Descartes, Frenicle, and subsequently Sorli, conjectured that $k = 1$, if $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$).

The inequality $q^k < n^2$ follows from a result of Pomerance. In particular, this implies that $n < q \Longrightarrow k = 1$ is true. (Note that, if $n < q$, then the *Euler prime* $q$ becomes the largest prime factor of the odd perfect number $N$. In comparison to the case of even perfect numbers, the Mersenne prime $2^p - 1$ is the largest prime factor with exponent $1$.)

Taking off from a recent MO question, we have the following related papers:

Descartes Numbers by Banks, et.al.

and

SPOOF ODD PERFECT NUMBERS by Dittmer

My question is this: Does anybody here know of *more recent* papers that tackle whether the quasi-Euler prime of a spoof odd perfect / Descartes number is the largest quasi-prime factor and also has exponent $1$? Quick searches via Google for the keywords "Descartes number" or "spoof odd perfect number" do not seem to yield much results.

I have likewise e-mailed Professors Banks and Dittmer, hoping to get some information.

[Edit (March 5, 2015): Just for the sake of comparison - In OEIS sequence A228059, T. D. Noe lists odd numbers of the form ${r^s}{t^2}$, where $r$ is prime with $r \equiv s \equiv 1 \pmod 4$ and $\gcd(r,t)=1$, that *are closer to being perfect than previous terms*. He notes that, coincidentally the first $9$ numbers in this sequence have the exponent $s=1$. Additionally, except for the first number in the sequence ($45$), all of the succeeding numbers listed in A228059 satisfy $r < t$.]