This hyperlink redirects to a Scribd document.

Here is the abstract:

On a Conjecture of Dris Regarding Odd Perfect Numbers

Dris conjectured (in his M.Sc. thesis) that the inequality $q^k < n$ always holds, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form. In this short note, we show that either of the two conditions $n < q^k$ or $\sigma(q)/n < \sigma(n)/q$ holds. This is achieved by first proving that $\sigma(q)/n \neq \sigma(n)/q^k$, where $\sigma(x)$ is the sum of the divisors of $x$. Hence, we show that the inequalities $q < n < q^k$ hold in four out of a total of six cases.

## 12.11.14

## 9.11.13

### Eureka Moments A Few Weeks Before My Birthday

Eureka Moment #1: The Abundancy Index of Divisors of Odd Perfect Numbers - Part III (Already uploaded to arXiv, pending a hold from an arXiv administrator) - This paper contains, among other things, a proof for the inequality q < n.

Eureka Moment #2: A Short Proof for Sorli's Conjecture on Odd Perfect Numbers (Also available via arXiv at http://arxiv.org/abs/1308.2156) - This paper contains a simple (and short) proof for Descartes' / Frenicle's / Sorli's conjecture on odd perfect numbers.

Together, these will then imply the conjecture q^k < n from this

M. Sc. thesis.

Related Stuff:

Euclid-Euler Heuristics for (Odd) Perfect Numbers

New Results for Sorli's Conjecture on Odd Perfect Numbers

The Abundancy Index of Divisors of Odd Perfect Numbers - Part II

The Abundancy Index of Divisors of Odd Perfect Numbers

Solving the Odd Perfect Number Problem: Some Old and New Approaches

OEIS sequence A228059 by Tony D. Noe, Aug 14 2013

Math Questions in MSE and MO:

#
On odd perfect numbers

http://math.stackexchange.com/questions/548528

Eureka Moment #2: A Short Proof for Sorli's Conjecture on Odd Perfect Numbers (Also available via arXiv at http://arxiv.org/abs/1308.2156) - This paper contains a simple (and short) proof for Descartes' / Frenicle's / Sorli's conjecture on odd perfect numbers.

Together, these will then imply the conjecture q^k < n from this

M. Sc. thesis.

Related Stuff:

Euclid-Euler Heuristics for (Odd) Perfect Numbers

New Results for Sorli's Conjecture on Odd Perfect Numbers

The Abundancy Index of Divisors of Odd Perfect Numbers - Part II

The Abundancy Index of Divisors of Odd Perfect Numbers

Solving the Odd Perfect Number Problem: Some Old and New Approaches

OEIS sequence A228059 by Tony D. Noe, Aug 14 2013

Math Questions in MSE and MO:

#
On odd perfect numbers N given in the Eulerian form N=qkn2

http://math.stackexchange.com/questions/548528# On J. T. Condict's Senior Thesis on Odd Perfect Numbers

http://mathoverflow.net/questions/83161## 11.9.13

### OPN Research - September 2013

In this post, we consider the problem of deriving bounds for the quantity

$$\frac{q^k}{n^2},$$

in terms of $n$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.

We will be using the bounds obtained in the post from July 2013.

Case I: $q^k < n$

Case I-A: $k = 1 \Longrightarrow q = q^k < n$

We have the chain of inequalities

$$\frac{1}{2}n < q^k < n < 2{q^k}$$

which implies that

$$\frac{1}{2n} < \frac{q^k}{n^2} < \frac{1}{n}.$$

Case I-B: $k \neq 1 \Longrightarrow q < q^k < n$

We have the chain of inequalities

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

which implies that

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

Case II: $n < q^k$

Case II-A:$k = 1 \Longrightarrow n < q = q^k$

We have the chain of inequalities

$$\frac{q^k}{\sqrt{3}} < n < \sqrt[4]{\frac{108}{125}}{q^k} < \sqrt[4]{\frac{108}{125}}{\sqrt{3}}{n}$$

which implies that

$$\sqrt[4]{\frac{125}{108}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{\sqrt{3}}{n}.$$

Case II-B:$k \neq 1 \Longrightarrow q < n < q^k$

We have the chain of inequalities

$$\frac{q^k}{2} < n < \sqrt[4]{\frac{125}{128}}{q^k} < 2{\sqrt[4]{\frac{125}{128}}}{n}$$

which implies that

$$\sqrt[4]{\frac{128}{125}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{2}{n}.$$

We have therefore proven the following theorem (as we already know, from the theorems $q^k < n^2$ [Dris, 2012] and $N = {q^k}{n^2} > {10}^{1500}$ [Ochem and Rao, 2012], that $n > {10}^{375}$):

Theorem: If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $q^k = \circ(n^2)$.

$$\frac{q^k}{n^2},$$

in terms of $n$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.

We will be using the bounds obtained in the post from July 2013.

Case I: $q^k < n$

Case I-A: $k = 1 \Longrightarrow q = q^k < n$

We have the chain of inequalities

$$\frac{1}{2}n < q^k < n < 2{q^k}$$

which implies that

$$\frac{1}{2n} < \frac{q^k}{n^2} < \frac{1}{n}.$$

Case I-B: $k \neq 1 \Longrightarrow q < q^k < n$

We have the chain of inequalities

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

which implies that

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

Case II: $n < q^k$

Case II-A:$k = 1 \Longrightarrow n < q = q^k$

We have the chain of inequalities

$$\frac{q^k}{\sqrt{3}} < n < \sqrt[4]{\frac{108}{125}}{q^k} < \sqrt[4]{\frac{108}{125}}{\sqrt{3}}{n}$$

which implies that

$$\sqrt[4]{\frac{125}{108}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{\sqrt{3}}{n}.$$

Case II-B:$k \neq 1 \Longrightarrow q < n < q^k$

We have the chain of inequalities

$$\frac{q^k}{2} < n < \sqrt[4]{\frac{125}{128}}{q^k} < 2{\sqrt[4]{\frac{125}{128}}}{n}$$

which implies that

$$\sqrt[4]{\frac{128}{125}}\frac{1}{n} < \frac{q^k}{n^2} < \frac{2}{n}.$$

We have therefore proven the following theorem (as we already know, from the theorems $q^k < n^2$ [Dris, 2012] and $N = {q^k}{n^2} > {10}^{1500}$ [Ochem and Rao, 2012], that $n > {10}^{375}$):

Theorem: If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then $q^k = \circ(n^2)$.

## 9.8.13

## 2.7.13

### OPN Research - July 2013

If $N = {q^k}{n^2}$ is an odd perfect number (OPN) given in Eulerian form, then since $\gcd(q, n) = \gcd(q^k, n) = 1$, we either have $q^k < n$ or $n < q^k$.

As of

If $k = 1$, then

$$\frac{1}{2} < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \frac{\sigma(q)}{n} < 1$$

$$1 < \frac{n}{q} < \frac{\sigma(n)}{\sigma(q)} < \frac{\sigma(n)}{q} < 4$$

Otherwise, if $k > 1$ then

$$\frac{1}{2} < \frac{\sigma(q^k)}{\sigma(n)} < \sqrt{\frac{125}{128}}$$

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

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

$$\sqrt{\frac{1}{2}} < \frac{q^k}{n} < 1 < \frac{n}{q^k} < \sqrt{2}$$

$$\sqrt{\frac{128}{125}} < \frac{\sigma(n)}{\sigma(q^k)} < \frac{\sigma(n)}{q^k} < 2$$

$$1 < \frac{\sigma(q^k)}{n} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(n)}{q^k} < 2$$

If $k = 1$, then

$$\sqrt{\frac{1}{3}} < \frac{n}{q} < \frac{\sigma(n)}{\sigma(q)} < \frac{\sigma(n)}{q} < 1$$

$$1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \sqrt{3}$$

$$1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \frac{\sigma(q)}{n} < 2\sqrt{3}$$

$$\sqrt{\frac{1}{3}} < \frac{n}{q} < \sqrt[4]{\frac{108}{125}} < \sqrt[4]{\frac{125}{108}} < \frac{q}{n} < \sqrt{3}$$

Otherwise, if $k > 1$ then

$$\frac{1}{2} < \frac{n}{q^k} < \sqrt[4]{\frac{125}{128}} < \sqrt[4]{\frac{128}{125}} < \frac{q^k}{n} < 2$$

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

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

$$\frac{4\left(1 + \sqrt{\frac{8}{5}}\right)}{13} < \frac{\sigma(n)}{\sigma(q^k)} < 1 < \frac{\sigma(q^k)}{\sigma(n)} < \frac{13}{4\left(1 + \sqrt{\frac{8}{5}}\right)}$$

$$1 < \frac{\sigma(n)}{q^k} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(q^k)}{n} < 2$$

As of

*July 2013*, the author is able to obtain the following bounds:**$q^k < n$**__Case 1:__If $k = 1$, then

$$\frac{1}{2} < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \frac{\sigma(q)}{n} < 1$$

$$1 < \frac{n}{q} < \frac{\sigma(n)}{\sigma(q)} < \frac{\sigma(n)}{q} < 4$$

Otherwise, if $k > 1$ then

$$\frac{1}{2} < \frac{\sigma(q^k)}{\sigma(n)} < \sqrt{\frac{125}{128}}$$

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

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

$$\sqrt{\frac{1}{2}} < \frac{q^k}{n} < 1 < \frac{n}{q^k} < \sqrt{2}$$

$$\sqrt{\frac{128}{125}} < \frac{\sigma(n)}{\sigma(q^k)} < \frac{\sigma(n)}{q^k} < 2$$

$$1 < \frac{\sigma(q^k)}{n} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(n)}{q^k} < 2$$

**$n < q^k$**__Case 2:__If $k = 1$, then

$$\sqrt{\frac{1}{3}} < \frac{n}{q} < \frac{\sigma(n)}{\sigma(q)} < \frac{\sigma(n)}{q} < 1$$

$$1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \sqrt{3}$$

$$1 < \frac{\sigma(q)}{\sigma(n)} < \frac{q}{n} < \frac{\sigma(q)}{n} < 2\sqrt{3}$$

$$\sqrt{\frac{1}{3}} < \frac{n}{q} < \sqrt[4]{\frac{108}{125}} < \sqrt[4]{\frac{125}{108}} < \frac{q}{n} < \sqrt{3}$$

Otherwise, if $k > 1$ then

$$\frac{1}{2} < \frac{n}{q^k} < \sqrt[4]{\frac{125}{128}} < \sqrt[4]{\frac{128}{125}} < \frac{q^k}{n} < 2$$

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

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

$$\frac{4\left(1 + \sqrt{\frac{8}{5}}\right)}{13} < \frac{\sigma(n)}{\sigma(q^k)} < 1 < \frac{\sigma(q^k)}{\sigma(n)} < \frac{13}{4\left(1 + \sqrt{\frac{8}{5}}\right)}$$

$$1 < \frac{\sigma(n)}{q^k} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(q^k)}{n} < 2$$

## 28.6.13

### From Yahoo! SHE - Dirty Little Secrets: 8 Disturbing Ingredients Hidden in Your Food

A lot has been said lately about knowing what is in our foods when grocery shopping. From dyes to chemicals, food manufacturers are filling our bellies with who knows what. For instance, did you know that the manufacturing process for those popular Greek yogurts produces millions of pounds of toxic acid waste annually? It really is such a lovely image. Well, at least the acid waste is not in the yogurt. The truth is, there are several "ingredients" hidden in the foods we eat every day.

- By Jessica Cohen

**Cochineal Extract**

You may have been warned to stay away from red dyes, but do you know what they are made of? Much of the red dye used in red and pink food products are made from cochineal extract, which is basically the bodies of crushed-up beetles. Doesn't that make you just want to run out and purchase a giant pack of licorice?

**Animal Tissue**

Gelatin is used in the majority of gummy products such as fruit chews and marshmallows. Gelatin is a pork derivative. That sounds harmless enough, right? What if I tell you that gelatin is made from boiled animal connective tissue? Yummy!

**Castoreum**

Castoreum is a "natural" flavor enhancer used in items such as ice creams. However, it is enhancing the flavor of your favorite scoop with the secretions of beavers. Yes, you read that right. The Food and Drug Administration regards castoreum extract as safe, so at least there's that.

**Sawdust**

The newest food scandal to come to the forefront is sawdust. It is used to prevent stickiness in items such as shredded cheeses and is more commonly known as cellulose. If you have ever purchased a natural, organic shredded cheese and noticed that it was a little clumpier than brand name packages, now you may know why. You're welcome.

**Propylene Glycol**

Another preservative found in our foods (and in many hair and body products) is propylene glycol, which is also used to make anti-freeze. Experts say that propylene glycol alters the structure of the skin by helping other chemicals reach your bloodstream. In foods it is used to make biscuits, cakes, sweets, and other baked goods.

**L-Cysteine**

Many of those baked goods and bread products you see on the shelves of the supermarket are also created using an ingredient called L-Cysteine. L-Cysteine is actually made from fibers which commonly come from human hair. Occasionally it can be derived from duck feathers. Quack.

**Brominated Vegetable Oil**

Be careful of those sports drinks you use to stay hydrated. They are often made with brominated vegetable oil (BVO), which is added to keep the ingredients from separating and contains bromine, an element also found in flame retardants. Experts worry about this ingredient which builds up in the body and may compete with iodine for receptor sites. It has been removed or banned from food and drinks in Europe and Japan but not here.

## 8.6.13

### OPN Research - June 2013

Version 5 of http://arxiv.org/abs/1302.5991 (New Results for Sorli's Conjecture on Odd Perfect Numbers) and version 3 of http://arxiv.org/abs/1303.2329 (New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II) are set to appear in the arXiv by Tue, 11 Jun 2013 00:00:00 GMT.

My profuse thanks to Math@StackExchange user Tharsis for "[admiring my] work on odd perfect numbers". I hope you like these revisions, too!