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Improving the lower bound for $I(n)$ where $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, with $n < q$

Per Will Jagy's answer (and a subsequent comment by Erick Wong) to the following MSE question, we have the (sharp?) bounds

$$1 \leq \frac{I(x^2)}{I(x)} \leq \prod_{p}{\frac{p^2 + p + 1}{p^2 + p}} = \frac{\zeta(2)}{\zeta(3)} \approx 1.3684327776\ldots$$

Now, let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.

From Theorem 4.2 [pages 14 to 15 of this paper], we have the following biconditional:

$$\frac{2n}{n + 1} < I(n^2) \Longleftrightarrow n < q.$$

In particular, if $n < q$ (combining the two results), we get

$$\frac{2n}{(n + 1)I(n)} < \frac{I(n^2)}{I(n)} \leq \frac{\zeta(2)}{\zeta(3)}.$$

It follows that

$$I(n) > \frac{\zeta(3)}{\zeta(2)}\cdot\frac{2n}{n + 1}.$$

But we have the lower bound $n > {10}^{375}$ from $q^k < n^2$ [Dris, 2012] and ${10}^{1500} < N = {q^k}{n^2}$ [Ochem and Rao, 2012]. Consequently, we have

$$I(n) > \frac{\zeta(3)}{\zeta(2)}\cdot\frac{2n}{n + 1} > \frac{\zeta(3)}{\zeta(2)}\cdot\frac{2\cdot{10}^{375}}{{10}^{375} + 1}.$$

Note that we have the rational approximation

$$\frac{\zeta(3)}{\zeta(2)}\cdot\frac{2\cdot{10}^{375}}{{10}^{375} + 1} \approx 1.4615259388\ldots$$