Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

This blog post aims to discuss results related to those contained in the preprint titled On a Conjecture of Dris Regarding Odd Perfect Numbers.

In particular, we amend the original conjecture $q^k < n$ in the author's thesis as follows:

**Conjecture**(Dris, 2017): If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $q^k < 3n$ holds.

**Heuristics**

First, note that $1 < I(q^k n)$, so that $\sigma(q^k)/n < 1$ and $\sigma(n)/q^k \leq 1$ cannot both be true.

We consider four (4) different cases:

**(a)**$\sigma(q^k)/n < 1 < \sigma(n)/q^k$

This implies that $q^k < n$. (Note that $q^k < 3n$ is a weaker inequality.)

**(b)**$1 < \sigma(q^k)/n < \sigma(n)/q^k < 2$

Since $I(q^k n) < 2$ holds (because $q^k n$ is a proper divisor of the [odd] perfect number $N = q^k n^2$), it follows that $\sigma(q^k)/n < \sqrt{2}$, from which we get $q^k < n\sqrt{2}$. (Notice again that $q^k < 3n$ is a weaker inequality.)

**(c)**$1 < \sigma(n)/q^k < \sigma(q^k)/n < 2$

This implies that $n < q^k < 2n$. (Note that $q^k < 3n$ is a weaker inequality.)

**(d)**$\sigma(n)/q^k \leq 1 < \sigma(q^k)/n$

We obtain $n < q^k$. Also, since $\sigma(q^k) \leq (2/3)n^2$ (see this paper for a proof), we get $\sigma(q^k)/n \leq (2/3)n$. It thus appears difficult to bound $\sigma(q^k)/n$ by a purely numerical limit from above, as $q^k < n^2$ implies (together with Ochem and Rao's lower bound of ${10}^{1500}$ for the magnitude of an odd perfect number) that $n > {10}^{375}$.

It would have been possible to show that $q^k < 3n$ by proving that $\sigma(q^k)/n < 3$ via the inequality

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

so that the biconditionals

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

hold.

This is because of the following equation:

$$\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)).$$

Notice that, if $q^k < 3n$, then since $I(q^k)=\sigma(q^k)/q^k < 5/4$, we have that

$$\frac{2n^2}{\sigma(q^k)} > \frac{8}{5}\cdot\frac{n^2}{q^k} > \frac{8}{5}\cdot\frac{1}{3}\cdot{n} = \frac{8}{15}\cdot{n} > \frac{1}{2}\cdot{{10}^{499}} = {5}\cdot{{10}^{498}}$$

since $q^k n^2 = N > {10}^{1500}$ by Ochem and Rao's result for a lower bound on the magnitude of an odd perfect number.

This will then imply that the quantities

$$\frac{\sigma(N/q^k)}{q^k},$$

$$2n^2 - \sigma(n^2),$$

and

$$\gcd(n^2,\sigma(n^2))$$

are all large.

It would have been possible to show that $q^k < 3n$ by proving that $\sigma(q^k)/n < 3$ via the inequality

$$\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k} < I(q^k)+I(n) < I(q^k)+I(n^2) < 3.$$

However, we know that

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

holds if and only if the biconditionals

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

hold. Notice that, under

**Case (d)**, we have$$\bigg(n < q^k\bigg) \land \bigg(\sigma(n) < \sigma(q^k)\bigg) \land \bigg(\frac{\sigma(n)}{q^k}<\frac{\sigma(q^k)}{n}\bigg)$$

so that the biconditionals

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

hold.

**Motivation****Why do we bother to come up with a proof for $q^k < 3n$?**

This is because of the following equation:

$$\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)).$$

Notice that, if $q^k < 3n$, then since $I(q^k)=\sigma(q^k)/q^k < 5/4$, we have that

$$\frac{2n^2}{\sigma(q^k)} > \frac{8}{5}\cdot\frac{n^2}{q^k} > \frac{8}{5}\cdot\frac{1}{3}\cdot{n} = \frac{8}{15}\cdot{n} > \frac{1}{2}\cdot{{10}^{499}} = {5}\cdot{{10}^{498}}$$

since $q^k n^2 = N > {10}^{1500}$ by Ochem and Rao's result for a lower bound on the magnitude of an odd perfect number.

This will then imply that the quantities

$$\frac{\sigma(N/q^k)}{q^k},$$

$$2n^2 - \sigma(n^2),$$

and

$$\gcd(n^2,\sigma(n^2))$$

are all large.