Vtgyjfy

Let $N$ be an odd (positive) integer. If $sigma(N)=2N$ where $sigma(N)$ is the sum of the divisors of $N$, then $N$ is called an odd perfect number. Let $I(N)=sigma(N)/N$ denote the abundancy index of $N$. Likewise, denote the deficiency of $x in mathbb{N}$ by $D(x)=2x-sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x)=sigma(x)-x$.

Euler showed that an odd perfect number, if one exists, must have the so-called Eulerian form $N = q^k n^2$, where $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$.

The question on the existence of an odd perfect number is one of the long-standing open problems of number theory.

Since $gcd(q,n)=1$ and the abundancy index function $I$ is multiplicative, we have
$$sigma(N)=2N iff I(N)=2=I(q^k)I(n^2) iff frac{sigma(q^k)}{q^k} = I(q^k) = frac{2}{I(n^2)} = frac{2n^2}{sigma(n^2)}.$$

Using the formula
$$frac{A}{B}=frac{C}{D}=frac{C-A}{D-B}$$
and the substitutions $A=sigma(q^k)$, $B=q^k$, $C=2n^2$, and $D=sigma(n^2)$, and noting that $C-A > 0$ and $D-B > 0$ (see this paper by Dris (2012) for more information), we obtain
$$frac{sigma(q^k)}{q^k} = I(q^k) = frac{2}{I(n^2)} = frac{2n^2}{sigma(n^2)} = frac{2n^2 - sigma(q^k)}{sigma(n^2) - q^k}.$$

Using the known bounds
$$frac{q+1}{q} leq I(q^k) < frac{q}{q-1}$$
we get the system of inequalities
$$(q+1)(sigma(n^2) - q^k) leq 2qn^2 - qsigma(q^k)$$
and
$$(q-1)(2n^2 - sigma(q^k)) < q(sigma(n^2) - q^k)$$
from which we get
$$qsigma(q^{k-2})= q(sigma(q^{k-1}) - q^{k-1}) = q(s(q^k) - q^{k-1}) = qs(q^k) - q^k leq qD(n^2) - sigma(n^2)$$
and
$$qD(n^2) - 2n^2 < qs(q^k) - sigma(q^k) = qsigma(q^{k-1}) - sigma(q^k) = -1,$$
of which the last inequality implies that
$$q < frac{2n^2 - 1}{D(n^2)} = frac{2n^2 - 1}{2n^2 - sigma(n^2)}.$$

The two resulting inequalities may be summarized as follows:

If $N=q^k n^2$ is an odd perfect number given in Eulerian form, then
$$qsigma(q^{k-2}) + sigma(n^2) leq qD(n^2) < 2n^2 - 1.$$

Consequently, we have
$$qsigma(q^{k-2}) + sigma(n^2) < 2n^2 - 1 implies D(n^2) > qsigma(q^{k-2}) + 1.$$

However, this last inequality appears to be trivial, as it is known that
$$frac{sigma(n^2)}{q^k}=frac{2n^2}{sigma(q^k)}=gcd(n^2,sigma(n^2))=frac{D(n^2)}{s(q^k)}=frac{2s(n^2)}{D(q^k)}$$
so that, if we are willing to assume the Descartes-Frenicle-Sorli Conjecture that $k=1$, and knowing that this conjecture implies $sigma(q^k) < n$ (see this paper by Dris (2017) for more information), then we obtain
$$frac{sigma(n^2)}{q^k}=gcd(n^2,sigma(n^2))=frac{D(n^2)}{s(q^k)}=frac{2s(n^2)}{D(q^k)}=frac{2n^2}{sigma(q^k)}>2n>sigma(n).$$

(Please see OEIS sequence A322154 for more information.)

Here is my question:

Is it possible to tweak the argument presented in this MSE post in order to come up with a lower bound better than $D(n^2) > 2n$?

I am guessing our best shot would have to emanate from the inequality
$$sigma(q^{k-2}) + frac{sigma(n^2)}{q} leq D(n^2)$$
where equality holds if and only if $k=1$ (whence $sigma(q^{k-2})=0$ since it would be an empty sum).

So suppose that $k>1$. It follows from our method that
$$sigma(q^{k-2}) + frac{sigma(n^2)}{q} < D(n^2) = s(q^k)frac{sigma(n^2)}{q^k} = sigma(q^{k-1})frac{sigma(n^2)}{q^k}$$
from which we obtain
$$frac{sigma(q^{k-2})}{sigma(q^{k-1})} + frac{sigma(n^2)}{qsigma(q^{k-1})} < frac{sigma(n^2)}{q^k}$$
which is trivial as
$$sigma(q^{k-2}) < sigma(q^{k-1})$$
while
$$q^k < qsigma(q^{k-1}) = q(1 + ldots + q^{k-1}) = q + ldots + q^k.$$