A Ramanujan-type supercongruence related to harmonic numbers

We prove that, for all primes $ p > 5 $, positive integers $ r $ and $ m $ with $ p\nmid m $ and $ p\nmid \binom{2mp^{r-1}}{mp^{r-1}} $, there holds$ \begin{align*} &\frac{1}{m^4p^{4r}{\binom{2mp^{r-1}}{mp^{r-1}}}^3}\bigg(\sum\limits_{k = 0}^{mp^r-1}(21k+8){\binom{2k}{k}}^3-p\sum\limits_{k = 0}^{mp^{r-1}-1}(21k+8){\binom{2k}{k}}^3\bigg)\\ &\quad \equiv-6\frac{H_{p-1}}{p^{2}}\ ({\rm{mod}}\ {p^{2}}), \end{align*} $where $ H_n = \sum_{j = 1}^{n}\frac{1}{j} $ ($ n = 0, 1, 2, \cdots $) are the usual harmonic numbers. This partially confirms a conjecture by Z.-W. Sun.

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Publication Details

Journal
AIMS Mathematics
Published
2025-01-01
DOI
https://doi.org/10.3934/math.2025521
Primary Topic
Advanced Mathematical Identities
Type
article
Field-Weighted Citation Impact
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article

A Ramanujan-type supercongruence related to harmonic numbers

Yong Zhang, Wenbin Zhang, Jiachen Wu
AIMS Mathematics
Advanced Mathematical Identities
article

A Ramanujan-type supercongruence related to harmonic numbers

Yong Zhang, Wenbin Zhang, Jiachen Wu
article en

Abstract

We prove that, for all primes $ p > 5 $, positive integers $ r $ and $ m $ with $ p\nmid m $ and $ p\nmid \binom{2mp^{r-1}}{mp^{r-1}} $, there holds$ \begin{align*} &\frac{1}{m^4p^{4r}{\binom{2mp^{r-1}}{mp^{r-1}}}^3}\bigg(\sum\limits_{k = 0}^{mp^r-1}(21k+8){\binom{2k}{k}}^3-p\sum\limits_{k = 0}^{mp^{r-1}-1}(21k+8){\binom{2k}{k}}^3\bigg)\\ &\quad \equiv-6\frac{H_{p-1}}{p^{2}}\ ({\rm{mod}}\ {p^{2}}), \end{align*} $where $ H_n = \sum_{j = 1}^{n}\frac{1}{j} $ ($ n = 0, 1, 2, \cdots $) are the usual harmonic numbers. This partially confirms a conjecture by Z.-W. Sun.

AIMS MathematicsVol. 10(5)
Guangdong University Of Finances and Economics (CN), Nanjing Institute of Technology (CN)
Openalex Percentile: Top 84%
Advanced Mathematical Identities
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