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Let $p>3$ be a prime, and let $q_p(2)=(2^{p-1}-1)/p$ be the Fermat quotient of $p$ to base 2. Recently, Z. H. Sun proved that \sum_{k=1}^{p-1}\frac{1}{k\cdot 2^k}\equiv q_p(2)-\frac{p}{2}q_p(2)^2 \pmod{p^2} which is a generalization of a…

数论 · 数学 2011-10-20 Romeo Mestrovic

Given an odd prime p, we prove a congruence on the p-residue of the Fermat quotient q_p(a) in base a with 0<a<p, which arises from a generalization of the Voronoi congruences and from some other congruences on sums and weighted sums of…

数论 · 数学 2021-08-31 Claire I. Levaillant

In this paper, we confirm some congruences conjectured by V.J.W. Guo and M.J. Schlosser recently. For example, we show that for primes $p>3$, $$…

数论 · 数学 2020-06-30 Chen Wang

In this paper, we mainly prove two congruence conjecture of Z.-W. Sun. Let $p\equiv3\pmod 4$ be a prime. Then $$\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{8^k}\equiv-\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{(-16)^k}\pmod{p^3}.$$ And for any odd prime…

数论 · 数学 2023-04-11 Guo-Shuai Mao

Let $p>5$ be a prime. Motivated by the known formulae $\sum_{k=1}^\infty(-1)^k/(k^3\binom{2k}{k})=-2\zeta(3)/5$ and $\sum_{k=0}^\infty \binom{2k}{k}^2/((2k+1)16^k)=4G/\pi$$ (where $G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2$ is the Catalan…

数论 · 数学 2013-12-04 Zhi-Wei Sun

We give q-analogues of the following congruences by Z.-W. Sun: \sum_{k=1}^{p-1}\frac{D_k}{k} \equiv -\frac{2^{p-1}-1}{p} \pmod p,\\ \sum_{k=1}^{p-1}\frac{H_k}{k 2^k}\equiv 0 \pmod{p},\quad p\geqslant 5, where p is a prime,…

数论 · 数学 2013-12-24 Victor J. W. Guo

We prove several supercongruences involving the harmonic number of order two $H_n^{(2)}:=\sum_{k=1}^n1/k^2$. For example, if $p>5$ is prime and $\alpha$ is $p$-integral, then we can completely determine $$…

数论 · 数学 2022-01-19 Guo-Shuai Mao , Hao Pan

We prove some polynomial identities from which we deduce congruences modulo $p^2$ for the Fermat quotient $\frac{2^p-2}{p}$ for any odd prime $p$ (Proposition 1 and Theorem 1). These congruences are simpler than the one obtained by…

数论 · 数学 2023-09-19 Takao Komatsu , B. Sury

Let $m>2$ and $q>0$ be integers with $m$ even or $q$ odd. We show the supercongruence $$\sum_{k=0}^{p-1}(-1)^{km}\binom{p/m-q}{k}^m\equiv0\pmod{p^3}.$$ for any prime $p>mq$. This confirms a conjecture of Sun.

数论 · 数学 2015-04-29 Xiang-Zi Meng , Zhi-Wei Sun

In this paper we establish some sophisticated congruences involving central binomial coefficients and Fibonacci numbers. For example, we show that if $p\not=2,5$ is a prime then $$\sum_{k=0}^{p-1}F_{2k}\binom{2k}{k}=(-1)^{[p/5]}(1-(p/5))…

数论 · 数学 2009-12-20 Zhi-Wei Sun

Recently the first author proved a congruence proposed in 2006 by Adamchuk: $\sum_{k=1}^{\lfloor\frac{2p}{3}\rfloor}\binom{2k}{k}\equiv 0\pmod{p^2}$ for any prime $p=1 \pmod{3}$. In this paper, we provide more examples (with proofs) of…

数论 · 数学 2020-07-21 Guo-Shuai Mao , Roberto Tauraso

We give a short proof of the following known congruence: for every odd prime $p$ $$\sum_{k=0}^{p-1}{2k\choose k}^2 16^{-k}\equiv (-1)^{{p-1\over 2}}\pmod{p^2}.$$ Moreover, we provide some new results connected with the above congruence.

数论 · 数学 2009-11-24 Roberto Tauraso

In this paper, we mainly prove a congruence conjecture of Z.-W. Sun \cite{Sjnt}: Let $p>5$ be a prime. Then $$ \sum_{k=(p+1)/2}^{p-1}\frac{\binom{2k}k^2}{k16^k}\equiv-\frac{21}2H_{p-1}\pmod{p^4}, $$ where $H_n$ denotes the $n$-th harmonic…

数论 · 数学 2026-01-26 Guo-Shuai Mao

Let $p$ be an odd prime. In the paper, by using the properties of Legendre polynomials we prove some congruences for $\sum_{k=0}^{\frac{p-1}2}\binom{2k}k^2m^{-k}\mod {p^2}$. In particular, we confirm several conjectures of Z.W. Sun. We also…

数论 · 数学 2010-12-20 Zhi-Hong Sun

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/n$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k^2}\equiv \sum_{k=1}^{p-1}\frac{H_k^2}{k}…

数论 · 数学 2018-04-10 Romeo Mestrovic

In this paper, we mainly prove the following conjectures of Z.-W. Sun \cite{S13}: Let $p>2$ be a prime. If $p=x^2+3y^2$ with $x,y\in\mathbb{Z}$ and $x\equiv1\pmod 3$, then $$x\equiv\frac14\sum_{k=0}^{p-1}(3k+4)\frac{f_k}…

数论 · 数学 2024-09-20 Guo-Shuai Mao , Yan Liu

This article deals with a conjecture, introduced in [GQ] (hereinafter $SFLT2$), which generalizes the second case of Fermat's Last Theorem: {\it Let $p>3$ be a prime. The diophantine equation $\frac{u^p+v^p}{u+v}=w_1^p$ with $u,v,u+v,…

数论 · 数学 2013-05-30 Roland Quême

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/k$ be the $n$th harmonic number. In this note we prove that for any prime $p\ge 7$, $$ \sum_{k=1}^{p-1}\frac{H_k^2}{k^2} \equiv4/5pB_{p-5}\pmod{p^2}, $$ which confirms the conjecture…

数论 · 数学 2012-03-27 Romeo Mestrovic

Let $p$ be a prime and let $S_2(\Gamma(p))$ be the space of weight $2$ cusp forms for the principal congruence subgroup $\Gamma(p)$. Then $\mathrm{SL}_2(\mathbb{F}_p)$ acts on $S_2(\Gamma(p))$ in a natural way. Around 1928, Hecke proved…

表示论 · 数学 2025-08-26 Zhe Chen , Yongqi Feng

The Franel numbers are defined by $ f_n=\sum_{k=0}^n {n\choose k}^3. $ Motivated by the recent work of Z.-W. Sun on Franel numbers, we prove that \begin{align*} \sum_{k=0}^{n-1}(3k+1)(-16)^{n-k-1} {2k\choose k} f_k &\equiv…

数论 · 数学 2014-04-29 Victor J. W. Guo
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