arXiv 1901.019932019-01-06math.NT

几个划分不等式的加细

Refinements of Some Partition Inequalities

James Mc Laughlin

备注：: 11 pages
发表信息：: INTEGERS: The Electronic Journal of Combinatorial Number Theory 16 (2016), A66, 11 pp

摘要：

本文首先研究了一类划分不等式，例如，如果$M\geq 5$是一个整数，整数$a$和$b$与$M$互素，且满足$1\leq a<b<M/2$，且$c(m,n)$由\[ \frac{1}{(sq^a,sq^{M-a};q^M)_{\infty}}-\frac{1}{(sq^b,sq^{M-b};q^M)_{\infty}}:=\sum_{m,n\geq 0} c(m,n)s^m q^n, \]定义，则对所有整数$m\geq 0, n\geq 0$都有$c(m, Mn)\geq 0$. %此外，如果$M$是偶数，则对所有整数$m\geq 0, n\geq 0$都是$c(m, Mn+M/2)\geq 0$。对于由\[ (-sq^a,-sq^{M-a};q^M)_{\infty}-(-sq^b,-sq^{M-b};q^M)_{\infty}:=\sum_{m,n\geq 0} d(m,n)s^m q^n. \]定义的整数$d(m,n)$证明了类似的结果。在每种情况下，根据整数划分都有明显的解释。例如，如果$p_{1,5}(m,n)$（分别为$p_{2,5}(m,n)$）表示将$n$划分为恰好$m$部分$\equiv \pm 1 (\mod 5)$（分别为$\equiv \pm 2 (\mod 5)$）的数目，则对于每个整数$n \geq 1$，\[ p_{1,5}(m,5n)\geq p_{2,5}(m,5n), \,\,\,1 \leq m \leq 5n. \]

英文摘要：

In the present paper we initiate the study of a certain kind of partition inequality, by showing, for example, that if $M\geq 5$ is an integer and the integers $a$ and $b$ are relatively prime to $M$ and satisfy $1\leq a<b<M/2$, and the $c(m,n)$ are defined by \[ \frac{1}{(sq^a,sq^{M-a};q^M)_{\infty}}-\frac{1}{(sq^b,sq^{M-b};q^M)_{\infty}}:=\sum_{m,n\geq 0} c(m,n)s^m q^n, \] then $c(m, Mn)\geq 0$ for all integers $m\geq 0, n\geq 0$. %If, in addition, $M$ is even, then $c(m, Mn+M/2)\geq 0$ for all integers $m\geq 0, n\geq 0$. A similar result is proved for the integers $d(m,n)$ defined by \[ (-sq^a,-sq^{M-a};q^M)_{\infty}-(-sq^b,-sq^{M-b};q^M)_{\infty}:=\sum_{m,n\geq 0} d(m,n)s^m q^n. \] In each case there are obvious interpretations in terms of integer partitions. For example, if $p_{1,5}(m,n)$ (respectively $p_{2,5}(m,n)$) denotes the number of partitions of $n$ into exactly $m$ parts $\equiv \pm 1 (\mod 5)$ (respectively $\equiv \pm 2 (\mod 5)$), then for each integer $n \geq 1$, \[ p_{1,5}(m,5n)\geq p_{2,5}(m,5n), \,\,\,1 \leq m \leq 5n. \]

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：

相关论文