31.7.17

On the Descartes-Frenicle-Sorli conjecture and the Euler prime of odd perfect numbers

(Preamble: My apologies for the somewhat long post - I merely wanted to include all the details that I had in mind for ease of reference later.)

This post is an offshoot of this earlier MSE question.

So to summarize:  We have an odd perfect number $N=q^k n^2$ with Euler prime $q$.  (That is, $q$ satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  In particular, note that $q \geq 5$.)

Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$.

The Descartes-Frenicle-Sorli conjecture predicts that $k=1$.

Since
$$I(n^2) \leq \frac{2q}{q+1},$$
if $q=5$ then $I(n^2) \leq 5/3$.  This then implies that, if $q=5$, then
$$I(q^k)+I(n^2) \leq \frac{43}{15}.$$
It turns out that we have the biconditional
$$q>5 \iff \bigg(I(q^k)+I(n^2) > \frac{43}{15}\bigg).$$
However, if $k=1$, we have
$$I(q^k)+I(n^2) \geq \frac{43}{15}.$$
By the contrapositive,
$$I(q^k)+I(n^2) < \frac{43}{15}$$
implies that $k>1$.
Lastly, we have the biconditional
$$\bigg(\bigg(k=1\bigg) \land \bigg(q=5\bigg)\bigg) \iff \bigg(I(q^k)+I(n^2) = \frac{43}{15}\bigg).$$

Summarizing, we have:

If $57/20 < I(q^k) + I(n^2) < 43/15$, then $q=5$ and $k>1$.

Under this case, $k \geq 5$, so that
$$1.24992 = \frac{3906}{3125} = \frac{5^6 - 1}{{5^5}(5 - 1)}= I(5^5) \leq I(q^k) < \frac{5}{4} = 1.25$$
$$1.6 = \frac{8}{5} < I(n^2) \leq \frac{3125}{1953} \approx 1.\overline{600102406554019457245263696876},$$
resulting in the improved upper bound
$$I(q^k) + I(n^2) \leq \frac{3125}{1953}+\frac{3906}{3125}=\frac{17394043}{6103125}=2.85002\overline{240655401945724526369687660010}$$

Thus, we have:

If $q=5$ and $k>1$, then $57/20 < I(q^k) + I(n^2) \leq 17394043/6103125$, and conversely.

If $I(q^k) + I(n^2) = 43/15$, then $q=5$ and $k=1$, and conversely.

If $I(q^k) + I(n^2) > 43/15$, then $q>5$, and conversely.

Suppose that $I(q^k) + I(n^2) > 43/15$.  It follows that $q > 5$, from which we obtain $q \geq 13$ (since $q$ is a prime satisfying $q \equiv 1 \pmod 4$).

Consequently,
$$I(q^k) < \frac{q}{q-1} \leq \frac{13}{12} = 1.08\overline{333},$$
and hence we have
$$I(n^2) = \frac{2}{I(q^k)} > \frac{24}{13} = 1.\overline{846153},$$
resulting in the improved lower bound
$$2.92\overline{948717} = \frac{457}{156} = \frac{24}{13} + \frac{13}{12} < I(q^k) + I(n^2).$$

We can improve on these estimates if $q>5$ and $k=1$.  We obtain
$$I(q^k) = I(q) = 1+\frac{1}{q} \leq 1+\frac{1}{13}=\frac{14}{13} = 1.0\overline{769230}$$
and
$$I(n^2) \geq \frac{13}{7} = 1.\overline{857142},$$
resulting in the improved lower bound
$$2.\overline{934065} = \frac{267}{91} = \frac{14}{13} + \frac{13}{7} \leq I(q^k) + I(n^2).$$

Hence, we have:

If $q>5$ and $k=1$, then $2.\overline{934065} = 267/91 = 14/13 + 13/7 \leq I(q^k) + I(n^2) < 3$.

If $q>5$ and $k>1$, then $2.92\overline{948717} = 457/156 = 24/13 + 13/12 < I(q^k) + I(n^2) < 3$.

Notice that we have sharper bounds when the Descartes-Frenicle-Sorli conjecture that $k=1$ is true.

Here is my question:

Is it possible to do better than these present bounds?