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23.5.15

OPN Research - May 2015

Here are the latest updates regarding my research on odd perfect numbers:


Lemma 1. If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, then Sorli's conjecture (i.e., $k = 1$) implies that 
$\gcd(n^2, \sigma(n^2)) = 2n^2 - \sigma(n^2)$.

Lemma 1 was proved by Jaycob Coleman in the following two MSE posts:

and

We also have:

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

The proof of Lemma 2 follows readily from the definition of perfect numbers.

Observe that, since $\sigma(n^2)$ is always odd, $2n^2 - \sigma(n^2)$ is likewise odd. Furthermore, as $q \equiv 1 \pmod 4$, $q + 1$ is even (but not divisible by $4$). We conclude, by Lemma 1, that $\gcd(n^2, \sigma(n^2)) \neq q + 1$.

We now claim the truth of the following statement:

Proposition 1. Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and let $k = 1$.  If $\gcd(n^2, \sigma(n^2)) > q + 1$, then $q < n\sqrt{2}$.

Proof. By Lemmas 2 and 1, 
$\frac{n^2}{\frac{q+1}{2}} = 2n^2 -\sigma(n^2) = \gcd(n^2,\sigma(n^2))$

Since $\gcd(n^2, \sigma(n^2)) > q + 1$, we get $q < q + 1 < n\sqrt{2}$.

We can also prove the following result:

Proposition 2. Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and let $k = 1$.  If $\frac{q + 1}{2}$ is prime, then $\gcd(n^2, \sigma(n^2)) \geq \frac{q + 1}{2}$.

Proof. If $\frac{q + 1}{2}$ is prime, then 
${\left(\frac{q + 1}{2}\right)}^2 \mid n^2$.

This implies that
$q{\left(\frac{q + 1}{2}\right)} \mid \sigma(n^2)$.

By Lemmas 2 and 1, these two divisibility constraints both imply that
$\gcd(n^2, \sigma(n^2)) \geq \frac{q + 1}{2}$.

It remains to rule out the case $\gcd(n^2, \sigma(n^2)) < q + 1$ in order to prove that $k = 1$ implies $q < n\sqrt{2}$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form. This can be very difficult to do, as it has been conjectured that the Euler prime $q$ might have to be the largest prime factor of the odd perfect number $N$, as discussed by Nielsen [Zeta-Flux] here. Dris has given a sufficient condition [i.e., $n < q$] for Sorli's conjecture, which further supports the conjecture that $q$ must be the largest prime divisor of $N$.

The interested reader is hereby referred to this MO post for another unsuccessful attempt of mine at improving Acquaah and Konyagin's estimate of $q < n\sqrt{3}$ to $q < n\sqrt{2}$.

Let me know via e-mail if you have any comments, questions or clarifications regarding this post. You can find my e-mail in my arXiv papers. Please send to the gmail account.

Added [05/23/2015 12:00 NN Manila time]:

It turns out that we can prove the following (stronger) claim:

Proposition 3. Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form. If $k = 1$, then we have:


(a) $\gcd(n^2, \sigma(n^2)) > 2(q + 1)$ implies $q < \sigma(q) < n$.
(b) $\gcd(n^2, \sigma(n^2)) < 2(q + 1)$ implies $n < q < \sigma(q)$.

Proof. The proof uses Lemmas 1 and 2, and the fact that the biconditional


$q < n \Longleftrightarrow \sigma(q) < n$

holds.

Added [05/24/2015 14:30 PM Manila time]:

If $N = q^k n^2$ is an odd perfect number given in Eulerian form, then


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

and

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

Now, since $\gcd(q^k, \sigma(q^k)/2) = 1$, it follows that

$\gcd(n^2, \sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$.

Note that the deficiency $D$ of $n^2$ (written simply as $D(n^2)$) is given by

$D(n^2) = 2n^2 - \sigma(n^2)$

so that

$\sigma(n^2) = 2n^2 - D(n^2)$.

In other words,

$\sigma(n^2) = 2n^2 - {{\sigma(q^{k-1})}{\gcd(n^2, \sigma(n^2))}}$.

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.