Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number

Question

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) is YES, 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 $k=1$.

Answer 1

Here is a quick way to further refine the improved lower bound for $I(n^2)$:

Write $$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg).$$ Now use, for instance, $$q^{k+1} - \frac{1}{q^2} > q^{k+1} - 1$$ to obtain $$I(n^2)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - \frac{1}{q^2}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\Bigg(1 + \frac{q^2}{q^{k+3} - 1}\Bigg).$$

Note that $$\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{q^2}{q^{k+3} - 1}\bigg) - \bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\frac{2(q-1)}{q^{k+2} (q^{k+3} - 1)}>0$$ since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.

In fact, this method shows that there are infinitely many ways to refine the improved lower bound $$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}+1}{q^{k+1}}\bigg).$$

It remains to be seen whether there is a refined (improved) lower bound that is independent of $k$ (and therefore expressed entirely in terms of $q$).

Answer 2

Here is a way to come up with an improved lower bound for $I(n^2)$, albeit in terms of $q$ and $n$:

We write $$I(n^2) - \frac{2(q - 1)}{q} = \frac{I(n^2)}{q^{k+1}} = \frac{\sigma(n^2)}{q^k}\cdot\frac{1}{qn^2} > \frac{1}{qn^2},$$ from which it follows that $$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{qn^2}.$$

This improved lower bound for $I(n^2)$, which does not contain $k$, can then be used to produce an upper bound for $k$.