20061119, 19:57  #1 
May 2005
2^{2}×3×5 Posts 
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 sfold 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) , 1393, 10001669 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 
20061119, 21:49  #2 
"Jason Goatcher"
Mar 2005
3·7·167 Posts 

20061120, 00:08  #3  
"William"
May 2003
New Haven
23×103 Posts 
Quote:
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. 

20061120, 00:28  #4 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
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 
20061121, 15:38  #5 
May 2005
2^{2}·3·5 Posts 
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 
20061123, 19:45  #6 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
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 
20061123, 21:40  #7 
May 2005
3C_{16} Posts 
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 
20061126, 13:29  #8 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Here's the remaining factorisation:
59122930024750227495169443662787547931316348401001823  163350799^191 I'll send you an email. Cheers, Alex 
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