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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…

数论 · 数学 2022-10-13 Sandro Mattarei , Roberto Tauraso

Let $p$ be a prime greater than $3$ and let $a$ be a rational p-adic integer. In this paper we try to determine $\sum_{k=1}^{[p/3]}\binom{3k}ka^k\pmod p$, and real the connection between cubic congruences and the sum…

数论 · 数学 2013-11-21 Zhi-Hong Sun

Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper we prove some supercongruences concerning $$\align &\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{3k}k}{54^k},\…

数论 · 数学 2011-01-06 Zhi-Hong Sun

For $n=0,1,2,\ldots$ let $W_n=\sum_{k=0}^{[n/3]}\binom{2k}k \binom{3k}k\binom n{3k}(-3)^{n-3k}$, where $[x]$ is the greatest integer not exceeding $x$. Then $\{W_n\}$ is an Ap\'ery-like sequence. In this paper we deduce many congruences…

数论 · 数学 2020-05-12 Zhi-Hong Sun

It is known that $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)4^k)=\pi/2$ and $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)16^k)=\pi/3$. In this paper we obtain their p-adic analogues such as…

数论 · 数学 2011-08-03 Zhi-Wei Sun

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…

数论 · 数学 2013-10-31 Zhi-Wei Sun

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},\…

数论 · 数学 2022-11-28 Zhi-Hong Sun

Let p be a prime and let a be a positive integer. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ for all d=0,...,p^a, where m is any integer not divisible by p.…

数论 · 数学 2010-04-02 Zhi-Wei Sun , Roberto Tauraso

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…

数论 · 数学 2022-02-15 Zhi-Hong Sun

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}…

数论 · 数学 2009-11-06 Roberto Tauraso

Let $p$ be an odd prime and let $x$ be a $p$-adic integer. In this paper, we establish supercongruences for $$ \sum_{k=0}^{p-1}\frac{\binom{x}{k}\binom{x+k}{k}(-4)^k}{(dk+1)\binom{2k}{k}}\pmod{p^2} $$ and $$…

数论 · 数学 2024-07-30 Chen Wang , Hui-Li Han

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

数论 · 数学 2015-05-18 Zhi-Wei Sun

The polynomials $d_n(x)$ are defined by \begin{align*} d_n(x) &= \sum_{k=0}^n{n\choose k}{x\choose k}2^k. \end{align*} We prove that, for any prime $p$, the following congruences hold modulo $p$: \begin{align*}…

数论 · 数学 2016-04-19 Song Guo , Victor J. W. Guo

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…

We primarily investigate congruences modulo $p$ for finite sums of the form $\sum_k\binom{rk}{k}x^k/k$ over the ranges $0<k<p$ and $0<k<p/r$, where $p$ is a prime larger than the positive integer $r$. Here $x$ is an indeterminate, thus…

数论 · 数学 2026-03-18 Sandro Mattarei , Roberto Tauraso

In this paper, we prove several supercongruences conjectured by Z.-W. Sun ten years ago via certain strange hypergeometric identities. For example, for any prime $p>3$, we show that…

数论 · 数学 2021-08-10 Chen Wang , Zhi-Wei Sun

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.

数论 · 数学 2009-12-20 Roberto Tauraso

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,…

数论 · 数学 2018-09-25 Guo-Shuai Mao , Zhi-Wei Sun

Let $p>3$ be a prime. For any $p$-adic integer $a$, we determine $$\sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k,\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k^{(2)},\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}k\frac{H_k^{(2)}}{2k+1}$$ modulo…

数论 · 数学 2024-01-11 Zhi-Wei Sun

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak…

数论 · 数学 2015-03-12 Zhi-Hong Sun
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