Recursively abundant and recursively perfect numbers

In this note, we investigate properties of the ratio $D(n)/n$, which we will call the deficiency index. We will discuss some concepts recast in the language of the deficiency index, based on similar considerations in terms of the abundancy index.

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 $F_{2k+1}$ are Fibonacci primes. Moreover, we obtain other interesting results and raise a new conjecture on perfect numbers.

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 $n$ is an even perfect number $\neq 2^{k-1}(2^k-1)$. Furthermore, if $n = 2^{\alpha-1}p^{\beta-1}$ for some $\beta>1$, then $n\ |\ \sigma_5(n)$ if and only if $n$ is an even perfect number $\neq 496$.

Abundancy index refers to the ratio of the sum of the divisors of a number to the number itself. It is a concept of great importance in defining friendly and perfect numbers. Here, we describe a suitable generalization of abundancy index to the ring of Gaussian integers ($\mathbb{Z}[i]$). We first show that this generalization possesses many of the useful properties of the traditional abundancy index in $\mathbb{Z}$. We then investigate $k$-powerful $\tau$-perfect numbers and...

The abundancy index of a positive integer is the ratio between the sum of its divisors and itself. We generalize previous results on abundancy indices by defining a two-variable abundancy index function as $I(x,n)\colon\mathbb{Z^+}\times\mathbb{Z^+}\to\mathbb{Q}$ where $I(x,n)=\frac{\sigma_x(n)}{n^x}$. Specifically, we extend limiting properties of the abundancy index and construct sufficient conditions for rationals greater than one that fail to be in the image of the functi...

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 not exist any odd integers with equality between $\sigma(N)$ and 2N. The existence of distinct prime divisors in the repunits in $\sigma(N)$ follows from a theorem on the primitive divisors of the Lucas sequences $U_{2\alpha_i+1}(q_i+1,q_i)...

We shall give an explicit upper bound for the smallest prime factor of multiperfect numbers of the form $N=p_1^{\alpha_1}\cdots p_s^{\alpha_s} q_1^{\beta_1}\cdots q_t^{\beta_t}$ with $\beta_1, \ldots, \beta_t$ bounded by a given constant. We shall also give similar results for quasiperfect numbers and relatively prime amicable pairs of opposite parity.

We call positive integer n a near-perfect number, if it is sum of all its proper divisors, except of one of them ("redundant divisor"). We prove an Euclid-like theorem for near-perfect numbers and obtain some other results for them.

Generalizing the concept of a perfect number, Sloane's sequences of integers A083207 lists the sequence of integers $n$ with the property: the positive factors of $n$ can be partitioned into two disjoint parts so that the sums of the two parts are equal. Following Clark et al., we shall call such integers, Zumkeller numbers. Generalizing this, Clark et al., call a number n a half-Zumkeller number if the positive proper factors of n can be partitioned into two disjoint parts s...

Let $q^k n^2$ be an odd perfect number with special prime $q$. Define the GCDs $$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$$ and $$I = \gcd\bigg(n,\sigma(n^2)\bigg).$$ We prove that $G \times H = I^2$. (Note that it is trivial to show that $G \mid I$ and $I \mid H$ both hold.) We then compute expressions for $G, H,$ and $I$ in terms of $\sigma(q^k)/2, n,$ and $\gcd\bigg(\sigma(q^k)/2,n\bigg)$. Afterwards, we prove that if $G = H = I$,...