20121005, 17:51  #551  
Feb 2012
3^{4}·5 Posts 
Quote:


20121005, 18:52  #552 
Aug 2010
Kansas
547 Posts 
Composite exponent + composite divisor= ?

20121006, 04:02  #553  
Jun 2003
3^{2}×19×29 Posts 
Quote:
Quote:
Last fiddled with by axn on 20121006 at 04:03 

20121009, 00:39  #554 
Sep 2002
789_{10} Posts 
P1 found a factor in stage #2, B1=530000, B2=9805000.
UID: Jwb52z/Clay, M58188989 has a factor: 55849484816777970918764473 85.530 bits. 
20121011, 16:45  #555 
Sep 2006
Odenton, MD, USA
2^{2}·41 Posts 
Two factors found with BrentSuyama:
P1 found a factor in stage #2, B1=260000, B2=6305000, E=12. M4037023 has a factor: 46132411290706485444839 k = 5713667136737453 = 17 × 31 × 53 × 1327 × 154,154,969 P1 found a factor in stage #2, B1=290000, B2=7395000, E=12. M4438789 has a factor: 1402524780745530895151 k = 157985069885675 = 5^2 × 73883 × 85,532,569 
20121013, 09:41  #556 
"Åke Tilander"
Apr 2011
Sandviken, Sweden
566_{10} Posts 
A big one
ANONYMOUS Manual testing 1019 FECM Oct 13 2012 2:05AM 0.0 0.0000 1140356877758679056056869944845540826402854641895928218298013381554156431441
249.334 bits Quote from http://www.mersenne.org/report_recent_cleared/ There are now 4 known factors of M1019 with a total size of 452.3 bits Factor was not found by me. It would be interesting to know who was the "Anonymous" this time? 13th biggest known factor of any Mp (not counting the biggest factors of fully factored Mps). Does anyone know wether the remaining 567bit factor is composite or not? OK Now I have found this post. Adding a question: Does "prp" in frmky:s log mean that the factors are probable primes, not proven to be primes? Last fiddled with by aketilander on 20121013 at 10:39 
20121013, 10:16  #557 
Apr 2012
993438: i1090
2·73 Posts 
Found by NFS@home
http://escatter11.fullerton.edu/nfs/...ead.php?id=386 The remaining 567bit factor is prime Last fiddled with by Jatheski on 20121013 at 10:16 
20121013, 17:34  #558  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1C35_{16} Posts 
Quote:


20121013, 18:13  #559  
"Åke Tilander"
Apr 2011
Sandviken, Sweden
2×283 Posts 
Quote:


20121014, 08:37  #560  
May 2007
Kansas; USA
2×3×11×157 Posts 
Quote:
Quote:
For what it's worth: M268435456 = 3 * 5 * 17 * 257 * 641 * 65537 * 6700417 * ?? Or more interestingly M(2^28) = (2^1+1) * (2^2+1) * (2^4+1) * (2^8+1) * (2^16+1) * (2^32+1) * ?? Note that 2^32+1 = 641 * 6700417. Edit: The question is: How long does this sequence continue? In other words are 2^64+1 and 2^128+1 factors also? Last fiddled with by gd_barnes on 20121014 at 09:36 Reason: edit 

20121015, 00:16  #561 
May 2007
Kansas; USA
10100001111010_{2} Posts 
After some analysis, I just answered my own question above. I'm sure many on here recognize this but I did not. I'll state it for others like me who did not know the following:
For any 2^(2^q)1 where q is sufficiently large, algebraic factors are: Code:
[2^(2^0)+1] * [2^(2^1)+1] * [2^(2^2)+1] * [2^(2^3)+1] * [2^(2^4)+1] * ..... * [2^(2^(q1))+1] Code:
[2^(2^0)+1] * [2^(2^1)+1] * [2^(2^2)+1] * [2^(2^3)+1] * [2^(2^4)+1] * ..... * [2^(2^27)+1] Last fiddled with by gd_barnes on 20121015 at 00:21 
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