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27.7.16

"Theorem": If $p$ is an odd prime, then $\frac{p+2}{p}$ is an abundancy outlaw.

We first show some preliminary lemmata.
Lemma 1.  If $p$ is odd and $\frac{p+2}{p}$ is the abundancy index of some integer $n$, then $n$ is deficient.

Proof.  Since $p$ is odd, $p > 2$.  Furthermore, since
$$I(n) = \frac{p+2}{p} = 1 + \frac{2}{p} < 1 + 1 = 2,$$
then $n$ is deficient.          QED

Lemma 2.  If $p$ is odd, then $\gcd(p, p+2)=1$.

Proof.
$$\gcd(p,p+2)=\gcd(p,(p+2)-p)=\gcd(p,2)=1$$
where the last equality uses the fact that $p$ is odd.          QED

Lemma 3.  If $p$ is an odd prime and $I(n)=\frac{p+2}{p}$, then $D(n) = 2n - \sigma(n) \neq 1$.

Proof.
Let $p$ be an odd prime.  Assume that $I(n)=\frac{p+2}{p}$.  Now, assume to the contrary that
$$D(n) = 2n - \sigma(n) = 1.$$
Note that this implies
$$\gcd(n,\sigma(n))=\gcd(n,2n-1)=1.$$
Additionally, since $p$ is odd, by Lemma 2 we obtain
$$\gcd(p,p+2)=1.$$
Consequently, since $I(n)=\frac{p+2}{p}$, we get
$$p\sigma(n)=(p+2)n.$$
But $\gcd(p,p+2)=1$ implies that $p \mid n$, and $\gcd(n,\sigma(n)=1$ implies that $n \mid p$.  Hence, it follows that $p=n$.

But since $p$ is an odd prime, $I(n)=I(p)=\frac{p+1}{p}$, which contradicts $I(n)=\frac{p+2}{p}$.

We conclude that $D(n) = 2n - \sigma(n) \neq 1$.          QED

The following lemma is actually a theorem from this preprint (to be updated soon).  (We omit the proof.)

Lemma 4.  If $n$ is a positive integer satisfying $D(n) = 2n - \sigma(n) > 1$, then we have the following bounds for the abundancy index of $n$ in terms of the deficiency of $n$:
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}.$$

We now attempt to prove the following "result":

"Theorem".   If $p$ is an odd prime, then $\frac{p+2}{p}$ is an abundancy outlaw.

"Proof Attempt".
Let $p$ be an odd prime, and assume to the contrary that $I(n)=\frac{p+2}{p}$.

By Lemma 1, $n$ is deficient (so that $D(n) \neq 0$).

By Lemma 3, $D(n) \neq 1$.

Thus, by Lemma 4, we have the bounds
$$\frac{2}{1 + \frac{D(n)}{n}}= \frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)} = \frac{2+\frac{D(n)}{n}}{1+\frac{D(n)}{n}}.$$

Now, $I(n) = \frac{p+2}{p}$ implies that
$$D(n) = p\left(\sigma(n) - n\right) - \sigma(n).$$

We compute:
$$n + D(n) = (p - 1)\left(\sigma(n) - n\right) \Longleftrightarrow 1 + \frac{D(n)}{n} = (p - 1)\left(I(n) - 1\right)$$
and
$$2n + D(n) = (p - 1)\left(\sigma(n) - n\right) + n \Longleftrightarrow 2 + \frac{D(n)}{n} = (p - 1)\left(I(n) - 1\right) + 1.$$

Thus, we have the lower bound
$$\frac{2}{(p - 1)\left(I(n) - 1\right)} < I(n)$$
which implies that (upon setting $I(n) = \frac{p+2}{p}$)

$$\frac{2}{(p - 1)\left(\frac{p+2}{p} - 1\right)} < \frac{p + 2}{p},$$
from which we obtain
$$2p^2 < 2(p - 1)(p + 2) = 2(p^2 + p - 2) = 2p^2 + 2p - 4.$$
Lastly, this implies
$$2 < p.$$
No contradictions so far.  We now consider the upper bound.

We have the upper bound
$$I(n) < \frac{(p - 1)\left(I(n) - 1\right) + 1}{(p - 1)\left(I(n) - 1\right)},$$
which implies that
$$I(n) - 1 < \frac{1}{(p - 1)\left(I(n) - 1\right)}.$$
On setting $I(n) = \frac{p+2}{p}$, we get
$$\frac{2}{p} < \frac{1}{(p - 1)\frac{2}{p}}.$$
This implies that
$$4(p - 1) < p^2.$$
Thus, $p^2 - 4p + 4 = (p - 2)^2 > 0$.  Still no contradictions.

It thus seems fruitful to try to improve on the (trivial?) bounds
$$\frac{2n}{n+D(n)} < I(n) < \frac{2n+D(n)}{n+D(n)}$$
for $n$ satisfying $D(n)>1$.

20.7.16

If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$ implies that $k=1$ and $q=5$.

(Note:  This blog post is essentially an elucidation of the answer to this MSE question.)

Here, we prove the following proposition.

Theorem.  If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, then $$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$$
implies that
$$k=1$$
and
$$q=5.$$


Proof.  Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

Since $N$ is perfect, we have
$$2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \left(\frac{q^{k+1} - 1}{q - 1}\right)\cdot\sigma(n^2).$$
It follows that
$$\frac{\sigma(n^2)}{n^2} = \frac{2{q^k}\left(q - 1\right)}{q^{k+1} - 1} = \frac{2q^{k+1} - 2q^k}{q^{k+1} - 1} = \frac{2q^{k+1} - 2}{q^{k+1} - 1} - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right)$$
$$= 2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right).$$

By assumption, we have
$$\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}.$$

This inequality is equivalent to
$$2 - 2\left(\frac{q^k - 1}{q^{k+1} - 1}\right) \geq 2 - \frac{5}{3q},$$
which, in turn, is equivalent to
$$6q\left(q^k - 1\right) \leq 5(q^{k+1} - 1).$$
This last inequality simplifies to
$$q^{k+1} - 6q + 5 \leq 0,$$
which cannot be true when $k > 1$.  Thus, we know that $k=1$.

Hence, we have
$$q^2 - 6q + 5 \leq 0.$$
But this inequality implies that
$$1 \leq q \leq 5,$$
which, together with $q \geq 5$, implies that $q=5$.

QED.

18.7.16

On conditions equivalent to the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

The Descartes-Frenicle-Sorli conjecture predicts that $k=1$ always holds.

From the equation $\sigma(N)=2N$, we get that
$$\left(q^k + \sigma(q^{k-1})\right)\sigma(n^2) = \sigma(q^k)\sigma(n^2) = 2{q^k}{n^2}$$
so that we obtain
$$\frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})} = 2n^2 - \sigma(n^2).$$
(Note that $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of $n^2$, and that
$$I(x) = \frac{\sigma(x)}{x}$$
is the abundancy index of $x$.)

Now suppose that
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
Trivially, we know that
$$\frac{\sigma(n^2)}{q} \mid \sigma(n^2).$$
Thus, we have
$$\frac{\sigma(n^2)}{q} \mid \left(2n^2 - \sigma(n^2)\right) = \frac{\sigma(n^2)}{q}\cdot{I(q^{k-1})}.$$
This implies that $I(q^{k-1})$ is an integer;  in other words, $k=1$.

The other direction
$$k=1 \Longrightarrow \frac{\sigma(n^2)}{q} \mid n^2$$
is trivial.

We therefore have the following lemma.

Lemma 1.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longleftrightarrow \frac{\sigma(n^2)}{q} \mid n^2.$$

Now assume that $k=1$.

By Lemma 1, we have
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
This implies that there exists an integer $d$ such that
$$n^2 = d \cdot \left(\frac{\sigma(n^2)}{q}\right).$$

Note that, from the equation $\sigma(N)=2N$, we obtain (upon setting $k=1$)
$$(q+1)\sigma(n^2) = \sigma(q)\sigma(n^2) = 2q{n^2}$$
from which we get
$$d = \frac{n^2}{\frac{\sigma(n^2)}{q}} = \frac{q + 1}{2}.$$

Notice that, when $k=1$, we can derive
$$\frac{5}{3} \leq I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} < 2$$
so that we have
$$\frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

We state this latest result as our second lemma.

Lemma 2.  Suppose that $N = q^k n^2$ is an odd perfect number with Euler prime $q$.  Then
$$k = 1 \Longrightarrow \frac{q}{2} < d = \frac{q}{I(n^2)} \leq \frac{3q}{5}.$$

Note that, when $k=1$, we have
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2q+2}{q+1} - \frac{2}{q+1} = 2 - \frac{1}{\frac{q+1}{2}} = 2 - \frac{1}{d}$$

However, notice that we know, by Lemma 2,
$$\frac{q}{2} < d \leq \frac{3q}{5} \Longrightarrow \frac{5}{3q} \leq \frac{1}{d} < \frac{2}{q} \Longrightarrow 2 - \frac{2}{q} < 2 - \frac{1}{d} = I(n^2) \leq 2 - \frac{5}{3q}.$$

Since we already have
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1}$$
this WolframAlpha computation validates that, indeed,
$$\frac{2q}{q+1} \leq 2 - \frac{5}{3q}$$
implies
$$q \geq 5.$$

Iterating the computations, we obtain
$$\frac{2 - \frac{2}{q}}{q} < \frac{1}{d} = \frac{I(n^2)}{q} \leq \frac{2 - \frac{5}{3q}}{q}$$
so that
$$2 - \frac{2 - \frac{5}{3q}}{q} \leq 2 - \frac{1}{d} = I(n^2) < 2 - \frac{2 - \frac{2}{q}}{q}.$$
But
$$2 - \frac{2 - \frac{5}{3q}}{q} = \frac{6q^2 - 6q + 5}{3q^2} \geq \frac{5}{3}$$
$$2 - \frac{2 - \frac{2}{q}}{q} = \frac{2q^2 - 2q + 2}{q^2} < 2.$$

Lastly, note that, after multiplying throughout
$$\frac{6q^2 - 6q + 5}{3q^2} \leq I(n^2) < \frac{2q^2 - 2q + 2}{q^2}$$
by $I(q) = \frac{q+1}{q}$, and asking WolframAlpha to solve the resulting inequality
$$\left(\frac{q+1}{q}\right)\cdot\left(\frac{6q^2 - 6q + 5}{3q^2}\right) \leq 2 = I(q)I(n^2) = I(qn^2) < \left(\frac{q+1}{q}\right)\cdot\left(\frac{2q^2 - 2q + 2}{q^2}\right)$$
we obtain
$$q \geq 5.$$

This blog post is currently a WORK IN PROGRESS.

14.7.16

The Abundancy Index of Divisors of Even Almost Perfect Numbers That Are Not Powers of Two

Antalan and Tagle showed that, if $M \neq 2^t$ ($t \geq 1$) and $\sigma(M) = 2M - 1$ (where $\sigma = \sigma_{1}$ is the (classical) sum-of-divisors function), then $M$ takes the form
$$M = 2^r b^2$$
where $b > 1$ is an odd composite.

In this preprint, Antalan and Dris proved the following bounds:

Theorem A.  If $2^r b^2$ is an almost perfect number with $r \geq 1$, $\gcd(2, b) = 1$ and $b > 1$, then
$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$

In this blog post, we will prove the following series of inequalities:

Theorem B.  If $2^r b^2$ is an almost perfect number with $r \geq 1$, $\gcd(2, b) = 1$ and $b > 1$, then
$$1 < \frac{b}{\sigma(2^r)} < \frac{\sigma(b)}{\sigma(2^r)} < \frac{b}{2^r} < \frac{\sigma(b)}{2^r} < \frac{b^2}{\sigma(2^r)} < \frac{\sigma(b^2)}{\sigma(2^r)} < \frac{b^2}{2^r} < \frac{\sigma(b^2)}{2^r}.$$ 

In other words, taking reciprocals, we should obtain:
$$\frac{2^r}{\sigma(b^2)} < \frac{2^r}{b^2} < \frac{\sigma(2^r)}{\sigma(b^2)} < \frac{\sigma(2^r)}{b^2} < \frac{2^r}{\sigma(b)} < \frac{2^r}{b} < \frac{\sigma(2^r)}{\sigma(b)} < \frac{\sigma(2^r)}{b} < 1.$$



Proof.  Since
$$\frac{\sigma(b)}{b} < \frac{\sigma(b^2)}{b^2} < \frac{4}{3},$$
by Theorem A, it suffices to prove the inequality in the middle, namely 
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$

Suppose to the contrary that
$$\frac{\sigma(b^2)}{\sigma(2^r)} \leq \frac{\sigma(b)}{2^r}.$$

It follows that
$$\frac{\sigma(b^2)}{\sigma(b)} \leq \frac{\sigma(2^r)}{2^r} < 2,$$
by Theorem A.  Since $\frac{\sigma(b)}{b} < \frac{4}{3}$ (again, by Theorem A), it follows that
$$\frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b}\right) < \left(\frac{b}{\sigma(b)}\right)\cdot\left(\frac{\sigma(b^2)}{b}\right) = \frac{\sigma(b^2)}{\sigma(b)} < 2.$$
Dividing through by $1 < b$, we obtain 
$$\frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b^2}\right) < \frac{2}{b}.$$
By Theorem A, we get 
$$\frac{3}{4} < \frac{3}{4}\cdot\left(\frac{\sigma(b^2)}{b^2}\right) < \frac{2}{b},$$
which finally implies that
$$b < \frac{8}{3}.$$
Since $1 < b$, this forces $b = 2$A contradiction, as $b$ has to be odd.

We therefore conclude that 
$$\frac{\sigma(b)}{2^r} < \frac{\sigma(b^2)}{\sigma(2^r)}.$$

The proof for 
$$\frac{\sigma(b)}{2^r} < \frac{b^2}{\sigma(2^r)}$$
is very similar, and we are done.

Odd Numbers Ruled Out by the Dagal-Dris Criterion on Almost Perfect Numbers (Using the Abundancy Index)

In this post, we consider odd numbers that are ruled out by the Dagal-Dris theorem, which gives the following criterion for almost perfect numbers in terms of the abundancy index:

Theorem.  (Dagal-Dris)  Let $N$ be a positive integer.  Then $\sigma(N) = 2N - 1$ if and only if
$$\frac{2N}{N + 1} \leq I(N) < \frac{2N + 1}{N + 1}.$$
Equality holds if and only if $N = 1$.

First, note that $N_0 = 2^r$ satisfies the inequality in the criterion, as WolframAlpha confirms that
$$\frac{2^{r+1}}{2^r + 1} \leq \frac{2^{r+1} - 1}{2^r} < \frac{2^{r+1} + 1}{2^r + 1}$$
is true for $r \geq 0$.  (Note that $1$ is almost perfect.)

Second, note that if $N$ is odd and $\sigma(N) = 2N - 1$, then $N$ must be a square.

A.  Suppose that $N_1 = p^k$, where $p$ is an odd prime and $k \geq 2$ is an even integer.  We want to show that $N_1$ is not almost perfect.

To this end, assume that
$$\frac{2{p^k}}{p^k + 1} < \frac{p^{k+1} - 1}{{p^k}(p - 1)} < \frac{2{p^k} + 1}{p^k + 1}.$$

WolframAlpha says that this inequality is equivalent to
$$2p^k < -\frac{p^{-k}}{p - 1} + \frac{p^{k+1}}{p - 1} + \frac{p}{p - 1} - \frac{1}{p - 1} < 2{p^k} + 1,$$
which, by a brief inspection, is not true.  (It suffices to check the inequality on the left.)

Since the inequality in the criterion is satisfied when $p = 2$, we now know that $N_1 = p^k$ is not almost perfect, when $p$ is an odd prime with $k \geq 2$.

B.  Suppose that $N_2 = p^k q^l$, where $p$ and $q$ are distinct odd primes, and $k \geq 2$, $l \geq 2$ are even integers.  We want to show that $N_2$ is not almost perfect.

To this end, assume that
$$\frac{2{p^k}{q^l}}{{p^k}{q^l} + 1}  < \frac{\left(p^{k+1} - 1\right)\left(q^{l+1} - 1\right)}{{p^k}{q^l}\left(p - 1\right)\left(q - 1\right)} < \frac{2{p^k}{q^l} + 1}{{p^k}{q^l} + 1}.$$

WolframAlpha says that this inequality is equivalent to
$$2{p^k}{q^l} < \frac{{p^{-k}}{q^{-l}}}{\left(p - 1\right)\left(q - 1\right)} + \frac{{p^{k+1}}{q^{l+1}}}{\left(p - 1\right)\left(q - 1\right)} - \frac{q{p^{-k}}}{\left(p - 1\right)\left(q - 1\right)} - \frac{p^{k+1}}{\left(p - 1\right)\left(q - 1\right)}$$
$$-\frac{p{q^{-l}}}{\left(p - 1\right)\left(q - 1\right)} - \frac{q^{l+1}}{\left(p - 1\right)\left(q - 1\right)} + \frac{pq}{\left(p - 1\right)\left(q - 1\right)} + \frac{1}{\left(p - 1\right)\left(q - 1\right)} < 2{p^k} + 1,$$
which, by a brief inspection, is not true.  (Again, it suffices to check the inequality on the left.)

We conclude that $N_2$ is not almost perfect.

(Moving forward, since the inequality on the right appears to be always satisfied, we will only consider the inequality on the left.)

An Aside:  Let $\omega(N)$ denote the number of distinct prime factors of $N$. If $N$ is odd and satisfies $\sigma(N) = 2N - 1$, then an easy way to prove that $\omega(N) \geq 3$ is as follows:

Since $\sigma(N) = 2N - 1$ and $N$ is odd, it follows that $N$ is a square.  

Suppose that $\omega(N)=1$.  Then $N$ takes the form $p^{2k}$, where $p$ is an odd prime and $k \geq 1$ is an integer.  Note that $p \geq 3$, from which it follows that $N = p^{2k} \geq p^2 \geq 9$.

We obtain 
$$I(N) = \frac{\sigma(N)}{N} = 2 - \frac{1}{N} \geq 2 - \frac{1}{9} = \frac{17}{9} = 1.\overline{888}.$$

But we also have 
$$I(N) = \frac{\sigma(N)}{N} = \frac{\sigma(p^{2k})}{p^{2k}} = \frac{p^{2k+1} - 1}{{p^{2k}}\left(p - 1\right)} < \frac{p^{2k+1}}{{p^{2k}}\left(p - 1\right)} = \frac{p}{p - 1} \leq \frac{3}{2} = 1.5,$$
since $p \geq 3$.

This results to the contradiction 
$$1.\overline{888} = \frac{17}{9} \leq I(N) < \frac{3}{2} = 1.5.$$

This means that $\omega(N) \geq 2$.  Now, suppose that $\omega(N) = 2$.  Again, as before, $N$ takes the form ${p^{2k}}{q^{2l}}$, where $p$ and $q$ are distinct odd primes (so that we may take $p < q$, without loss of generality) and $k \geq 1$, $l \geq 1$ are integers.  Note that we may take $3 \leq p$ and $5 \leq q$, from which it follows that 
$$N = {p^{2k}}{q^{2l}} \geq {p^2}{q^2} = (pq)^2 \geq ({3}\cdot{5})^2 = {15}^2 = 225.$$

As before, we obtain 
$$I(N) = \frac{\sigma(N)}{N} = 2 - \frac{1}{N} \geq 2 - \frac{1}{225} = \frac{449}{225} = 1.99\overline{555},$$
since $p \geq 3$ and $q \geq 5$.

But we also have 
$$I(N) = \frac{\sigma(N)}{N} = \frac{\sigma({p^{2k}}{q^{2l}})}{{p^{2k}}{q^{2l}}} = \frac{\left(p^{2k+1} - 1\right)\left(q^{2l + 1} - 1\right)}{{p^{2k}}{q^{2l}}\left(p - 1\right)\left(q - 1\right)}$$
$$< \frac{{p^{2k+1}}{q^{2l+1}}}{{p^{2k}}{q^{2l}}\left(p - 1\right)\left(q - 1\right)} = \left(\frac{p}{p - 1}\right)\cdot\left(\frac{q}{q - 1}\right) \leq \left(\frac{3}{2}\right)\cdot\left(\frac{5}{4}\right) = \frac{15}{8} = 1.875$$
since $p \geq 3$ and $q \geq 5$.
 
This results to the contradiction$$1.99\overline{555} = \frac{449}{225} \leq I(N) < \frac{15}{8} = 1.875.$$

We therefore conclude that $\omega(N) \geq 3$.  QED.

C.  Suppose that $N_3 = p^k q^l r^m$, where $p$, $q$ and $r$ are distinct odd primes, and $k \geq 2$, $l \geq 2$, $m \geq 2$ are even integers.  We want to show that $N_3$ is not almost perfect.

To this end, assume that
$$\frac{2{p^k}{q^l}{r^m}}{{p^k}{q^l}{r^m} + 1}  < \frac{\left(p^{k+1} - 1\right)\left(q^{l+1} - 1\right)\left(r^{m+1} - 1\right)}{{p^k}{q^l}{r^m}\left(p - 1\right)\left(q - 1\right)\left(r - 1\right)} < \frac{2{p^k}{q^l}{r^m} + 1}{{p^k}{q^l}{r^m} + 1}.$$

As indicated earlier, we will only be considering the inequality on the left.  This inequality implies
$$2{p^k}{q^l}{r^m} < \left({p^k}{q^l}{r^m} + 1\right)\cdot\left(\frac{p}{p - 1}\cdot\frac{q}{q - 1}\cdot\frac{r}{r - 1}\right),$$
which again, by a brief inspection, is not true.

Generalizing to the case of a fixed $\omega(N)$, we now show that indeed, the following "theorem" holds:

"Theorem".  No odd almost perfect numbers exist apart from $1$.

"Proof".  It suffices to consider the case when we have an odd $N > 1$ and $\omega(N) = M$, where $M$ is fixed.

Assume to the contrary that $N = \prod_{i=1}^{M}{{p_i}^{\alpha_i}}$ is almost perfect.  (Note that the $\alpha_i$'s are all even, but it will not matter in this argument.)  By the Dagal-Dris criterion, we have 
$$\frac{2N}{N + 1} < I(N) = \frac{\sigma(N)}{N} < \prod_{i=1}^{M}{\frac{p_i}{p_i - 1}}$$
so that 
$$2{\prod_{i=1}^{M}{{p_i}^{\alpha_i}}}  < \left(\left({\prod_{i=1}^{M}{{p_i}^{\alpha_i}}}\right) + 1\right)\cdot\left(\prod_{i=1}^{M}{\frac{p_i}{p_i - 1}}\right).$$
 
Similar to the reasoning above, by a brief inspection, this inequality cannot be true.

We therefore conclude that $N$ cannot be almost perfect.  Since $M = \omega(N)$ was arbitrary, it follows that there are no odd almost perfect numbers, apart from $1$.

"QED". 

Update (Added July 14 2016):  There is a serious flaw in the arguments beginning Part C.  However, by utilizing a result of Kishore that an odd almost perfect number has to have at least six (6) distinct prime factors, it is possible to prove the following result instead:

Theorem.  If $N > 1$ is an odd almost perfect number with $\omega(N)=6$, then $N$ must be divisible by $3$.

Proof.  The details are in this MSE post. 

8.7.16

A proof of the Descartes-Frenicle-Sorli conjecture on odd perfect numbers?

(Note:  This has been cross-posted from MathOverflow.)


Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.  Denote the abundancy index of the positive integer $x$ by
$$I(x) = \frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the sum of the divisors of $x$.

Suppose that the Descartes-Frenicle-Sorli conjecture on odd perfect numbers is false.  (That is, assume that $k > 1$.)

Then since we have
$$\frac{q+1}{q} = I(q) < I(q^k) < \frac{q}{q-1} < \frac{5}{4}$$
$$\frac{8}{5} < \frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} < \frac{2}{I(q)} = \frac{2q}{q+1},$$
we consider the product
$$\left(I(q^k) - \frac{q+1}{q}\right)\left(I(n^2) - \frac{q+1}{q}\right)$$
which is positive.  Therefore,
$$2 + \left(\frac{q+1}{q}\right)^2 = I(q^k)I(n^2) + \left(\frac{q+1}{q}\right)^2 > \frac{q+1}{q}\cdot\left(I(q^k) + I(n^2)\right)$$
which implies that
$$\frac{2q}{q+1} + \frac{q+1}{q} > I(q^k) + I(n^2).$$
But we have
$$I(q) + I(n^2) = I(q) + \frac{2}{I(n^2)} = \frac{q+1}{q} + \frac{2q}{q+1}.$$
We therefore have
$$I(q) + I(n^2) = \frac{q+1}{q} + \frac{2q}{q+1} > I(q^k) + I(n^2).$$
This results to the contradiction $I(q) > I(q^k)$.  Therefore, $k = 1$.

My question is:  What are the consequences of the truth of the Descartes-Frenicle-Sorli conjecture on odd perfect numbers?

Added on July 08, 2016
MathOverflow user Michael Renardy found a flaw with the reasoning above.  He says:

You seem to prove $I(n^2) < 2q/(q+1)$ (second line of inequalities), but a few lines onward you set $I(n^2) = 2q/(q+1)$.  You cannot have it both ways.

1.7.16

If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, are the following statements known to hold in general?

ORIGINAL QUESTION

(Note: This question has been cross-posted from MSE.)

Let $\sigma = \sigma_{1}$ be the classical sum-of-divisors function.

A number is said to be perfect if $\sigma(N)=2N$.

If $q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), are the following statements known to hold in general?

(a) $\gcd(n^2, \sigma(n^2))$ is large.

(b) The deficiency $D(n^2) = 2n^2 - \sigma(n^2)$ is large.

(c) The index $i(q^k) = \sigma(N/q^k)/q^k$ is large.

Using the trivial relationships:

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

then if the Descartes-Frenicle-Sorli conjecture that $k = 1$ is true, it is possible to show that a lower bound for the quantities (a), (b) and (c) is given by $n/\sqrt{3}$.  (Here, I have used Acquaah and Konyagin's estimate $q < n\sqrt{3}$. The inequality $q^k < n^2$ then gives the desired large numerical bound if we use known lower bounds for the odd perfect number $N = q^k n^2,$ latest of which are by Ochem and Rao.)

What happens when $k > 1$?  I do know that 
$$\frac{\sigma(N/q^k)}{q^k} = \sigma(n^2)/q^k \geq 315$$
by using a result of Broughan, Delbourgo, and Zhou.

Is it possible to do better than this, apart from attempting a proof of (obviously) $q^k < n$?

RECENTLY POSTED ANSWER
Let $N = q^k n^2$ be an odd perfect number with Euler prime $q$.

In a recent preprint, Brown claims a complete proof for $q < n$, and a partial proof that the inequality $q^k < n$ holds under many cases.  (See arXiv.)

In particular, since $q < q^k < n$ holds if Brown's proofs are correct (and completed), then the resulting lower bound is
$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = i(q^k) = \frac{2n^2}{\sigma(q^k)} > \frac{8}{5}\cdot\frac{n^2}{q^k} > \frac{8}{5}\cdot{n}.$$

Notice that Brown's proofs hold unconditionally (i.e., even if $k=1$).

We then have the desired large numerical lower bound
$$\gcd(n^2, \sigma(n^2)) = \frac{D(n^2)}{\sigma(q^{k-1})} = \frac{\sigma(N/q^k)}{q^k} > \frac{8}{5}\cdot{n} > \frac{8}{5}\cdot{{10}^{500}},$$

which is an easy consequence of Ochem and Rao's $N > {10}^{1500}$ and the inequality $q^k < n$.

30.6.16

On the conjectured nonexistence of even almost perfect numbers (other than powers of two) and odd perfect numbers

(Note:  This question has been cross-posted from MSE.)

Let $\sigma(a) = \sigma_{1}(a)$ be the sum of the divisors of the positive integer $a$.

A number $M$ is called almost perfect if $\sigma(M) = 2M - 1$.  $N$ is called perfect if $\sigma(N) = 2N$.

$M_s = 2^s$ where $s \geq 0$ are almost perfect, with $M_0 = 1$ being the only odd almost perfect number that is currently known.  If $M \neq 2^t$ is an even almost perfect number, then Antalan and Tagle showed that $M$ must have the form
$$M = {b^2}{2^r}$$
where $r \geq 1$ and $b$ is an odd composite.

On the other hand, only $49$ even perfect numbers have been discovered, and they are of the form 
$$N_e = 2^{p-1}\left(2^p - 1\right)$$
where $2^p - 1$ and $p$ are primes.  It is currently unknown whether there are any odd perfect numbers, but Euler showed that they are of the form
$$N_o = n^2 q^c$$
where $q$ is prime with $q \equiv c \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$.

Notice the following:

For numbers that we know exist

Even Perfect Numbers

Assuming $N_e \neq 6$ (because it is squarefree),
$$\frac{\sigma(2^{(p-1)/2})}{2^p - 1} = \frac{2^{(p+1)/2} - 1}{2^p - 1}< 1 < 4 \le 2^{(p+1)/2} = \frac{\sigma(2^p - 1)}{2^{(p-1)/2}}.$$

Note that
$$\frac{\sigma(2^p - 1)}{2^p - 1} \leq \frac{8}{7} < \sqrt{\frac{7}{4}} < \frac{\sigma(2^{(p-1)/2})}{2^{(p-1)/2}}.$$

Almost Perfect Numbers (Powers of Two)

Since $s \geq 0$,
$$\frac{\sigma(\sqrt{1})}{2^s} \leq 1 \leq 2^{s+1} - 1 = \frac{\sigma(2^s)}{\sqrt{1}}.$$

Note that
$$\frac{\sigma(\sqrt{1})}{\sqrt{1}} = 1 \leq \frac{\sigma(2^s)}{2^s}.$$


In Descartes' example $D = km$, we have
$$m = 22021 = {{19}^2}\cdot{61}$$
and
$$\sqrt{k} = {3}\cdot{7}\cdot{11}\cdot{13}.$$

Note that
$$\frac{\sigma(\sqrt{k})}{m} = \frac{5376}{22021} < 1 < \frac{22022}{3003} < \frac{m+1}{\sqrt{k}}$$
and
$$\frac{m+1}{m} = \frac{22022}{22021} < \frac{5376}{3003} = \frac{\sigma(\sqrt{k})}{\sqrt{k}}.$$

For numbers that are conjectured not to exist

Even Almost Perfect Numbers (Other Than Powers of Two)

$$\frac{\sigma(2^r)}{b} < 1 < 2 < \frac{\sigma(b)}{2^r}$$

Note that
$$\frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r}.$$

Odd Perfect Numbers

The following inequalities are conjectured in New Results for Sorli's Conjecture on Odd Perfect Numbers - Part II:

$$\frac{\sigma(q^c)}{n} < 1 < \sqrt{\frac{8}{5}} < \frac{\sigma(n)}{q^c}.$$

Note that
$$\frac{\sigma(q^c)}{q^c} < \frac{5}{4} < \sqrt{\frac{8}{5}} < \frac{\sigma(n)}{n}.$$

(Added June 27 2016 - In fact, in a recent preprint, Brown claims a partial proof for $q^c < n$, which would be consistent with the conjecture here.)

Here are my questions:

(1) If $K = {x^2}{y^z}$ is a (hypothetical) number satisfying $\sigma(K) = 2K + \alpha$ (with $y$ prime, $\gcd(x,y)=1$, and where $\alpha$ could be zero or negative), might there be a specific reason why the inequalities
$$\frac{\sigma(x)}{y^z} \leq 1 \leq \frac{\sigma(y^z)}{x}$$
seem to guarantee existence of such numbers $K$?

(2) If $L = {u^2}{v^w}$ is a (hypothetical) number satisfying $\sigma(L) = 2L + \beta$ (with $v$ prime, $\gcd(u,v)=1$ and where $\beta$ could be zero or negative), might there be a specific reason why the inequalities
$$\frac{\sigma(v^w)}{u} < 1 < \frac{\sigma(u)}{v^w}$$
seem to predict nonexistence of such numbers $L$?

Question (1) is illustrated (as detailed above) in the case of even perfect numbers, almost perfect numbers which are powers of two, and spoof odd perfect numbers (otherwise known in the literature as Descartes numbers).

Question (2) is illustrated (as detailed above) in the case of even almost perfect numbers which are not powers of two, and odd perfect numbers.