3
$\begingroup$

Let $\sigma(n)$ denote the sum of the divisors of the natural number $n$.

Denote the deficiency of $n$ as $D(n) = 2n - \sigma(n)$.

Here is my question:

What is the asymptotic density of natural numbers $n$ such that $D(n) \mid n$?

Numbers $n$ such that $D(n) \mid n$ are called deficient-perfect.

In particular, do the deficient-perfect numbers have positive asymptotic density among the natural numbers?

Added December 21 2016 It is known that for $n=2^k$ and $k \geq 0$, $D(n)=1$ so that $D(n) \mid n$. This means that there is at least one infinite family of natural numbers $n$ such that $D(n) \mid n$. Can this fact then be used to prove that the deficient-perfect numbers have positive asymptotic density among the natural numbers?

$\endgroup$
  • 1
    $\begingroup$ We can only say that they are infinitely many. $\endgroup$ – Crostul Dec 21 '16 at 11:21
  • 1
    $\begingroup$ Primes or squares are examples with infinitely many numbers but asymptotic density equal $0$. $\endgroup$ – Sil Dec 21 '16 at 11:29
  • 1
    $\begingroup$ The striking behavior of the logarithmic scatterplot for this sequence (A271816 on OEIS, to which you supplied a link) suggests there are very few such integers. It seems likely the deficient-perfect numbers have density zero in the positive integers. $\endgroup$ – Kyle Dec 22 '16 at 4:01
1
$\begingroup$

Here is a sketch of a proof that the deficient-perfect numbers have density zero.

As above, we have $D(n) := 2n - \sigma(n)$. We wish to show that the sum $$\sum_{\substack{n \leq x \\ D(n)|n}} 1$$ is $o(x)$. We break the integers $n$ under consideration into several sets.

First, if $n$ is a power of two then there are $O(\log x)$ such numbers $\leq x$.

Second, suppose that $n$ is odd. Then the equation $D(n)|n$ implies $\sigma(n)$ must be odd. If $p$ is an odd prime then $\sigma(p^k)$ is odd if and only if $k$ is even, so we deduce that $n$ must be a square. It follows that $$\sum_{\substack{n \leq x \\ n \text{ odd} \\ D(n)|n}} 1 \ll x^{1/2}.$$

The third, and hardest, case is when $n = 2^a m$, where $a \geq 1$ and $m > 1$ is odd. We break into two further cases, depending on the $2$-adic valuation of $n$. The number of $n \leq x$ divisible by $2^a$ with $a>T$ is $$\ll \frac{x}{2^T}.$$

It therefore remains to estimate $$\sum_{1 \leq a \leq T} \sum_{\substack{m \leq 2^{-a}x \\ m \text{ odd} \\ D(2^a m)|2^am}} 1.$$

Turning to the equation $D(2^a m) | 2^a m$, we see that the highest power of $2$ dividing $\sigma(m)$ is at most $a$. Again using the fact that $\sigma(p^k)$ is odd if and only if $k$ is even, it follows that each such $m$ can be written as $$m = qs,$$ where $q$ is squarefree with $\omega(q) \leq a$ ($\omega(n)$ is the number of distinct prime factors of $n$), and $s$ is squarefull. Thus $$\sum_{\substack{m \leq y \\ m \text{ odd} \\ D(2^a m)|2^am}} 1 \leq \mathop{\sum \sum}_{\substack{qs \leq y \\ \omega(q) \leq a \\ s \ \square-\text{full}}} \mu^2(q) \ll y^{1/2} \sum_{\substack{q \leq y \\ \omega(q) \leq a}} \frac{\mu^2(q)}{q^{1/2}},$$ the second inequality following since the number of squarefull integers up to $z$ is $O(z^{1/2})$. Using partial summation and the fact that $$\sum_{\substack{q \leq y \\ \omega(q) \leq a}} \mu^2(q) \ll_a \frac{y}{\log y}(\log \log y)^{a-1},$$ we deduce $$\sum_{\substack{m \leq 2^{-a}x \\ m \text{ odd} \\ D(2^a m)|2^am}} 1 \ll_a \frac{x}{2^a} \frac{(\log \log x)^{a-1}}{\log x}.$$ Upon summing over $a$ and collecting all of our bounds we obtain $$\sum_{\substack{n \leq x \\ D(n)|n}} 1 \ll_T x \frac{(\log \log x)^{T-1}}{\log x} + \frac{x}{2^T},$$ and the desired result follows by letting $T$ tend slowly to infinity.

$\endgroup$
0
$\begingroup$

According to the online encyclopedia of integer sequences, the smallest odd deficient-perfect number is 9018009.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.