Let $N = {q^k}{n^2}$ be an odd perfect number (OPN) given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$). Note that $q \neq n$ and $q^k \neq n$.

In his Ph.D. thesis completed in the year 2003, Ronald Maurice Sorli conjectured that $k = 1$.

In the author's paper titled "The Abundancy Index of Divisors of Odd Perfect Numbers" published in the Journal of Integer Sequences in September 2012, he proved that $n < q$ implies $k = 1$. He likewise conjectured that $q^k < n$, on the basis of the related inequality $I(q^k) < \sqrt[3]{2} < I(n)$, where $I(x) = \frac{\sigma(x)}{x}$ is the abundancy index of the (positive) integer $x$.

It is now natural to ask as to whether the converse "$k = 1$ implies $n < q$" holds. It is easy to see that this is true if $n < q^k$. From earlier blog posts here, we have shown that if $q^k < n$, then $k \geq 1$ if and only if $q < n$. Consequently, to prove the biconditional "$k = 1$ if and only if $n < q$", it suffices to rule out the conjunction $q^k < n$ and $k = 1$.

Update (February 25, 2013) - I have decided instead to upload a paper containing the details for some new results that I have obtained related to Sorli's conjecture. The paper is set to appear in the arXiv in a few days.

Update (February 26, 2013 - 9:00 AM MNL time) - The paper has just appeared in the arXiv.

Update (March 27, 2013 - 10:30 PM MNL time) - A sequel to the first paper uploaded last February 26 is now available in the arXiv.

Update (March 29, 2013 - 12:15 AM MNL time) - The third installment to this series of papers for Feb-Mar 2013 will appear in the arXiv at around 9:00 AM MNL time today.

Update (March 30, 2013 - 11:30 AM MNL time) - There is some delay in the posting of the third installment to arXiv. I have temporarily uploaded a copy to Scribd.