arXivDaily arXiv每日学术速递 周一至周五更新
北美退伍军人节(Veterans Day),12 日无更新
arXiv 1002.2054math.COmath.AC
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具有k个逆的置换的个数

The number of permutations with k inversions

Gábor Hegedüs

摘要:

令$n\geq 1$、$0\leq t\leq {n \choose 2}$为任意整数。将数字$I_n(t)$定义为具有$t$反转的$[n]$的排列数。令$n,d\geq 1$和$0\leq t\leq (d-1)n$为任意整数。将{ \em多项式系数} $H(n,d,t)$定义为具有至多$n$部分的$t$的组合物的数目,其中没有一个大于$d-1$。本文利用Gröbner基和自由分解的理论给出了数$I_n(t)$和$H(n,d,t)$的显式公式。

英文摘要:

Let $n\geq 1$, $0\leq t\leq {n \choose 2}$ be arbitrary integers. Define the numbers $I_n(t)$ as the number of permutations of $[n]$ with $t$ inversions. Let $n,d\geq 1$ and $0\leq t\leq (d-1)n$ be arbitrary integers. Define {\em the polynomial coefficients} $H(n,d,t)$ as the numbers of compositions of $t$ with at most $n$ parts, no one of which is greater than $d-1$. In our article we give explicit formulas for the numbers $I_n(t)$ and $H(n,d,t)$ using the theory of Gröbner bases and free resolutions.

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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