## 7.3.15

### The Abundancy Index of Divisors of Spoof Odd Perfect Numbers

I have a new paper out there (currently already in Scribd).

To summarize:  I extended the results that I have obtained in my previous papers on odd perfect numbers, to the case of spoof odd perfect numbers, also known as Descartes numbers in the literature.

Happy reading everyone!  =)

## 12.2.15

### Eureka for the Month of Hearts, Year 2015!

The biconditional $k = 1 \Longleftrightarrow n < q$ is indeed true, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form.

See the following MSE post for more details:

## 24.1.15

### Improving the lower bound for $I(n)$ where $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, with $n < q$

Per Will Jagy's answer (and a subsequent comment by Erick Wong) to the following MSE question, we have the (sharp?) bounds

$1 \leq \frac{I(x^2)}{I(x)} \leq \prod_{p}{\frac{p^2 + p + 1}{p^2 + p}} = \frac{\zeta(2)}{\zeta(3)} \approx 1.3684327776\ldots$

Now, let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.

From Theorem 4.2 [pages 14 to 15 of this paper], we have the following biconditional:

$\frac{2n}{n + 1} < I(n^2) \Longleftrightarrow n < q.$

In particular, if $n < q$ (combining the two results), we get

$\frac{2n}{(n + 1)I(n)} < \frac{I(n^2)}{I(n)} \leq \frac{\zeta(2)}{\zeta(3)}.$

It follows that

$I(n) > \frac{\zeta(3)}{\zeta(2)}\cdot\frac{2n}{n + 1}.$

But we have the lower bound $n > {10}^{375}$ from $q^k < n^2$ [Dris, 2012] and ${10}^{1500} < N = {q^k}{n^2}$ [Ochem and Rao, 2012].  Consequently, we have

$I(n) > \frac{\zeta(3)}{\zeta(2)}\cdot\frac{2n}{n + 1} > \frac{\zeta(3)}{\zeta(2)}\cdot\frac{2\cdot{10}^{375}}{{10}^{375} + 1}.$

Note that we have the rational approximation

$\frac{\zeta(3)}{\zeta(2)}\cdot\frac{2\cdot{10}^{375}}{{10}^{375} + 1} \approx 1.4615259388\ldots$

## 30.12.14

### Improving the bound $q < n\sqrt{3}$ for an odd perfect number $N = {q^k}{n^2}$ given in Eulerian form

http://mathoverflow.net/questions/188831

http://math.stackexchange.com/questions/1009929

## 12.11.14

### On a Conjecture of Dris Regarding Odd Perfect Numbers

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.

## 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

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 $q^k = \circ(n^2)$.