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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 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 $q^k = \circ(n^2)$.

9.8.13

August 2013 - On Some Recent Math@StackExchange Questions


Does the following inequality hold if and only if N is an odd deficient number?

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 July 2013, the author is able to obtain the following bounds:

Case 1:  $q^k < n$

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

$$\frac{1}{2} < \frac{\sigma(q^k)}{\sigma(n)} < \sqrt{\frac{125}{128}}$$
$$\frac{1}{2} < \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < 1$$
$$1 < \frac{\sigma(q^k)}{n} < \frac{n}{q^k} < \frac{\sigma(n)}{\sigma(q^k)} < \frac{\sigma(n)}{q^k} < 2$$
$$\sqrt{\frac{1}{2}} < \frac{q^k}{n} < 1 < \frac{n}{q^k} < \sqrt{2}$$
$$\sqrt{\frac{128}{125}} < \frac{\sigma(n)}{\sigma(q^k)} < \frac{\sigma(n)}{q^k} < 2$$
$$1 < \frac{\sigma(q^k)}{n} < I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n) < \frac{\sigma(n)}{q^k} < 2$$

Case 2:  $n < q^k$

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!  

5.6.13

OPN Research - May 2013

First, we prove the following result:

Lemma 1.  Let $N = {q^k}{n^2}$ be an odd perfect number (OPN) given in Eulerian form.  If $k = 1$ implies $\sigma(n) < q^k$, then $k = 1$ if and only if $n < q$.

Proof.
Suppose $N = {q^k}{n^2}$ is an odd perfect number (OPN) given in Eulerian form.

If $k = 1$ implies $\sigma(n) < q^k$, then we have:

$k = 1 \Longrightarrow \sigma(n) < q^k \Longrightarrow n < \sigma(n) < q^k = q.$

Thus, we have: $k = 1 \Longrightarrow n < q$.

But we also have: $n < q \Longrightarrow k = 1$ [Dris, 2012].

Therefore, $k = 1$ if and only if $n < q$.
QED.

Remark 1.  Notice that, in fact if we have $n < q^k$, then $k = 1 \Longleftrightarrow n < q$ would follow.  Furthermore, note that if $n < q$, then we have $n < q^k$ (since it is true that $k \geq 1$, in general).  Consequently, to prove the biconditional $k = 1 \Longleftrightarrow n < q$, it suffices to prove the inequality $n < q$.

Next, we prove the following proposition:

Lemma 2.  Let $N = {q^k}{n^2}$ be an odd perfect number (OPN) given in Eulerian form.  If $k = 1$ implies $\sigma(q^k) < n$, then $q < n$.

Proof.
Suppose $N = {q^k}{n^2}$ is an odd perfect number (OPN) given in Eulerian form.


If $k = 1$ implies $\sigma(q^k) < n$, then we have:


$k = 1 \Longrightarrow \sigma(q^k) < n \Longrightarrow q < \sigma(q) = \sigma(q^k) < n$.

Therefore, we have: $k = 1 \Longrightarrow q < n$.

But we also have: $n < q \Longrightarrow k = 1$ [Dris, 2012].  By the contrapositive:

$k \neq 1 \Longrightarrow q < n$.

Consequently, we conclude that $q < n$.
QED.

Remark 2.  If $\sigma(q^k) < n$, then $q^k < n$.  (This last inequality was conjectured in the M.Sc. thesis [Dris, 2008] and in the published article [Dris, 2012]).  Note that, if $q^k < n$, then it is trivial to prove that $k \geq 1 \Longleftrightarrow q < n$.

This blog post is currently a WORK IN PROGRESS.  Please do check back from time to time for updates!