What is the asymptotic density of deficient-perfect numbers? 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?
 A: 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.
A: According to the online encyclopedia of integer sequences, the smallest odd deficient-perfect number is 9018009. 
