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Old 2006-09-07, 19:45   #1
ewmayer
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Default Official Odd Perfect Number thread

First, I owe some folks an apology:

Earlier today I merged several older threads on the subject of OPNs, but at some point during my tidying-up I must've clicked the wrong link or button, because I inadvertently wound up hard-deleting the whole result. Mike (a.k.a. Xyzzy) says it is theoretically possible to reconstruct the original threads from server backups of the SQL database, but as that would be a major headache, and since most of the meaningful content in the old threads was links to OPN references, unless someone has a massive objection, I'd like to just move on - lesson for me is in future to never hard-delete anything, especially since the forum now has a nifty merge-threads-with-timed-expire-of-redirects feature for the moderators (you can currently see several redirects in the Math forum to threads I've moved which have this timed-expire feature). I apologize for any inconvenience this may cause, and to the folks who took the time to post to the deleted threads. The perils of housecleaning - sometimes good stuff inadvertently winds up going out ith the garbage.

So, please feel free to re-post links to any of your favorite OPN references and continue/reconstruct the deleted discussion, which can more or less be summarized as "discussion of the web-of-constraints evolution w.r.to OPNs, the current proven lower limit on the number of factors an OPN must have, and the utility of numerical search for OPNs."

I'll start with some related links:

Odd Perfect Number -- from Wolfram MathWorld

oddperfect.org -- Odd Perfect Number Search

Odd Perfect Numbers - A Factoring Challenge

=========================================

...and a recent e-mail exchange from the Number-Theory mailing list:

Quote:
For N, perfect number (PN), we have:

(1) h(N) = S(N)/N = 2,

where S(N) := number of all divisors of N. According to Euler for n, odd PN (OPN), we have:

(2) n = p1^k * m^2,

where p1 := prime, p1 == 1 (mod 4), k == 1 (mod 4). May be there are no OPN, but exist n*, approximately OPN (AOPN), complying with (2) and with h(n*), which very slightly differs from 2. I found the next examples (r is number of different prime divisors of AOPN):

r=6:
h(3^2*7^10*11^8*13^1*541^4*291457^2) = 2,0000000015..., p1=13;

r=7:
h(3^4*5^1*11^18*73^10*2647^6*348031^4*24133257823^2) = 2,000000000000000000026..., p1=5;

r=8:
h(3^2*7^32*11^26*13^1*541^10*291619^6*4475618963^2*399988521576257^2) = 2,000000000000000000000000000125..., p1=13.

HYPOTHESIS: There are infinitely many AOPN n* such, that with increasing r, lim h(n*) = 2.

I would like to have any information about the hypothesis.
Quote:
Subject: Re: Odd perfect numbers
From: Carl Pomerance
Date: Tue, 5 Sep 2006 09:53:48 -0400

The question is whether 2 is a limit point of the sequence h(n), where h(n) = sigma(n)/n = sum of the reciprocals of the divisors of n, and n is restricted to odd numbers of the form p^k *m^2, with p a prime that is 1 mod 4, with k = 1 mod 4, and p not dividing m. (These last are the conditions for sigma(n) = 2 mod 4.) In fact, h(n) is dense in [1,infinity) with n so restricted, so yes, 2 is a limit point. The proof uses the Erdos--Wintner theorem applied to the additive function g(m) = log h(m^2) for m odd, g(m) = 0 for m even. One immediately gets that not only is the sequence g(m) dense in [0,infinity), but there is a continuous distribution (in natural density). This almost solves the problem. To completely solve it, let p(m) be a prime that is 1 mod 4 and greater than m. Then the sequence h(p(m)*m^2) is dense in [1,infinity) and satisfies the conditions that are called for.

Carl Pomerance
Quote:
Subject:
Re: Odd perfect numbers
From: Charles Greathouse
Date: Tue, 5 Sep 2006 09:53:49 -0400

It is possible to construct an odd perfect number with abundance arbitrarily close to 2 with careful choice of factors. In fact, if a number has abundance less than 2, it is always possible to make the abundance greater without making it exceed 2; the unique sequence formed by choosing the lowest unused prime (starting at 3) proves your hypothesis, assuming I understand correctly: 3, 5, 7, 11, 389, 29959, 128194589, ...

Hmm, I don't see that on Sloane's list. Maybe I'll add it.

Charles Greathouse

Last fiddled with by ewmayer on 2006-09-13 at 22:22 Reason: Added URLs
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Old 2006-09-08, 00:49   #2
philmoore
 
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I see I must have had quite a few posts deleted in that process, but at least we probably won't miss all the heated exchanges about whether OPN research is even worthwhile...

On the other hand, I wouldn't be too quick to merge threads, as I remember discussions about several distinct topics that probably deserve separate threads, i.e., number of factors, size of an OPN, heuristics, etc. Fortunately, some of the more interesting discussions actually occurred in the factoring thread, but if anyone remembers anything interesting about what was deleted, feel free to ask about it and perhaps if I can't come up with references, Pace Nielsen or Joseph Chein or Kevin Hare, all contributors to this forum who have also published papers on odd perfect numbers, can help us out. William Lipp (oddperfect.org) is also a good resource.
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Old 2006-11-10, 08:41   #3
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Default On some recent results re: OPNs

Hi all. I was able to show recently that, given an odd perfect number
N = (p^k)*(m^2) with special prime factor p, then it must be true that
p^k < (2/3)*(m^2). I will post here the details of the proof as soon as I have the paper published in a local journal. Meanwhile, this result shows in particular that m^2 > (sqrt(6)/2)*(10^250) (since N > 10^500, according to http://www.oddperfect.org), which somehow strengthens Cohen's 1987 result that (p_sub{i})^(2*alpha_sub{i}) > 10^20 for some i. (Here (p_sub{i}) divides m; see http://en.wikipedia.org/wiki/Perfect_number.)

Best regards,

Jose
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Old 2006-11-10, 15:12   #4
Zeta-Flux
 
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It is always good to see someone else interested in OPN's. Keep it up.

Not to discourage you, but your result doesn't seem that difficult. I think it was Pomerance who proved that for an odd perfect number, there is no decomposition N=AB with gcd(A,B)=1 and \sigma(A)=2B, \sigma(B)=A. In particular, \sigma(m^2)\neq p^k. But we do know p^k |\sigma(m^2). Therefore, 3p^k <= \sigma(m^2) <= 2m^2; which recovers your result. (That is, if I didn't make a mistake.)
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Old 2006-11-10, 21:55   #5
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I didn't understand a bit of the last two posts, but I still think OPNs are way cool if they exist.

/me plans on doing a few more curves on wblipp's servers, even though he doesn't understand one bit of the math involved.

Hey, guys, if we find an OPN, maybe we could mail a hard copy of it and all it's factors to Mr. Silverman? Can we, PLEEEEEAAAAASSSSEEE? Huh, huh, can we? Huh?

Last fiddled with by jasong on 2006-11-10 at 21:58
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Old 2006-11-11, 07:46   #6
akruppa
 
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Quote:
Originally Posted by jasong View Post
Hey, guys, if we find an OPN, maybe we could mail a hard copy of it and all it's factors to Mr. Silverman? Can we, PLEEEEEAAAAASSSSEEE? Huh, huh, can we? Huh?
We sure could. Somehow I doubt he'll be holding his breath, though. It really isn't very likely that any exist.

Alex
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Old 2006-11-11, 19:13   #7
jasong
 
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Quote:
Originally Posted by akruppa View Post
It really isn't very likely that any exist.

Alex
Maybe, but it isn't like my question of whether or not 2^n-1 and 2^n+1 could be simultaneously prime(I later discovered it's only true for n=2). Unless it's definitively proven, it's always going to be possible.
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Old 2006-11-11, 19:20   #8
jasong
 
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While we're on the subject of perfect numbers:

If we did find an OPN, wouldn't it be possible that there are even perfect numbers that don't correspond to Mersenne primes? (I apologize if this question isn't mathematically robust, I'm not sure how to change it)
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Old 2006-11-11, 19:27   #9
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It's been proven by Euler that every even perfect number is of the form 2n-1*(2n-1), where 2n-1 must be prime. It's been known since Euclid's times that every Mersenne prime (although they didn't call them "Mersenne" as Father Mersenne's work was still two millenia off) leads to a perfect number, so even numbers are perfect if and only if they are of the stated form.

Alex

Last fiddled with by akruppa on 2006-11-11 at 19:29
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Old 2006-11-11, 23:55   #10
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Heh... there is a new result on odd perfect numbers: "Odd perfect numbers have a prime factor exceeding 10^8" by Takeshi Goto and Yasuo Ohno. This research brought to you by the E. W. Dijkstra foundation...

Alex
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Old 2006-11-18, 23:59   #11
jasong
 
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Quote:
Originally Posted by jasong View Post
/me plans on doing a few more curves on wblipp's servers, even though he doesn't understand one bit of the math involved.
Sorry if this is considered thread-crapping, but I just wanted to say that the "he" reference in the quote is supposed to refer to myself.

Wblipp seems like a nice guy, I'm not interested in offending him, accidentally or otherwise.
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