This question is an offshoot from the following [MSE post](https://math.stackexchange.com/questions/1205692). I hope that it is appropriate for this site.


Let $\sigma(x)$ be the sum of the divisors of $x$.


An integer $a$ is said to be solitary if there does not exist another integer $b \neq a$ such that
$$\frac{\sigma(a)}{a}=\frac{\sigma(b)}{b}.$$


An integer $N$ is said to be almost perfect if $\sigma(N)=2N-1$.


My question is this: Does there exist an integer that is both solitary and almost perfect, apart from $2^r, r \geq 0$? I guess my main question is that: If a number is an odd almost perfect number, can it then be solitary?


This inquiry arises out of trying to rule out the condition $k < m$ for a [Descartes number](http://arxiv.org/abs/1503.03860) $n = km$, where $\sigma(k)(m+1)=2n$. This condition is known to be equivalent to $k$ being almost perfect.


**Update (September 18 2017)**: In the following [preprint](https://arxiv.org/pdf/1503.03860.pdf), the author appears to have proved that $m < k$, if $n = km$ is a Descartes number (where $\sigma(k)(m+1)=2n$). The proof is dependent on the validity of a reasonable assumption regarding the unboundedness of the function $f(x)=x+(1/x)$, for "general" $x > 0$.



Does there exist an integer that is both solitary and almost perfect?
An elementary sufficient condition for $n$ to be solitary is that $\gcd(n,\sigma(n))=1$. Indeed, suppose $\gcd(n,\sigma(n))=1$ and $\sigma(n)/n = \sigma(m)/m$. Since the left-hand fraction is in lowest terms, $n\mid m$. But then $\sigma(m)/m \ge \sigma(n)/n$ unless $n=m$. 


Now if $n$ is almost perfect, then $\gcd(\sigma(n),n) = \gcd(2n-1,n) = 1$. So every almost perfect number is solitary.


So your question is just the question of whether there are almost perfect numbers other than the powers of $2$. As mentioned above, this is open.


