(Note: This has been cross-posted to MO.)

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$).

Therefore, $q \neq n$. It follows that either $q < n$ or $n < q$.

Dris [2012] proved that $n < q$ implies Sorli's conjecture that $k = 1$. By the contrapositive, $k > 1$ implies that $q < n$.

Acquaah and Konyagin [2012] showed that all the prime factors $r$ of $N$ satisfy $r < (3N)^{1/3}$. In particular, if $k = 1$, then

$$q < (3N)^{1/3} \Longrightarrow q^3 < 3N = 3qn^2 \Longrightarrow q < n\sqrt{3}.$$

Therefore, regardless of the status of Sorli's conjecture, we know that $$q < n\sqrt{3}$$ must be true.

Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$. Let $$I(x) = \sigma(x)/x$$ be the abundancy index of $x$.

We claim that

$$\frac{\sigma(q)}{n} \neq \frac{\sigma(n)}{q}.$$

Suppose to the contrary that

$$\frac{\sigma(q)}{n} = \frac{\sigma(n)}{q}.$$

It follows that

$$q\sigma(q) = n\sigma(n).$$

Since $\gcd(q, n) = 1$, then we have $$n \mid \sigma(q)$$ and $$q \mid \sigma(n)$$ so that $$\frac{\sigma(q)}{n}, \frac{\sigma(n)}{q} \in \mathbb{N}.$$

Since $1 < qn$ is a proper factor of the (odd) perfect number $N = {q^k}{n^2}$, then $qn$ is deficient, and we have $$1 < I(qn) = \frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q} < 2.$$ This is a contradiction.

Now, in an e-mail sent by Ochem to Dris in April of 2013, the following result was communicated:

If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $$I(n) > \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9}} \approx 1.44440557369828207.$$

Since $q < n\sqrt{3}$ (in general), then ( since $n > {10}^{375}$ follows from $q^k < n^2$ (Dris [2012]) and $N > {10}^{1500}$ (Ochem and Rao [2012]) ) $$\sigma(q) = q + 1 << n\sqrt{3}$$ $$\frac{\sigma(q)}{n} << \sqrt{3}$$ $$\frac{\sigma(n)}{q} > \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9}}\frac{n}{q} > {\frac{\sqrt{3}}{3}}\cdot\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9}} \approx 0.83392794679369899.$$

In particular, if $$\frac{\sigma(q)}{n} < \frac{\sigma(n)}{q},$$ so that $$\frac{\sigma(q)}{n} < \sqrt{2},$$ then we have $$\frac{\sigma(n)}{q} > \sqrt{\frac{\sigma(q)}{n}\cdot\frac{\sigma(n)}{q}} = \sqrt{I(q)I(n)} > \sqrt{I(n)} > \sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}}.$$

But $$\sqrt{\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}} \approx 1.20183425383797497745284556 > I(5).$$

My question is this -- Does anybody here have any bright ideas on how to improve the following bounds:

$$\frac{\sigma(q)}{n} << \sqrt{3}$$

$$\frac{\sigma(n)}{q} > {\frac{\sqrt{3}}{3}}\cdot\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}}$$

Basically, I'm trying to get an upper bound for $\sigma(q)/n$ that's smaller than the lower bound for $\sigma(n)/q$, in order to force $\sigma(q)/n < \sigma(n)/q$. As you can see, it's still a long way to go. (For more details on the motivation behind pursuing a proof for this inequality, the interested reader is hereby referred to this preprint.)

Thanks everyone, and my apologies for the somewhat very long post. =)

Update - July 19 2016

Brown has announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds "in many cases", here.


Update - July 19 2016

Brown has announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds "in many cases", here.

The following occurred to me just now. This is not a complete answer to my question, but I merely wanted to collect my thoughts into a single location:

Since $\sigma(q^k) \equiv k + 1 \pmod 4$, $k \equiv 1 \pmod 4$ and $n$ is odd, then $\sigma(q^k) \neq n$. We consider two cases:

$$\sigma(q^k) < n$$


$$n < \sigma(q^k).$$

The first implies $q^k < n$, which further implies $q < n$.

Under the second case, we consider two sub-cases:

Sub-Case (1) : $$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

In this subcase, $\sigma(q^k) < \sqrt{2}n$. This implies that $q < \sqrt{2}n$.

Additionally, note that under this subcase:

$$\frac{\sigma(q)}{n} \leq \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \leq \frac{\sigma(n)}{q}$$

Sub-Case (2) : $$\frac{\sigma(n)}{q^k} < \frac{\sigma(q^k)}{n}$$

This implies that $n < q^k$, which implies that the biconditional $k = 1 \Longleftrightarrow n < q$ is true.

Under Sub-Case (2):

(a) If $k > 1$, then $q < n$.

(b) If $k = 1$, then (by Acquaah and Konyagin's estimate) $q < \sqrt{3}n$.

This is my best shot at this, so far.

  • $\begingroup$ It seems to me that the next step will be to try to prove that $k = 1 \Longrightarrow q < \sqrt{2}n$ in general. $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 7 '14 at 2:08
  • $\begingroup$ Lastly, note that: $$I(q^k) < \sqrt[3]{2} < I(n).$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 7 '14 at 2:14
  • $\begingroup$ There is also a partial answer to this same question in MO. $\endgroup$ – Jose Arnaldo Bebita-Dris Dec 12 '14 at 2:10

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