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Related papers: A congruence modulo $n^3$ involving two consecutiv…

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We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

Number Theory · Mathematics 2009-11-06 Roberto Tauraso

In this paper we deduce some new supercongruences modulo powers of a prime $p>3$. Let $d\in\{0,1,\ldots,(p-1)/2\}$. We show that $$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{8^k}\equiv 0\ (\mbox{mod}\ p)\ \ \ \mbox{if}\ d\equiv…

Number Theory · Mathematics 2013-10-31 Zhi-Wei Sun

Let $p>5$ be a prime. We prove congruences modulo $p^{3-d}$ for sums of the general form $\sum_{k=0}^{(p-3)/2}\binom{2k}{k}t^k/(2k+1)^{d+1}$ and $\sum_{k=1}^{(p-1)/2}\binom{2k}{k}t^k/k^d$ with $d=0,1$. We also consider the special case…

Number Theory · Mathematics 2012-10-09 Khodabakhsh Hessami Pilehrood , Tatiana Hessami Pilehrood , Roberto Tauraso

Let $\overline{p}_{k}(n)$ denote the number of overpartition $k$-tuples of $n$. In 2023, Saikia \cite{saikia} conjectured the following congruences: \begin{align*} \overline{p}_{q}(8n+2)& \equiv 0 \pmod{4},\quad \overline{p}_{q}(8n+3)\equiv…

Number Theory · Mathematics 2025-09-23 G. Kavya Keerthana , S. Ananya , Ranganatha D

Let $p>3$ be a prime, and let $a$ be a rational $p$-adic integer, using WZ method we establish the congruences modulo $p^3$ for $$\sum_{k=0}^{p-1} \binom ak\binom{-1-a}k\binom{2k}k\frac {w(k)}{4^k},$$ where $$w(k)=1,\frac 1{k+1},\frac…

Number Theory · Mathematics 2022-02-15 Zhi-Hong Sun

The following congruence for power sums, $S_n(p)$, is well known and has many applications: $1^n+2^n +\dots +p^n \equiv\begin{cases} -1 \text{ mod } p, & \text{ if } \ p-1 \ | \ n; 0 \text{ mod } p, & \text{ if } \ p-1 \ \not| \ n,…

Number Theory · Mathematics 2018-01-08 Nicholas J. Newsome , Maria S. Nogin , Adnan H. Sabuwala

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence, which is $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n$-th Bernoulli number. In this paper, we will…

Number Theory · Mathematics 2025-12-03 Jiaqi Wang , Rong Ma

In this paper, we establish the following two congruences: \begin{gather*} \sum_{k=0}^{(p+1)/2}(3k-1)\frac{\left(-\frac{1}{2}\right)_k^2\left(\frac{1}{2}\right)_k4^k}{k!^3}\equiv…

Number Theory · Mathematics 2020-06-30 Chen Wang

Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3),$$ $$\sum_{k=1}^{(p-1)/2}\binom{2k}{k}/k=(-1)^{(p+1)/2}8/3*pE_{p-3} (mod p^2),$$…

Number Theory · Mathematics 2015-05-18 Zhi-Wei Sun

In this paper we prove the supercongruence $$\sum_{n=0}^{(p-1)/2}\frac{6n+1}{256^n}\binom{2n}n^3\equiv p(-1)^{(p-1)/2}+(-1)^{(p-1)/2}\frac{7}{24}p^4B_{p-3}\pmod{p^5}$$ for any prime $p>3$, which was conjectured by Sun in 2019.

Number Theory · Mathematics 2021-09-22 Guo-Shuai Mao , Zhi-Wei Sun

We produce congruences modulo a prime $p>3$ for sums $\sum_k\binom{3k}{k}x^k$ over ranges $0\le k<q$ and $0\le k<q/3$, where $q$ is a power of $p$. Here $x$ equals either $c^2/(1-c)^3$, or $4s^2/\bigl(27(s^2-1)\bigr)$, where $c$ and $s$ are…

Number Theory · Mathematics 2022-10-13 Sandro Mattarei , Roberto Tauraso

We present some congruences modulo $p^{6-d}$ for sums of the type $\sum_{k=0}^{(p-3)/2}x^k{2k\choose k}/(2k+1)^d$, for $d=1,2,3$ where $p>5$ is a prime.

Number Theory · Mathematics 2011-11-01 Roberto Tauraso

We give a family of congruences for the binomial coefficients ${kp-1\choose p-1}$ in terms of multiple harmonic sums, a generalization of the harmonic numbers. Each congruence in this family (which depends on an additional parameter $n$)…

Number Theory · Mathematics 2018-10-16 Julian Rosen

Let $\overline{p}_{j,k}(n)$ denotes the number of $(j,k)$-regular overpartitions of a positive integer $n$ such that none of the parts is congruent to $j$ modulo $k$. Naika et. al. (2021) proved infinite families of congruences modulo…

Number Theory · Mathematics 2021-09-16 Riyajur Rahman , Nipen Saikia

In 1950 G. Giuga studied the congruence $\sum_{j=1}^{n-1} j^{n-1} \equiv -1$ (mod $n$) and conjectured that it was only satisfied by prime numbers. In this work we generalize Giuga's ideas considering, for each $k \in \mathbb{N}$, the…

Number Theory · Mathematics 2011-03-18 José María Grau , Antonio M. Oller-Marcén

Let $p>3$ be a prime and let $a$ be a positive integer. We show that if $p\equiv1\pmod 4$ or $a>1$ then $$\sum_{k=0}^{\lfloor\frac34p^a\rfloor}\frac{\binom{2k}k^2}{16^k}\equiv\l(\frac{-1}{p^a}\r)\pmod{p^3}$$ with $(-)$ the Jacobi symbol,…

Number Theory · Mathematics 2018-09-25 Guo-Shuai Mao , Zhi-Wei Sun

In this work we consider the congruence $\sum_{j=1}^{n-1} j^{k(n-1)} \equiv -1 \pmod n$ for each $k \in \mathbb{N}$, thus extending Giuga's ideas for $k=1$. In particular, it is proved that a pair $(n,k)\in \mathbb{N}^2$ satisfies this…

Number Theory · Mathematics 2013-11-15 Antonio M. Oller-Marcén , José María Grau

We present several congruences modulo a power of prime $p$ concerning sums of the following type $\sum_{k=1}^{p-1}{m^k\over k^r}{2k\choose k}^{-1}$ which reveal some interesting connections with the analogous infinite series.

Number Theory · Mathematics 2009-12-20 Roberto Tauraso

For integer $k \geq 1$, let $S_k(n)$ denote the sum of the $k$th powers of the first $n$ positive integers. In this paper, we derive a new formula expressing $2^{2k}$ times $S_{2k}(n)$ as a sum of $k$ terms involving the numbers in the…

General Mathematics · Mathematics 2025-01-27 José L. Cereceda

Let $p>3$ be a prime. In this paper, we obtain the congruences for $$\sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^3}{(-8)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{3k}k}{(-192)^k},\…

Number Theory · Mathematics 2022-11-28 Zhi-Hong Sun
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