In this blog post, we present some recent improvements on the results contained in the paper titled "**New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II**".

We refer the interested reader to the following arXiv preprint for the latest version of this paper.

**Preliminaries**

Denote the sum of the divisors of the positive integer $x$ by $\sigma(x)$. For example, $\sigma(6)=1+2+3+6=12$. A number $M$ is called *perfect* when $\sigma(M)=2M$. Euler showed that an *odd perfect number* must have the form $q^k n^2$ where $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.

Denote the **abundancy index*** ***I** by

$$I(x)=\sigma(x)/x.$$

Note that, if $M$ is perfect, then

$$I(M)=\sigma(M)/M=2M/M=2.$$

Recall that $\sigma$ is *weakly multiplicative*, that is, it satisfies

$$\sigma(yz)=\sigma(y)\sigma(z)$$

if $\gcd(y,z)=1$. In particular, note that the abundancy index $I$ is also weakly multiplicative, so that if $q^k n^2$ is an odd perfect number, then we have the equation

$$I(q^k)I(n^2) = I({q^k}{n^2}) = 2.$$

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

Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds. Notice that $\gcd(q, n) = 1$ implies that $q \neq n$ and $q^k \neq n$. Also, we have the bounds $1 < I(qn) \leq I({q^k}n) < 2$, which follows from the fact that ${q^k}n$ is *deficient*, being a proper divisor of the perfect number $N = q^k n^2$. It follows that we have the following inequations:

$$\sigma(q)/n \neq \sigma(n)/q$$

and

$$\sigma(q^k)/n \neq \sigma(n)/q^k.$$

We will be using the following ("general") bounds from this publication:
$$1 < 1 + \frac{1}{q} = I(q) \leq I(q^k) < \frac{5}{4} < \sqrt[3]{2} < \sqrt{\frac{8}{5}} < I(n) < I(n^2) < 2,$$

or when $k=1$ is known to hold, we will utilize the ("slightly") stronger bounds

$$1 < I(q) = 1 + \frac{1}{q} \leq \frac{6}{5} < \sqrt[3]{2} < \sqrt{\frac{5}{3}} < I(n) < I(n^2) < 2.$$

(Hereinafter, both sets of bounds will be abbreviated as $I(q^k) < \sqrt[3]{2} < I(n)$.)

**Improved Lower Bound for** $I(n)$

The following result was communicated to Arnie Dris (the author of this blog post) by *Pascal Ochem* via e-mail, on April 17 2013:

**THEOREM P**

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

In particular, note that

$$\left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} > \sqrt{2}.$$

A proof of this claimed lower bound for $I(n)$ was likewise communicated by Ochem to Dris (in the same e-mail), and the details of the proof are presented in this preprint.

We will utilize **Theorem P** in getting improvements on the results contained in this preprint.

**Lemmas**

In light of this more recent preprint, we have the following unconditional result:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, then $q^k < n$ is true if and only if the biconditional

$$q^k < n \Longleftrightarrow \left(q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \sigma(q^k)/n < \sigma(n)/q^k\right)$$

holds.

From **Remark 2.2 (page 2)** of this preprint, we have the biconditional
$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longleftrightarrow \left(\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right).$$

Here is a brief outline of the proof (taken from this preprint, to appear in JANTA):

First, we show that

$$\frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k} \Longrightarrow \left(\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right).$$

Since we have the inequality

$$\frac{\sigma(q^k)}{q^k} = I(q^k) < \sqrt[3]{2} < I(n) = \frac{\sigma(n)}{n},$$

it follows that

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

By assumption, we have

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

which is equivalent to

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

Putting everything together, we obtain

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

Now consider the product

$$\left(\frac{\sigma(q^k)}{\sigma(n)} - \frac{q^k}{n}\right)\cdot\left(\frac{\sigma(n)}{\sigma(q^k)} - \frac{q^k}{n}\right),$$

which should be negative. Thus, we have

$$0 > \left(\frac{\sigma(q^k)}{\sigma(n)} - \frac{q^k}{n}\right)\cdot\left(\frac{\sigma(n)}{\sigma(q^k)} - \frac{q^k}{n}\right) = 1 + \left(\frac{q^k}{n}\right)^2 - {\frac{q^k}{n}}\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right),$$

which implies that

$${\frac{q^k}{n}}\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right) > 1 + \left(\frac{q^k}{n}\right)^2.$$

Finally, we obtain

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

The proof for the other direction

$$\left(\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right) \Longrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

is very similar, and is left as an exercise for the interested reader.

**Main Results**

By considerations in the preceding paragraphs, we know that $q^k \neq n$. We therefore need to cover two cases:

__Case A__ $q^k < n$

This means that the biconditionals

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longrightarrow \sigma(q^k)/n < \sigma(n)/q^k$$

and

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

both hold.

Since $q^k < n$ is true by assumption, all of the inequalities

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

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

and

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

simultaneously hold. Note further that we also have

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

It follows that

$$\left(\frac{q^k}{n}\right)^2 < \frac{I(n)}{I(q^k)} < 2.$$

Now, note that

$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < \sqrt{2}$$

and

$$\frac{1}{\sqrt{2}} < \frac{n}{q^k} < \frac{\sigma(n)}{\sigma(q^k)}.$$

Taking square roots, we get

$$\sqrt{\frac{\sigma(q^k)}{\sigma(n)}} < \sqrt[4]{2}$$

and

$$\frac{1}{\sqrt[4]{2}} < \sqrt{\frac{\sigma(n)}{\sigma(q^k)}}.$$

Subtracting, we obtain

$$\sqrt{\frac{\sigma(q^k)}{\sigma(n)}} - \sqrt{\frac{\sigma(n)}{\sigma(q^k)}} < \sqrt[4]{2} - \frac{1}{\sqrt[4]{2}}.$$

Squaring both sides, we now have

$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} - 2 = \left(\sqrt{\frac{\sigma(q^k)}{\sigma(n)}} - \sqrt{\frac{\sigma(n)}{\sigma(q^k)}}\right)^2 < \left(\sqrt[4]{2} - \frac{1}{\sqrt[4]{2}}\right)^2 = \sqrt{2} + \frac{1}{\sqrt{2}} - 2,$$

so that

$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}.$$

Note the rational approximation

$$\frac{3}{\sqrt{2}} \approx 2.12132.$$

Now, because of the fact that $\sigma(q^k) \neq n$ (since $\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4$ and $n$ is odd), we have two further sub-cases to analyze:

__Sub-case A-1__ $q^k < \sigma(q^k) < n < \sigma(n)$

Under this **Sub-case A-1**, we have

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

from which we obtain the lower bound

$$I(q^k)I(n) + 1 < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k},$$

which we get by considering the *negative* product

$$\left(\sigma(q^k)/n - 1\right)\left(\sigma(n)/q^k - 1\right).$$

Thus, we have

$$1 + \left(\frac{8}{5}\right)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(q^k)I(n) + 1 < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}.$$

This validates the (*trivial*!) inequality

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

We now compute a lower bound for the quantity

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

Note that a trivial lower bound is given by the Arithmetic Mean-Geometric Mean Inequality, as follows:

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

since $\sigma(q^k) < \sigma(n)$. Note further that $\sigma(n)/\sigma(q^k) > 1$.

We now attempt to get an improved upper bound for $\sigma(q^k)/\sigma(n)$:

$$\sigma(q^k)/n < 1 < \sigma(q^k)/q^k < 5/4 < (8/5)^{\ln(4/3)/\ln(13/9)} < \sigma(n)/n < \sigma(n)/q^k$$

$$\frac{\sigma(q^k)}{\sigma(n)}=\frac{\sigma(q^k)/n + \sigma(q^k)/q^k}{\sigma(n)/n + \sigma(n)/q^k} < \frac{1 + (5/4)}{2\cdot{(8/5)^{\ln(4/3)/\ln(13/9)}}} = \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 0.778867$$

This translates to the improved lower bound

$$\frac{\sigma(n)}{\sigma(q^k)} > \frac{8}{9}\cdot{\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}} \approx 1.283916$$

for $\sigma(n)/\sigma(q^k)$.

From another perspective, since

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

and because

$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 0.778867,$$

then we get

$$\frac{\sigma(n)}{\sigma(q^k)} > 2 - \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 1.2211.$$

(Notice that, in general, multiplicative estimates are better/tighter than additive estimates.)

Let

$$\theta_1 := \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} \approx 0.778867.$$

Then

$$\frac{1}{\theta_1} = \frac{8}{9}\cdot{\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}} \approx 1.283916.$$

Note that the product

$$\bigg(\frac{\sigma(q^k)}{\sigma(n)} - \theta_1\bigg)\bigg(\frac{\sigma(n)}{\sigma(q^k)} - \theta_1\bigg)$$

is negative since $\sigma(q^k)/\sigma(n) < \theta_1 < \sigma(n)/\sigma(q^k)$. In particular, we finally get the (improved) lower bound

$$1 + {\theta_1}^2 < {\theta_1}\cdot\bigg(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\bigg),$$

which implies that

$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} > \theta_1 + \frac{1}{\theta_1} \approx 2.062783179.$$

We now try to derive an improved estimate for ${q^k}/n$.

$$\frac{q^k}{n}=\frac{{q^k}/\sigma(q^k) + {q^k}/\sigma(n)}{n/\sigma(q^k) + n/\sigma(n)}<\frac{1+(5/8)^{\ln(4/3)/\ln(13/9)}}{1+(1/2)}=\frac{2}{3}\cdot\bigg(1 + \left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}\bigg) \approx 1.1282,$$

which is trivial as compared to $q^k < n$.

Lastly, since we have

$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{3}{\sqrt{2}},$$

and because $n/{q^k} > 1$, we obtain

$$\frac{q^k}{n} < \frac{3}{\sqrt{2}} - 1 = \frac{3\sqrt{2} - 2}{2} \approx 1.12132,$$

which again is trivial in comparison to $q^k < n$. Hence, it appears that it will not be possible to improve on the upper bound

$$\frac{q^k}{n} + \frac{n}{q^k} < \frac{3}{\sqrt{2}}$$

using our current method.

Here is a summary of the results that we have obtained for

Sub-case A-1:

$$2 < \frac{q^k}{n} + \frac{n}{q^k} < \frac{3}{\sqrt{2}} \approx 2.12132$$

$$2.062783179 \approx \frac{9}{8}\cdot{\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}} + \frac{8}{9}\cdot{\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{3}{\sqrt{2}}.$$

__Sub-case A-2__ $q^k < n < \sigma(q^k) < \sigma(n)$

Note that **Sub-case A-2** implies that $k>1$.

Under this **Sub-case A-2**, we have

$$1 <\sigma(q^k)/n < \sigma(q^k)/q^k < 5/4 < (8/5)^{\ln(4/3)/\ln(13/9)} < \sigma(n)/n < \sigma(n)/q^k < 2,$$
from which we obtain
$$\frac{1}{2}=\frac{1+1}{2+2}<\frac{\sigma(q^k)}{\sigma(n)}=\frac{\sigma(q^k)/n + \sigma(q^k)/q^k}{\sigma(n)/n + \sigma(n)/q^k}<\frac{(5/4)+(5/4)}{2\cdot{(8/5)^{\ln(4/3)/\ln(13/9)}}}=\frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)} \approx 0.8654.$$
It follows that
$$2 > \frac{\sigma(n)}{\sigma(q^k)}>\frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)} \approx 1.1555.$$

Let
$$\theta_2 := \frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)}$$
and
$$\frac{1}{\theta_2} = \frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)}.$$
Since the product
$$\bigg(\frac{\sigma(q^k)}{\sigma(n)} - \theta_2\bigg)\cdot\bigg(\frac{\sigma(n)}{\sigma(q^k)} - \theta_2\bigg)$$
is negative, we have
$$1 + {\theta_2}^2 < {\theta_2}\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right),$$
which implies that
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} > \theta_2 + \frac{1}{\theta_2} = \frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)} + \frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)} \approx 2.02093.$$

Similarly, since the product
$$\bigg(\frac{\sigma(q^k)}{\sigma(n)} - 2\bigg)\cdot\bigg(\frac{\sigma(n)}{\sigma(q^k)} - 2\bigg)$$
is positive, we obtain
$$1 + 4 > 2\cdot\left(\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)}\right),$$
which implies that
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{5}{2}.$$
This last upper bound is trivial when compared to the earlier bound
$$\frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{3}{\sqrt{2}} \approx 2.12132.$$
Here is a summary of the results that we have obtained for
Sub-case A-2:
$$2 < \frac{q^k}{n}+\frac{n}{q^k} < \frac{3}{\sqrt{2}} \approx 2.12132$$
$$2.02093 \approx \frac{5}{4}\cdot\left(\frac{5}{8}\right)^{\ln(4/3)/\ln(13/9)} + \frac{4}{5}\cdot\left(\frac{8}{5}\right)^{\ln(4/3)/\ln(13/9)} < \frac{\sigma(q^k)}{\sigma(n)} + \frac{\sigma(n)}{\sigma(q^k)} < \frac{3}{\sqrt{2}} \approx 2.12132$$

__Case B__ $n < q^k$

It remains to consider three remaining sub-cases under **Case B**:

__Sub-case B-1__ $n < q^k < \sigma(q^k) \leq \sigma(n)$

__Sub-case B-2__ $n < q^k \leq \sigma(n) < \sigma(q^k)$

__Sub-case B-3__ $n < \sigma(n) < q^k < \sigma(q^k)$

We remark that work is underway to fill in the gaps in Brown's proof for $q^k < n$. (The interested reader is hereby referred to this preprint for more details.)

**THIS POST IS CURRENTLY A WORK IN PROGRESS.**