20060506, 21:05  #12 
Mar 2005
Internet; Ukraine, Kiev
11×37 Posts 
biwema,
I'm now at p=20T. I'll start another sieving of [5e9; 25e9] and when it reaches 20T, I will merge them. I will start just now, so there is not much work to catch up. 
20060512, 19:10  #13 
Mar 2005
Internet; Ukraine, Kiev
11×37 Posts 
I have merged [150e6; 5e9] and [5e9; 25e9] this morning. Now the whole range [150e6; 25e9] is at p=21.4T, 10,947,000 k's left, rate is about 1 k/sec.

20060513, 02:43  #14  
Apprentice Crank
Mar 2006
1C6_{16} Posts 
Quote:


20060516, 16:18  #15 
Mar 2005
Internet; Ukraine, Kiev
11·37 Posts 
I have released [150e6; 200e6] for LLR at p=30.6T. Now the range [200e6; 25e9] is at p=30.6T, 10,667,000 k's left, rate is ~1.41.6 k/sec.
MooooMoo's edit: The ranges are now in the "pre sieved range reservation thread". Last fiddled with by MooooMoo on 20060517 at 07:34 
20060530, 18:31  #16 
Mar 2005
Internet; Ukraine, Kiev
627_{8} Posts 
Now the range [250e6; 25e9] is at p=51.7T, 10,334,553 k's left, rate is ~2.52.7 k/sec.

20060603, 13:08  #17 
Mar 2005
Internet; Ukraine, Kiev
11×37 Posts 
Now the range [250e6; 25e9] is at p=61.0T, 10,226,786 k's left, rate is ~2.52.7 k/sec.

20060611, 08:55  #18  
"Robert Gerbicz"
Oct 2005
Hungary
5BC_{16} Posts 
Quote:
Code:
T=4.0;forprime(p=3,10^5,T*=(p2)/p/(11/p)^2);p=T*(195000*log(2))^2;\ forstep(i=5,50,5,print("k=",i," G ","chance of finding a twin: ",1(1p)^(i*10^9/2))) k=5 G chance of finding a twin: 0.3032663771235724704662848430 k=10 G chance of finding a twin: 0.5145622587534880610868001205 k=15 G chance of finding a twin: 0.6617792038603679416316561856 k=20 G chance of finding a twin: 0.7643501993734845214281243748 k=25 G chance of finding a twin: 0.8358148606793800287773442252 k=30 G chance of finding a twin: 0.8856066930586734521235834760 k=35 G chance of finding a twin: 0.9202983368219543694155290908 k=40 G chance of finding a twin: 0.9444691714646835051627448017 k=45 G chance of finding a twin: 0.9613098046532592367695271155 k=50 G chance of finding a twin: 0.9730432400262686098144673574 

20060611, 10:31  #19 
Mar 2004
3×127 Posts 
Nice.
My Approach was slightly different: I was a bit afraid on how much the thiw is dependant on each other. Therefor I sieved a range of 10G to K=1T, and did the probability calculation with the number of remaining candidates to minimize the dependency. Ofcourse, this number has also a statistical distribution. 
20060613, 21:46  #20 
Mar 2005
Internet; Ukraine, Kiev
11·37 Posts 
The range [300e6; 25e9] is at p=83.0T, 10,005,865 k's left, rate is ~5.08.0 sec/k.
Please excuse me, in posts #15, #16 and #17 I made a mistake: I wrote "k/sec", while it is "sec/k". 
20060614, 23:15  #21  
Mar 2004
381_{10} Posts 
hi gribozavr,
Could it be, that you are quite unlucky? Quote:
Quote:
10667000 * (24.7G / 24.8G) * (Log2(30.6T) / Log2(83T))^2 = 9972751 Candidates, so there are 28500 Factors too few. Candidates * New Range * Newfactoring Depth ratio^twin Is there a flaw in my calculations? Edit: More detailed calculations with excel give: 3.06E+13 24.8 10667000 5.17E+13 24.75 10334553 10294806 39747 6.10E+13 24.75 10226768 10227121 353 8.30E+13 24.7 10005867 10010909 5042 from 30.6T to 51.7T 39747 too few factors; from 51.7T to 61T approx. correct from 61T to 83T 5000 too many Last fiddled with by biwema on 20060614 at 23:32 

20060624, 20:05  #22 
Mar 2005
Internet; Ukraine, Kiev
11·37 Posts 
The range [400e6; 25e9] is at p=101.1T, 9,849,791 k's left, rate is ~5.010.0 sec/k.
Sieving has passed 100T biwema, I'm intrested in this math, but I don't understand the "Newfactoring Depth ratio" part. Can you explain it to me, please? (I undersatnd that log2(x) is the number of bits in x) 
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