This question is pretty basic, so I apologize in advance if it is unsuitable for MO. If so, please do let me know and I will migrate it over to MSE.


Essentially, by work of Kanold, we know that the natural density of the perfect numbers $N$ for which
$$\sigma(N) = 2N$$
is $0$. Now my question is: Is it known that a similar result holds for almost perfect numbers $M$, whereby
$$\sigma(M) = 2M - 1$$
holds?


I tried searching for an answer via Google, but it did not return any relevant results.


If this problem is currently open, I would appreciate it very much if you could point me to recent literature on research work done in this area.


Thanks!



On the natural density of almost perfect numbers
The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim\_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Fix any $k \in \mathbb{Z}$. Considering any $\epsilon>0$ we see that for $n>\frac{|k|}{\epsilon}$ the natural density of integers $n$ such that $\sigma(n)=2n+k$ is at most the density of integers $n$ with $$-\epsilon+2<\frac{\sigma(n)}{n}\leq 2$$ which equals $\Phi(2)-\Phi(2-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish.
Obviously for a specific choice of $k$ one can do better but the approach here does in fact provide density $0$ in very general situations.  

EDIT:(references) According to [this paper](http://projecteuclid.org/euclid.pjm/1102912212) of Erdős the first published proof was given by Davenport in a 1934 paper not listed in Mathscinet but the method was identically the same as Schoenberg's earlier result that $\phi(n)/n$ has a continuous distribution.


