## 14.6.15

### OPN Research - June 2015

"Maybe this is the case that needs to be eliminated:
$$N = {q}{p^{2a}}{m^2}$$
where
$$\sigma(m^2) = p^{2a},$$
$$\sigma(p^{2a}) = q,$$

and

$$q + 1 = 2{m^2}.$$

It would seem that an odd perfect $N$ of this form would contradict [the] conjecture that $q < n$.  $\ldots$ maybe this is the only problem case."
[From an e-mail communication of Professor Carl Pomerance dated June 1, 2015]

Preliminary Results

By using the fact that the abundancy index $I(x) = \sigma(x)/x$ satisfies $1 < I(x)$ for all $x > 1$, then we have the inequalities
$$m^2 < p^{2a} < q.$$
In particular, note that we have
$$m < p^a,$$
and also that $q$ is the largest prime divisor of $N$.

We now establish the following claims:

Claim 1:
$I(q) = I(p^{2a}) = I(m^2)$ is false.

Proof of Claim 1.
Suppose to the contrary that
$$I(q) = I(p^{2a}) = I(m^2)$$ is true.

Then
$$2 = I(q)I(p^{2a})I(m^2) = (I(q))^3 = (I(p^{2a}))^3 = (I(m^2))^3.$$

This implies that

$$\sqrt[3]{2} = I(q) = I(p^{2a}) = I(m^2).$$

The number $\sqrt[3]{2}$ is irrational, while all of the abundancy indices $I(q)$, $I(p^{2a})$, and $I(m^2)$ are rational.

QED.

Claim 2:
$I(q) \neq I(m^2)$ is true.

Proof of Claim 2.
Suppose to the contrary that $I(q) = I(m^2)$.

This then contradicts the fact that the prime $q$ is solitary.

QED.

The proof of the following claim is similar to that of Claim 2.

Claim 3:
$I(p^{2a}) \neq I(m^2)$ is true.

Proof of Claim 3.
Use the fact that the prime-power $p^{2a}$ is solitary.

Claim 4:
$I(q) \neq I(p^{2a})$ is true.

Proof of Claim 4.
Note that $q$ is prime and $p^{2a}$ is a prime power.

Main Results

Assume that $p = 3$.

We then have:

$$\sigma(m^2) = 3^{2a}$$
$$\sigma(3^{2a}) = \frac{3^{2a + 1} - 1}{2} = q$$
$$2m^2 = q + 1.$$

This then implies that
$$3\sigma(m^2) - 1 = 2q = 2(2m^2 - 1).$$

This finally gives:
$$\frac{\sigma(m^2)}{m^2} + \frac{1}{3m^2} = \frac{4}{3}.$$

Consequently:
$$I(m^2) < \frac{4}{3}.$$

We are now able to improve the following bounds:
$$\frac{10}{9} < I(m^2) = \frac{2{p^{2a}}}{q + 1} < 2$$
$$1 < I(q) = \frac{2m^2}{\sigma(p^{2a})} < 1 + {10}^{-500}$$
$$1 < I(p^{2a}) = \frac{q}{\sigma(m^2)} < \frac{3}{2}$$

to
$$\frac{10}{9} < I(m^2) < \frac{4}{3}$$
$$1 < I(q) < 1 + {10}^{-500}$$
$$\frac{5}{4} < I(p^{2a}) < \frac{3}{2}.$$

In particular, we obtain the estimates
$$\frac{\sqrt{10}}{3} < I(m)$$
$$1 < I(q)$$
$$\frac{\sqrt{5}}{2} < I(p^a).$$

By the Arithmetic Mean-Geometric Mean Inequality:
$$\sqrt[3]{2} = \sqrt[3]{I(q)I(p^{2a})I(m^2)} < \frac{I(q) + I(p^{2a}) + I(m^2)}{3}.$$

This then gives the (trivial [?]) lower bound

$$3\sqrt[3]{2} < I(q) + I(p^{2a}) + I(m^2).$$

(Note the rational approximation $3\sqrt[3]{2} \approx 3.779763$.)

But we have the (nontrivial) upper bounds
$$I(m^2) < \frac{4}{3}$$
$$I(q) < 1 + {10}^{-500}$$

$$I(p^{2a}) < \frac{3}{2}$$

from which we obtain the upper bound
$$I(q) + I(p^{2a}) + I(m^2) < \frac{4}{3} + \frac{3}{2} + 1 + {10}^{-500} \approx 3.8\bar{3}.$$

Further Results

Using the (trivial) estimate $I(x^2) < {(I(x))}^2$ (which is true for all $x > 1$), we get the bounds
$$\sqrt{\frac{10}{9}} < I(m) < \frac{4}{3}$$
$$1 < I(q) < 1 + {10}^{-500}$$
$$\sqrt{\frac{5}{4}} < I(p^a) < \frac{3}{2}.$$

### The "Proof"

In light of recent events, I wish to refer everyone to the following hyperlinks:

http://arxiv.org/abs/1206.3230 - On the Components of an Odd Perfect Number

http://arxiv.org/abs/1204.1450 - Solving the Odd Perfect Number Problem: Some Old and New Approaches

http://arxiv.org/abs/1206.1548 - Solving the Odd Perfect Number Problem: Some New Approaches

http://arxiv.org/abs/1103.1090 - The Abundancy Index of Divisors of Odd Perfect Numbers

http://arxiv.org/abs/1302.5991 - New Results for Sorli's Conjecture on Odd Perfect Numbers

http://arxiv.org/abs/1303.2329 - New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II

http://arxiv.org/abs/1310.5616 - Euclid-Euler Heuristics for (Odd) Perfect Numbers

http://arxiv.org/abs/1503.03860 - The Abundancy Index of Divisors of Spoof Odd Perfect Numbers

http://math.stackexchange.com/questions/548528 - On odd perfect numbers $N$ given in the Eulerian form $N = {q^k}{n^2}$

http://mathoverflow.net/questions/188831 - Improving the bound $q < n\sqrt{3}$ for an odd perfect number $N = {q^k}{n^2}$ given in Eulerian form

http://arxiv.org/abs/1308.6767 - A Criterion for Almost Perfect Numbers Using the Abundancy Index

http://arxiv.org/abs/1312.6001 - The Abundancy Index of Divisors of Odd Perfect Numbers - Part II

http://math.stackexchange.com/questions/1030760 - Prove a property of the divisor function (Part 2)

http://math.stackexchange.com/questions/1097803 - Improving the inequality $x\sigma(x) \leq \sigma(x^2)$ for $x \in \mathbb{N}$

http://arxiv.org/abs/1103.1437 - A note on odd perfect numbers

http://math.stackexchange.com/questions/1300684 - If ${q^k}{n^2}$ is an odd perfect number with Euler prime $q$, are the following statements known to hold in general?

http://math.stackexchange.com/questions/1294399 - Mathematical terminology for primes $(q+1)/2$ such that $q$ is also prime

http://math.stackexchange.com/questions/1191510 - Computational verification request

http://math.stackexchange.com/questions/1188446 - Is the Euler prime of an odd perfect number a palindrome (in base $10$), or otherwise?

http://math.stackexchange.com/questions/1188405 - Is the Euler prime of an odd perfect number a repunit, or otherwise?

http://math.stackexchange.com/questions/1293668 - Is there a solution to this system of equations?

and

http://mathoverflow.net/questions/201634 - On Odd Perfect Numbers and Spoof Odd Perfect Numbers

## 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.

## 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.