Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Define the *abundancy index*
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical *sum of divisors* of $x$.

Since $q$ is prime, we have the bounds $$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$ which implies, since $N$ is perfect, that $$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$

We now prove the following claim:

CLAIM:$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$

**PROOF:** We know that
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k}=\gcd(n^2,\sigma(n^2)),$$
(since $\gcd(q^k,\sigma(q^k))=1$).

However, we have
$$\sigma(q^k) - q^k = 1 + q + \ldots + q^{k-1} = \frac{q^k - 1}{q - 1},$$
so that we obtain
$$\frac{\sigma(n^2)}{q^k}=\frac{\bigg(q - 1\bigg)\bigg(2n^2 - \sigma(n^2)\bigg)}{q^k - 1}=\sigma(n^2) - \bigg(q - 1\bigg)\bigg(2n^2 - \sigma(n^2)\bigg)$$
$$=q\sigma(n^2) - 2(q - 1)n^2.$$
Dividing both sides by $qn^2$, we get
$$I(n^2) - \frac{2(q-1)}{q} = \frac{I(n^2)}{q^{k+1}} > \frac{1}{q^{k+1}}\cdot\frac{2(q-1)}{q},$$
which implies that
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
**QED.**

To illustrate the improved bound:

**(1)** Unconditionally, we have
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5} = 1.6.$$

**(2)** Under the assumption that $k=1$:
$$I(n^2) > 2\bigg(1 - \frac{1}{q}\bigg)\bigg(1 + \left(\frac{1}{q}\right)^2\bigg) \geq \frac{208}{125} = 1.664.$$

**(3)** However, it is known that under the assumption $k=1$, we actually have
$$I(q^k) = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} \geq \frac{5}{3} = 1.\overline{666}.$$

Here are my questions:

(A)Is it possible to improve further on the unconditional lower bound for $I(n^2)$?

(B)If the answer to Question(A)isYES, my next question is "How"?

I did notice that $$\frac{2(q-1)}{q}+\frac{2}{q(q+1)}=I(n^2)=\frac{2q}{q+1}$$ when $k=1$.