相关论文: On Sinha's note on perfect numbers

A perfect number is a number whose divisors add up to twice the number itself. The existence of odd perfect numbers is a millennia-old unsolved problem. This note proposes a proof of the nonexistence of odd perfect numbers. More generally,…

While the general form of even perfect numbers is well-known, the existence or non-existence of odd perfect numbers is still an open problem. We address this problem and prove that if a natural number is odd, then it's not perfect.

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…

Euler showed that if an odd perfect number exists, it must be of the form $N = p^\alpha q_{1}^{2\beta_{1}}$ $\ldots$ $q_{k}^{2\beta_{k}}$, where $p, q_{1}, \ldots, q_k$ are distinct odd primes, $\alpha$, $\beta_{i} \geq 1$, for $1 \leq i…

Let $\sigma(n)$ be the sum of the positive divisors of $n$. A number $n$ is said to be 2-near perfect if $\sigma(n) = 2n +d_1 +d_2 $, where $d_1$ and $d_2$ are distinct positive divisors of $n$. We give a complete description of those $n$…

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.

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…

In this paper, we prove the conjecture that if there is an odd perfect number, then there are infinitely many of them.

In this paper, we introduce the concept of $F$-perfect number, which is a positive integer $n$ such that $\sum_{d|n,d<n}d^2=3n$. We prove that all the $F$-perfect numbers are of the form $n=F_{2k-1}F_{2k+1}$, where both $F_{2k-1}$ and…

We show that for every length of form $4^k-1$, there exists a binary $1$-perfect code that does not include any Preparata-like code.

Let $N$ be an odd perfect number. Then, Euler proved that there exist some integers $n, \alpha$ and a prime $q$ such that $N = n^{2}q^{\alpha}$, $q \nmid n$, and $q \equiv \alpha \equiv 1 \bmod 4$. In this note, we prove that the ratio…

It is sufficient to prove that there is an excess of prime factors in the product of repunits with odd prime bases defined by the sum of divisors of the integer $N=(4k+1)^{4m+1}\prod_{i=1}^\ell ~ q_i^{2\alpha_i}$ to establish that there do…

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 $\sigma(n)$ to be the sum of the positive divisors of $n$. A number is non-deficient if $\sigma(n) \geq 2n$. We establish new lower bounds for the number of distinct prime factors of an odd non-deficient number in terms of its second…

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.

We mainly introduce two new kinds of numbers given by $$R_n=\sum_{k=0}^n\binom nk\binom{n+k}k\frac1{2k-1}\quad\ (n=0,1,2,...)$$ and $$S_n=\sum_{k=0}^n\binom nk^2\binom{2k}k(2k+1)\quad\ (n=0,1,2,...).$$ We find that such numbers have many…

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…

A natural number $n$ is called {\it multiperfect} or {\it$k$-perfect} for integer $k\ge2$ if $\sigma(n)=kn$, where $\sigma(n)$ is the sum of the positive divisors of $n$. In this paper, we establish the structure theorem of odd multiperfect…

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…

In this short paper we prove that the square of an odd prime number cannot be a very perfect number.