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7.7.19

If $N = q^k n^2$ is an odd perfect number with special/Euler prime $q$, then $q=5$.

MSE QUESTION

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.  That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

From a comment underneath this earlier question, we have the equation (and corresponding inequalities)
$$1 < \frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1} \leq \frac{5}{4}$$
since $q$ is prime with $q \equiv 1 \pmod 4$ implies that $q \geq 5$.
This implies that
$$\frac{4}{5} \leq \frac{\frac{\varphi(N)}{N}}{\frac{\varphi(n)}{n}} = \frac{q-1}{q} < 1.$$

But from the following source:
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}.$$

However, we also have
$$\frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}.$$
Notice that
$$\frac{4}{5} \leq \frac{\varphi(q^k)}{q^k} = \frac{q^k \bigg(1 - \frac{1}{q}\bigg)}{q^k} = \frac{q - 1}{q} < 1.$$ 
Therefore, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{\varphi(n)}{n},$$
and
$$\frac{4}{5}\cdot\frac{\varphi(n)}{n} \leq \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{1}{2},$$
which implies that
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}.$$

WolframAlpha gives the rational approximation
$$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$

Here is my question:
Is it possible to improve on the bounds for $\varphi(N)/N$, if $N = q^k n^2$ is an odd perfect number with special prime $q$?

MOTIVATION FOR THE INQUIRY

It can be shown that the equation
$$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$
together with the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}$$
and
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}$$
imply
$$0.92 \approx \frac{\frac{120}{217\zeta(3)}}{\frac{1}{2}} < \frac{q}{q-1} < \frac{\frac{5}{8}}{\frac{120}{217\zeta(3)}} \approx 1.358574729,$$
from which we obtain trivial bounds.

Nonetheless, it can be shown that the equation
$$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$
together with the upper bound $\varphi(N)/N < 1/2$ implies that
$$q < \frac{x}{x-1}$$
where
$$x = \frac{2\varphi(n)}{n}.$$
Thus, if we can improve the upper bound for $\varphi(N)/N$ to something smaller than $1/2$ (say $1/2 - \varepsilon$ for some tiny $\varepsilon > 0$), then we can improve the coefficient of $\frac{\varphi(n)}{n}$ in $x$ to some number bigger than $2$.  Likewise, if we can get a better lower bound for $\varphi(N)/N$, then we will be able to get an improved lower bound for $\varphi(n)/n$.  Together, they would translate (hopefully!) to a numerical upper bound for the special/Euler prime $q$!


Eureka!!!

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

From the equation and lower bound for $\varphi(N)/N$
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}$$
and the equation
$$\frac{\varphi(q^k)}{q^k} = \frac{q - 1}{q},$$
we get the lower bound
$$2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1} < \frac{2\varphi(n)}{n} = x.$$
This implies that we have the upper bound
$$q < \frac{x}{x-1} < \frac{2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}}{\bigg(2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}\bigg) - 1}$$
which can be solved using WolframAlpha, yielding the upper bound
$$q < \frac{217\zeta(3)}{217\zeta(3) - 240} \approx 12.5128,$$
from which it follows that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.  

QED

REVISED BOUNDS FOR THE ABUNDANCY INDICES OF DIVISORS OF ODD PERFECT NUMBERS

Since $q=5$ holds, then since $q \mid q^k$ (for all positive integers $k$), then
$$\frac{q+1}{q} = I(q) \leq I(q^k) < \frac{q}{q - 1} < \frac{2(q - 1)}{q} < \frac{2}{I(q^k)} = I(n^2) \leq \frac{2q}{q + 1},$$
where $I(x)=\sigma(x)/x$ is the abundancy index of the positive integer $x$ (and $\sigma(x)$ is the sum of divisors of $x$).  We therefore have the revised bounds
$$\frac{6}{5} \leq I(q^k) < \frac{5}{4} < \frac{8}{5} < I(n^2) \leq \frac{5}{3}.$$


Note that we then have

$$\bigg(I(q^k) - \frac{6}{5}\bigg)\bigg(I(n^2) - \frac{6}{5}\bigg) \geq 0$$

which implies
$$I(q^k)I(n^2) - \frac{6}{5}\bigg(I(q^k) + I(n^2)\bigg) + \bigg(\frac{6}{5}\bigg)^2 \geq 0$$
from which it follows that
$$\frac{43}{15} = \frac{5}{3} + \frac{6}{5} = 2\cdot\frac{5}{6} + \frac{6}{5} \geq I(q^k) + I(n^2).$$
We also have the lower bound
$$I(q^k) + I(n^2) > \frac{57}{20}.$$

Better values/bounds are known when the Descartes-Frenicle-Sorli Conjecture that $k=1$ is assumed true (or otherwise), given that $q=5$ holds.  (See the following MSE question for more information: On the Descartes-Frenicle-Sorli conjecture and the Euler prime of odd perfect numbers.)

18.4.19

Arnie Dris's Publications - 1st Quarter, 2019

A note on the OEIS sequence A228059 (co-authored with Doli-Jane Uvales Tejada)

Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers?

Let
$$\sigma(x) = \sum_{e \mid x}{e}$$
denote the sum of divisors of the positive integer $x$.  Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$.  A positive integer $N$ is said to be deficient-perfect if $D(N) \mid N$.

Here is my question:
Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers $N > 1$?
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$

(Note that the inequality
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$
is true if and only if $N$ is deficient.)

References


Posted Answer #1

ILLUSTRATING VIA A TOY EXAMPLE

Let $M$ be an odd perfect number given in the so-called Eulerian form
$$M = p^k m^2$$
(i.e. $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$).

It is known that the non-Euler part $m^2$ is deficient-perfect if and only if the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.  (See this paper for a proof of this fact.)

So, suppose that $k=1$.  Then $m^2$ is deficient-perfect.

In particular, $m^2$ is deficient, so that the criterion in this paper applies.

We have
$$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}.$$

Under the hypothesis that $k=1$, $m^2$ is deficient-perfect, with deficiency 
$$D(m^2) = \frac{m^2}{(p+1)/2}.$$

We also have
$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p+1}.$$

Putting these all together, we have
$$\frac{m^2}{D(m^2)} = \frac{p+1}{2}$$
$$\frac{2p}{p+1} = I(m^2) > \frac{2m^2}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg)}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg)}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+1}{\frac{p+3}{2}} = \frac{2(p+1)}{p+3}$$
which implies that
$$p^2 + 3p = p(p+3) > (p+1)^2 = p^2 + 2p + 1$$
$$p > 1$$
(This last inequality is trivial as $p$ is prime with $p \equiv 1 \pmod 4$ implies that $p \geq 5$.)
$$\frac{2p}{p+1} = I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg) + 1}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg) + 1}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+2}{\frac{p+3}{2}} = \frac{2(p+2)}{p+3}$$
which implies that
$$p^2 + 3p = p(p+3) < (p+1)(p+2) = p^2 + 3p + 2$$
$$0 < 2.$$


This example illustrates my interest in improvements to the bounds in terms of the abundancy index and deficiency functions of $N$, when $N > 1$ is deficient-perfect.

Posted Answer #2

Suppose that $N > 1$ is deficient-perfect.  Since $N$ is deficient, then
$$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}.$$

I think that, since $D(N) \mid N$ when $N$ is deficient-perfect, then $N/D(N)$ is an integer, so that we have (since $\frac{N}{D(N)} \mid N$)
$$I\bigg(\frac{N}{D(N)}\bigg) \leq I(N) < \frac{2N + D(N)}{N + D(N)} = \frac{2\bigg(\frac{N}{D(N)}\bigg) + 1}{\bigg(\frac{N}{D(N)}\bigg) + 1}.$$

CLAIM

$$\frac{2\bigg(\frac{N}{D(N)}\bigg)}{\bigg(\frac{N}{D(N)}\bigg) + 1} < I\bigg(\frac{N}{D(N)}\bigg)$$

This claim, if true, would prove that all deficient-perfect numbers $N$ correspond to almost perfect numbers $N/D(N)$.

Added April 18 2019 (6:13 PM - Manila time)

The claim is false.  A counterexample is given by 
$$N = \bigg({3}\cdot{7}\cdot{11}\cdot{13}\bigg)^2.$$

Added April 18 2019 (6:17 PM - Manila time)


It appears that the claim is true when $D(N)=1$.

10.3.19

Breaking the barriers at $q=5$ and $q=13$ for $q^k n^2$ an odd perfect number with special prime $q$

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


Let $\sigma(x)$ be the sum of divisors of the positive integer $x$.  If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.

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

Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $\sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.

So now suppose that $\sigma(n^2)/q^k$ is a square.  This implies that $k=1$, and also that $\sigma(n^2) \equiv 1 \pmod 4$, since $\sigma(n^2)/q^k$ is odd and $q \equiv k \equiv 1 \pmod 4$.

The congruence $\sigma(n^2) \equiv 1 \pmod 4$ then implies that $q \equiv k \pmod 8$.  (See this MO post for the details.)  Substituting $k=1$, we obtain
$$q \equiv 1 \pmod 8.$$

This implies that the lowest possible value for the special prime $q$ is $17$.  (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $\sigma(n^2)/q^k$ is a square.)  Note that, if $q=17$, then $(q+1)/2 = 3^2 \mid n^2$.

Here is my question:
Can we push the lowest possible value from $q \geq 17$, to say, $q \geq 41$ or even $q \geq 97$, using the ideas in this post, and possibly more?


POSTED ANSWER


Note that if
$$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}$$
is a square, then $k=1$ and $\sigma(q^k)/2 = (q+1)/2$ is also a square.

The possible values for the special prime satisfying $q < 100$ and $q \equiv 1 \pmod 8$ are $17$, $41$, $73$, $89$, and $97$.

For each of these values:
$$\frac{q_1 + 1}{2} = \frac{17 + 1}{2} = 9 = 3^2$$
$$\frac{q_2 + 1}{2} = \frac{41 + 1}{2} = 21 \text{ which is not a square.}$$
$$\frac{q_3 + 1}{2} = \frac{73 + 1}{2} = 37 \text{ which is not a square.}$$
$$\frac{q_4 + 1}{2} = \frac{89 + 1}{2} = 45 \text{ which is not a square.}$$
$$\frac{q_5 + 1}{2} = \frac{97 + 1}{2} = 49 = 7^2$$

Thus, if $\sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q \geq 97$.

23.2.19

On the golden ratio and odd perfect numbers

(Note:  The following post was copied verbatim from this MSE question.)


On the golden ratio and odd perfect numbers

Here is my question:
Is $I(n^2) - 1 > 1/I(n^2)$ true when $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $k>1$?

My Attempt

If $k>1$, then since $q$ is the special prime, then $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  In particular, we know that $q \geq 5$ and $k \geq 5$.

We know that
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} \leq \frac{5}{4}.$$

It follows that
$$I(n^2) = \frac{2}{I(q^k)} > \frac{2(q - 1)}{q} \geq \frac{8}{5}.$$

Thus,
$$I(n^2) - 1 > \frac{2(q - 1)}{q} - 1 = \frac{(2q - 2) - q}{q} = \frac{q - 2}{q} > \frac{q}{2(q - 1)} > \frac{1}{I(n^2)}$$
where the inequality
$$\frac{q - 2}{q} > \frac{q}{2(q - 1)}$$
holds provided $q > 3+\sqrt{5} \approx 5.23607$.

However, the resulting inequality for $I(n^2)$ from
$$I(n^2) - 1 > \frac{1}{I(n^2)}$$
together with the following upper bound for $I(n^2)$ (which holds when $k>1$)
$$\frac{2q}{q+1} > I(n^2)$$
only yields
$$\frac{2q}{q+1} > I(n^2) > \frac{\sqrt{5}+1}{2}$$
thereby giving
$$q > \frac{1+\sqrt{5}}{3-\sqrt{5}} = 2+\sqrt{5} \approx 4.23607.$$

Comments

Note that there is no discrepancy when $k=1$, as then we have
$$I(n^2) - 1 \geq \frac{2}{3} > \frac{3}{5} \geq \frac{1}{I(n^2)}$$
yielding the lower bound $q > 2+\sqrt{5} \approx 4.23607$ from
$$\frac{2q}{q+1}=I(n^2) > \frac{\sqrt{5}+1}{2}.$$

When $k>1$, we get
$$I(n^2) - 1 > \frac{3}{5} \not\gt \frac{5}{8} > \frac{1}{I(n^2)}.$$

4.1.19

On Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College

(Note:  This blog post was lifted verbatim from this MSE question and the answer contained therein.)


THE QUESTION

In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$.  Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)$.  Finally, let $s(X)=\sigma(X)-X$ denote the sum of the aliquot divisors of $X$.

This is a question about Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College, titled Characterization of the Positive Integers with Abundancy Index of the Form $(2x-1)/x$.  (A copy of the poster presentation is available via the following hyperlink.)

In the abstract of the paper, it is stated in the fourth sentence that
Rational numbers of the form $(2x-1)/x$ are important since both even and odd perfect numbers have a divisor with abundancy index of this form.

Let $M = 2^{p-1}(2^p - 1)$ be an even perfect number, and let $N = q^k n^2$ be an odd perfect number.

Clearly,
$$I(2^{p-1}) = \frac{2^p - 1}{2^{p-1}} = \frac{2x_1 - 1}{x_1}$$
where $x_1 = 2^{p-1}$.  (In other words, $2^{p-1}$ is an even almost perfect number, since it is a power of two.)

However,
$$I(p^k) = \frac{p^{k+1} - 1}{p^{k+1} - p^k}$$
and
$$I(n^2) = \frac{2}{I(p^k)} = \frac{2(p^{k+1} - p^k)}{p^{k+1} - 1}$$
so clearly $p^k$ is not almost perfect (since $p$ must be odd).

Additionally, since
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)} \geq 3$$
(see the paper [Dris, 2012]), then clearly $n^2$ is likewise not almost perfect.  (Similarly, it can be proved that $n$ and $q^k n$ are not almost perfect.)

So I think the trivial divisor $1$ of an odd perfect number has the required abundancy index
$$I(1) = 1 = \frac{2\cdot{1} - 1}{1} = \frac{2x_2 - 1}{x_2}$$
where $x_2 = 1$.

Here is my question:
Is there any other divisor $m > 1$ of an odd perfect number $N = q^k n^2$ such that
$$I(m) = \frac{2x - 1}{x}$$
for some positive integer $x$?

POSTED ANSWER

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

Suppose that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.

Then
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = 2 - \frac{1}{(q+1)/2} = \frac{2((q+1)/2) - 1}{(q+1)/2},$$
where $(q+1)/2$ is an integer (since $q \equiv 1 \pmod 4$).

It remains to consider the case when $k>1$.  (Note that $k \equiv 1 \pmod 4$.)

24.12.18

Arnie Dris's Publications - 4th Quarter, 2018

Revisiting some old results on odd perfect numbers (co-authored with Doli-Jane Uvales Tejada)

10.10.18

Arnie Dris's Publications - 3rd Quarter, 2018

Conditions equivalent to the Descartes–Frenicle–Sorli Conjecture on odd perfect numbers – Part II (co-authored with Doli-Jane Uvales Tejada)