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A positive integer $N$ is said to be $k$-multiperfect if

$$\sigma(N) = kN$$

where $\sigma(x)$ is the sum of the divisors of $x$ and $k$ is a positive integer.

(The case $k = 2$ reduces to the original notion of perfect numbers.)

Now my question is the following: Are all known $k$-multiperfect numbers (for $k > 2$) not squarefree?

For the case $k = 2$, the only known exception is $N = 6 = 2\cdot3$.

Update [October 06 2013 - Manila time] :: This question has been cross-posted to MathOverflow here.

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  • $\begingroup$ Chen and Luo proves in Odd Multiperfect Numbers (preprint available via arXiv) that odd multiperfect numbers of a certain shape have a square part $M^2$ and are therefore not squarefree. $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 2 '13 at 23:42
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    $\begingroup$ Note that every even perfect number is a practical number. I've confirmed that this is also true of the $6$ known $3$-multiperfect numbers, the first $13$ $4$-multiperfect numbers, and the first $4$ $5$-multiperfect numbers. $\endgroup$ – Jaycob Coleman Oct 11 '13 at 6:00
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    $\begingroup$ I think it's a distinct possibility. I considered the question recently while trying to determine whether every known harmonic divisor number is a practical number. I've confirmed the latter for the first $700$ harmonic divisor numbers so far, but I'm currently working on extending this to the $937$th, $98079457512960$. I'll see if I can manage to confirm some larger multiperfect numbers as well. The relationship of this observation to your question is loose, so I'm glad it's been resolved by other means. To your observation in comments; this implies $4$ or $6 \mid N$. $\endgroup$ – Jaycob Coleman Oct 12 '13 at 1:03
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    $\begingroup$ I forgot that I had intended to update you on my results from further testing. Perhaps this isn't an appropriate place, but considering the question was resolved on MathOverflow perhaps it should be deleted anyway? Using the factorizations from Flammenkamp it takes only seconds to confirm that the known multiply perfect numbers are practical. You can find the conjecture at oeis. A generalization, which I've not tested nearly as far, is that if the odd part of $n$ divides $\sigma(n)$, $n$ is practical. Equivalently, if there exist $m,k$ such that $\sigma(n)=\dfrac{nk}{2^m}$, $n$ is practical. $\endgroup$ – Jaycob Coleman Nov 9 '13 at 13:10
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    $\begingroup$ Perhaps you could submit as your answer the link to MO answer. The conjectures are A007691 for multiply-perfect, A001599 for Ore. I was recently introduced to abundancy outlaws actually, but was unaware of the constructive result. I have emailed Kurt Ludwick about obtaining a copy of the paper. Thanks for the information. $\endgroup$ – Jaycob Coleman Nov 10 '13 at 0:36
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This question has been answered in MathOverflow here.

In particular, the answer confirms that all known $k$-multiperfect numbers (for $k > 2$) are not squarefree.

Additionally, $1$ and $6$ are the only squarefree $k$-perfect numbers (allowing for $k = 1$, and where $k = 2$ corresponds to the classical notion of perfect numbers).

These results were further validated by Flammenkamp (in a response to an e-mail inquiry), who maintains a database of $k$-multiperfect numbers.

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