# Does there exist an integer that is both solitary and almost perfect?

This question is an offshoot from the following MSE post. 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 $n = km$, 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, 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" $x > 0$.

• oeis.org/A000079 [Comment from Max Alekseyev, Jan 26 2005: All the powers of 2 are least deficient numbers but it is not known if there exists a least deficient number that is not a power of 2.]... least deficient == almost perfect – joro Mar 25 '15 at 10:38
• @joro, thank you for your comment. I am already aware of the fact that it is currently an open problem to determine whether powers of two are the only examples of almost perfect / least deficient numbers. What I am asking in this post is: If we know that a number is an odd almost perfect number, can it then be solitary? Note that powers of two are both (even) almost perfect and solitary. – Arnie Bebita-Dris Mar 25 '15 at 10:40
• Yes, but, Arnie, you should have pointed out that powers of two are the only known almost perfect numbers. – Gerry Myerson Mar 25 '15 at 11:53

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.