On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$ The topic of odd perfect numbers likely needs no introduction.
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.  Let
$$D(x) = 2x - \sigma(x)$$
denote the deficiency of $x$, and let
$$s(x) = \sigma(x) - x$$
denote the sum of aliquot/proper divisors of $x$.  Finally, denote the abundancy index of $x$ by
$$I(x) = \frac{\sigma(x)}{x}.$$
Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $q$ is prime and $N$ is perfect, we obtain
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}$$
so that we get
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Using the identity
$$\frac{D(x)}{x} = 2 - I(x)$$
we obtain the bounds
$${q^k}\bigg(\frac{q-2}{q-1}\bigg) < D(q^k) \leq {q^{k-1}}\bigg(q-1\bigg)$$
and
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q}.$$
This implies that
$${2q^k n^2}\cdot\bigg(\frac{q-2}{(q-1)(q+1)}\bigg) < D(q^k)D(n^2) < {2q^k n^2}\cdot\bigg(\frac{q-1}{q^2}\bigg).$$
Dividing both sides of the last inequality by $2q^k n^2$, we get
$$\frac{q-2}{(q-1)(q+1)} < \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-1}{q^2}.$$
But we know that
$$D(q^k)D(n^2)=2s(q^k)s(n^2),$$
which can be verified by a direct, brute-force computation.
Thus, the fraction in the middle of the last inequality simplifies to
$$\frac{D(q^k)D(n^2)}{2q^k n^2}=\bigg(\frac{s(q^k)}{q^k}\bigg)\bigg(\frac{s(n^2)}{n^2}\bigg)=\bigg(I(q^k) - 1\bigg)\bigg(I(n^2) - 1\bigg) = 3 - \bigg(I(q^k) + I(n^2)\bigg).$$
We therefore finally have the bounds
$$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2) < 3 - \bigg(\frac{q-2}{(q-1)(q+1)}\bigg)$$
which does not improve on the known bounds
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
(See this paper for a proof.)
Here are my:

QUESTIONS Is it possible to improve on the bounds for $D(q^k)$ and $D(n^2)$ (where $q^k n^2$ is an odd perfect number with special prime $q$) to hopefully produce stronger bounds for $I(q^k) + I(n^2)$?  If so, how could this be done?

 A: On OP's request, I am converting my comment into an answer.
Let $f(k):=I(q^k) + I(n^2)$.
Then, we have
$$f'(k)=\frac{-( q^{2 k + 2}- 4 q^{2 k + 1}+2 q^{k + 1}   + 2 q^{2 k} - 1) \log(q)}{(q - 1)q^k (q^{k + 1} - 1)^2}$$
which is negative, so we see that $f(k)$ is decreasing.
It follows that
$$\lim_{k\to\infty}f(k)\lt f(k)\le f(1)\tag1$$
which is the known bounds
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg)\tag2$$
So, I think that this means the following :

*

*If you prove that $k\not=1$, then you can get an improved upper bound for $f(k)$.


*If you get an improved upper bound for $f(k)$, then you can say that $k\not=1$.


*If you prove that there is an integer $a$ such that $k\le a$, then you can get an improved lower bound for $f(k)$.


*If you get an improved lower bound for $f(k)$, then you can say that there is an integer $a$ such that $k\le a$.
(I'm not saying that if you can't improve the bounds for $k$, you can't improve the bounds for $f(k)$.)
A: Let me try to work backwards from
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
This can be rewritten as
$$\frac{q-1}{q(q+1)} \leq 3 - \bigg(I(q^k) + I(n^2)\bigg) = \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-2}{q(q-1)}.$$
We also have
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q},$$
which we can rewrite as
$$q < \frac{2n^2}{D(n^2)} \leq q + 1.$$
We therefore obtain
$$\frac{q-1}{q+1} = \frac{q(q-1)}{q(q+1)} < \bigg(\frac{q-1}{q(q+1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{D(q^k)}{q^k} < \bigg(\frac{q-2}{q(q-1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{(q+1)(q-2)}{q(q-1)},$$
which implies that
$$\frac{q-1}{q+1} < 2 - I(q^k) < \frac{(q+1)(q-2)}{q(q-1)},$$
which simplifies to
$$\frac{q^2 - q + 2}{q^2 - q} = 2 - \bigg(\frac{(q+1)(q-2)}{q(q-1)}\bigg) < I(q^k) < 2 - \bigg(\frac{q-1}{q+1}\bigg) = \frac{q+3}{q+1}.$$
Note that both bounds do not improve on the currently known
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}.$$
