20221206, 23:44  #1 
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Aug 2009
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CRUS sieving questions
rogue suggested I come here because I like to sieve. His srsieve2cl and the RTX 2080 I have do quite well. I have read most of the pages in this topic. There is a lot more to read in the primary topic.

20221207, 01:01  #2 
"Gary"
May 2007
Overland Park, KS
10111000001111_{2} Posts 
Great! Feel free to contribute however you want. Mark (Rogue) is the residential sieving expert on many different projects and writes/maintains many sieving programs. He will be able to answer whatever questions you have regarding sieving.

20221207, 01:28  #3 
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Thank you for the reply! Is there a table for something similar where a person can pick what they wish to run?

20221207, 03:09  #4  
"Mark"
Apr 2003
Between here and the
2·7^{2}·71 Posts 
Quote:
Look for a conjecture that isn't reserved. Then go to http://www.noprimeleftbehind.net/cru...ereserves.htm or http://www.noprimeleftbehind.net/cru...ereserves.htm to see if there is a sieve file for the an unreserved range. You will then need to look in this thread to see if anything is reserved for sieving. To save you time, look at conjectures with fewer than 1000 k remaining. I have or am sieving most of the ones with more than that number of k. Sieve from n to 2n. You will need to run a test with llr to determine target removal rate. I typically run a test for k*b^(1.7n)+/1. 

20221207, 05:44  #5  
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Below is a snip from the first link you provided. How do these column headers relate to k*2^n1. The second link appears to be finished sieves when any zip file on the right is selected. 

20221207, 06:29  #6  
"Gary"
May 2007
Overland Park, KS
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Quote:
Because of this, I'd suggest picking something with 30 to 100 k's remaining (that is not base 2) to get your feet wet. I'll recommend Riesel base 60 or R60 as we call it here. It has 33 k's remaining. See the k's remaining here: http://www.noprimeleftbehind.net/cru...60reserve.htm Riesel is the 1 side and Sierpinski is the +1 side. So you would sieve 36*60^n1, 1770*60^n1, 4708*60^n1, etc. with all of the 33 kvalues shown on that page. You can see on the page that R60 has already been tested to n=100K and it does not already have a sieve file. I would suggest sieving n=100K250K or you could sieve n=100K500K but the sieving job would be much longer. To determine an optimum sieving rate, you'd need to be able to run a partial LLR test to estimate how long that test would take. Nash weight is meaningless when sieving multiple k's together. Software changes quite often these days. New and faster sieves are created, better versions of LLR are created, etc. Each time that happens, the optimum sieve depth can change. That's why the speed of an LLR test must be compared to the speed of your sieve to determine the optimum rate. As per your final comment, if you see a sieve file on a base, it means it's already been sieved so of course you should choose something else. Last fiddled with by gd_barnes on 20221207 at 06:33 

20221207, 07:18  #7  
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20221207, 15:04  #8  
"Gary"
May 2007
Overland Park, KS
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Quote:
Just checking: You will be sieving all of the k's at one time. Is that correct? There are no formal reservations for sieving. Just state here in this thread that you will be sieving R60 for n=100K250K. 

20221207, 16:25  #9  
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I started out using a P value is 100e9. I will probably increase this to get the final term count down more. Of course, raising this increases the time required. The k=36 sieve I ran last night took 40 minutes and ended with 6,091 terms. I am running it again with P at 250e9. I would rather they take longer. Not so much machine interaction. I will leave these in .abcd format. It is the default and far more compact than .npg. I won't post the sieves until all are complete. Do you have a recommendation for a final P value? 

20221207, 17:25  #10  
"Mark"
Apr 2003
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20221207, 18:58  #11  
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Note: I think you meant 250,000 above. 

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