On Basak’s “Bounds On Factors Of Odd Perfect Numbers”

Question

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In what follows, we denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$, where $\sigma(x)$ is the sum of the divisors of $x$.

In Case 1 under Remark 3.1 on page 4 of Basak's Bounds On Factors Of Odd Perfect Numbers, it is proven that

$$\frac{16}{7\zeta(3)} < \frac{16q^3}{7\zeta(3)(q^3 – 1)} < \bigg(1 + \frac{1}{q}\bigg)\prod_{p \mid n, \hspace{0.05in} p \neq q}\bigg(1 + \frac{1}{p} + \frac{1}{p^2}\bigg).$$

But we also have
$$\bigg(1 + \frac{1}{q}\bigg)\prod_{p \mid n, \hspace{0.05in} p \neq q}\bigg(1 + \frac{1}{p} + \frac{1}{p^2}\bigg) < I(q)I(n^2) \leq \frac{6I(n^2)}{5},$$
since $q$ is prime with $q \equiv 1 \pmod 4$ implies that $q \geq 5$.

(Note that this last inequality is unconditional on the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$.)

This implies that
$$\frac{16}{7\zeta(3)} < \frac{6I(n^2)}{5}$$
from which it follows that
$$I(n^2) > \frac{5}{6}\cdot\frac{16}{7\zeta(3)} = \frac{40}{21\zeta(3)} \approx 1.58458547158229994034881195966.$$

But then this resulting numerical lower bound for $I(n^2)$ is trivial, as it is known that
$$I(q^k) < \frac{q}{q – 1} \leq \frac{5}{4} < \frac{8}{5} \leq \frac{2(q – 1)}{q} < I(n^2),$$
so that we already know, unconditionally, that $I(n^2) > 1.6$.

Here is my question:

Would it be possible to tweak Basak's argument in order to come up with an improved lower bound for
$$\bigg(1 + \frac{1}{q}\bigg)\prod_{p \mid n, \hspace{0.05in} p \neq q}\bigg(1 + \frac{1}{p} + \frac{1}{p^2}\bigg)?$$

Update (May 1, 2020 – 13:45 PM Manila time)

I re-read Basak's argument in Case 1 under Remark 3.1 on page 4 of the hyperlinked paper in arXiv, it appears that Basak does make use of the assumption that $k=1$.

Answer 1

We get $$\frac{16q^3}{7\zeta(3)(q^3 - 1)} < I(q)I(n^2)$$ from which we have

$$I(n^2)\gt\frac{16}{7\zeta(3)}f(q)$$ where $$f(q)=\frac{q^4}{(q^3-1)(q+1)}$$

Now, we have, for $q\ge 5$, $$f'(q)=\frac{ q^3 (q(q^2-3)-4)}{(q + 1)^2 (q^3 - 1)^2}\gt 0$$ from which we have

$$I(n^2)\gt\frac{16}{7\zeta(3)}f(q)\ge \frac{16}{7\zeta(3)}f(5)=\frac{1250}{651\zeta(3)}\approx 1.597364$$ which is better than $1.584585$, but still smaller than $1.6$.