## 11.11.15

### OPN Research - 11/11/2015

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

We denote the abundancy index of $x$ as
$$I(x) = \frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the sum of the divisors of $x$.

First, we prove the following claim:

Proposition 1.
$$I(n^2) \neq \frac{9}{5}$$

Proof.
Suppose that $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, and assume further that $I(n^2) = 9/5$.

This implies that $I(q^k) = 2/I(n^2) = 10/9$.  Since $9 < 10 < \sigma(9) = 13$ and $\gcd(10, 9) = 1$, then the fraction $10/9$ is an abundancy outlaw. This contradicts $I(q^k) = 10/9$.

QED.

By Proposition 1, either $I(n^2) < 9/5$ or $I(n^2) > 9/5$ is true.

If $I(n^2) < 9/5$, we have the following result.

Proposition 2.
$$I(n^2) < \frac{9}{5} \Longleftrightarrow q = 5$$

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

First, we show that $I(n^2) < 9/5 \Longrightarrow q = 5$:
$$\frac{2}{I(q^k)} = I(n^2) < \frac{9}{5} \Longrightarrow \frac{10}{9} < I(q^k) < \frac{q}{q - 1} \Longrightarrow q < 10.$$
That $q = 5$ follows from the fact that $q$ is the Euler prime (i.e., $q \equiv 1 \pmod 4$).

Next, we show that $I(n^2) > 9/5 \Longrightarrow q \geq 13$:
$$\frac{9}{5} < I(n^2) = \frac{2}{I(q^k)} \Longrightarrow 1 + \frac{1}{q} \leq I(q^k) < \frac{10}{9} \Longrightarrow q > 9.$$
That $q \geq 13$ follows from the fact that $q$ is the Euler prime.

In particular, we have shown that
$$I(n^2) < \frac{9}{5} \Longleftrightarrow q = 5.$$

QED.

The following result appears as Lemma 12 in "The third largest prime divisor of an odd perfect number exceeds one hundred" by Iannucci.

Proposition 3.
$$q = 5 \Longrightarrow k = 1$$

(This post is currently a WORK IN PROGRESS.)

## 13.10.15

### Revision to one of my arXiv papers has been uploaded

A revision to my joint paper with Keneth Adrian P. Dagal has been uploaded to arXiv and is now titled as Criteria for Almost Perfect and Deficient Numbers.

In the revised paper, we extend our previous criterion to show that numbers $m$ satisfying $\sigma(m) = 2m - p$ (where $\sigma$ is the sum-of-divisors function, and $p > 1$) also satisfy the inequalities

$$\frac{2m}{3m - \sigma(m)} < \frac{\sigma(m)}{m} < \frac{4m - \sigma(m)}{3m - \sigma(m)},$$
and vice-versa.

## 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]

In particular, note that
$$p\sigma(m^2) - 1 = p^{2a+1} - 1 = (p - 1)\sigma(p^{2a}) = (p - 1)q = (p - 1)(2m^2 - 1),$$
from which we obtain
$$\sigma(m^2) = 2m^2 - 1 - \left(\frac{2(m^2 - 1)}{p}\right),$$
so that we have
$$D(m^2) = 1 + \frac{2(m^2 - 1)}{p}$$
where $D(x) = 2x - \sigma(x)$ is the deficiency of $x$.

Divisibility Constraints

From the equation
$$D(m^2) = 1 + \frac{2(m^2 - 1)}{p}$$
we have the condition
$$p \mid (m^2 - 1),$$
since $\gcd(2, p) = 1$.

$$p\sigma(m^2) - 1 = (p - 1)q$$
we get
$$p\left(\sigma(m^2) - 1\right) = (p - 1)(q - 1).$$
Since $\gcd(p, p - 1) = 1$, we have the condition
$$p \mid (q - 1).$$

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

Note that:
$$1 < I(q) = \frac{2m^2}{\sigma(p^{2a})} < 1 + {10}^{-500}$$

Since $p$ and $q$ are primes, then we have
$$I(p^{2a}) < I(q)$$
since $p^{2a} < q$.

Note that $I(q) < I(m^2)$, since if $I(m^2) < I(q)$, then
$$I(p^{2a}) < I(q) < 1 + {10}^{-500}$$
and
$$I(m^2) < I(q) < 1 + {10}^{-500}.$$

By the Arithmetic Mean-Geometric Mean Inequality:
$$\sqrt[3]{2} = \sqrt[3]{I(q)I(p^{2a})I(m^2)} < \frac{I(q) + I(p^{2a}) + I(m^2)}{3} < 1 + {10}^{-500},$$

Consequently,
$$I(p^{2a}) < I(q) < I(m^2).$$

Recall that, under the assumption $p = 3$, then
$$3\sqrt[3]{2} < I(q) + I(p^{2a}) + I(m^2) < 2 + 2\cdot{10}^{-500} + \frac{4}{3} \approx \frac{10}{3}.$$

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

Consequently, $p \neq 3$.

(In fact, it is possible to do significantly better than this, since
$$I(p^2) \leq I(p^{2a}) < I(q) < 1 + {10}^{-500}$$
implies that the prime $p$ is large.)

## 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 and arXiv).

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.

## 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