11.6.17

A question on the Euler prime of odd perfect numbers

(Note:  This post was copied verbatim from this MSE link.)

A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function.  For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect.  (Note that $6$ is even.)

It is currently unknown whether there are any odd perfect numbers.  Euler proved that an odd perfect number, if any exists, must take the form $N = q^k n^2$ where $q$ is prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  We call $q$ the Euler prime of the odd perfect number $N$.

In what follows, we denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.

MOTIVATION

Since the non-Euler part $n^2$ is deficient but not almost perfect, we know (from this preprint) that
$$I(n^2) < \frac{2n^2 + D(n^2)}{n^2 + D(n^2)},$$
where $D(n^2)={2n^2}-\sigma(n^2)$ is the deficiency of $n^2$.

When the Descartes-Frenicle-Sorli conjecture that $k=1$ is true, we also know that we have the lower bound
$$\frac{5}{3} \leq I(n^2),$$
whereupon we obtain, from the last two inequalities, the following lower bound
$$\frac{n^2}{D(n^2)} > 2.$$
(When $k=1$, ${n^2}/D(n^2)=(q+1)/2$, so that ideally we should have ${n^2}/D(n^2) \geq 3$ since $q \geq 5$.)

From the following Answer 1 and Answer 2 to a related MSE question, we obtain improved upper bounds for $I(n^2)$ by successively computing the mediant of
$$I(n^2) = \frac{2n^2 - D(n^2)}{n^2}=I_0$$
and the upper bounds $u_i$ (for $I(n^2)$) given by
$$u_0 = \frac{2n^2 + D(n^2)}{n^2 + D(n^2)}$$
$$u_1 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(2n^2 + D(n^2)\bigg)}{n^2 + \bigg(n^2 + D(n^2)\bigg)} = \frac{4n^2}{2n^2 + D(n^2)}$$
$$u_2 = \frac{\bigg(2n^2 - D(n^2)\bigg)+4n^2}{n^2 + \bigg(2n^2 + D(n^2)\bigg)}=\frac{6n^2 - D(n^2)}{3n^2 + D(n^2)}$$
$$u_3 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(6n^2 - D(n^2)\bigg)}{n^2 + \bigg(3n^2 + D(n^2)\bigg)} = \frac{8n^2 - 2D(n^2)}{4n^2 + D(n^2)}$$
$$u_4 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(8n^2 - 2D(n^2)\bigg)}{n^2 + \bigg(4n^2 + D(n^2)\bigg)} = \frac{10n^2 - 3D(n^2)}{5n^2 + D(n^2)}$$
$$\ldots \ldots \ldots$$
$$\ldots \ldots \ldots$$
$$\ldots \ldots \ldots$$
(That is, $u_i = \text{mediant}(u_{i-1},I_0)$.)

Note that it is easy to prove that $u_i$ (for $i=0, 1, 2, \ldots$) is a strictly decreasing sequence.

The resulting lower bounds for ${n^2}/D(n^2)$ (on assuming $k=1$) are:

$$u_0 : \frac{n^2}{D(n^2)} > 2$$
$$u_1 : \frac{n^2}{D(n^2)} > \frac{5}{2}$$
$$u_2 : \frac{n^2}{D(n^2)} > \frac{8}{3}$$
$$u_3 : \frac{n^2}{D(n^2)} > \frac{11}{4}$$
$$u_4 : \frac{n^2}{D(n^2)} > \frac{14}{5}$$
$$\ldots : \ldots$$

QUESTIONS

(1)   Do the resulting lower bounds for ${n^2}/D(n^2)$ (using this procedure) have a closed form?

(2)  Does the lower bound ever reach $3$, or is it possible to prove that it does not?

POSTED ANSWER

It appears that the numerators of the lower bounds form OEIS sequence A016789.  Hence, the lower bounds have the closed form
$$\frac{3i + 2}{i + 1}.$$

Consequently,
$$\frac{3i + 2}{i + 1}=3-\frac{1}{i + 1},$$

so that the lower bounds never reach $3$.