1
$\begingroup$

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" $x > 0$.

$\endgroup$
3
  • 3
    $\begingroup$ 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 $\endgroup$ – joro Mar 25 '15 at 10:38
  • $\begingroup$ @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. $\endgroup$ – Arnie Bebita-Dris Mar 25 '15 at 10:40
  • 6
    $\begingroup$ Yes, but, Arnie, you should have pointed out that powers of two are the only known almost perfect numbers. $\endgroup$ – Gerry Myerson Mar 25 '15 at 11:53
10
$\begingroup$

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.

$\endgroup$
0

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.