On Sorli's Conjecture Re: OPNs (Circa 2003) - MathOverflow

On Sorli's Conjecture Re: OPNs (Circa 2003)

2010-12-03T18:29:09Z

In the PhD dissertation titled "Algorithms in the Study of Multiperfect and Odd Perfect Numbers" (hyperlinked here) and completed in 2003, Ronald Sorli conjectured that the exponent $k$ on the Euler prime $p$ for an odd perfect number $N = (p^k)(m^2)$ is one (i.e. we can drop $k$).

Assuming Sorli's conjecture is true, does anyone know if there exist (any) "effective" results (pardon my use of the term, I just could not think of a better word) in the literature, particularly with respect to relations between the Euler prime $p$, the exponent $k$ and the number $\sqrt{\frac{N}{p^k}}$? I have, so far, only been able to get hold of Paolo Starni's article titled "Odd Perfect Numbers: A Divisor Related to the Euler′s Factor".

In particular, note that Sorli's conjecture implies the following relations:

$$I(p^k) = I(p) = \frac{p+1}{p}$$

$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p + 1}$$

which, in turn, gives the (trivial) algebraic identity:

$$\frac{1}{p} = \frac{1}{p+1} + \frac{1}{p}\left(\frac{1}{p+1}\right)$$

where $p$ is the Euler prime (i.e. $p^k$ is the Euler's Factor) and $$I(x) = \frac{\sigma(x)}{x}$$ is the abundancy index of $x$.

Answer by Pace Nielsen for On Sorli's Conjecture Re: OPNs (Circa 2003)

2010-12-04T00:12:46Z

As far as I know, there are no such effective bounds. In fact, even if $p=5$ and $k=1$, there are no known effective bounds on $N$. (There are bounds on $N$ in terms of the number of distinct factors.)

Answer by Jose Arnaldo Dris for On Sorli's Conjecture Re: OPNs (Circa 2003)

2011-04-03T07:24:34Z

Based from this preprint, in order to prove Sorli's conjecture, it suffices to show that $m < p$, from which it follows that the Euler prime $p$ is the largest prime factor of the odd perfect number $N$. One even has much leeway up to showing $m < p^2$ and Sorli's conjecture that $k = 1$ would still follow, since $k \equiv 1 \pmod 4$.

In the same preprint, I present numerical evidence for my 2008 conjecture that $p^k < m$, which is a relatively big improvement over my previous result $\sigma(p^k) < \frac{2m^2}{3}$. The key ingredient in my method is to establish an equivalence between (to use the same notation in the preprint): $$\rho_3 = \frac{\sigma(p^k)}{m} < \frac{\sigma(m)}{p^k} = \mu_4$$ and $p^k < m$. The last stumbling block to this method is mentioned in Remark $4.3$ from pages $17$ to $18$.