Step Two: Almost Perfect Numbers... (#OPNResearch #May2023)

Let $N = p^k m^2$ be an odd perfect number with special prime $p$.  Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x) = \sigma_1(x)$ is the classical sum of divisors of $x$.  Additionally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$.

I claim that the estimate $m < p$ holds if and only if $D(m) = 1$.

Recall the following material facts:

• $$\left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)} \approx 1.4912789897463723558019369442648428792691131693617462298521097731$$
• $$k = 1 \Rightarrow I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)} \geq \left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)}$$
• [Dagal, Dris (NNTDM, 06/2021) - The Abundancy Index of Divisors of Odd Perfect Numbers - Part II]
• We have the (generic/sharp [?]) upper bound $$\frac{I(m^2)}{I(m)} \leq \frac{\zeta(2)}{\zeta(3)}.$$
• Reference: Comment by Erick Wong underneath an answer by Will Jagy to MSE question https://math.stackexchange.com/questions/1097803.
• If $I(m) < 2$, then $$\frac{2m}{m+D(m)} \leq I(m) < \frac{2m+D(m)}{m+D(m)},$$ where equality holds if and only if $I(m)=D(m)=1$.
• Dris [JANTA (02/2018) - Lemma 2.2 (page 2), Theorem 3.3 (page 5)]
• The inequality $p^k < m^2$, obtained by Dris in 09/2012 [JIS (2012)], means we can infer that the implication $m < p \Rightarrow k = 1$ holds.

First, we prove the following claims:

LEMMA 1.  $I(m^2) \neq 2m/(m+1)$

Proof: Suppose to the contrary that $I(m^2) = 2m/(m+1)$.  Since 

$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1},$$ then we have

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

This then implies that $m < p$ (since $\gcd(p,m) = 1$).  But then $m < p$ implies $k = 1$.  Consequently, we obtain

$$\frac{2m}{m+1} = I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(p)} = \frac{2p}{p+1},$$

which then gives $m = p$.  This contradicts $\gcd(p,m)=1$.

LEMMA 2.  $I(m^2) < 2m/(m+1)$ if and only if $p < m$.

Proof.  By Lemma 1, we know that $I(m^2) \neq 2m/(m+1)$.  

It is obvious that $2m/(m+1) < I(m^2)$ implies $m < p$, since 

$$I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$

We want to show that $I(m^2) < 2m/(m+1)$ implies $p < m$.

To this end, suppose that the inequality $I(m^2) < 2m/(m+1)$ is true. Assume that the estimate $m < p$ holds. This implies that $k = 1$, whereupon we obtain

$$\frac{2p}{p+1} = \frac{2}{I(p)} = \frac{2}{I(p^k)} = I(m^2) < \frac{2m}{m+1}.$$

Thus, we have $2p/(p+1) < 2m/(m+1)$, which implies that $p < m$, contradicting our earlier assumption that $m < p$.

It follows that $I(m^2) < 2m/(m+1) \iff p < m$, and we are done.

It is known that $D(m) = 1 \Rightarrow m < p$, since $m > 1$ is deficient, whence by a criterion for deficient numbers by Dris:

$$D(m) = 1 \iff \frac{2m}{m+1} = \frac{2m}{m+D(m)} < I(m) < \frac{2m+D(m)}{m+D(m)} = \frac{2m+1}{m+1}$$

(Use the $LHS$ inequality for $I(m)$.)

$$\bigg(\frac{2m}{m+1} < I(m) < I(m^2) \leq \frac{2p}{p+1}\bigg) \Rightarrow m < p$$

However, it is also known (by Dris [JIS (09/2012)]) that $m < p$ implies $k = 1$ (since $p^k < m^2$).

Consequently,

$$\left(D(m) = 1\right) \Rightarrow \left(m < p\right) \Rightarrow \left(k = 1\right).$$

We want to determine whether the implication $m < p \Rightarrow D(m) = 1$ is true.

Assume that $m < p$.  Suppose to the contrary that $D(m) > 1$.

We consider three cases:

(1)  $$\left(r < I(m) < I(m^2)\right) \Rightarrow \left((D(m) = 1) \land (m < p)\right) \Rightarrow m \text{ is almost perfect.}$$

(2)  $$\left(I(m) < r < I(m^2)\right) \Rightarrow \left((D(m) > 1) \land (m < p)\right) \Rightarrow m \text{ is almost perfect.}$$

(3)  $$\left(I(m) < I(m^2) < r\right) \Rightarrow \left((D(m) > 1) \land (p < m)\right) \Rightarrow m \text{ is not almost perfect.}$$

Note that Case (3) is ruled out under the assumption $m < p$.

We consider the remaining problematic case:

Case (2): Under this case, $I(m) < r < I(m^2)$:

Since $m < p$ is true under Case (2), then we infer that (mainly because $k = 1$),

$$I(m) < (I(m^2))^{\ln(13/9)/\ln(4/3)}$$

$$\frac{I(m)}{I(m^2)} < \left(I(m^2)\right)^{\ln(13/12)/\ln(4/3)}$$

where $\ln(13/12)/\ln(4/3)$ is approximately $0.2782332141567583199541292544159705$; consequently, we obtain

$$\frac{I(m^2)}{I(m)} > \left(I(m^2)\right)^{\ln(4/3)/\ln(13/12)},$$

where $\ln(4/3)/\ln(13/12)$ is approximately $3.5941072061820548592874747274664451171$.

But since $k = 1$, then $p$ is the special prime implies that $p \geq 5$, so that

$$I(p^k) = I(p) = \frac{p+1}{p} = 1+\frac{1}{p} \leq \frac{6}{5}$$

$$I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(p)} = \frac{2p}{p+1} \geq \frac{5}{3}$$

Therefore, 

$$\frac{I(m^2)}{I(m)} > \left(I(m^2)\right)^{\ln(4/3)/\ln(13/12)} \geq (5/3)^{\ln(4/3)/\ln(13/12)} \approx 6.271164452351745712932500473652648.$$

This contradicts
$$\frac{I(m^2)}{I(m)} \leq \frac{\zeta(2)}{\zeta(3)} \approx 1.36843277762020587573676.$$

Consequently, only Case (1) remains: 

Case (1):  $$r < I(m) < I(m^2) \Rightarrow \left((D(m) = 1) \land (m < p)\right) \Rightarrow m \text{ is almost perfect}.$$

In particular, $m$ is almost perfect if and only if $m < p$. (In other words, $m < p$ if and only if $D(m) = 1$.)