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

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 $x$.

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Consider the sum $I(p^k)+I(m^2)$. Then we have the (sharp) bounds $L(p)=\frac{3p^2−4p+2}{p(p−1)}$ and $U(p)=\frac{3p^2+2p+1}{p(p+1)}$ with $L(p) \lt I(p^k)+I(m^2) \le U(p)$. If Sorli's conjecture is proved, then $I(p^k)+I(m^2)=U(p)$. – Jose Arnaldo Dris Dec 5 2010 at 7:02
Additionally, the derivative $U'(p) \gt 0$ and, while $U(p)$ has no maximum value on the interval $[5, \infty)$, it does have a least upper bound of $\lim_{p\to\infty}{U(p)} = 3$. As remarked by Joshua Zelinksy a few years back: "Any improvement on the upper bound of $3$ would have similar implications for all arbitrarily large primes and thus would be a very major result." (e.g. $U(p) < 2.99$ implies $p \le 97$.) Thus, (assuming Sorli's conjecture), "heuristically" we can have the following approach for the OPN problem: Fix an upper bound for $u = U(p)$, and then use factor chains... – Jose Arnaldo Dris Dec 5 2010 at 7:30
... to loop until you get the contradiction $U^{-1}(p) < p$. – Jose Arnaldo Dris Dec 5 2010 at 7:33
Update: If Sorli's conjecture is indeed true, then there are no odd perfect numbers. The proof proceeds via reductio ad absurdum, and is replete with all sorts of contradictions at every (succeeding) step. This reminds me of James Joseph Sylvester's quote from 1888: "… a prolonged meditation on the subject has satisfied me that the existence of any one such [OPN] — its escape, so to say, from the complex web of conditions which hem it in on all sides — would be little short of a miracle." – Jose Arnaldo Dris Dec 7 2010 at 14:18
Arnie, that is a bold claim. You really should back it up. To claim that if $k=1$ there are no OPN's would be a very big deal. – Pace Nielsen Dec 7 2010 at 17:49

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.)

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 @Pace, thanks a lot! What about if we consider the sum $$I(p^k) + I(m^2)$$ with this additional information from Sorli's conjecture? Can you comment on that one? – Jose Arnaldo Dris Dec 4 2010 at 4:18 I usually think of I as a multiplicative function. But I imagine that you can use the identities you derived above to get a lower bound on this thing. But again, if $p=5$ and $k=1$ this sum is just $6/5 + 5/3$, and there isn't much to say. – Pace Nielsen Dec 6 2010 at 23:36 More about this topic here: arnienumbers.blogspot.com/2010/12/… – Jose Arnaldo Dris Dec 7 2010 at 14:21 Lastly, I noticed that the numerators of $L(p)$ and $U(p)$ are irreducible quadratic polynomials over $\mathbb{Z}$. I don't know if that would have a bearing though, on the remaining case $p^k \lt m$, $k \ge 5$. – Jose Arnaldo Dris Dec 7 2010 at 14:21

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 $18$.

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Dear Arnie Dris, MathOverflow is not a bulletin board for news regarding ongoing mathematical projects. It is meant for mathematical questions and answers. Please confine your announcements of progress to your blog. – S. Carnahan Apr 3 2011 at 8:36
Okay, thanks for the advice. – Jose Arnaldo Dris Apr 3 2011 at 10:47