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Old 2006-11-19, 19:57   #1
jchein1
 
May 2005

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Default Lower bounds for odd multiperfect numbers.

Happy thanksgiving everyone,

Let sigma(n) denote the sum of the positive divisors of n and let Ns be a hypothetical odd s-fold perfect number, i.e. sigma(Ns) = s*Ns. It’s known that BCR (1991) established N2 > 10^300 and Cohen & Hagis (1985) proved that Ns > 10^70. Recently, I finished a paper “There is no odd multiperfect numbers < 10^300 except possibly for N4” which proved that

N3 > 10^300; // Sorli(2003) proved that N3 > 10^128
N4 > 10^100;
N4 > 10^300 if 3 || N4;
N5 > 10^300; // McCarthy(1970) proved that N5 > 10^138
N6 > 10^700;
N7 > 10^2163;
N8 > 10^6816.

I left 5 uncompleted factorizations. Although it did not affect the proof, they are essential if one wishes to extend N3:

1) c117 | sigma(25249969^16) , 2282 curves completed
2) c114 | sigma(163350799^16) , 1-393, 1000-1669 curves completed
3) c127 | sigma(163350799^18) , 533 curves completed
4) c118 | sigma(242980751471069070767002375553611^4), 2190 curves completed
5) 4898: c116 | sigma(6294091^18) // 2215 curves completed (I bet wblipp has this one; 4898 is line number from original BCR proof)

Remarks:
1) The proof for Ns (s >= 6) is strictly number theory. In particular, I improved w(N6) from 142 (Reidlinger in 1983) to 144, where w(n) is the number of distinct factors of n.

2) What I will do next is try to use the BCR (N2 > 10^300) proof to show that N3 > 10^300. Therefore the N3 proof is heavily dependent on the BCR proof. It will be difficult to understand this if you don’t know BCR.

Suppose N3 < 10^300. Since N3 is a perfect square and contains no special primes, this implies all BCR trees (factor chains) contains no special prime and has been terminated with B1 or B3 bound can be reused for N3, where B1 and B3 are running bounds used in both N2 and N3 but there is a slight difference. For example, if 3^2*7^3*467^2*2801 | N2 then N2 >= 3^2*7^4*467^2*2801 > 10^12 (exponents adjusted in accordance Euler’s form), then B1 = 12. If 3^2*7^3*467^2*2801 | N3 then N3 >= 3^2*7^4*467^2*2801^2 > 10^16 (exponents adjusted accordance N3 is a perfect square), thus B1 = 16. This show by example implies that if same tree belong to both N2 and N3, then B1 (for N3) >= B1 (for N2).

3) I proved next that if s = 3,4,5 and if q^k || Ns, where q is an odd prime then Ns > q^2k. This fact takes care of N2 trees involving no special prime and terminated with B2 bound.

The total lines of N3 proof is closed to 4000, 3500 lines borrowed from BCR and I added 500 lines to resolve N2 trees terminated with abundant (i.e. sigma(N2) > 2*N2).

Since N3 proof involved no abundant logic, it’s easy to see that N5 proof is identical to N3 proof.

4) It seemed to me that BCR approach doesn’t treat N4 well because of small prime overlapping badly. I improved w(N4) from 23 (Reidlinger in 1983) to 24, then and I came up with an alternate approach (approximately 900 total lines), proving that N4 > 10^100. N4 is most difficult one to deal with. We might need a better approach to handle N4.

5) I sincerely hope wblipp will publish his “N2 > 10^500” proof soon, can wblipp show us some of his smaller roadblocks which lie in between 10^140 and 10^275 ?

I was working on the BCR’s “lifting” algorithms for quite some time. I am certain it’s feasible but difficult to push N2 > 10^700. We definitely need “lifting” algorithms to overcome many potential roadblocks. Thanks to alpertron and jasong for their excellent tools, which enabled me to complete this paper.


Joseph E.Z. Chein
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Old 2006-11-19, 21:49   #2
jasong
 
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"Jason Goatcher"
Mar 2005

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Quote:
Originally Posted by jchein1 View Post
Thanks to alpertron and jasong for their excellent tools, which enabled me to complete this paper.


Joseph E.Z. Chein
I think you mean jasonp, since my mathematical education isn't even close to his.
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Old 2006-11-20, 00:08   #3
wblipp
 
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May 2003
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Quote:
Originally Posted by jchein1 View Post
1) c117 | sigma(25249969^16) , 2282 curves completed
2) c114 | sigma(163350799^16) , 1-393, 1000-1669 curves completed
3) c127 | sigma(163350799^18) , 533 curves completed
4) c118 | sigma(242980751471069070767002375553611^4), 2190 curves completed
5) 4898: c116 | sigma(6294091^18) // 2215 curves completed (I bet wblipp has this one; 4898 is line number from original BCR proof)
I don't have any of these factorizations, although I'll be glad to collect them when available. Let me know if there are any other factorizations of interest that I might have. At some point we will want to consolidate our factorization lists.

Quote:
Originally Posted by jchein1 View Post
5) I sincerely hope wblipp will publish his “N2 > 10^500” proof soon, can wblipp show us some of his smaller roadblocks which lie in between 10^140 and 10^275 ?
The continuing time crunch at work has been a real surprise to me - I wouldn't have shared my intentions if I had known I would continue to have so little time for this.
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Old 2006-11-20, 00:28   #4
akruppa
 
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Aug 2002
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Joseph,

these five numbers are suitable for SNFS and can be factored very quickly. I've started sigma(25249969^16), although it probably isn't even worthwhile to post progress as the whole batch should be done quite soon.

Alex
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Old 2006-11-21, 15:38   #5
jchein1
 
May 2005

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Wblipp,

Thank for your quick response. Since the BCR proof has 12564 lines, I guess your proof is at least several times bigger. My gut feeling is that your N2 > 10^500 proof may be much more difficult and complicated than people think. Please give me a brief description of your project if you can. If there is any way I can help please let me know.


Happy Thanksgiving Day


Joseph
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Old 2006-11-23, 19:45   #6
akruppa
 
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1) 6238111573895116928785756590919735630276379 | c117
2) 805093815816265024108813825580038113018833 | c114
3) not quite done yet, should be there tomorrow
4) 2368569500780882215999847855758811 | c118
5) 12006895985805536119334711015880094863231210125088463127 | c116

This took much longer than I had expected, mostly because I got the sieving parameters wrong and ended up sieving far more than I should have...

Alex
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Old 2006-11-23, 21:40   #7
jchein1
 
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Alex,

Thank you very much for your help. I am working on a plan. Hopefully we can work together to push “N3, N5 > 10^500 in the shortest possible time. I have a couple of aces otherwise I wouldn’t invite you. I firmly believe that factorizations alone are rare solve math problems. I am looking forward to working with you. If you are interested please let me know.


Happy Thanksgiving Day


Joseph
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Old 2006-11-26, 13:29   #8
akruppa
 
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Here's the remaining factorisation:

59122930024750227495169443662787547931316348401001823 | 163350799^19-1

I'll send you an email.

Cheers,
Alex
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