On Sorli's Conjecture Re: OPNs (Circa 2003) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:48:43Z http://mathoverflow.net/feeds/question/48203 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48203/on-sorlis-conjecture-re-opns-circa-2003 On Sorli's Conjecture Re: OPNs (Circa 2003) Jose Arnaldo Dris 2010-12-03T18:29:09Z 2011-04-03T07:24:34Z <p>In the PhD dissertation titled "Algorithms in the Study of Multiperfect and Odd Perfect Numbers" (<a href="http://utsescholarship.lib.uts.edu.au/dspace/handle/2100/275?show=full" rel="nofollow">hyperlinked here</a>) 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 $k$).</p> <p>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".</p> <p>In particular, note that Sorli's conjecture implies the following relations:</p> <p>$$I(p^k) = I(p) = \frac{p+1}{p}$$</p> <p>$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p + 1}$$</p> <p>which, in turn, gives the (trivial) algebraic identity:</p> <p>$$\frac{1}{p} = \frac{1}{p+1} + \frac{1}{p}\left(\frac{1}{p+1}\right)$$</p> <p>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 $x$.</p> http://mathoverflow.net/questions/48203/on-sorlis-conjecture-re-opns-circa-2003/48236#48236 Answer by Pace Nielsen for On Sorli's Conjecture Re: OPNs (Circa 2003) Pace Nielsen 2010-12-04T00:12:46Z 2010-12-04T00:12:46Z <p>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.)</p> http://mathoverflow.net/questions/48203/on-sorlis-conjecture-re-opns-circa-2003/60413#60413 Answer by Jose Arnaldo Dris for On Sorli's Conjecture Re: OPNs (Circa 2003) Jose Arnaldo Dris 2011-04-03T07:24:34Z 2011-04-03T07:24:34Z <p>Based from this <a href="http://www.scribd.com/doc/46766721/OPNPaper3" rel="nofollow">preprint</a>, in order to prove Sorli's conjecture, it suffices to show that $m &lt; p$, 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 $m &lt; p^2$ and Sorli's conjecture that $k = 1$ would still follow, since $k \equiv 1 \pmod 4$.</p> <p>In the same <a href="http://www.scribd.com/doc/46766721/OPNPaper3" rel="nofollow">preprint</a>, I present numerical evidence for my 2008 conjecture that $p^k &lt; m$, which is a relatively big improvement over my previous result $\sigma(p^k) &lt; \frac{2m^2}{3}$. The key ingredient in my method is to establish an equivalence between (to use the same notation in the preprint): $$\rho_3 = \frac{\sigma(p^k)}{m} &lt; \frac{\sigma(m)}{p^k} = \mu_4$$ and $p^k &lt; m$. The last stumbling block to this method is mentioned in Remark $4.3$ from pages $17$ to $18$.</p>