Does the following approach (or something near it) exist in the number theory literature?

I will provide some motivation for $\omega(p^n - 1)$ as $n \rightarrow \infty$ and for this question. First, I recall that $\sigma(m)$ is the sum of positive divisors of the natural number $m$, and that $m$ is an $n/d$-multiperfect number when for coprime natural numbers $n \gt d $ one has $nm=d\sigma(m)$. Perfect numbers are the case $n=2$ and $d=1$.

Now here is an attack on the Odd Perfect Number Conjecture, with perhaps an extension to some odd $n/d$-multiperfects. Most perfect numbers are mostly powerful, where I mean all but one prime factor of the perfect number occurs with multiplicity greater than one. (6 is provably the sole exception to this characterization.) Indeed, all known existing perfect numbers $m$ satisfy $\log(m) \lt 2\omega(m)$. (Here as in the linked post above, $\omega$ counts prime factors with multiplicity, so $\omega(75)=3$.) While we may not expect the same inequality for odd perfect numbers, because they are mostly powerful, they should have more prime factors than most numbers of roughly the same size.

If we write the equation above for perfect numbers, using a prime factorization for $m$, we get that $2\prod p_i^{e_i} = \prod\sigma(p_i^{e_i})$, or in terms of $\omega$, $1 + \sum e_i = \sum \omega(\sigma(p_i^{e_i}))$. Now there is very little control over $\omega(\sigma(p_i^{e_i}))$ when $e_i=1$, but there are things to say when $e_i \gt 1$, and when the factors of $\sigma(p_i^{e_i})$ are limited to a finite set, this says a lot about $e_i$, and in particular may imply very few factors and a very special relationship between $p_i$ and the other prime factors.

We may not be able to dispose of the conjecture entirely, since there may be one factor of $m$ which is one less than a largely composite number, but we might be able to constrain that one prime and use some other information to rule out the existence of $m$.

I am aware of (but not well versed in) Kevin Hare's lower bounds on $\omega(m)$ for odd perfect $m$. Is there any other work on $\omega$ and multiperfects, even if they aren't trying to rule out odd perfect numbers? I would especially appreciate a paper in the odd perfect literature that entertains the above approach.

Comments or suggestions for improving/focusing the question are welcome, as are downvotes with accompanying rationale.

  • $\begingroup$ This question appears to be off-topic because it's asking us to check his approach to some old open problem. $\endgroup$ Jun 26, 2013 at 20:00
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    $\begingroup$ The intent is not to check the approach. The intent is to check the collective memory to see if someone remembers an approach similar to or even mildly resembling this. If someone says "this reminds me of paper X", that remark may prove useful. I will do the research regardless, but if someone has a useful pointer, that will speed things up for me. Or is what I am asking not a reference request? $\endgroup$ Jun 26, 2013 at 21:18
  • $\begingroup$ You say - "Most perfect numbers are mostly powerful, where I mean all but one prime factor of the perfect number occurs with multiplicity greater than one." - The definition of powerful numbers (as they are currently used in the literature) are in Wikipedia and Wolfram MathWorld, so I think you are using the terminology "powerful" in a different context here. Nonetheless, I would like to refer you to the following MSE post. $\endgroup$ Aug 11, 2013 at 1:07
  • $\begingroup$ Additionally, it is $\Omega(75)=\Omega({3}\cdot{5^2})=1+2=3$, not $\omega(m)$, which gives the number of prime factors of $m$, counting multiplicities. (In fact, $\omega(m)$ actually gives the number of distinct prime factors of $m$.) $\endgroup$ Aug 11, 2013 at 1:11
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    $\begingroup$ Jose, thank you for your comments and the reference. If you would like, perhaps we could try a chat room. I think your computation of $\Omega(q^kn^2)$ above has an error; but I point out the inequality to emphasize that perfect numbers have lots of factors, and if I were more careful I would say Euler proved that odd perfect numbers are either powerful or mostly powerful, if they exist at all. $\endgroup$ Aug 11, 2013 at 3:21


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