Related papers: Generalizing Giuga's conjecture

In the present paper, we determine the sums $\sum_{j=1}^{p-1}\frac{H_j^{(s_1)}H_j^{(s_3)}}{j^{s_2}}$ and $\sum_{j=1}^{p-1}\frac{H_j^{(s_1)}H_j^{(s_3)}H_j^{(s_4)}}{j^{s_2}}$ modulo $p$ and modulo $p^2$ in certain cases. This is done by using…

Let $p\equiv3\pmod{4}$ be a prime and $r$ a positive integer. We show that $$ \prod_{k=1}^{(p^{2r}-1)/2}\frac{4k-1}{4k+1}\equiv1\pmod{p^2}. $$ This confirms a recent conjecture of Guo.

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

In 1971 Cusick proved that every real number $x\in[0,1]$ can be expressed as a sum of two continued fractions with no partial quotients equal to $1$. In other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{ for all }…

Given a prime $p\geq 5$, we reduce modulo p a convolution of order p-1 of powers of two weighted Bernoulli numbers with Bernoulli numbers in terms of harmonic numbers and generalized harmonic numbers. Our proof is based on studying the…

Given k>1, let a_n be the sequence defined by the recurrence a_n=c_1a_{n-1}+c_2a_{n-2}+...+c_ka_{n-k} for n>=k, with initial values a_0=a_1=...=a_{k-2}=0 and a_{k-1}= 1. We show under a couple of assumptions concerning the constants c_i…

This paper presents both a method and a result. The result presents a closed formula for the sum of the first $m+1,m \ge 0,$ squares of the sequence $F^{(k)}$ where each member is the sum of the previous $k$ members and with initial…

We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer…

For every nonzero integer $m$ and every integer $n \ge 1$, the $n$\textsuperscript{th} harmonic number $H_n = 1 + \tfrac12 + \dots + \tfrac1n$ satisfies the identity \[ H_n \;=\; \frac{1}{m}\,\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}\, \binom{m…

It is shown that for any prime $p$ and any natural numbers $\ell, m,$ and $s$ such that $0<s<p$, the three following congruences \begin{align*}\sum_{i\ge \ell+1}(-1)^{m-i} {m \choose i}{m+s-1+i(p-1) \choose m+s-1+\ell(p-1)} &\equiv 0 \bmod…

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.…

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…

Consider the linear congruence equation $${a_1^{s}x_1+\ldots+a_k^{s} x_k \equiv b\,(\text{mod } n^s)}\text { where } a_i,b\in\mathbb{Z},s\in\mathbb{N}$$ Denote by $(a,b)_s$ the largest $l^s\in\mathbb{N}$ which divides $a$ and $b$…

We will study the solution of a congruence, $x \equiv g^{(1/2)\omega_g(2^n)} \bmod 2^n$, depending on the integers $g$ and $n$, where $\omega_g(2^n)$ denotes the order of $g$ modulo $2^n$. Moreover, we introduce an application of the above…

In 1851 Prouhet showed that when $N=j^{k+1}$ where $j$ and $k$ are positive integers, $j \geq 2$, the first $N$ consecutive positive integers can be separated into $j$ sets, each set containing $j^k$ integers, such that the sum of the…

In this paper we determine the congruence of Jacobi sums $J(1, 1)_{49}$ of order 49 over a field $\mathbb{F}_p$. We also show that simpler congruences hold for $J(1, 1)_{49}$ in the case of artiad and hyperartiad primes.

For a non-negative integer $m$, let $S(m)$ denote the sum given by $$S(m):=\sum_{n=0}^{m}\frac{(-1)^n(8n+1)}{n!^3}\left(\frac{1}{4}\right)_n^3.$$ Using the powerful WZ-method, for a prime $p\equiv 3$ $($mod $4)$ and an odd integer $r>1$, we…

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…

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…

Inspired by Cohen and te Riele~\cite{Cohen1996}, who computationally verified that for every $n \leq 400$ there exists $k$ such that $\sigma^k(n) \equiv 0 \pmod{n}$ (where $\sigma^k$ denotes the $k$-fold iteration of the sum-of-divisors…