Related papers: On a problem of De Koninck

Let $q$ be a prime. We classify the odd primes $p\neq q$ such that the equation $x^2\equiv q\pmod{p}$ has a solution, concretely, we find a subgroup $\mathbb{L}_{4q}$ of the multiplicative group $\mathbb{U}_{4q}$ of integers relatively…

Let $\gamma(S_n)$ be the minimum number of proper subgroups $H_i$ of the symmetric group $S_n$ such that each element in $S_n$ lies in some conjugate of one of the $H_i.$ In this paper we conjecture that…

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…

Let p(n, k) denote the number of partitions of n into parts less than or equal to k. We show several properties of this function modulo 2. First, we prove that for fixed positive integers k and m, p(n,k) is periodic modulo m. Using this, we…

Let $\lambda(n)$ denote the exponent of the multiplicative group modulo $n$. We show that when $q$ is odd, each coprime residue class modulo $q$ is hit equally often by $\lambda(n)$ as $n$ varies. Under the stronger assumption that…

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…

A conjecture predicting an injective and surjective mapping $X = \displaystyle\frac{\sigma(p^k)}{p^k}, Y = \displaystyle\frac{\sigma(m^2)}{m^2}$ between OPNs $N = {p^k}{m^2}$ (with Euler factor $p^k$) and rational points on the hyperbolic…

We show that, modulo some odd prime p, the powers of two weighted sum of the first p-2 divided Bernoulli numbers equals the Agoh-Giuga quotient plus twice the number of permutations on p-2 letters with an even number of ascents and distinct…

Let $p$ be an odd prime and $r\geq 1$. Suppose that $\alpha$ is a $p$-adic integer with $\alpha\equiv2a\pmod p$ for some $1\leq a<(p+r)/(2r+1)$. We confirm a conjecture of Sun and prove that…

We prove that if $q$ is a power of a prime $p$ and $p^k$ divides $a$, with $k\ge 0$, then \[ 1+(q-1)\sum_{0\le b(q-1)<a} \binom{a}{b(q-1)}\equiv 0\pmod{p^{k+1}}. \] The special case of this congruence where $q=p$ was proved by Carlitz in…

Let $n,k\in\mathbb{N}$ and let $p_{n}$ denote the $n$th prime number. We define $p_{n}^{(k)}$ recursively as $p_{n}^{(1)}:=p_{n}$ and $p_{n}^{(k)}=p_{p_{n}^{(k-1)}}$, that is, $p_{n}^{(k)}$ is the $p_{n}^{(k-1)}$th prime. In this note we…

Carmichael showed for sufficiently large $L$, that $F_L$ has at least one prime divisor that is $\pm 1({\rm mod}\, L)$. For a given $F_L$, we will show that a product of distinct odd prime divisors with that congruence condition is a…

Let \sigma(n) = \sum_{d \mid n}d be the usual sum-of-divisors function. In 1933, Davenport showed that that n/\sigma(n) possesses a continuous distribution function. In other words, the limit D(u):= \lim_{x\to\infty} \frac{1}{x}\sum_{n \leq…

We study sets $E(\Sigma,q)=\left\{\sum_{i=1}^\infty \sigma_iq^i\colon(\sigma_i)\in\Sigma^{\mathbb N}\right\}$ for a finite set $\Sigma\subset \mathbb R$ and $q\in(0,1)$. Under the assumption $q|\Sigma|=1$ we prove several new equivalent…

Dris conjectured in his masters thesis that the inequality $q^k < n$ always holds, if $N = {q^k}{n^2}$ is an odd perfect number with special prime $q$. In this note, we initially show that either of the two conditions $n < q^k$ or…

We introduce a variant of the large sieve and give an example of its use in a sieving problem. Take the interval [N] = {1,...,N} and, for each odd prime p <= N^{1/2}, remove or ``sieve out'' by all n whose reduction mod p lies in some…

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))…

We show that for every $r \geq 1$, and all $r$ distinct (sufficiently large) primes $p_1,..., p_r > p_0(r)$, there exist infinitely many integers $n$ such that ${2n \choose n}$ is divisible by these primes to only low multiplicity. From a…

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

We define A_n=\sum_{i=1}^n (-1)^i\frac{1}{i} and we show that, for every prime p, there exists a number n such that A_n\equiv 0 (mod p).