20071113, 01:15  #1 
Apr 2007
Spessart/Germany
10100010_{2} Posts 
43 digit primefactor with P1 of GMPECM
hi,
never saw such a prime detected with P1: GMPECM 6.1.2 [powered by GMP 4.1.4] [P1] Input number is (151^1391)/150/2503/295411610933/52665989/114307715297 (268 digits) Using B1=10000000, B2=880276332, polynomial x^24, x0=294542479 Step 1 took 52953ms Step 2 took 15031ms ********** Factor found in step 2: 6020018307223318401319089049820274622700173 Found probable prime factor of 43 digits: 6020018307223318401319089049820274622700173 Composite cofactor ((151^1391)/150/2503/295411610933/52665989/114307715297)/6020018307223318401319089049820274622700173 has 225 digits worth to be reported? greetings Matthias edit: GMPECM in subject, sorry Last fiddled with by MatWurS530113 on 20071113 at 01:17 
20071113, 02:20  #2  
Jun 2005
lehigh.edu
2^{10} Posts 
Quote:
link on the page for this section, GMPECM, you'll find the current p1 records; in particular, at http://www.loria.fr/%7Ezimmerma/records/Pminus1.html there are 10 factors above p50, with p66 as the current record; pending a current focussed effort  with improved step2 development code  reported elsewhere, with the recent new p+1 record. Bruce 

20071113, 02:25  #3 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
The group order is remarkably smooth, the largest prime factors are only 1943657 and 43270121.
Alex 
20071114, 05:07  #4 
"William"
May 2003
New Haven
23·103 Posts 
Congratulations on the nice factor!
Richard Brent publishes factors of a^n ± 1 with a and n both < 10,000. http://wwwmaths.anu.edu.au/~brent/factors.html I'll add this to my factors of p^q1 with p an odd prime and q a prime for oddperfect.org. I'd enjoy learning about any other such factors you have found. William 
20071115, 02:34  #5 
Apr 2007
Spessart/Germany
2·3^{4} Posts 
Hello,
thanks for your replies. Of course I know Richard Brent's page with the extension of the Cunningham numbers. And Zimmermann's page with the champs and factorization records, too (his pages are in french... I can't speak or read any word french...). I try to factor RepUnits to bases which depends on the exponent. f.e. f(p, k)=((p+k)^p1)/(p+k1) with p prime and 2p <= k <= p+1 the factor is out of the series with k = 12. For k = 2p the series is the Mersenne series. And my special series is for k = 1 f(p) = ((p1)^p1)/(p2) this series is something like a 'twinseries' to the Mersenne numbers. f.e. proven is: if the exponent p of a Mersenne number is a SG3 (Sophie Germain prime congruent 3 mod 4) then 2p+1 divides M(p) for this series I suppose (testet with SG1's up to ~ 10000): if the exponent is a SG1 (SG congruent 1 mod 4) then 2p+1 divides f(p) divisors are congruent +1 or +3 mod 8 divisors are congruent 1 mod 2p the numbers of the series seem to be relative prime to each other (like the Mersenne numbers) the series seems to be squarefree (as supposed for the Mersenne numbers, too) @ Bruce: a p50+ I only saw at the end of a factorization @ Alex: I think the group order (or the P1 factoring as I always say) must be so smooth. I think I can get the factor only if B1 > 2nd largest primefactor of P1 and B2 > largest primefactor of P1. Or is there a fault in my thoughts? @ William: yes, I have some factors which are not in Brent's lists. And they are not in the list with factors found by GMPECM, which is available at Brent's page, too. I try to make my own lists for repunits with base <= 257 and prime exponents <= 257. But I miss an update of the factors which are already known... btw: I run msieve on a 116 digit composite remaining from 89^1191 In Brent's list this number is not solved, but the last update was from february or so. Maybe someone know the status of this number? If it is already done I don't need to continue msieve (it will last some weeks, I only can run 1 core ~ 8 hours a day on it). greetings Matthias 
20071116, 00:30  #6 
Nov 2007
home
5^{2} Posts 
I factored 2^30001 and found a factor of 71 digits (8877945148742945001146041439025147034098690503591013177336356694416517527310181938001) How can I check if this is a new factor? In 2^200001 I found a prime (It passed the deterministic prime test) factor of (408 digits) 155386 735065 302807 657338 888324 840574 967976 392677 934613 718122 647287 544656 765233 262721 341393 606016 744531 000112 436610 774942 779955 600519 736335 602844 010769 786254 387975 001963 076667 172256 354309 736758 381224 102266 657673 701281 341390 405297 835922 866481 748603 417931 493488 599294 075617 908954 452734 945612 088800 522580 496750 320104 713305 287483 233447 455701 091443 252058 311861 464889 310635 462175 309145 469354 631868 486251 948453 328001
Last fiddled with by vector on 20071116 at 01:20 
20071116, 01:15  #7 
Einyen
Dec 2003
Denmark
2·1,601 Posts 
Sorry to say this was found as a factor of 2^10001:
http://www.jjj.de/mathdata/mersennefactors.txt Very nice find though Last fiddled with by ATH on 20071116 at 01:16 
20071116, 10:24  #8  
Feb 2006
Denmark
2·5·23 Posts 
Quote:
The 408digit factor of 2^200001 is a known cofactor of (2^2000+1)/(2^400+1). http://www.euronet.nl/users/bota/mediumpeven4k.txt doesn't give the decimal expansion but it's 2^2000+1 divided by all the listed factors. 

20071116, 15:06  #9  
Nov 2003
1D24_{16} Posts 
Quote:
primitive prime factor and an algebraic factor. A good intro to this subject is the Cunningham book itself. You should read it before continuing your pursuits. See Sam Wagstaff's web pages. 

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