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OPN Research - Status as of Q4 - 2010

Hello World! Hello Philippines! :-)

Some good news: I think I may have proven that the inequality
m < p^k holds true in general. (Given an OPN N of the form N = (p^k)(m^2) where p is the Euler prime and gcd(p, m) = 1.)

If the proof is correct, then this means the following:
1. Refer to this post for a related result In particular, my latest result will imply that k = 1. This resolves a conjecture by Ronald Sorli in his Ph. D. thesis (2003).
2. It also follows that we may drop k ALTOGETHER in our OPN computations. (Isn't that cute? Now we only have to deal with sigma(p) and NO LONGER with sigma(p^k).)
3. Since it also follows that m < p (by dropping k), we now know that the Euler prime p is the largest prime factor of an OPN. (Since it is the largest, it definitely cannot be the smallest -- this latter assertion has been known for some time.)
4. Note that we also already know that p^k < (2/3)*(m^2). Dropping k ( :-) ), we have
p < (2/3)*(m^2), and thus, we have suitable (both) upper and lower bounds for:
(a) m in terms of p : (1/2)*sqrt(6p) < m < p
(b) p in terms of m : m < p < (2/3)*(m^2)
5. The restrictions on the structure of an OPN would now be so tight, that I am inclined to think that we may be able to see a conclusive proof that "An OPN does not exist" before the year 2010 ends. @hooray@

Will post the proof for the inequality "m < p^k" as soon as I have it completely re-worded.


Arnie Dris said...

The assumption m < p^k leads to a contradiction.

m < p^k
--> k = 1
--> m < p
--> m < p < Sigma(p) < Sigma(m) < 2m
--> 1 < Sigma(p)/m < 2

This last inequality contradicts:

Sigma(p^k)/m = Sigma(p)/m <= 2/3

(Please see for more details.)

Arnie Dris said...

Sigma(p^k)/m = Sigma(p)/m <= 2/3  --> retracted

What is true, however, is that:

Sigma(p^k)/m^2 <= 2/3