Related papers: Solving the Odd Perfect Number Problem: Some New A…

Using a sieve-theoretic argument, we show that almost all gaps $(p_n, p_{n+1})$ between consecutive primes $p_n, p_{n+1}$ contain a natural number $m$ whose least prime factor $p(m)$ is at least the length $p_{n+1} - p_n$ of the gap,…

If $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form, then the Descartes-Frenicle-Sorli conjecture predicts that $k=1$. Brown has recently announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds…

Some new results concerning the equation $\sigma(N)=aM, \sigma(M)=bN$ are proved. As a corollary, there are only finitely many odd superperfect numbers with a fixed number of distinct prime factors.

For a normal projective variety $X$, the $\bf Q$-factoriality defect $\sigma(X)$ is defined to be the rank of the quotient of the group of Weil divisors by the subgroup of Cartier ones. We prove a slight improvement of a topological formula…

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

Let $\sigma(n)$ be the sum of the positive divisors of $n$. A positive integer $n$ is said to be $2$-near perfect when $\sigma(n)=2n+d_1+d_2$, where $d_1$ and $d_2$ are distinct positive divisors of $n$. We show that there are no odd…

Let $k\ge2$ be an integer. A natural number $n$ is called $k$-perfect if $\sigma(n)=kn.$ For any integer $r\ge1$ we prove that the number of odd $k$-perfect numbers with at most $r$ distinct prime factors is bounded by $k4^{r^3}$.

If $N={q^k}{n^2}$ is an odd perfect number given in Eulerian form, then the Descartes-Frenicle-Sorli conjecture predicts that $k=\nu_{q}(N)=1$. In this article, we give a short proof for this conjecture.

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…

For $n \geq 3$, an asymptotic formula is derived for the number of representations of a sufficiently large natural number $N$ in the form $p_1+p_2+m^n=N$, where $p_1$, $p_2$ $-$ prime numbers, $m$ $-$ natural number satisfying the…

A classic question in analytic number theory is to find asymptotics for $\sigma_{k}(x)$ and $\pi_{k}(x)$, the number of integers $n\leq x$ with exactly $k$ prime factors, where $\pi_{k}(x)$ has the added constraint that all the factors are…

We study some divisibility properties of multiperfect numbers. Our main result is: if $N=p_1^{\alpha_1}... p_s^{\alpha_s} q_1^{2\beta_1}... q_t^{2\beta_t}$ with $\beta_1, ..., \beta_t$ in some finite set S satisfies…

We consider several old problems involving the number of prime divisors function $\omega(n)$, as well as the related functions $\Omega(n)$ and $\tau(n)$. Firstly, we show that there are infinitely many positive integers $n$ such 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 show that if $N$ is an odd perfect number and $q^{\alpha}$ is some prime power exactly dividing it, then $\sigma(N/q^{\alpha})/q^{\alpha}>5$. In general, we also show that if $\sigma(N/q^{\alpha})/q^{\alpha}<K$, where $K$…

Let $p$ be an odd prime and let $d$ be an integer not divisible by $p$. We prove that $$ \prod_{1\le m,n\le p-1\atop p\nmid m^2-dn^2}\ (x-(m+n\sqrt{d})) \equiv \begin{cases}\sum_{k=1}^{p-2}\frac{k(k+1)}2x^{(k-1)(p-1)}\pmod p &\text{if}\…

Let $N$ be an odd perfect number. Let $\omega(N)$ be the number of distinct prime factors of $N$ and let $\Omega(N)$ be the total number of prime factors of $N$. We prove that if $(3,N)=1$, then $ \frac{302}{113}\omega - \frac{286}{113}…

In this paper, we partly prove a supercongruence conjectured by Z.-W. Sun in 2013. Let $p$ be an odd prime and let $a\in\mathbb{Z}^{+}$. Then if $p\equiv1\pmod3$, we have \begin{align*}…

In 2003, Zhao discovered a curious congruence involving harmonic series and Bernoulli numbers: for any odd prime $p$, $$\sum_{\substack{i,j,k\ge 1\\\gcd(ijk,p)=1\\i+j+k=p}}\frac{1}{ijk}\equiv -2B_{p-3} \pmod{p},$$ where $B_n$ is the $n$-th…

Let $p$ be an odd prime. In 2008 E. Mortenson proved van Hamme's following conjecture: $$\sum_{k=0}^{(p-1)/2}(4k+1)\binom{-1/2}k^3\equiv (-1)^{(p-1)/2}p\pmod{p^3}.$$ In this paper we show further that…