Related papers: Proof of a conjectural supercongruence modulo $p^5…

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 $n$)…

Let the numbers $\alpha_n,\beta_n$ and $\gamma_n$ denote \begin{align*} \alpha_n=\sum_{k=0}^{n-1}{2k\choose k},\quad \beta_n=\sum_{k=0}^{n-1}{2k\choose k}\frac{1}{k+1}\quad\text{and}\quad \gamma_n=\sum_{k=0}^{n-1}{2k\choose…

In this paper, we employ the Wilf-Zeilberger (WZ) method to prove a supercongruence conjecture posed by Z.-W. Sun: for any prime $p$, \begin{align*} \sum_{k=0}^{\frac{p-3}{2}}\frac{92k^2+61k+9}{(2k+1)64^k}{2k \choose k}{3k \choose k}{4k…

In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that…

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer. Let $\{B_n(x)\}$ and $\{E_n(x)\}$ denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that $$\sum_{k=0}^{p-1}\binom…

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))…

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},\…

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…

Let $p$ be an odd prime. Define the Gaussian power sum \[ G_n(p)=\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}(a+bi)^n\in\mathbb Z[i]. \] We determine $G_p(p)$ modulo high powers of $p$: if $p\equiv 1\pmod 4$ then $$G_p(p)\equiv p^2(1+i)\pmod{p^3},$$…

We prove a supercongruence modulo $p^3$ between the $p$th Fourier coefficient of a weight 6 modular form and a truncated ${}_6F_5$-hypergeometric series. Novel ingredients in the proof are the comparison of two rational approximations to…

In this paper, we mainly establish a congruence for a sum involving Ap\'{e}ry numbers, which was conjectured by Z.-W. Sun. Namely, for any prime $p>3$ and positive odd integer $m$, we prove that there is a $p$-adic integer $c_m$ only…

In this paper, we prove the following result conjectured by Z.-W. Sun: $$ (2n-1){3n\choose n}| \sum_{k=0}^{n}{6k\choose 3k}{3k\choose k}{6(n-k)\choose 3(n-k)}{3(n-k)\choose n-k}. $$ by showing that the left-hand side divides each summand on…

We establish some supercongruences related to a supercongruence of Van Hamme, such as \begin{align*} \sum_{k=0}^{(p+1)/2} (-1)^k (4k-1)\frac{(-\frac{1}{2})_k^3}{k!^3} &\equiv p(-1)^{(p+1)/2}+p^3(2-E_{p-3})\pmod{p^{4}},\\…

We prove a strengthened form of a conjecture of Sun on a determinant attached to a binary quadratic form. Let $n>3$ and let $c,d\in\Z$. If $n$ is composite, then \[ \det\big[(i^2+cij+dj^2)^{n-2}\big]_{0\leq i,j\leq n-1}\equiv 0\pmod {n^2}…

Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{\deg f}\sum_{k=r(mod p^a)}\binom{n}{k}(-1)^k f((k-r)/p^a) =0 (mod p^{\sum_{i=a}^{\infty}[n/p^i]})$$ (motivated by…

The Ap\'ery numbers $A_n$ and the Franel numbers $f_n$ are defined by $$A_n=\sum_{k=0}^{n}{\binom{n+k}{2k}}^2{\binom{2k}{k}}^2\ \ \ \ \ {\rm and }\ \ \ \ \ \ f_n=\sum_{k=0}^{n}{\binom{n}{k}}^3(n=0, 1, \cdots,).$$ In this paper, we prove…

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…

We prove congruences, modulo a power of a prime p, for certain finite sums involving central binomial coefficients $\binom{2k}{k}$.

Let n be a positive odd integer and let p>n+1 be a prime. We mainly derive the following congruence: $$\sum_{0<i_1<...<i_n<p}(i_1/3)(-1)^{i_1}/(i_1...i_n)=0 (mod p).$$

In this paper, we pose lots of challenging conjectures on congruences for the sums involving binomial coefficients and Ap\'ery-like numbers modulo $p^3$, where $p$ is an odd prime.