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.



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 $r < t$.]

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    $\begingroup$ Sam Dittmer was my undergraduate research student when he wrote that paper. He is currently a first year PhD student at UCLA, and so I believe he is thinking about other topics at the moment! $\endgroup$ – Pace Nielsen Mar 5 '15 at 5:34
  • $\begingroup$ Thanks for the heads-up, @PaceNielsen! I guess what I'm really asking in this MO question is that: Do results for odd perfect numbers carry over to the Descartes / spoof odd perfect numbers? (e.g., if $n = km$ is a Descartes / spoof odd perfect number such that $\sigma(k)(m + 1) = 2n = 2km$, I get the following: (1) $k$ must be an odd square. (2) $m$ must be $\equiv 1 \pmod 4$. (3) $5/3 \leq I(k) = \sigma(k)/k < 2$ and $1 < (m+1)/m \leq 6/5$. So it does seem that the Descartes-Frenicle-Sorli conjecture holds for spoof odd perfect numbers.) $\endgroup$ – Arnie Bebita-Dris Mar 5 '15 at 6:12
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    $\begingroup$ "Do results for odd perfect numbers carry over to the Descartes / spoof odd perfect numbers?" The answer is usually yes. In my experience, the only time the answer is no is when an OPN result depends on numerical computations involving prime factorizations. $\endgroup$ – Pace Nielsen Mar 5 '15 at 16:58
  • $\begingroup$ arnienumbers.blogspot.com/2017/08/… $\endgroup$ – Arnie Bebita-Dris Aug 25 '17 at 16:43

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