In what follows, we will denote the classical sum of divisors of the positive integer $x$ by
$$\sigma(x)=\sigma_1(x).$$
We will also denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
We start with a minor technical lemma:
Lemma 1: If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then $I(m^2) < \zeta(2)$ if and only if $p = 5$ and $k \neq 1$.
Proof:
Suppose that $I(m^2) < \zeta(2)$. Since $2(p - 1)/p < I(m^2)$, then it follows that
$$\frac{2(p - 1)}{p} < \zeta(2).$$
Hence, we derive
$$p < \frac{12}{12 - {\pi}^2} < 6,$$
from which we infer that $p = 5$ (since $p$ is a prime satisfying $p \equiv 1 \pmod 4$).
Now, since we already have $p = 5$, then
$$I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(5^k)} = \frac{2\cdot{5^k}\cdot(5 - 1)}{5^{k+1} - 1} < \zeta(2).$$
Solving for $k$, we obtain
$$k > 2\log_{5}{\pi} - \log_{5}\left(5{\pi}^2 - 48\right) \approx 1.23 > 1.$$
For the other direction, assume that $p = 5$ and $k \neq 1$.
Suppose to the contrary that
$$I(m^2) > \zeta(2) = \frac{{\pi}^2}{6} \approx 1.6449.$$
Since $k \equiv 1 \pmod 4$, it follows that $k \geq 5$. Hence, under the assumption that $p=5$, we have $k \geq 5$.
It follows that
$$I(m^2)=\frac{2}{I(p^k)}=\frac{2}{I(5^k)} \leq \frac{2}{I(5^5)} = \frac{3125}{1953}$$
where
$$\frac{3125}{1953} \approx 1.6001.$$
However, this contradicts our assumption that $I(m^2) > \zeta(2)$.
This concludes the proof of Lemma 1.
QED.
Remark 2: Note that we then have (either of) the improved bounds
$$\Bigg(I(m^2) > \zeta(2) \iff \left((p \geq 13) \lor (k = 1)\right)\Bigg) \oplus \Bigg(I(m^2) < \zeta(2) \iff \left((p = 5) \land (k \neq 1)\right)\Bigg).$$
Note that we have the implications
$$k = 1 \Rightarrow I(m^2)=\frac{2}{I(p^k)}=\frac{2}{I(p)}=\frac{2p}{(p+1)} \geq \frac{5}{3} = 1.\overline{666}$$
and
$$p \geq 13 \Rightarrow I(m^2) > \frac{2(p-1)}{p} \geq \frac{24}{13} \approx 1.\overline{846153}.$$
We thus have
$$k = 1 \Rightarrow I(m^2) > \zeta(2)$$
and
$$p \geq 13 \Rightarrow I(m^2) > \zeta(2)$$
Hence, we have the following proposition.
Theorem 3. Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Then the following implications hold:
- If $k=1$, then $I(m^2) > \zeta(2)$.
- If $p=5$, then we have the two cases:
- If $k = 1$, then $I(m^2) = 5/3 > \zeta(2)$.
- If $k \neq 1$, then $I(m^2) < \zeta(2)$.
(This blog post is currently a Work In Progress/Under Construction.)
