## 25.9.16

### Some Recent Improvements on Results Contained in the Paper Titled "New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II"

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.

## 8.9.16

### An Elementary Proof (???) of the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers

EUREKA???

Introduction

If $N=q^k n^2$ is an odd perfect number with Euler prime $q$, then the Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

In this blog post, we will attempt to give an elementary proof for this conjecture. We will also derive the possible value(s) for the Euler prime $q$!

We denote the abundancy index of the positive integer $x$ by
$$I(x) = \frac{\sigma(x)}{x}.$$

Preliminaries

First, we restate the following results from these earlier posts in this blog:

Proposition 1
Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$.  Then $k=1$ implies
$$I(n^2) \leq 2 - \frac{5}{3q}.$$

Proposition 2
Suppose that $N=q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$I(n^2) \geq 2 - \frac{5}{3q}$$
implies that $k=1$ and $q=5$.

Main Results

Theorem
Suppose that  $N=q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$I(n^2) \leq 2 - \frac{5}{3q}.$$
Additionally, $I(n^2) = 2 - \frac{5}{3q}$ if and only if $k=1$ and $q=5$.

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

By the contrapositive of Proposition 1,
$$I(n^2) > 2 - \frac{5}{3q} \Rightarrow k \neq 1.$$

By Proposition 2, we have
$$I(n^2) \geq 2 - \frac{5}{3q} \Rightarrow \left\{k = 1 \land q = 5\right\}.$$

Now, assume that
$$I(n^2) \geq 2 - \frac{5}{3q}$$
is true.  Then we have $k=1$ and $q=5$.  In particular, we are sure that $k=1$ must hold.

Consequently,
$$I(n^2) \geq 2 - \frac{5}{3q} \Rightarrow k = 1$$
holds.

Note that
$$I(n^2) > 2 - \frac{5}{3q} \Rightarrow k = 1$$
also holds.

Putting it all together,
$$\left\{I(n^2) > 2 - \frac{5}{3q} \Rightarrow k \neq 1\right\} \land \left\{I(n^2) > 2 - \frac{5}{3q} \Rightarrow k = 1\right\} \Rightarrow I(n^2) \leq 2 - \frac{5}{3q}.$$

Notice that
$$I(n^2) = 2 - \frac{5}{3q} \Rightarrow \left\{k=1 \land q=5\right\}$$
holds.

Thus, it remains to rule out the case
$$I(n^2) < 2 - \frac{5}{3q}.$$

It is easy to show that if
$$I(n^2) < 2 - \frac{5}{3q}$$
is true, then it cannot happen that both $k=1$ and $q=5$ are true.  Thus, either $k \neq 1$ or $q \neq 5$ is true.  This means that the implication
$$k = 1 \Rightarrow q \neq 5$$
is true.

However, Iannucci (1999) [Lemma 12, page 873] proved that
$$q=5$$
implies
$$k=1.$$

Consequently, we have that $q \neq 5$.  (Note that this implies that $q \geq 13$.)

(THIS BLOG POST IS CURRENTLY A WORK IN PROGRESS.)

## 27.7.16

### When $p$ is an odd prime, is $\frac{p+2}{p}$ an outlaw or an index?

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the abundancy index $I$ of $x$ by
$$I(x)=\frac{\sigma(x)}{x}$$
and the deficiency $D$ of $y$ by
$$D(y)=2y - \sigma(y).$$

We first show some preliminary lemmata.

Lemma 1.  If $p$ is an odd prime, and $\frac{p+2}{p}$ is the abundancy index of some integer $n$, then $n$ is deficient.
Proof.  Since $p$ is an odd prime, $p > 2$.  Furthermore, since

$$I(n) = \frac{p+2}{p} = 1 + \frac{2}{p} < 1 + 1 = 2,$$
then $n$ is deficient.          QED

Lemma 2.  If $p$ is odd, then $\gcd(p, p+2)=1$.
Proof.
$$\gcd(p,p+2)=\gcd(p,(p+2)-p)=\gcd(p,2)=1$$
where the last equality uses the fact that $p$ is odd.          QED

Lemma 3.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $D(n) = 2n - \sigma(n) \neq 1$.
Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Now, assume to the contrary that
$$D(n) = 2n - \sigma(n) = 1.$$
Note that this implies
$$\gcd(n,\sigma(n))=\gcd(n,2n-1)=1.$$
Additionally, since $p$ is odd, by Lemma 2 we obtain
$$\gcd(p,p+2)=1.$$
Consequently, since $I(n)=\frac{p+2}{p}$, we get
$$p\sigma(n)=(p+2)n.$$
But $\gcd(p,p+2)=1$ implies that $p \mid n$, and $\gcd(n,\sigma(n))=1$ implies that $n \mid p$.  Hence, it follows that $p=n$.

But since $p$ is an odd prime, $I(n)=I(p)=\frac{p+1}{p}$, which contradicts $I(n)=\frac{p+2}{p}$.

We conclude that $D(n) = 2n - \sigma(n) \neq 1$.          QED

The following lemmas show that $p < n$, if $I(n)=\frac{p+2}{p}$ is true.  In particular, $p < n$ follows from Lemma 4 and $p \mid n$.

Lemma 4.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $I(p) < I(n)$.
Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.

We have
$$I(p)=\frac{p+1}{p}<\frac{p+2}{p}=I(n).$$

In fact, we have
$$I(n) - I(p) > I(n) - \frac{p}{p-1} = \left(\frac{1}{p-1}\right)\cdot\left(\frac{D(n)}{n}\right).$$
QED

Lemma 5.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $\gcd(n,\sigma(n)) \neq 1$.
Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Suppose to the contrary that we have $\gcd(n,\sigma(n))=1$.  Following the proof of Lemma 3, since $\gcd(p,p+2)=1$ (by Lemma 2), we have
$$n = p$$
and
$$\sigma(n) = p+2.$$
Thus,
$$p+1=\sigma(p)=\sigma(n)=p+2.$$

Remarks.
In fact, one can show that, if $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then
$$\sigma(n)=\left(\frac{n}{p}\right)\cdot(p+2)$$
and
$$n=\left(\frac{\sigma(n)}{p+2}\right)\cdot{p}.$$
(Note that
$$\frac{n}{p}$$
and
$$\frac{\sigma(n)}{p+2}$$
are integers because of Lemma 2.)  Consequently,
$$\gcd\left(n,\sigma(n)\right) = \frac{n}{p} = \frac{\sigma(n)}{p+2}.$$
More is actually true.  One can also show that
$$p\left(2n-\sigma(n)\right)=(p-2)n$$
so that
$$D(n)=\left(p-2\right)\cdot\left(\frac{n}{p}\right)=\left(p-2\right)\cdot\left(\frac{\sigma(n)}{p+2}\right) = \left(p-2\right)\cdot\gcd\left(n,\sigma(n)\right).$$
We therefore conclude that
$$\frac{D(n)}{n}=\frac{p-2}{p}=\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$
To facilitate easier reference later, we compute:
$$1+\frac{D(n)}{n}=2\cdot\left(\frac{p-1}{p}\right)=1+\left(\frac{p-2}{p+2}\right)\cdot{I(n)},$$
and
$$2+\frac{D(n)}{n}=\frac{3p-2}{p}=2+\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$

The following lemma is actually a theorem from this preprint.  (We omit the proof.)

Lemma 6.  If $n$ is a positive integer satisfying $D(n) = 2n - \sigma(n) > 1$, then we have the following bounds for the abundancy index of $n$ in terms of the deficiency of $n$:
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}.$$

We now attempt to prove the following "result":

"Conjecture".   If $p$ is an odd prime, then $\frac{p+2}{p}$ is an abundancy outlaw.
"Proof Attempt".
Let $p$ be an odd prime, and assume to the contrary that $I(n)=\frac{p+2}{p}$.

By Lemma 1, $n$ is deficient (so that $D(n) \neq 0$).

By Lemma 3, $D(n) \neq 1$.

Thus, by Lemma 6, we have the bounds
$$\frac{2}{1 + \frac{D(n)}{n}}= \frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}}.$$

By the Remarks, $I(n) = \frac{p+2}{p}$ implies that
$$1+\frac{D(n)}{n}=2\cdot\left(\frac{p-1}{p}\right)=1+\left(\frac{p-2}{p+2}\right)\cdot{I(n)},$$
and
$$2+\frac{D(n)}{n}=\frac{3p-2}{p}=2+\left(\frac{p-2}{p+2}\right)\cdot{I(n)}.$$

We compute:
$$I(n) > \frac{2n}{n+D(n)} = \frac{2}{1 + \frac{D(n)}{n}} = \frac{p}{p-1} > I(p),$$
$$I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}} = \frac{\left(\frac{3p-2}{p}\right)}{\left(\frac{2(p-1)}{p}\right)} = \frac{3p-2}{2p-2} = \frac{4p-4}{2p-2} - \frac{p-2}{2(p-1)} = 2 - {\frac{1}{2}}\cdot\left(\frac{p-2}{p-1}\right),$$
(still no contradictions, per this WolframAlpha computational verification).

It thus seems fruitful to try to improve on the (trivial?) bounds
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}$$
for $n$ satisfying $D(n)>1$.

## 20.7.16

### If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$ implies that $k=1$ and $q=5$.

(Note:  This blog post is essentially an elucidation of the answer to this MSE question.)

Here, we prove the following proposition.

Theorem.  If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$$
implies that
$$k=1$$
and
$$q=5.$$

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

Since $N$ is perfect, we have
$$2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \left(\frac{q^{k+1} - 1}{q - 1}\right)\cdot\sigma(n^2).$$
It follows that
$$\frac{\sigma(n^2)}{n^2} = \frac{2{q^k}\left(q - 1\right)}{q^{k+1} - 1} = \frac{2q^{k+1} - 2q^k}{q^{k+1} - 1} = \frac{2q^{k+1} - 2}{q^{k+1} - 1} - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right)$$
$$= 2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right).$$

By assumption, we have
$$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}.$$

This inequality is equivalent to
$$2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right) \geq 2 - \frac{5}{3q},$$
which, in turn, is equivalent to
$$6q\left(q^k - 1\right) \leq 5(q^{k+1} - 1).$$
This last inequality simplifies to
$$q^{k+1} - 6q + 5 \leq 0,$$
which cannot be true when $k > 1$.  Thus, we know that $k=1$.

Hence, we have
$$q^2 - 6q + 5 \leq 0.$$
But this inequality implies that
$$1 \leq q \leq 5,$$
which, together with $q \geq 5$, implies that $q=5$.

QED.

## 18.7.16

### On conditions equivalent to the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

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

The Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

From the equation $\sigma(N)=2N$, we get that
$$\left(q^k + \sigma(q^{k-1})\right)\sigma(n^2) = \sigma(q^k)\sigma(n^2) = 2{q^k}{n^2}$$
so that we obtain
$$\frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})} = 2n^2 - \sigma(n^2).$$
(Note that $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of $n^2$, and that
$$I(x) = \frac{\sigma(x)}{x}$$
is the abundancy index of $x$.)

Now suppose that
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
Trivially, we know that
$$\frac{\sigma(n^2)}{q} \mid \sigma(n^2).$$
Thus, we have
$$\frac{\sigma(n^2)}{q} \mid \left(2n^2 - \sigma(n^2)\right) = \frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})}.$$
This implies that $I(q^{k-1})$ is an integer;  in other words, $k=1$.

The other direction
$$k=1 \Longrightarrow \frac{\sigma(n^2)}{q} \mid n^2$$
is trivial.

We therefore have the following lemma.

Lemma 1.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longleftrightarrow \frac{\sigma(n^2)}{q} \mid n^2.$$

Now assume that $k=1$.

By Lemma 1, we have
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
This implies that there exists an integer $d$ such that
$$n^2 = d \cdot \left(\frac{\sigma(n^2)}{q}\right).$$

Note that, from the equation $\sigma(N)=2N$, we obtain (upon setting $k=1$)
$$(q+1)\sigma(n^2) = \sigma(q)\sigma(n^2) = 2q{n^2}$$
from which we get
$$d = \frac{n^2}{\frac{\sigma(n^2)}{q}} = \frac{q + 1}{2}.$$

Notice that, when $k=1$, we can derive
$$\frac{5}{3} \leq I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} < 2$$
so that we have
$$\frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

We state this latest result as our second lemma.

Lemma 2.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longrightarrow \frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

Note that, when $k=1$, we have
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2q+2}{q+1} - \frac{2}{q+1} = 2 - \frac{1}{\frac{q+1}{2}} = 2 - \frac{1}{d}$$

However, notice that we know, by Lemma 2,
$$\frac{q}{2} < d \leq \frac{3q}{5} \Longrightarrow \frac{5}{3q} \leq \frac{1}{d} < \frac{2}{q} \Longrightarrow 2 - \frac{2}{q} < 2 - \frac{1}{d} = I(n^2) \leq 2 - \frac{5}{3q}.$$

$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1}$$
this WolframAlpha computation validates that, indeed,
$$\frac{2q}{q+1} \leq 2 - \frac{5}{3q}$$
implies
$$q \geq 5.$$

Iterating the computations, we obtain
$$\frac{2 - \frac{2}{q}}{q} < \frac{1}{d} = \frac{I(n^2)}{q} \leq \frac{2 - \frac{5}{3q}}{q}$$
so that
$$2 - \frac{2 - \frac{5}{3q}}{q} \leq 2 - \frac{1}{d} = I(n^2) < 2 - \frac{2 - \frac{2}{q}}{q}.$$
But
$$2 - \frac{2 - \frac{5}{3q}}{q} = \frac{6q^2 - 6q + 5}{3q^2} \geq \frac{5}{3}$$
$$2 - \frac{2 - \frac{2}{q}}{q} = \frac{2q^2 - 2q + 2}{q^2} < 2.$$

Lastly, note that, after multiplying throughout
$$\frac{6q^2 - 6q + 5}{3q^2} \leq I(n^2) < \frac{2q^2 - 2q + 2}{q^2}$$
by $I(q) = \frac{q+1}{q}$, and asking WolframAlpha to solve the resulting inequality
$$\left(\frac{q+1}{q}\right)\cdot\left(\frac{6q^2 - 6q + 5}{3q^2}\right) \leq 2 = I(q)I(n^2) = I(qn^2) < \left(\frac{q+1}{q}\right)\cdot\left(\frac{2q^2 - 2q + 2}{q^2}\right)$$
we obtain
$$q \geq 5.$$

This blog post is currently a WORK IN PROGRESS.

## 14.7.16

### The Abundancy Index of Divisors of Even Almost Perfect Numbers That Are Not Powers of Two

Antalan and Tagle showed that, if $M \neq 2^t$ ($t \geq 1$) and $\sigma(M) = 2M - 1$ (where $\sigma = \sigma_{1}$ is the (classical) sum-of-divisors function), then $M$ takes the form
$$M = 2^r b^2$$
where $b > 1$ is an odd composite.

In this preprint, Antalan and Dris proved the following bounds:

Theorem A.  If $2^r b^2$ is an almost perfect number with $r \geq 1$, $\gcd(2, b) = 1$ and $b > 1$, then
$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$

In this blog post, we will prove the following series of inequalities:

Theorem B.  If $2^r b^2$ is an almost perfect number with $r \geq 1$, $\gcd(2, b) = 1$ and $b > 1$, then
$$1 < \frac{b}{\sigma(2^r)} < \frac{\sigma(b)}{\sigma(2^r)} < \frac{b}{2^r} < \frac{\sigma(b)}{2^r} < \frac{b^2}{\sigma(2^r)} < \frac{\sigma(b^2)}{\sigma(2^r)} < \frac{b^2}{2^r} < \frac{\sigma(b^2)}{2^r}.$$

In other words, taking reciprocals, we should obtain:
$$\frac{2^r}{\sigma(b^2)} < \frac{2^r}{b^2} < \frac{\sigma(2^r)}{\sigma(b^2)} < \frac{\sigma(2^r)}{b^2} < \frac{2^r}{\sigma(b)} < \frac{2^r}{b} < \frac{\sigma(2^r)}{\sigma(b)} < \frac{\sigma(2^r)}{b} < 1.$$

Proof.  Since
$$\frac{\sigma(b)}{b} < \frac{\sigma(b^2)}{b^2} < \frac{4}{3},$$
by Theorem A, it suffices to prove the inequality in the middle, namely
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$

Suppose to the contrary that
$$\frac{\sigma(b^2)}{\sigma(2^r)} \leq \frac{\sigma(b)}{2^r}.$$

It follows that
$$\frac{\sigma(b^2)}{\sigma(b)} \leq \frac{\sigma(2^r)}{2^r} < 2,$$
by Theorem A.  Since $\frac{\sigma(b)}{b} < \frac{4}{3}$ (again, by Theorem A), it follows that
$$\frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b}\right) < \left(\frac{b}{\sigma(b)}\right)\cdot\left(\frac{\sigma(b^2)}{b}\right) = \frac{\sigma(b^2)}{\sigma(b)} < 2.$$
Dividing through by $1 < b$, we obtain
$$\frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b^2}\right) < \frac{2}{b}.$$
By Theorem A, we get
$$\frac{3}{4} < \frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b^2}\right) < \frac{2}{b},$$
which finally implies that
$$b < \frac{8}{3}.$$
Since $1 < b$, this forces $b = 2$A contradiction, as $b$ has to be odd.

We therefore conclude that
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$

The proof for
$$\frac{\sigma(b)}{2^r} < \frac{b^2}{\sigma(2^r)}$$
is very similar, and we are done.