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31.7.17

On the Descartes-Frenicle-Sorli conjecture and the Euler prime of odd perfect numbers

(Preamble: My apologies for the somewhat long post - I merely wanted to include all the details that I had in mind for ease of reference later.)

This post is an offshoot of this earlier MSE question.

So to summarize:  We have an odd perfect number $N=q^k n^2$ with Euler prime $q$.  (That is, $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  In particular, note that $q \geq 5$.)

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.

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

Since
$$I(n^2) \leq \frac{2q}{q+1},$$
if $q=5$ then $I(n^2) \leq 5/3$.  This then implies that, if $q=5$, then
$$I(q^k)+I(n^2) \leq \frac{43}{15}.$$
It turns out that we have the biconditional
$$q>5 \iff \bigg(I(q^k)+I(n^2) > \frac{43}{15}\bigg).$$
However, if $k=1$, we have
$$I(q^k)+I(n^2) \geq \frac{43}{15}.$$
By the contrapositive,
$$I(q^k)+I(n^2) < \frac{43}{15}$$
implies that $k>1$.
Lastly, we have the biconditional
$$\bigg(\bigg(k=1\bigg) \land \bigg(q=5\bigg)\bigg) \iff \bigg(I(q^k)+I(n^2) = \frac{43}{15}\bigg).$$

Summarizing, we have:

If $57/20 < I(q^k) + I(n^2) < 43/15$, then $q=5$ and $k>1$.

Under this case, $k \geq 5$, so that 
$$1.24992 = \frac{3906}{3125} = \frac{5^6 - 1}{{5^5}(5 - 1)}= I(5^5) \leq I(q^k) < \frac{5}{4} = 1.25$$
$$1.6 = \frac{8}{5} < I(n^2) \leq \frac{3125}{1953} \approx 1.\overline{600102406554019457245263696876},$$
resulting in the improved upper bound
$$I(q^k) + I(n^2) \leq \frac{3125}{1953}+\frac{3906}{3125}=\frac{17394043}{6103125}=2.85002\overline{240655401945724526369687660010}$$

Thus, we have:

If $q=5$ and $k>1$, then $57/20 < I(q^k) + I(n^2) \leq 17394043/6103125$, and conversely.

If $I(q^k) + I(n^2) = 43/15$, then $q=5$ and $k=1$, and conversely.

If $I(q^k) + I(n^2) > 43/15$, then $q>5$, and conversely.

Suppose that $I(q^k) + I(n^2) > 43/15$.  It follows that $q > 5$, from which we obtain $q \geq 13$ (since $q$ is a prime satisfying $q \equiv 1 \pmod 4$).

Consequently,
$$I(q^k) < \frac{q}{q-1} \leq \frac{13}{12} = 1.08\overline{333},$$
and hence we have
$$I(n^2) = \frac{2}{I(q^k)} > \frac{24}{13} = 1.\overline{846153},$$
resulting in the improved lower bound
$$2.92\overline{948717} = \frac{457}{156} = \frac{24}{13} + \frac{13}{12} < I(q^k) + I(n^2).$$

We can improve on these estimates if $q>5$ and $k=1$.  We obtain
$$I(q^k) = I(q) = 1+\frac{1}{q} \leq 1+\frac{1}{13}=\frac{14}{13} = 1.0\overline{769230}$$
and
$$I(n^2) \geq \frac{13}{7} = 1.\overline{857142},$$
resulting in the improved lower bound
$$2.\overline{934065} = \frac{267}{91} = \frac{14}{13} + \frac{13}{7} \leq I(q^k) + I(n^2).$$

Hence, we have:

If $q>5$ and $k=1$, then $2.\overline{934065} = 267/91 = 14/13 + 13/7 \leq I(q^k) + I(n^2) < 3$.

If $q>5$ and $k>1$, then $2.92\overline{948717} = 457/156 = 24/13 + 13/12 < I(q^k) + I(n^2) < 3$.

Notice that we have sharper bounds when the Descartes-Frenicle-Sorli conjecture that $k=1$ is true.

Here is my question:

Is it possible to do better than these present bounds?

4.7.17

A curious inequality involving divisors of odd perfect numbers

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

Since $q^k n \mid N$ and $N$ is perfect, then $q^k n$ is deficient.  Therefore, $I(q^k n) < 2$, where $I(x)=\sigma(x)/x$ is the abundancy index of $x \in \mathbb{N}$.

It follows that
$$\frac{1}{2}\cdot\frac{\sigma(q^k)}{n} < \frac{q^k}{\sigma(n)}$$
and
$$\frac{1}{2}\cdot\frac{\sigma(n)}{q^k} < \frac{n}{\sigma(q^k)}.$$
Adding the last two inequalities, we get
$$\frac{\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}}{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$

Since the arithmetic mean is never less than the harmonic mean, and since
$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k},$$
then we have
$$\frac{2}{\frac{1}{\sigma(q^k)/n}+\frac{1}{\sigma(n)/q^k}}< \frac{\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}}{2},$$
so that we obtain
$$\frac{2}{\frac{q^k}{\sigma(n)}+\frac{n}{\sigma(q^k)}}< \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$
This last inequality implies that
$$2 < \bigg(\frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}\bigg)^2$$
which further means that
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}.$$

Now, we also have that (for the case of two summands/factors)
$$\text{ Arithmetic Mean }\cdot\text{ Harmonic Mean } = \bigg(\text{ Geometric Mean }\bigg)^2.$$
(See this hyperlink for a proof.)

Therefore, we have
$$\frac{\frac{\sigma(q^k)}{n}+\frac{\sigma(n)}{q^k}}{\frac{n}{\sigma(q^k)}+\frac{q^k}{\sigma(n)}}=\bigg(\sqrt{\frac{\sigma(q^k)}{n}\cdot\frac{\sigma(n)}{q^k}}\bigg)^2 = I({q^k}n),$$
an equation which can be readily verified by an inspection of the complex fraction on the LHS.  We shall return to this relationship later.

From the inequality
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}$$
we claim that either
$$\frac{\sigma(q^k)}{n} < \sqrt{2} < \frac{\sigma(n)}{q^k}$$
or
$$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n}$$
holds.
It suffices to prove one inequality, as the proof for the other one is very similar.

To this end, assume that
$$\sqrt{2} < \frac{\sigma(q^k)}{n}.$$
This implies that
$$\frac{n}{\sigma(q^k)} < \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
which further means that
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)} < \frac{q^k}{\sigma(n)} + \frac{\sqrt{2}}{2}.$$
We obtain
$$\frac{\sqrt{2}}{2} < \frac{q^k}{\sigma(n)},$$
from which we get
$$\frac{\sigma(n)}{q^k} < \sqrt{2}.$$
It follows that
$$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n},$$
and the proof is done.

We state this result as a theorem in this blog post.

Theorem 1.  Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.  Then we have the unconditional result
$$\sqrt{2} < \frac{q^k}{\sigma(n)} + \frac{n}{\sigma(q^k)}$$
from which it follows that either one of the following conditions hold:
(a)  $$\frac{\sigma(q^k)}{n} < \sqrt{2} < \frac{\sigma(n)}{q^k}$$
(b)  $$\frac{\sigma(n)}{q^k} < \sqrt{2} < \frac{\sigma(q^k)}{n}$$

As before, let $N=q^k n^2$ be an odd perfect number with Euler prime $q$. Suppose that the Descartes-Frenicle-Sorli conjecture that $k=1$ is true. From pages 7 to 8 of this preprint, under the assumption $k=1$, we have the following cases to consider:

$$\text{ Case I: } \sigma(q)/n < 1 < \sqrt{2} < \sigma(n)/q \Longrightarrow q < n$$
$$\text{ Case II: } 1 < \sigma(q)/n < \sqrt{2} < \sigma(n)/q < 2 \Longrightarrow n < q < n\sqrt{2}$$
$$\text{ Case III: } \sigma(n)/q < \sqrt{2} < \sigma(q)/n < \sqrt{3}+{10}^{-375} \Longrightarrow n < q < n\sqrt{3}$$

THIS BLOG POST IS CURRENTLY A WORK IN PROGRESS.

24.6.17

If $N=q^k n^2$ is an odd perfect number and $q = k$, why does this bound not imply $q > 5$?

Let $\mathbb{N}$ denote the set of natural numbers (i.e., positive integers).

A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum of divisors.  For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect.  (Note that $6$ is even.)  Denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.

Euler proved that an odd perfect number $N$, if any exists, must take the form $N=q^k n^2$, where $q$ is prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Suppose that $k=q$.  Since $q$ is prime and $q \equiv 1 \pmod 4$, this implies that $k \geq 5$.  (In particular, $k \neq 1$, so that the Descartes-Frenicle-Sorli conjecture is false in this case.)

Using WolframAlpha, we get the upper bound
$$I(q^k)=I(q^q)=\frac{q^{q+1}-1}{{q^q}(q-1)} \leq \frac{3906}{3125} = 1.24992$$
which corresponds to the lower bound
$$I(n^2)=\frac{2}{I(q^k)} \geq \frac{3125}{1953} \approx 1.6001\ldots.$$
Consider the product
$$\bigg(I(q^q) - \frac{3906}{3125}\bigg)\bigg(I(n^2) - \frac{3906}{3125}\bigg).$$
This product is nonpositive.  Therefore,
$$I(q^q)I(n^2) + \bigg(\frac{3906}{3125}\bigg)^2 \leq \frac{3906}{3125}\cdot\bigg(I(q^q) + I(n^2)\bigg).$$
Since $N=q^k n^2$ is perfect with $q=k$, then $I(q^k)I(n^2)=I(q^q)I(n^2)=2$, so that
$$I(q^q) + I(n^2) \geq \frac{3906}{3125} + \frac{3125}{1953} = \frac{17394043}{6103125} \approx 2.850022406554\ldots.$$

But in the paper [Dris, 2012 (pages 4 to 5)], it is proved that
$$I(q^k) + I(n^2) \leq \frac{3q^2 + 2q + 1}{q(q+1)} = 3 - \frac{q-1}{q(q+1)}$$
with equality occurring if and only if $k=1$.

In our case, since $k = q \geq 5$, we obtain
$$\frac{17394043}{6103125} \leq I(q^q) + I(n^2) = I(q^k) + I(n^2) < 3 - \frac{q-1}{q(q+1)}$$
$$q > \frac{3125}{781} \approx 4.00128\ldots.$$

Here is my question:
Why does the bound 
$$I(q^q) + I(n^2) \geq \frac{3906}{3125} + \frac{3125}{1953} = \frac{17394043}{6103125} \approx 2.850022406554\ldots$$
not imply that $q > 5$?

I am thinking along the lines that:

(1) $57/20 < I(q^k) + I(n^2) < 3$ is best-possible.

(2) Improving the upper bound $3$ would result in a finite upper bound for the Euler prime $q$.

(3) Therefore, improving the lower bound $57/20$ would result in a lower bound for $q$ better than the currently known $q \geq 5$.

REFERENCES

POSTED ANSWER

I am guessing that it has got something to do with the interaction between the conditions $k=1$ and $q=5$.

When $k=1$, we have the bounds
$$I(q^k)=I(q)=1+\frac{1}{q} \leq \frac{6}{5}$$
and
$$I(n^2)=\frac{2}{I(q)} \geq \frac{5}{3}.$$

When $q=5$, we have the bounds
$$I(n^2) \leq 2 - \frac{5}{3q} = \frac{5}{3}$$
and
$$I(q^k) \geq \frac{6}{5}.$$

Note that, when $k=1$, we have the lower bound
$$I(q^k) + I(n^2) \geq \frac{43}{15} = 2.8\overline{666} > 2.85$$
Note further that, when $q=5$, we have the upper bound

$$I(q^k) + I(n^2) \leq \frac{43}{15}.$$

ADDITIONAL NOTES

It is easy to show that
$$\bigg(q = 5\bigg) \land \bigg(k = 1\bigg) \Longrightarrow I(q^k) + I(n^2) = \frac{43}{15}.$$

I am currently in the process of writing up a proof for the reverse implication
$$I(q^k) + I(n^2) = \frac{43}{15} \Longrightarrow \bigg(\left(q = 5\right) \land \left(k = 1\right)\bigg).$$

11.6.17

A question on the Euler prime of odd perfect numbers

(Note:  This post was copied verbatim from this MSE link.)


A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function.  For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect.  (Note that $6$ is even.)

It is currently unknown whether there are any odd perfect numbers.  Euler proved that an odd perfect number, if any exists, must take the form $N = q^k n^2$ where $q$ is prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  We call $q$ the Euler prime of the odd perfect number $N$.

In what follows, we denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.

MOTIVATION

Since the non-Euler part $n^2$ is deficient but not almost perfect, we know (from this preprint) that
$$I(n^2) < \frac{2n^2 + D(n^2)}{n^2 + D(n^2)},$$
where $D(n^2)={2n^2}-\sigma(n^2)$ is the deficiency of $n^2$.

When the Descartes-Frenicle-Sorli conjecture that $k=1$ is true, we also know that we have the lower bound
$$\frac{5}{3} \leq I(n^2),$$
whereupon we obtain, from the last two inequalities, the following lower bound
$$\frac{n^2}{D(n^2)} > 2.$$
(When $k=1$, ${n^2}/D(n^2)=(q+1)/2$, so that ideally we should have ${n^2}/D(n^2) \geq 3$ since $q \geq 5$.)

From the following Answer 1 and Answer 2 to a related MSE question, we obtain improved upper bounds for $I(n^2)$ by successively computing the mediant of
$$I(n^2) = \frac{2n^2 - D(n^2)}{n^2}=I_0$$
and the upper bounds $u_i$ (for $I(n^2)$) given by
$$u_0 = \frac{2n^2 + D(n^2)}{n^2 + D(n^2)}$$
$$u_1 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(2n^2 + D(n^2)\bigg)}{n^2 + \bigg(n^2 + D(n^2)\bigg)} = \frac{4n^2}{2n^2 + D(n^2)}$$
$$u_2 = \frac{\bigg(2n^2 - D(n^2)\bigg)+4n^2}{n^2 + \bigg(2n^2 + D(n^2)\bigg)}=\frac{6n^2 - D(n^2)}{3n^2 + D(n^2)}$$
$$u_3 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(6n^2 - D(n^2)\bigg)}{n^2 + \bigg(3n^2 + D(n^2)\bigg)} = \frac{8n^2 - 2D(n^2)}{4n^2 + D(n^2)}$$
$$u_4 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(8n^2 - 2D(n^2)\bigg)}{n^2 + \bigg(4n^2 + D(n^2)\bigg)} = \frac{10n^2 - 3D(n^2)}{5n^2 + D(n^2)}$$
$$\ldots \ldots \ldots$$
$$\ldots \ldots \ldots$$
$$\ldots \ldots \ldots$$
(That is, $u_i = \text{mediant}(u_{i-1},I_0)$.)

Note that it is easy to prove that $u_i$ (for $i=0, 1, 2, \ldots$) is a strictly decreasing sequence.

The resulting lower bounds for ${n^2}/D(n^2)$ (on assuming $k=1$) are:

$$u_0 : \frac{n^2}{D(n^2)} > 2$$
$$u_1 : \frac{n^2}{D(n^2)} > \frac{5}{2}$$
$$u_2 : \frac{n^2}{D(n^2)} > \frac{8}{3}$$
$$u_3 : \frac{n^2}{D(n^2)} > \frac{11}{4}$$
$$u_4 : \frac{n^2}{D(n^2)} > \frac{14}{5}$$
$$\ldots : \ldots$$

QUESTIONS

(1)   Do the resulting lower bounds for ${n^2}/D(n^2)$ (using this procedure) have a closed form?

(2)  Does the lower bound ever reach $3$, or is it possible to prove that it does not?

POSTED ANSWER

It appears that the numerators of the lower bounds form OEIS sequence A016789.  Hence, the lower bounds have the closed form
$$\frac{3i + 2}{i + 1}.$$

Consequently,
$$\frac{3i + 2}{i + 1}=3-\frac{1}{i + 1},$$

so that the lower bounds never reach $3$.

Paolo Starni's "On Dris Conjecture about Odd Perfect Numbers"

Paolo Starni recently uploaded the preprint titled "On Dris Conjecture about Odd Perfect Numbers", with details below:

Abstract
The Euler's form of odd perfect numbers, if any, is $n = {{\pi}^{\alpha}} N^2$, where $\pi$ is prime, $\gcd(\pi, N) = 1$ and $\pi \equiv \alpha \equiv 1 \pmod 4$.  Dris conjecture states that $N > {\pi}^{\alpha}$.  We find that $N^2 > \frac{1}{2}{{\pi}^{\gamma}}$, with $\gamma = \max\{\omega(n) - 1, \alpha\}$; $\omega(n) \geq 9$ is the number of distinct prime factors of $n$.

3.6.17

On the condition $n < \sigma(q^k)$ where $N = q^k n^2$ is an odd perfect number given in Eulerian form

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

Brown recently announced a proof for the inequality $q < n$, and a partial proof that $q^k < n$ holds "in many cases".  The inequality $q^k < n$ was conjectured by Dris in his M. Sc. thesis.

Note that when the proof for $q^k < n$ is completed, it would follow that the following implication holds:
$$n < \sigma(q^k) \Longrightarrow k > 1.$$
(This is because $q$ and $\sigma(q)=q+1$ are consecutive integers.)

In this post, we will derive bounds for the quantities
$$\frac{\sigma(q^k)}{\sigma(n)}$$
and
$$\frac{q^k}{n}.$$
Notice that, already we have
$$\frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n}$$
since
$$I(q^k) < \frac{5}{4} < \bigg(\frac{8}{5}\bigg)^{\frac{\ln(4/3)}{\ln(13/9)}} < I(n),$$
where
$$I(x) = \frac{\sigma(x)}{x}$$
is the abundancy index of $x \in \mathbb{N}$.

First, we dispose of the following lemmas:

Lemma 1.
$$\frac{\sigma(q^k)}{\sigma(n)} = \frac{\frac{\sigma(q^k)}{q^k} + \frac{\sigma(q^k)}{n}}{\frac{\sigma(n)}{q^k} + \frac{\sigma(n)}{n}}$$

Lemma 2.
$$\frac{q^k}{n} = \frac{\frac{q^k}{\sigma(q^k)} + \frac{q^k}{\sigma(n)}}{\frac{n}{\sigma(q^k)} + \frac{n}{\sigma(n)}}$$

The proofs of Lemmas 1 and 2 are trivial.

(This blog post is currently a WORK IN PROGRESS.)

Verification of an equation from a recent preprint on odd perfect numbers (to appear in NNTDM)

The paper titled "Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers" has finally been accepted and is set to appear in the journal Notes on Number Theory and Discrete Mathematics (Volume 23, Number 2) this June or July, 2017.

The latest preprint for this paper is available online via the arXiv.

In this blog post, we will verify the equation
$$I(n^2)=2-\frac{5}{3q}=\frac{6q-5}{3q}$$
which is true if and only if $q=5$ and $k=1$, where $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $N=q^k n^2$ is an odd perfect number with Euler prime $q$.

To this end, suppose that
$$I(n^2)=2-\frac{5}{3q}=\frac{6q-5}{3q}.$$
This implies that
$$I(q^k)=\frac{2}{I(n^2)}=\frac{6q}{6q-5},$$
and that
$$q = 5 \land k = 1.$$
But $k=1$ implies that
$$I(q^k)+I(n^2)=I(q)+\frac{2}{I(q)}=\frac{q+1}{q}+\frac{2q}{q+1}$$
so that
$$I(q^k)+I(n^2)=\frac{(q+1)^2 + 2q^2}{q(q+1)}=\frac{3q^2 + 2q + 1}{q(q+1)}=3-\frac{q-1}{q(q+1)}.$$
Plugging in the values for $I(q^k)$ and $I(n^2)$ from above, and then solving for $q$, we obtain
$$\frac{6q}{6q-5}+\frac{6q-5}{3q}=\frac{3q^2 + 2q + 1}{q(q+1)}$$
$$\bigg(q = \frac{8}{3}\bigg) \lor \bigg(q = 5\bigg),$$
Since $q \geq 5$, $q = 5$ holds, which validates the result.

As an additional validation, when $q=5$ and $k=1$, we get
$$I(q^k)+I(n^2)=I(q)+\frac{2}{I(q)}=I(5)+\frac{2}{I(5)}=\frac{6}{5}+\frac{5}{3}=\frac{43}{15}.$$
Solving the equation
$$\frac{6q}{6q-5}+\frac{6q-5}{3q}=\frac{43}{15}$$
gives
$$\bigg(q = \frac{25}{12}\bigg) \lor \bigg(q = 5\bigg),$$
Since $q \geq 5$, $q = 5$ holds, which again validates the result.

2.6.17

Building on work from previous MSE question 2306650 (Re: Odd Perfect Numbers)

(Note:  This post builds on work from this previous MSE question.)

Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$.  If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number.  Denote the abundancy index $I$ of $N$ as $I(N)=\sigma(N)/N$.

Euler proved that every odd perfect number $N$ has to have the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In this paper, Dris proved that we have the bounds
$$L(q) = 3 - \frac{q-2}{q(q-1)} < I_{+} \leq 3 - \frac{q-1}{q(q+1)} = U(q)$$
for the sum of the abundancy indices $I_{+}=I(q^k)+I(n^2)$.  In particular, if $I_{+} \leq 3 - \epsilon$, then we will have $L(q) < 3 - \epsilon$, so that (using WolframAlpha) we get
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
if $0 < \epsilon < 3 - 2\sqrt{2}$.

Here is the problem:

PROBLEM STATEMENT

Does the converse hold?  That is, if $N=q^k n^2$ is an odd perfect number, does
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon},$$
with $0 < \epsilon < 3 - 2\sqrt{2}$, imply that $I(q^k) + I(n^2) \leq 3 - \epsilon$?

MY ATTEMPT

Assume that $$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon},$$
with $0 < \epsilon < 3 - 2\sqrt{2}$, and suppose to the contrary that
$$I(q^k) + I(n^2) > 3 - \epsilon.$$

This means that $U(q) > 3 - \epsilon$, so that (by using WolframAlpha again) we obtain (for $0 < \epsilon < 3 - 2\sqrt{2}$) either
$$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon}$$
or
$$0 < q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}.$$

Here is my question:  Do the resulting inequalities
$$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon}$$
or
$$0 < q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}},$$
for $0 < \epsilon < 3 - 2\sqrt{2}$, contradict the assumption
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}?$$

POSTED ANSWER


The upper bound

$$q < \frac{1-\epsilon}{2\epsilon} - \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}$$
is inconsistent with the lower bound
$$\frac{\epsilon + 1}{2\epsilon}-\frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}} < q$$
because of this WolframAlpha computation, where I have set $\epsilon=x$.

However, the lower bound
$$q > \frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon}$$
is not inconsistent with the upper bound
$$q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
because the inequality
$$\frac{\bigg(\sqrt{\frac{1}{{\epsilon}^2} - \frac{6}{\epsilon} + 1} - 1\bigg)\epsilon + 1}{2\epsilon} < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
is satisfied for $\epsilon > 0$, per this WolframAlpha computation, where again I have set $\epsilon = x$.

We therefore conclude that, if $N=q^k n^2$ is an odd perfect number with Euler prime $q$, then given $0 < \epsilon < 3 - 2\sqrt{2}$, we have the biconditional
$$q < \frac{1}{2}\sqrt{\frac{{\epsilon}^2 - 6\epsilon + 1}{{\epsilon}^2}}+\frac{\epsilon + 1}{2\epsilon}$$
if and only if $I(q^k) + I(n^2) \leq 3 - \epsilon$.

The case $\epsilon = 0.01$ verifies this to be the case, as in this MSE question.

If $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

(Note:  This blog post was copied verbatim from this MSE question.)

Let $\sigma(N)$ denote the sum of the divisors of the natural number $N$.  If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number.  Denote the abundancy index $I$ of $N$ as $I(N)=\sigma(N)/N$.

Euler proved that every odd perfect number $N$ has to have the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In this paper, Dris proved that we have the bounds
$$L(q) = 3 - \frac{q-2}{q(q-1)} < I_{+} \leq 3 - \frac{q-1}{q(q+1)} = U(q)$$
for the sum of the abundancy indices $I_{+}=I(q^k)+I(n^2)$.  In particular, if $I_{+} \leq 2.99$, then we will have $L(q) < 2.99$, so that (using WolframAlpha) we get
$$q < \frac{101 + \sqrt{9401}}{2} \approx 98.9794,$$
(since $q$ satisfies $q \geq 5$) from which we obtain $q \leq 97$ (since $q$ is prime).

Here is the problem:

PROBLEM STATEMENT

Does the converse hold?  That is, if $N=q^k n^2$ is an odd perfect number, does $q \leq 97$ imply that $I(q^k) + I(n^2) \leq 2.99$?

MY ATTEMPT

Since $N=q^k n^2$ is perfect and $\gcd(q,n)=1$, then $I(q^k)I(n^2)=2$, from which we have
$$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(q)} = \frac{2q}{q+1} = \frac{2}{1 + \frac{1}{q}}.$$

By assumption, $q \leq 97$, so that we get
$$\frac{1}{q} \geq \frac{1}{97} \Longrightarrow 1 + \frac{1}{q} \geq \frac{98}{97} \Longrightarrow \frac{2}{1 + \frac{1}{q}} \leq \frac{97}{49} \approx 1.97959\ldots.$$

However, this does not match up with the known upper bound
$$I(q^k) < \frac{5}{4} = 1.25.$$

ANOTHER APPROACH

Assume that $q \leq 97$, and suppose to the contrary that
$$I(q^k) + I(n^2) > 2.99.$$

This means that $U(q) > 2.99$, so that (by using WolframAlpha again) we obtain
$$q > \frac{99 + \sqrt{9401}}{2} \approx 97.9794,$$
contradicting $q \leq 97$. 

Here is my question:  Is this proof in the ANOTHER APPROACH section correct?

28.5.17

If $q^k n^2$ is an odd perfect number, then $k$ is bounded.

If $k$ is arbitrarily large in the equations
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
then
$$\lim_{k \to \infty}{\frac{\sigma(n^2)}{q^k}}=\lim_{k \to \infty}{\frac{2n^2}{\sigma(q^k)}}=\lim_{k \to \infty}{\bigg(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\bigg)}=0$$
while
$$\gcd(n^2,\sigma(n^2))>1$$
for a fixed Euler part $n^2$. This implies that $k$ is bounded.

UPDATE (June 11, 2017)

This claim is flawed.  Unconditionally, we know that $q^k < n^2$.  This implies that $k$ and $n$ are dependent.

Paper accepted in Notes on Number Theory and Discrete Mathematics

My paper titled "Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers" has finally been accepted and is set to appear in the journal Notes on Number Theory and Discrete Mathematics (Volume 23, Number 2) this June or July, 2017.


The latest preprint for this paper is available online via the arXiv.

If you have any comments and/or questions regarding this paper, you may e-mail me at the address indicated in the arXiv preprint.

Happy reading everyone!  😊