普通分拆的spt-Crank
The spt-Crank for Ordinary Partitions
摘要：
spt函数$spt(n)$是由Andrews引入的，作为$n$关于最小部分出现次数的划分的加权计数。Andrews，Garvan和Liang定义了$S$ -划分的spt-crank，它导致了$spt(n)$ mod 5和7的同余的组合解释。令$N_S(m,n)$表示具有spt-crank $m$的$n$的$S$ -分区的净数目。Andrews、Garvan和Liang证明了$N_S(m,n)$对所有整数$m$和正整数$n$都是非负的，他们提出了寻找$N_S(m,n)$的组合解释的问题。本文引入了双标记划分的结构，定义了双标记划分的spt-曲柄。我们表明，$N_S(m,n)$可以被解释为双标记分区的$n$与spt-crank $m$的数量。此外，我们还建立了$n$的标记划分和$n$的双重标记划分之间的双射。标记划分是由Andrews，Dyson和Rhoades定义的，它是一个划分，其中恰好有一个最小的部分被标记。他们认为这是一个挑战，找到一个定义的spt-crank的一个标记的分区，使一组标记的分区的$5n+4$和$7n+5$可以分为五个和七个equinumerous类。定义了双标记划分的spt-crank和双标记划分与标记划分之间的双射，从而解决了Andrews，Dyson和Rhoades的问题.
英文摘要：
The spt-function $spt(n)$ was introduced by Andrews as the weighted counting of partitions of $n$ with respect to the number of occurrences of the smallest part. Andrews, Garvan and Liang defined the spt-crank of an $S$-partition which leads to combinatorial interpretations of the congruences of $spt(n)$ mod 5 and 7. Let $N_S(m,n)$ denote the net number of $S$-partitions of $n$ with spt-crank $m$. Andrews, Garvan and Liang showed that $N_S(m,n)$ is nonnegative for all integers $m$ and positive integers $n$, and they asked the question of finding a combinatorial interpretation of $N_S(m,n)$. In this paper, we introduce the structure of doubly marked partitions and define the spt-crank of a doubly marked partition. We show that $N_S(m,n)$ can be interpreted as the number of doubly marked partitions of $n$ with spt-crank $m$. Moreover, we establish a bijection between marked partitions of $n$ and doubly marked partitions of $n$. A marked partition is defined by Andrews, Dyson and Rhoades as a partition with exactly one of the smallest parts marked. They consider it a challenge to find a definition of the spt-crank of a marked partition so that the set of marked partitions of $5n+4$ and $7n+5$ can be divided into five and seven equinumerous classes. The definition of spt-crank for doubly marked partitions and the bijection between the marked partitions and doubly marked partitions leads to a solution to the problem of Andrews, Dyson and Rhoades.
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