相关论文: On the quantity $m^2-p^k$ where $p^k m^2$ is an od…

If $N = {p^k}{m^2}$ is an odd perfect number with special prime factor $p$, then it is proved that ${p^k} < (2/3){m^2}$. Numerical results on the abundancy indices $\frac{\sigma(p^k)}{p^k}$ and $\frac{\sigma(m^2)}{m^2}$, and the ratios…

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…

Let $p^k m^2$ be an odd perfect number with special prime $p$. In this article, we provide an alternative proof for the biconditional that $\sigma(m^2) \equiv 1 \pmod 4$ holds if and only if $p \equiv k \pmod 8$. We then give an application…

Acquaah and Konyagin showed that if $N$ is an odd perfect number where $N= p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$ where $p_1 < p_2 \cdots < p_k$ then one must have $p_k < 3^{1/3}N^{1/3}$. Using methods similar to theirs, we show that…

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…

In this note, we continue an approach pursued in an earlier paper of the second author and thereby attempt to produce an improved lower bound for the sum $I(q^k) + I(n^2)$, where $q^k n^2$ is an odd perfect number with special prime $q$ and…

A perfect number is a positive integer $N$ such that the sum of all the positive divisors of $N$ equals $2N$, denoted by $\sigma(N) = 2N$. The question of the existence of odd perfect numbers (OPNs) is one of the longest unsolved problems…

The existence of a perfect odd number is an old open problem of number theory. An Euler's theorem states that if an odd integer $ n $ is perfect, then $ n $ is written as $ n = p ^ rm ^ 2 $, where $ r, m $ are odd numbers, $ p $ is a prime…

Let $h_{(m,k)}$ be the class number of $\mathbb{Q}(\sqrt{1-2m^k}).$ We prove that for any odd natural number $k,$ there exists $m_0$ such that $k \mid h_{(m,k)}$ for all odd $m > m_0.$ We also prove that for any odd $m \geq 3,$ $k \mid…

In this note, we present some new results on even almost perfect numbers which are not powers of two. In particular, we show that $2^{r+1} < b$, if ${2^r}{b^2}$ is an even almost perfect number.

Let $k>2$ be a prime such that $2^k-1$ is a Mersenne prime. Let $n = 2^{\alpha-1}p$, where $\alpha>1$ and $p<3\cdot 2^{\alpha-1}-1$ is an odd prime. Continuing the work of Cai et al. and Jiang, we prove that $n\ |\ \sigma_k(n)$ if and only…

If $N = {q^k}{n^2}$ is an odd perfect number, where $q$ is the Euler prime, then we show that $n < q$ is sufficient for Sorli's conjecture that $k = \nu_{q}(N) = 1$ to hold. We also prove that $q^k < 2/3{n^2}$, and that $I(q^k) < I(n)$,…

Let $p$ be an odd prime and let $a,m$ be integers with $a>0$ and $m \not\equiv0\pmod p$. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ mod $p^2$ for $d=0,1$; for example,…

We give a short proof of the following known congruence: for every odd prime $p$ $$\sum_{k=0}^{p-1}{2k\choose k}^2 16^{-k}\equiv (-1)^{{p-1\over 2}}\pmod{p^2}.$$ Moreover, we provide some new results connected with the above congruence.

We investigate the implications of a curious biconditional involving divisors of odd perfect numbers, if Dris conjecture that $q^k < n$ holds, where $q^k n^2$ is an odd perfect number with Euler prime $q$. We then show that this…

Let $p>3$ be a prime, and let $m$ be an integer with $p\nmid m$. In the paper, by using the work of Ishii and Deuring's theorem for elliptic curves with complex multiplication we solve some conjectures of Zhi-Wei Sun concerning…

An odd perfect number $N$ is said to be given in Eulerian form if $N = {q^k}{n^2}$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$. Similarly, an even perfect number $M$ is said to be given in Euclidean form if $M…

Recently, Z.-W. Sun made the following conjecture: for any odd prime $p$ and odd integer $m$, $$ \frac{1}{m^2{m-1\choose (m-1)/2}}\Bigg(\sum_{k=0}^{(pm-1)/2}\frac{{2k\choose k}}{8^k}…

Let $p$ be an odd prime. In the paper, by using the properties of Legendre polynomials we prove some congruences for $\sum_{k=0}^{\frac{p-1}2}\binom{2k}k^2m^{-k}\mod {p^2}$. In particular, we confirm several conjectures of Z.W. Sun. We also…

If $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form, then Sorli's conjecture predicts that $k=\nu_{q}(N)=1$. In this article, we give some further results related to this conjecture and those contained in the papers…