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

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 $$k=1$$.

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 $$q$$).

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 $$k$$.

• I think $$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{qn^2}$$ also improves on $$I(q^k) + I(n^2) > \frac{57}{20},$$ which by Lemma II.2, page 1 of this conference paper, "would be equivalent to showing that there are no odd perfect numbers of the form $5n^2$, which would be a very major result". Commented Jun 3, 2022 at 7:09
• Details for the argument in the preceding comment are in this recent MSE question. Commented Jun 3, 2022 at 8:20