20060624, 21:39  #23  
"Robert Gerbicz"
Oct 2005
Hungary
5·17^{2} Posts 
Quote:
if you want an estimation for the number of survivors sieving by primes up to H and sieving k from Nmin to Nmax ( sieve only odd numbers ) then there will be about prodprime(p=3,H,(p2)/p)*(NmaxNmin)/2 survivors because there are two forbidden resideu classes for every p>2 prime and primes are "independent". So we need only to approximate prodprime(p=3,H,(p2)/p). Known limits: limit(H>infinity,prodprime(p=3,H,p*(p2)/(p1)^2))=C2=0.6601618158 where C2 is the twin prime constant. limit(H>infinity,prodprime(p=2,H,(p1)/p)*log(H))=exp(gamma) where gamma is the EulerMascheroni=0.5772156649. From these limits we can get: limit(H>infinity,prodprime(p=3,H,(p2)/p)*log(H)^2)=4*exp(2*gamma)*C2. So a good approximation for the number of survivors: 2*exp(2*gamma)*C2*(NmaxNmin)/log(H)^(2), here log is the natural logarithm. Using your latest data: Nmax=25e9 and Nmin=400e6 and H=101.1T, so the estimation: 2*exp(2*0.5772156649)*0.6601618158*(25*10^94*10^8)*log(101.1*10^12)^2=9846234 Very good prediction because your correct data was 9849791 

20060625, 22:22  #24  
Mar 2004
17D_{16} Posts 
Quote:
The probability of not finding a factor of a number is log(beginningfactorrange)/log(endfactorrange) That is independant of the logarithm base (if you take the same base in the nominator and denominator. I squared this number bevcause we are searching for twins (after sieving to 10^12 they are independant enough). If you multiply it with the number of candidates, you get the remaining number of candidates. the (24.7G / 24.8G) is just the proportional shotening if the range. My predictions just say, that there are many too few candidates removed between 30.6T and 51.7T. Maybe your information was rounded or your initial data is taken from a range starting at 250M instead of 200M.  You can see that I'm a typical engineer (as you know it from many jokes): Take some input (your statistical data on sieving) and predict the future with as few mathematics as possible. R. Gerbicz is a typical mathematician: Take the whole equation with all constants etc. It is amazing how accurate these numbers are (if you consider that the impact of the statistical variation is bigger at the beginning of sieving) 

20060719, 19:18  #25 
Mar 2005
Internet; Ukraine, Kiev
11·37 Posts 
An Athlon64 machine sieved 150T200T in parallel with Celeron sieving 100150T.
Range [550e6; 25e9] at p=150.1T had 9,553,770 k's left. Now the range [550e6; 25e9] is at p=215.6T, 9,345,423 k's left. 
20060806, 11:48  #26 
Mar 2005
Internet; Ukraine, Kiev
11·37 Posts 
The range [550e6; 25e9] is at p=271.1T, 9,216,786 k's left.

20060820, 17:56  #27 
Mar 2005
Internet; Ukraine, Kiev
407_{10} Posts 
Again, I sieved in parallel:
Range [550e6; 25e9] at p=300.0T had 9,160,689 k's left. Now the range [550e6; 25e9] is at p=352.4T, 9,073,142 k's left. 
20060827, 11:56  #28 
Mar 2005
Internet; Ukraine, Kiev
11×37 Posts 
I've got my hands on an Athlon XP 2500+ @1800Mhz and got a sieving rate of 915 sec/k. LLR takes about 150200 sec, deending on the CPU. Maybe we should postpone LLR and start distributed sieving? With a couple of Athlon boxes we could sieve to 1000T quite easily.
The data file is 170 Mb uncompressed, about 3540 Mb compressed. I have hosting/bandwidth for it. I wonder, is there some ready software, that would allow to report factors only without uploading the whole data file. 
20060914, 05:04  #29 
Mar 2005
Internet; Ukraine, Kiev
627_{8} Posts 
Range [600e6; 25e9] at p=400.4T had 8,985,536 k's left.
Now the range [600e6; 25e9] is at p=550.8T, 8,878,628 k's left. 
20061012, 23:13  #30 
"Jason Goatcher"
Mar 2005
5×701 Posts 
Would it be beneficial for my 2.8GHz PentiumD to do some sieving? I know Pentiums are significantly slower, but I thought the help would be appreciated anyway.

20061022, 00:34  #31  
"Jason Goatcher"
Mar 2005
6661_{8} Posts 
I was doing a little surfing and discovered the following theorem about twin primes:
Quote:
Edit: if the "!" means factorial, then this formula will most definitely NOT help sieving, but just in case it means something different, I'll go ahead and post this. Last fiddled with by jasong on 20061022 at 00:37 

20061022, 02:26  #32 
2^{2}·3·7^{3} Posts 

20061022, 19:37  #33  
"Jason Goatcher"
Mar 2005
5×701 Posts 
Quote:
Would it be possible to find patterns in the (n1)! part of the equation? Maybe a situation where the last 6 to some odd digits are calculated correctly? Even a partial pattern might be helpful. I know there are plenty of people out there who have better skills than I do! 

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