20040505, 03:22  #1 
Mar 2003
New Zealand
13×89 Posts 
Where is P1, P+1 effort recorded?
Is there any record of how much P1 and P+1 effort has been applied to the Cunningham numbers?
If not, does anyone have a guess? Is it reasonable to assume the base two numbers have all had P1 done at least to the Prime95 limit B1=4.29e9? 
20040505, 08:34  #2  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×3×1,753 Posts 
Quote:
Paul 

20040505, 11:04  #3  
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Most of the P1 and all of the P+1 I've done myself was either on Fibonacci/Lucasnumbers which Blair Kelly keeps track of on his web site, or mostly on Cunninghams before factoring them with NFS.
AFAIK Paul Zimmermann and Torbjorn Granlund have done a lot of P1 and P+1 on Cunningham numbers, Torbjorn on base 10 numbers and Paul on other bases afaik. Paul recently told me in an email: Quote:
Alex 

20040505, 13:34  #4  
Nov 2003
2^{2}×5×373 Posts 
Quote:
used a lower B1 limit (10^7), with a much higher B2 limit (10^13). I was testing out my FFT implementation of Step 2 on an Alliant FX8 using 4 of its 68020 processors. I also ran the smaller unfactored Fermat numbers. 

20040505, 18:41  #5  
Jul 2003
UK
33_{16} Posts 
Quote:
According to PMINUS1.TXT (see bottom of the page of http://www.mersenne.org/status.htm) all base two numbers <10000 (those with no known factor anyway) have been tested to B1=4.29E9 

20040507, 08:14  #6 
Mar 2003
New Zealand
485_{16} Posts 
Thanks for the information, I don't think I will do any P1 or P+1 on the Cunningham numbers at this time, but just continue with ECM.

20040507, 08:36  #7 
Mar 2003
New Zealand
13×89 Posts 
Another thought about P1 testing:
Say I was to do a very large P1 stage one, say on M(2^19) with B1=4.29e9 which would take me about 2 months. If the test finished and the final GCD found all the expected known factors (those known P where the largest factor of P1 is less than 4.29e9) would that be a good enough check that no error had occured during the test, or is there some way that an incorrect residue could still lead to the known factors being found? 
20040510, 15:45  #8 
"Nancy"
Aug 2002
Alexandria
100110100011_{2} Posts 
Yes, that would be a very powerful check to ensure that the residue did not get corrupted during the computation. In fact, it should suffice to leave one resonably sized (and smooth enough) prime factor out of low[mp].txt and check that that one is found.
A factor p is found in stage 1 if the order of the starting element of P1 has only prime factors that are included in stage 1, i.e. that are <=B1. If an error occurs somewhere during the computation, we can look at it as another P1 factoring attempt with a random starting element which gets exponentiated by the stage 1 primes remaining at that point. So if the error occurs while we are processing the prime q, the factor will still be found if the corrupted residue happens to be a kth power where k is the product of all prime factors of p1 smaller than q. The probability of a random residue being such a power is 1/k. So unless the error occurs right at the start of the computation, k will usually be large enough to make an "accidental" discovery of p in spite of the error very unlikely. Alex Last fiddled with by akruppa on 20040510 at 15:45 
20040510, 23:13  #9 
"Phil"
Sep 2002
Tracktown, U.S.A.
2×13×43 Posts 
Actually, I have had apparent errors doing P1 on large, highly composite exponents, most recently on M9699690. For example, in October 2003 (this was with version 23.4.1) I ran stage 1 to a bound of B1=2000 and stage 2 to B2=200,000. (I found a number of factors at the same time which I then factored using Dario Alpern's Java applet.) But then I ran stage 1 to B1=100,000 and found the following two factors:
320731441801 = 2^3 * 3 * 5^2 * 7 * 11 * 17 * 19 * 21493 + 1 119609902307611 = 2 * 3^2 * 5 * 11 * 13 * 17^2 * 19 * 59 * 28687 + 1 My question is, why didn't these factors show up earlier when I ran stage 2 to 200,000? Then again, I continued running stage 2 to B2=10,000,000 and turned up: 150747052048811 = 2 * 5 * 7 * 17 * 19 * 307 * 509 * 42667 + 1 So why did this factor only turn up at the end of stage 2 when it seems like it should have turned up at the end of stage 1? I've had this on the todo list for quite awhile without getting to it, but since factors have been missed in both stage 1 and stage 2, I have wondered if it could be a GCD problem instead. This particular exponent requires a lowm.txt file of over 400 entries, so perhaps it isn't a problem that normally arises with typical (prime) exponents, but I have been wondering about it. In which case, geoff's testing idea might be a good one. 
20040511, 09:28  #10 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
The factor
M( 19427 )C: 53089736370439 which is 1 (mod 28687) is not being found with B1=2000, B2=200000, either. I didn't find another good test case that's 1 (mod 21493). It looks almost as if Prime95 skipped a few primes in stage 2. Which ones apparantly depends on the choice of B2 and perhaps B1. Maybe this has something to do with the new duplicationpreventing code that was introduced in, afaik, v23. It does pairing of primes similar as described in Montgomery's "Speeding" paper, but also cancels smaller primes p if p(mD+d) where mDd is a prime included in stage 2. Perhaps it cancels those small primes a little too eagerly, i.e. even in cases where (mD+D) ends up not being included in stage 2 after all. It not too many primes are skipped this way, it's probably not a big deal, but it is unfortunate that smallish primes are lost which have a good probability of dividing the group order. Alex 
20040511, 21:57  #11  
P90 years forever!
Aug 2002
Yeehaw, FL
1C93_{16} Posts 
Quote:
2 * 3 * 15877 * 19427 * 28687 P1 shouldn't find that factor with B1=2000 

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