n+1$$ 份的分割钻石上的新全等式

Yongqiang Chen, Eric H. Liu, Olivia X. M. Yao
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引用次数: 0

摘要

最近,Andrews 和 Paule 引入了分治函数 PDN1(N),该函数计算了具有 n 的 \(n+1\) 副本的分治菱形的数量,其中将链接处的部分相加得到 N。在论文的最后,安德鲁斯和波尔询问 PDN1(n) 是否存在其他类型的全等关系。受他们工作的启发,我们利用 Chern 和 Tang 的一些同余式证明了 PDN1(n) 的一些新的同余式 modulo 125 和 625。特别是,我们发现了 PDN1(n) modulo 625 的一系列奇特同余。例如,我们证明了对于(k\ge 0\),$$\begin{aligned}(开始{aligned})。PDN1 leave( 5^7 \cdot 7^{8k}+frac{ 19\cdot 5^7\cdot 7^{8k}+1 }{24} \right) \equiv 5^3 \pmod {5^4}.\end{aligned}$$
本文章由计算机程序翻译,如有差异,请以英文原文为准。

New congruences on partition diamonds with $$n+1$$ copies of n

New congruences on partition diamonds with $$n+1$$ copies of n

Recently, Andrews and Paule introduced a partition function PDN1(N) which counts the number of partition diamonds with \(n+1\) copies of n where summing the parts at the links gives N. They also established the generating function of PDN1(n) and proved congruences modulo 5,7,25,49 for PDN1(n). At the end of their paper, Andrews and Paule asked for the existence of other types of congruence relations for PDN1(n). Motivated by their work, we prove some new congruences modulo 125 and 625 for PDN1(n) by using some identities due to Chern and Tang. In particular, we discover a family of strange congruences modulo 625 for PDN1(n). For example, we prove that for \(k\ge 0\),

$$\begin{aligned} PDN1\left( 5^7 \cdot 7^{8k}+\frac{ 19\cdot 5^7\cdot 7^{8k}+1 }{24} \right) \equiv 5^3 \pmod {5^4}. \end{aligned}$$
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