20090324, 15:40  #166 
May 2007
Kansas; USA
2^{2}×5×11×47 Posts 
Problem with PFGW?...
That is VERY strange! Can any technical person help us with this apparent PFGW problem here?
If I don't get a response within a couple of days, I'll create a separate thread for this and perhaps PM a few more knowledgeable folks. Gary 
20090417, 06:50  #167 
May 2007
Kansas; USA
2^{2}·5·11·47 Posts 
I have not had time to do the follow ups with the particular issue involving PFGW. It is very strange for such a small test. The problem test was:
200212800*3^4+1, which factors to 19*31*41*61*101*109. Perhaps the large # of factors (i.e. smoothness) contributes to PFGW's problem. Therefore, I needed to check the k for myself using Alpertron's site and found that: 200212800*3^13+1 is prime. For k=200M300M and n<=25K, KEP went ahead and sent me his primes and k's remaining for n>2500 quite a while ago. I'll add k's remaining to the pages and show the range complete here shortly. Gary Last fiddled with by gd_barnes on 20090417 at 06:55 
20090417, 06:58  #168 
May 2007
Kansas; USA
10340_{10} Posts 
For historical balancing reference as sent to me by KEP:
k=200M300M has 715 k's remaining at n=25K. Subtract 246 k's that are divisible by 3. Add back 11 k's that are divisible by 3 where k+1 is prime. Total 480 k's remaining at n=25K. Gary 
20090417, 10:35  #169 
Quasi Admin Thing
May 2005
2^{2}·241 Posts 
@ Gary regarding my base 3 reservation:
Glad that you decided to accept the range as complete. And I guess you actually may be on to something, regarding the amount of factors, because maybe it reaches a limitation in an allowed length of factors found (or how to describe it). Don't know Off topic here, I can tell you, that I followed you advice on looking at the patterns on the riesel side, and I've actually been able to come up with a k == "remainder" mod "divisor" (divisor) answer in stead of just writing the apparent general factor for each k remaining. So in the next batches of completion I'll hand over to you, you'll get the modulos (or whatever you call it) in stead of just the factor I can btw not tell when I will hand over the next batch of completed bases to n=25K or proven, but I can tell that I'll hand over 84 bases which you can add in your own pace :) Also it is actually quite educational for me to do the investigation myself and having to do a lot of the thinking myself, so I actually thinks that my skills will improve drasticly as I move down the list of bases Take care everyone. Kenneth! 
20090417, 11:39  #170 
May 2007
Kansas; USA
2^{2}·5·11·47 Posts 
You haven't been specific enough to show that you understand which k's have trivial factors.
The pattern of the trivial factors for each base is as follows: 1. Find the prime factors of the (base minus 1). Let's say they are A and B. 2. The k's with trivial factors for your base are: (a) For Riesel, k==(1 mod A) and k==(1 mod B). (b) For Sierp, k==(A1 mod A) and k==(B1 mod B). Example for base 46: 1. The prime factors of 45 are 3*3*5. 2. The k's with trivial factors for base 46 are: (a) For Riesel, k==(1 mod 3) and k==(1 mod 5). (b) for Sierp, k==(2 mod 3) and k==(4 mod 5). For bases where the base1 is prime, it's simply k==(1 mod b1) for Riesel and k==(b2 mod b1) for Sierp. Example: base 48. Riesel has k's with trivial factors for k==(1 mod 47) and Sierp has k's with trivial factors for k==(46 mod 47). That is why base 2 is the only base with no k's that have trivial factors. If you subtract 1 from the base you get 1 but 1 is not considered prime and therefore cannot be a prime factor. Hence there are no base 2 k's with trivial factors. A more complex example is base 211. 210 factors to 2*3*5*7, therefore many k's have trivial factors for base 211. For Riesel, it's k==1mod2, 1mod3, 1mod5, and 1mod7. For Sierp, it's k==1mod2, 2mod3, 4mod5, and 6mod7. Does that make sense? It's quite simple when you see the pattern. You don't have to use srsieve or Alperton's site to eliminate k's or come up with factors. k's with algebraic factors on the Riesel side have a more complex, yet easily discernable, pattern once you are able to see it. Before I was able to pick it up, the project had to expand beyond base 32. Once it got near base 5060, I saw 12 recurring patterns and then was able to "extrapolate" several more, which subsequent testing bore out. Gary 
20090417, 12:45  #171 
Quasi Admin Thing
May 2005
964_{10} Posts 
Actually it does make sence. Thanks for the explanaition, as soon as I'm done replying here, I'll copy this information to a notepad document and then I'll use the information, to doublecheck weather or not I infact has come up with all k's with trivial factors. Also you're right, I was not quite clear enough on what I understood, so in short term, I understood at the time I posted that the trivial factors appears in a pattern i.e. k = = 1 mod 31 (31) will appear for bases 32, 63, 94, 125, 156, 187, 218, 249 etc. etc. etc.
But now I think that I'll prepare a spreadsheet that can help me find the primefactors of b1 Thanks for the very educational and informative knowledge. Regards Kenneth 
20090825, 01:45  #172 
May 2008
Wilmington, DE
2^{2}·23·31 Posts 
Sierp Base 3
Sierp Base 3
k=200M300M (480k's) Testing from n=25K50K 
20091027, 19:04  #173 
May 2008
Wilmington, DE
2^{2}·23·31 Posts 
Sierp Base 3
Sierp base 3 k=200M300Mn=25K80K complete
322 primes found and proven  see attached list I will continue to n=100M Last fiddled with by MyDogBuster on 20140902 at 09:16 
20091027, 23:52  #174  
May 2007
Kansas; USA
2^{2}×5×11×47 Posts 
Quote:
For balancing purposes for k=200M300M at n=80K, we have 480  322 = 158 k's remaining. Dang! Gonna be crunching a few years? You sure you'll live that long? If you're going to test that high, I'd suggest doing k<100M first. Last fiddled with by gd_barnes on 20091027 at 23:58 

20091028, 00:29  #175  
May 2008
Wilmington, DE
2852_{10} Posts 
Quote:


20091201, 23:42  #176  
May 2008
Wilmington, DE
2^{2}·23·31 Posts 
Sierp base 3 k=200M300Mn=80K100K complete
21 primes found and proven  see attached list Releasing range Quote:


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