Research article

A Ramanujan-type supercongruence related to harmonic numbers

  • Published: 20 May 2025
  • MSC : Primary: 11A07, 11B75; Secondary: 05A10, 11B65

  • 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 $H_n=\sum_{j=1}^{n}\frac{1}{j}$ ($n=1, 2, \cdots$) are the usual harmonic numbers. This partially confirms a conjecture by Z.-W. Sun.

    Citation: Wenbin Zhang, Yong Zhang, Jiachen Wu. A Ramanujan-type supercongruence related to harmonic numbers[J]. AIMS Mathematics, 2025, 10(5): 11444-11464. doi: 10.3934/math.2025521

    Related Papers:

  • 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 $H_n=\sum_{j=1}^{n}\frac{1}{j}$ ($n=1, 2, \cdots$) are the usual harmonic numbers. This partially confirms a conjecture by Z.-W. Sun.



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