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Here is my question:

Is $I(n^2) - 1 > 1/I(n^2)$ true when $I(n^2)=\sigma(n^2)/n^2$ is the abundancy index of $n^2$ and $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $k>1$?

My Attempt

If $k>1$, then since $q$ is the special prime, then $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In particular, we know that $q \geq 5$ and $k \geq 5$.

We know that $$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1} \leq \frac{5}{4}.$$

It follows that $$I(n^2) = \frac{2}{I(q^k)} > \frac{2(q - 1)}{q} \geq \frac{8}{5}.$$

Thus, $$I(n^2) - 1 > \frac{2(q - 1)}{q} - 1 = \frac{(2q - 2) - q}{q} = \frac{q - 2}{q} > \frac{q}{2(q - 1)} > \frac{1}{I(n^2)}$$ where the inequality $$\frac{q - 2}{q} > \frac{q}{2(q - 1)}$$ holds provided $q > 3+\sqrt{5} \approx 5.23607$.

However, the resulting inequality for $I(n^2)$ from $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ together with the following upper bound for $I(n^2)$ (which holds when $k>1$) $$\frac{2q}{q+1} > I(n^2)$$ only yields $$\frac{2q}{q+1} > I(n^2) > \frac{\sqrt{5}+1}{2}$$ thereby giving $$q > \frac{1+\sqrt{5}}{3-\sqrt{5}} = 2+\sqrt{5} \approx 4.23607.$$

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  • $\begingroup$ Note that there is no discrepancy when $k=1$, as then we have $$I(n^2) - 1 \geq \frac{2}{3} > \frac{3}{5} \geq \frac{1}{I(n^2)}$$ yielding the lower bound $q > 2 + \sqrt{5} \approx 4.23607$ from $$\frac{2q}{q+1}=I(n^2) > \frac{\sqrt{5}+1}{2}.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 23 at 10:18
  • $\begingroup$ When $k>1$, we get $$I(n^2) - 1 > \frac{3}{5} \not\gt \frac{5}{8} > \frac{1}{I(n^2)}.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 23 at 10:19
  • $\begingroup$ Consider the case when $k > 1$. If $q > 5$, then since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, it follows that $q \geq 13$, from which we obtain $$I(q^k) < \frac{q}{q - 1} \leq \frac{13}{12},$$ so that we get $$I(n^2) = \frac{2}{I(q^k)} > \frac{24}{13} = 1.\overline{846153},$$ which vastly improves on $$I(n^2) > \frac{\sqrt{5}+1}{2} \approx 1.61803.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 23 at 10:57
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Here is a partial answer for what happens when $k=1$.

We initially have $$I(n^2) - 1 = \frac{2q}{q+1} - 1 = \frac{q-1}{q+1} > \frac{q+1}{2q} = \frac{1}{I(n^2)},$$ where the inequality $$\frac{q-1}{q+1} > \frac{q+1}{2q}$$ holds when $$q > 2+\sqrt{5} \approx 4.23607.$$

In particular, the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ holds when $q \geq 5$.

Unfortunately, the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ implies that $I(n^2) > \frac{\sqrt{5}+1}{2} \approx 1.61803$, whereas $k=1$ implies that $$I(q^k)=I(q)=\frac{\sigma(q)}{q}=\frac{q+1}{q}=1+\frac{1}{q} \leq 1+\frac{1}{5}=\frac{6}{5}$$ (since $q \geq 5$), from which we obtain $$I(n^2)=\frac{2}{I(q^k)}=\frac{2}{I(q)}\geq \frac{5}{3} = 1.\overline{666}.$$

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Here is a full answer to the original question.

Let $q^k n^2$ be an odd perfect number with special prime $q$. Furthermore, let $I(x)=\sigma(x)/x$ be the abundancy index of the positive integer $x$. (Note that $\sigma(x)$ is the sum of divisors of $x$.)

Since the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ holds when $q>5$ or when $k=1$, it suffices to show that the inequality does not hold when $q=5$ and $k>1$.

First, note that the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ implies that $$I(n^2) > \frac{\sqrt{5}+1}{2} \approx 1.61803.$$

Next, from the following answer to a related MSE question, we have:

Cohen and Sorli ruled out $5^5$ as a possible Eulerian component $q^k$ for an odd perfect number in page 4 of their paper titled On Odd Perfect Numbers and Even 3-Perfect Numbers.

Thus, under the assumption $q=5$ and $k>1$, we have that $k \geq 9$ (since $k \equiv 1 \pmod 4$), whereupon we obtain $$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(5^9)} \leq \frac{1953125}{1220703} \approx 1.60000016384,$$ resulting in a contradiction.

Hence, the inequality $$I(n^2) - 1 > \frac{1}{I(n^2)}$$ does not hold when $q=5$ and $k>1$.

QED

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