δ_{ij}+\binom {m+i+j}j\big)$, where $\om$ is any 6th root of unity. These determinant evaluations are variations of a famous result due to Andrews (Invent. Math. 53 (1979), 193--225), which corresponds to $\om=1$.

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。

摘要：

英文摘要：

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