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20.9.17

Some remarks on the Descartes-Frenicle-Sorli conjecture on odd perfect numbers


In what follows, let $a, b, c \in \mathbb{N}$ and let $x, y \in \mathbb{Z}$.  Denote the abundancy index of $z \in \mathbb{N}$ by $I(z)=\sigma(z)/z$, and the deficiency of $z$ by $D(z)=2z-\sigma(z)$, where $\sigma(z)$ is the sum of the divisors of $z$.

This blog post is an offshoot of the following MSE post.  Essentially, we will be using the equation
$$\mathscr{A}\text{: }ax + by= c=\gcd(a,b)=\gcd(a,c)=\gcd(c,b)$$
in what follows.  Note that the values $x$ and $y$ in (Equation $\mathscr{A}$) are not unique.  (As hinted by Bill Dubuque in several comments to the MSE question mentioned, this equation can be proved in many ways, one of which is via the GCD Distributive Law.)

Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$.  (That is, $q$ is a prime that satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  Thus, $q, k, n, N \in \mathbb{N}$.)

Now, recall from this NNTDM paper that we have the equations
$$\mathscr{B}\text{: } \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)),$$
where $\sigma(x)$ is the sum of the divisors of $x$ and $D(y)=2y-\sigma(y)$ is the deficiency of $y$.

In particular, note from (Equation $\mathscr{A}$) (and the fact that both $n^2$ and $\sigma(n^2)$ are odd) that
$$\mathscr{C}\text{: } \gcd(n^2,\sigma(n^2))=\gcd(2n^2-\sigma(n^2),n^2)=\gcd(D(n^2),n^2)$$
and
$$\mathscr{D}\text{: } \gcd(n^2,\sigma(n^2))=\gcd(2n^2 - \sigma(n^2),\sigma(n^2))=\gcd(D(n^2),\sigma(n^2)).$$
But notice that (from (Equation $\mathscr{B}$) and (Equation $\mathscr{C}$)) we have
$$\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\gcd(n^2,\sigma(n^2))=\gcd(D(n^2),n^2),$$
and that (from (Equation $\mathscr{B}$) and (Equation $\mathscr{D}$)) we also have
$$\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2))=\gcd(D(n^2),\sigma(n^2)).$$
These last two equations imply that
$$\frac{\sigma(n^2)}{q^k} \text{   divides   } D(n^2)$$
and
$$\frac{D(n^2)}{\sigma(q^{k-1})} \text{   divides   } \sigma(n^2).$$
It follows from the last two divisibility constraints that
$$\frac{\sigma(n^2)}{{q^k}\sigma(q^{k-1})} \mid \frac{D(n^2)}{\sigma(q^{k-1})} \mid \sigma(n^2)$$
and also that
$$\frac{D(n^2)}{{q^k}\sigma(q^{k-1})} \mid \frac{\sigma(n^2)}{q^k} \mid D(n^2).$$

We therefore conclude that $\sigma(q^{k-1})$ divides $\sigma(n^2)$ and that $q^k$ divides $D(n^2)$.

This last divisibility constraint appears to result in a contradiction, as it implies that
$$\frac{D(n^2)}{q^k}=\frac{2n^2 - \sigma(n^2)}{q^k}$$
is an integer, which further means that 
$$\frac{2n^2}{q^k}$$
must also be an integer, since $\sigma(n^2)/q^k$ is an integer.  This contradicts $\gcd(q,n)=\gcd(q^k,n^2)=1$.

However, I think I may have used the assumption $k=1$ implicitly in the above proof.

Update (September 20, 2017, 7:00 AM [Manila time])
The assumption $k=1$ was not implicitly used in the argument above.  It turned out that
$$\frac{D(n^2)}{{q^k}\sigma(q^{k-1})} \not\in \mathbb{N}.$$

From (Equation $\mathscr{B}$), we also obtain
$$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)),$$
from which we have
$$\frac{\sigma(n^2)-n^2}{q^k - \sigma(q^k)/2}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)).$$
But we can simplify the first fraction in the last equation as follows
$$\frac{\sigma(n^2)-n^2}{q^k - \sigma(q^k)/2} = \frac{\sigma(n^2)-n^2}{D(q^k)/2} = 2\cdot\frac{\sigma(n^2)-n^2}{D(q^k)}.$$
In particular, by using (Equation $\mathscr{B}$) again, we get
$$D(q^k)D(n^2) = 2(\sigma(n^2) - n^2)\sigma(q^{k-1}).$$

This implies that the deficiency function $D$ is not multiplicative since, in particular, $\gcd(q^k, n^2) = 1$ but
$$D(q^k n^2) = 0 < 2(\sigma(n^2) - n^2)\sigma(q^{k-1}) = D(q^k)D(n^2).$$
(In general, if $\gcd(X,Y)=1$, then the deficiency function satisfies $D(XY) \leq D(X)D(Y)$.  See this article for a proof.)

Note that
$$\sigma(n^2) - n^2$$
is also called the sum of the aliquot parts of $n^2$.

Here are some relevant OEIS hyperlinks for the number sequences used in this blog post:

OEIS sequence A033879 - Deficiency of $z$, or $2z - \sigma(z)$

OEIS sequence A001065 - Sum of proper divisors (or aliquot parts) of n: $\sigma(z) - z$

Note the following (trivial!) relationships between these two sequences:

$$A033879 + 2\cdot(A001065) = \sigma(z)$$

$$A033879 + A001065 = z$$

Lastly, notice that the inequality $1 < I(z) < 2$ follows immediately from
$$\frac{A033879 + A001065}{A033879 + A001065} < \frac{\sigma(z)}{z} := \frac{A033879 + 2\cdot(A001065)}{A033879 + A001065} < \frac{2\cdot(A033879) + 2\cdot(A001065)}{A033879 + A001065}.$$

Comments from the readers of this blog are most welcome.  Please feel free to shoot me an e-mail.

12.9.17

On computing $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ when $q^k n^2$ is an odd perfect number with Euler prime $q$

In this blog post, we compute
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right)$$
when $N = q^k n^2$ is an odd perfect number with Euler prime $q$.

From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$

In particular, we know that the index $i(q)$ is an integer greater than $5$ [Dris, Luca (2016)] (a copy is available in arXiv).

We now attempt to compute an expression for $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ in terms of $i(q)$.

First, since we have 
$$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$
we obtain
$$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$
and
$$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$
so that we get
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$


Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) =  1$ and $i(q)$ is odd, we get
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$



Hence, we conclude that
$$G := \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$


I posed the following question in MathOverflow:

I seem to recall that somebody (was it Pomerance [?] et. al) proved that
$$G \neq 1.$$  Does anybody here happen to know a reference?  Additionally, does
$G \neq 1$ imply that $G = i(q)$?

11.9.17

Revisiting some old results on odd perfect numbers


I have a new paper out there (temporarily in Scribd), titled Revisiting some old results on odd perfect numbers.  It contains (among other things) new and shorter proofs for some old results on odd perfect numbers.  (It is set to appear in arXiv by tomorrow by 0:00 GMT.)

27.8.17

Some notes on the deficiency function

(Note:  This blog post was copied verbatim from this MSE question, and the subsequent answer.)

Let $\sigma(x)$ be the sum of the divisors of a (positive) integer $x$.  (For example, $\sigma(2) = 1 + 2 = 3$.)

Define the deficiency function $D(x)$ to be the number
$$D(x) = 2x - \sigma(x).$$

Let $y$ be a (positive) integer.  Now I compute the difference:
$$D(xy) - D(x)D(y) = 2xy - \sigma(xy) - (2x - \sigma(x))(2y - \sigma(y))$$
$$=2xy - \sigma(xy) - 4xy + 2y\sigma(x) + 2x\sigma(y) - \sigma(x)\sigma(y)$$
$$=-2xy - 2\sigma(x)\sigma(y) + 2y\sigma(x) + 2x\sigma(y) + (\sigma(x)\sigma(y) - \sigma(xy))$$
$$=2(x - \sigma(x))(\sigma(y) - y) + (\sigma(x)\sigma(y) - \sigma(xy)).$$

This is because I want to compare $D(xy)$ and $D(x)D(y)$.  Note that, in general we have
$$2(x - \sigma(x))(\sigma(y) - y) \leq 0$$
and
$$\sigma(x)\sigma(y) - \sigma(xy) \geq 0.$$

Lastly, observe that if we have $D(xy) = D(x)D(y)$, then it follows that
$$2(\sigma(x) - x)(\sigma(y) - y) = \sigma(x)\sigma(y) - \sigma(xy).$$

If $x$ and $y$ are relatively prime (i.e. $\gcd(x, y)=1$) then we obtain $D(xy) \leq D(x)D(y)$.

Answer by MSE user mixedmath

I ran some quick numerics to see what sort of things happen.

You have two questions. Firstly, you conjecture

> $D(xy) = D(x)D(y) \Longrightarrow x$ or $y = 1$, or both $x,y$ are powers of $2$.

I have no intuition for whether this should or shouldn't be true. But the smallest counterexample I found was $15$ and $3$. Note that $D(15) = 30 - (1 + 3 + 5 + 15) = 6$ and $D(3) = 6 - (1 + 3) = 2$. On the other hand, $D(45) = 90 - (1 + 3 + 5 + 9 + 15 + 45) = 12$.

So $D(45) = D(3)D(15)$, and this conjecture is false.

Secondly, you consider the inequalities

> Can we say that $D(xy) \geq D(x)D(y)$ or $D(xy) \leq D(x)D(y)$ in a meaningful way?

After the raw numerics, it seems that $D(xy) < D(x)D(y)$ more often. We might expect this, as we know that
$$ \sum_{n \leq X} \sigma(n) = \frac{\zeta(2)}{2}X^2 + O(X),$$

leading to the (very loose) heuristic that the average value of $\sigma(n)$ is about $n\zeta(2)/2$. But $D(xy) > D(x)D(y)$ very often, with few discernible patterns (at first glance). It will take (perhaps many or specific) additional constraints to say when either inequality holds.

Conjecture (August 2017): $q^k n^2$ is an OPN implies that $\sigma(n^2)/q^k > q$

It is easy to prove that $\sigma(n^2)/q^k \neq q$. For suppose to the contrary that
$$q=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$
This means that
$$2n^2 = q\sigma(q^k)$$
which implies that $q \mid n^2$, contradicting $\gcd(q,n)=1$.

If the conjecture $q^k < n$ is true (see here (page 117), here, and here), then it follows that $q < n$, so that if $\sigma(n^2)/q^k < q$, then we obtain
$$\sigma(n^2) < q\cdot{q^k} < n\cdot{n} = n^2$$
which is a contradiction.

Also, if the Descartes-Frenicle-Sorli conjecture that $k=1$ is true, then it follows that $q < n$, so that if $\sigma(n^2)/q^k < q$, then we have
$$\sigma(n^2) < q\cdot{q^k} = q\cdot{q} < n\cdot{n} = n^2$$
which again is a contradiction.

Hence, conjecturally we expect
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right) > q$$
to be true.

25.8.17

On a curious biconditional involving divisors of spoof odd perfect numbers

This blog post is an elucidation of some of the recent discoveries/advances in the preprint titled "On a Curious Biconditional Involving the Divisors of Odd Perfect Numbers" as applied to the case of spoof odd perfect numbers, with details in the older preprint titled "The Abundancy Index of Divisors of Spoof Odd Perfect Numbers".

Recall that we call $n$ a spoof odd perfect number if $n$ is odd and $n=km$ for two integers $k, m > 1$, such that $\sigma(k)(m+1)=2n=2km$.

We begin with the following very useful lemmas.  (In what follows, we take $\sigma(x)$ to be the sum of the divisors of $x$, and denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$, where $\mathbb{N}$ is the set of natural numbers or positive integers.)

Lemma 1.  Let $n = km$ be a spoof odd perfect number.  Then
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} \text{ is bounded } \iff \frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m} \text{ is bounded }.$$

Remark 1.1.  Note that, in Lemma 1, we define
$$\sigma(m) := m+1.$$

Proof of Lemma 1.  The claimed result follows from the inequalities
$$m < \sigma(m) = m + 1 < 2m$$
$$\sqrt{k} < \sigma(\sqrt{k}) < 2\sqrt{k}$$
(since $\sqrt{k}$ is a proper divisor of the deficient number $k$). Consequently,
$$\frac{m}{\sqrt{k}} < \frac{\sigma(m)}{\sqrt{k}} < 2\cdot\frac{m}{\sqrt{k}}$$
$$\frac{\sqrt{k}}{m} < \frac{\sigma(\sqrt{k})}{m} < 2\cdot\frac{\sqrt{k}}{m}$$
from which it follows that
$$\frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m} < \frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} < 2\cdot\bigg(\frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m}\bigg).$$
We therefore conclude that
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} \text{ is bounded } \iff \frac{m}{\sqrt{k}}+\frac{\sqrt{k}}{m} \text{ is bounded },$$
as desired.

Remark 1.2.  In general, the function
$$f(z) := z + \frac{1}{z}$$
is not bounded from above.  (It suffices to consider the cases $z \to 0^{+}$ and $z \to +\infty$.)  This means that we do not expect the sum
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m}$$
to be bounded from above.

Corollary 1.1.  Let $n = km$ be a spoof odd perfect number.  Then the following conditions hold:
(a)  $\sigma(\sqrt{k}) \neq m+1=\sigma(m)$
(b)  $\sigma(\sqrt{k}) \neq m$

Proof of Corollary 1.1.
(a)  Suppose that $n=km$ is a spoof satisfying the condition
$$\sigma(\sqrt{k}) = \sigma(m).$$
It follows that
$$\frac{\sigma(\sqrt{k})}{\sqrt{k}} = \frac{\sigma(m)}{\sqrt{k}}$$
and
$$\frac{\sigma(\sqrt{k})}{m} = \frac{\sigma(m)}{m}$$
from which we obtain
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m}=\frac{\sigma(\sqrt{k})}{\sqrt{k}}+\frac{\sigma(m)}{m}.$$
But we also have
$$\frac{\sigma(\sqrt{k})}{\sqrt{k}}+\frac{\sigma(m)}{m} < \frac{\sigma(k)}{k}+\frac{\sigma(m)}{m} = I(k) + \frac{m+1}{m} < 2 + \frac{10}{9} = \frac{28}{9}.$$
Finally, we get
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} < \frac{28}{9},$$
which contradicts Lemma 1 and Remark 1.2.  We conclude that
$$\sigma(\sqrt{k}) \neq m+1=\sigma(m).$$

(b)  Suppose that $n=km$ is a spoof satisfying the condition
$$\sigma(\sqrt{k}) = m.$$
It follows that
$$\frac{\sigma(\sqrt{k})}{m}=1.$$
But
$$1 < \frac{\sigma(\sqrt{k})}{\sqrt{k}}\cdot\frac{\sigma(m)}{m} = \frac{\sigma(\sqrt{k})}{m}\cdot\frac{\sigma(m)}{\sqrt{k}} < 2.$$
This implies that
$$1 < \frac{\sigma(m)}{\sqrt{k}} < 2.$$
In particular,
$$\frac{\sigma(\sqrt{k})}{m}+\frac{\sigma(m)}{\sqrt{k}}=1+\frac{\sigma(m)}{\sqrt{k}} < 1+2 = 3.$$
This contradicts Lemma 1 and Remark 1.2.  We conclude that
$$\sigma(\sqrt{k}) \neq m.$$

Lemma 2.  Let $a, b \in \mathbb{N}$.

(a)  If $I(a) + I(b) < \sigma(a)/b + \sigma(b)/a$, then $a < b \iff \sigma(a) < \sigma(b)$ holds.
(b)  If $\sigma(a)/b + \sigma(b)/a < I(a) + I(b)$, then $a < b \iff \sigma(b) < \sigma(a)$ holds.
(c)  If $I(a) + I(b) = \sigma(a)/b + \sigma(b)/a$ holds, then either $a = b$ or $\sigma(a) = \sigma(b)$ is true.

Proof of Lemma 2.  We refer the interested reader to a proof of (b) in page 6 of this preprint.  The proofs for parts (a) and (c) are very similar.

Remark 2.1.  Note that if we let $a = m$ and $b = \sqrt{k}$ in Lemma 2, and if we make the additional assumption that $\gcd(a, b) = \gcd(m, k) = 1$, then case (c) is immediately ruled out, as $\gcd(m, \sqrt{k}) = 1$ implies that $m \neq \sqrt{k}$.  Additionally, note that $\sigma(m) \neq \sigma(\sqrt{k})$ per Corollary 1.1 (a).

Likewise, note that Lemma 1 and Remark 1.2 rules out case (b), as it implies that
$$\frac{\sigma(m)}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m} < I(m) + I(\sqrt{k}) < \frac{m+1}{m} + I(k) < \frac{10}{9}+2=\frac{28}{9},$$
a contradiction.

Hence, we are left with the scenario under part (a):
$$I(m)+I(\sqrt{k}) = \frac{m+1}{m}+\frac{\sigma(\sqrt{k})}{\sqrt{k}} < \frac{m+1}{\sqrt{k}}+\frac{\sigma(\sqrt{k})}{m},$$
which per Lemma 2 implies that
$$m < \sqrt{k} \iff m+1<\sigma(\sqrt{k}).$$

The considerations in Remark 2.1 prove the following proposition.

Theorem 3.  Let $n = km$ be a spoof odd perfect number.  Then the series of biconditionals
$$m < \sqrt{k} \iff m+1<\sigma(\sqrt{k}) \iff \frac{m+1}{\sqrt{k}}<\frac{\sigma(\sqrt{k})}{m}$$
hold.

Proof of Theorem 3.  Trivial.

Remark 3.1.  Note that
$$\frac{m+1}{\sqrt{k}} \neq \frac{\sigma(\sqrt{k})}{m}$$
is in general true under the assumption $\gcd(m, \sqrt{k})=1$, since it follows from the fact that
$$1 < \frac{m+1}{\sqrt{k}}\cdot\frac{\sigma(\sqrt{k})}{m}=\frac{m+1}{m}\cdot\frac{\sigma(\sqrt{k})}{\sqrt{k}} < 2.$$

Also, note that since $m \neq \sqrt{k}$ (because $\gcd(m, \sqrt{k})=1$), and $m+1 = \sigma(m) \neq \sigma(\sqrt{k})$ (by Corollary 1.1 (a)), then equivalently, we have the series of biconditionals
$$\sqrt{k} < m \iff \sigma(\sqrt{k}) < m+1 \iff \frac{\sigma(\sqrt{k})}{m} < \frac{m+1}{\sqrt{k}}.$$

Remark 3.2.  Let us double-check the findings of Theorem 3 (and Remark 3.1) using the only spoof that we know of, as a test case.

In the Descartes spoof,
$$m = 22021 = {{19}^2}\cdot{61}$$
and
$$\sqrt{k} = {3}\cdot{7}\cdot{11}\cdot{13} = 3003$$
so that we obtain
$$m + 1 = 22022 = 2\cdot{11011}$$
and
$$\sigma(\sqrt{k}) = (3+1)\cdot(7+1)\cdot({11}+1)\cdot({13}+1) = 5376 = {2^8}\cdot{3}\cdot{7}.$$
Notice that we then have
$$\sqrt{k} < m,$$
$$\sigma(\sqrt{k}) < m+1,$$
and
$$\frac{\sigma(\sqrt{k})}{m} = \frac{5376}{22021} < 1 < \frac{22022}{3003} = \frac{m+1}{\sqrt{k}},$$
in perfect agreement with the results in Theorem 3.

Remark 3.3.  Using Theorem 3, we list down all allowable permutations of the set
$$\left\{m, m+1, \sqrt{k}, \sigma(\sqrt{k})\right\}$$
(following the usual ordering on $\mathbb{N}$) subject to the biconditional
$$m < \sqrt{k} \iff m+1<\sigma(\sqrt{k}) \iff \frac{m+1}{\sqrt{k}}<\frac{\sigma(\sqrt{k})}{m}$$
and the constraints
$$m < m + 1, \text{   } \sqrt{k} < \sigma(\sqrt{k}).$$

We have the following cases to consider:

Case A:  $m < m+1 := \sigma(m) < \sqrt{k} < \sigma(\sqrt{k})$

Under this case, $m < \sqrt{k} < k$ which is true if and only if $k$ is not an odd almost perfect number (see page 6 of this preprint).  That is, 
$$D(k) := 2k - \sigma(k) = \frac{\sigma(k)}{m} > \frac{\sigma(\sqrt{k})}{m} > 1,$$
whence there is no contradiction.

Case B:  $\sqrt{k} < \sigma(\sqrt{k}) < m < m + 1$

Noting that, for the Descartes spoof, we have
$$D(k) = D({3003}^2) = 2{3003}^2 - \sigma({3003}^2) = 819 > 1,$$
so that $k$ is not almost perfect, it would seem prudent to try to establish the inequality $m < k$ for Case B.

To do so, one would need to repeat the methodology starting in Lemma 2, but this time with $a = m$ and $b = k$.  Again, we are left with the scenario under part (a):
$$I(m)+I(k) = \frac{m+1}{m}+\frac{\sigma(k)}{k} < \frac{m+1}{k}+\frac{\sigma(k)}{m},$$
which per Lemma 2 implies that
$$m < k \iff m+1<\sigma(k).$$

Thus we see that we have the following corollary to Theorem 3.

Corollary 3.1.  Let $n = km$ be a spoof odd perfect number.  Then the series of biconditionals
$$m < k \iff m+1<\sigma(k) \iff \frac{m+1}{k}<\frac{\sigma(k)}{m}$$
hold.

Remark 3.4.  Note that
$$\frac{m+1}{k} \neq \frac{\sigma(k)}{m}$$
is in general true under the assumption $\gcd(m, k)=1$, since it follows from the fact that
$$1 < \frac{m+1}{k}\cdot\frac{\sigma(k)}{m}=\frac{m+1}{m}\cdot\frac{\sigma(k)}{k} = 2.$$
Otherwise, we have $2 = (m+1)/k$ (which implies that $m=2k-1$) and $\sigma(k)/m=1$, so that $D(k) = 2k - \sigma(k) = 1$, which is equivalent to $k < m$.  (This is not true under Case A above, so it suffices to show that $m < k$ directly under Case B above.)

Also, note that since $m \neq k$ (because $\gcd(m, k)=1$), and $m+1 = \sigma(m) \neq \sigma(k)$ (by a similar result as proved in Corollary 1.1 (a)), then equivalently, we have the series of biconditionals
$$k < m \iff \sigma(k) < m+1 \iff \frac{\sigma(k)}{m} < \frac{m+1}{k}.$$
Lastly, note that $\sigma(k) \neq m$ (by a similar result as proved in Corollary 1.1 (b)).

Remark 3.5.  Using Corollary 3.1 (and the considerations in Remark 3.4), we list down all allowable permutations of the set
$$\left\{m, m+1, k, \sigma(k)\right\}$$
(following the usual ordering on $\mathbb{N}$) subject to the biconditional
$$m < k \iff m+1<\sigma(k) \iff \frac{m+1}{k}<\frac{\sigma(k)}{m}$$
and the constraints
$$m < m + 1, \text{   } k < \sigma(k).$$

Since $m < k$ already holds under Case A above, we consider the following subcases under Case B above:

Case B.1:  $\sqrt{k} < \sigma(\sqrt{k}) < m < m + 1 < k < \sigma(k)$

Under this subcase,
$$m < k \iff \frac{\sigma(k)}{m} > 1 \iff D(k) > 1 \iff k \text{ is not an odd almost perfect number }.$$
Since we ultimately want to prove $m < k$, this case is OK.

Case B.2:  $\sqrt{k} < \sigma(\sqrt{k}) < k < \sigma(k) < m < m + 1$

Note that, under this subcase,
$$2k - \sigma(k) = D(k) = \frac{\sigma(k)}{m} < 1,$$
forcing $k$ perfect.  This contradicts the fact that $k$ is a square.

Hence, we now conclude that:

CONCLUSION

MAIN THEOREM.  Let $n = km$ be a spoof odd perfect number.  Then $m < k$, which is true if and only if $k$ is not an odd almost perfect number.

Remark 3.6.  Note that the MAIN THEOREM is an analogue of a similar result proved by Dris for the case of odd perfect numbers, i.e. the Euler factor $Q^K$ of an odd perfect number $Q^K N^2$ is less than the non-Euler part $N^2$.

Remark 3.7. Virtually everything written in this blog post (for spoofs) also apply to the case of the usual odd perfect numbers, except for the fact that in the case of spoofs, the quasi-Euler prime $m$ is tacitly assumed to have exponent $1$.  (In the language of the usual odd perfect numbers, the Descartes-Frenicle-Sorli conjecture is a given for spoofs.)

ACKNOWLEDGMENTS.  Arnie Dris thanks Professor Pace Nielsen for hinting over MathOverflow that most non-computational results on odd perfect numbers carry over to the case of spoofs, and vice-versa.  (Non-computational results means OPN results that do not depend on numerical computations involving prime factorizations.)

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.