20120427, 18:33  #67 
Jun 2009
2^{2}·173 Posts 
I simply calculate d^k where d=number of decimal digits and k=length of tuple. I miss your numbers by a factor of ~45. Do you aim for a certain probability of the range containing a tuple, say 90%?

20120427, 20:01  #68  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
1731_{16} Posts 
Quote:
I changed hash_bits to 20 which took me down to 20 minutes per 100T. I am now doing 4 lots of 5P overnight. This will take me upto 22P. 730P is the estimate. 

20120428, 05:49  #69 
Jun 2009
2^{2}·173 Posts 
20 minutes per 100T? Now that's what I call impressive! Using NewPGen we would be talking about weeks...
Last fiddled with by PuzzlePeter on 20120428 at 05:50 
20120430, 12:21  #70  
Mar 2004
3×127 Posts 
Quote:
Yesterday I took time to do that, but lost everything after I accidently hit the ‘page back’ key. Trying again… After sieving to deep limits, the primes in a tuple are independent. It is possible to multiply the probabilities of being prime. Example: henrzz, 6 Tuple with 1000 Bits (I don’t know the exact number) and 40 bit Mantissa. Sieving to 27 bit (131M); Probability of a prime is [Bits Sieving depth] / ([Size of Candidate]/2). Probability (prime): 27 / 520 = 1 in 19.25 or 5.2%. Twin: 1 in 371 3Tuple: 1 in 7143 4Tuple: 1 in 137580 5Tuple: 1 in 2.65 Million 6Tuple: 1 in 51 Million According to the Number of candidates in the sieved array it is possible to extrapolate the array, which is needed. Of course it is just a probability: After searching that range the probability of success is 11/e. (If the range is twice as big, the probability is 11/e². When sieving to low limits, the candidates are not independent: Having a NewPGen array of k*2^x for 6 Tuple, k odd. One out of 105 candidates is remaining after sieving 3, 5 and 7. Optimized sieves use an array which just use these numbers. When sieving p for NTuple, (pN)/p candidates are remaining. 6tuple: sieving 11, 13, 17, 19, 23 removes 5/11 * 7*13 * 11/17 * 13/19 * 17/23 = 8% remaining. After sieving up to 1023 less than 0.1% of the candidates are remaining. (<0.025% for 7tuple; sieving to 1048000 : 6tuple 0.0014%, 7tuple: 0.000175%) The reducing number of candidates makes it difficult to keep the sieve efficient. Up to 4Tuple, proth (k*2^n) numbers are more efficient than primordial (k*p#). 6 – 10(?) tuple, primorial numbers are faster. About 5tuple it depends on the tools used, which is faster. 15+ tuple need completely different algorithms. 

20120430, 20:01  #71 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5937_{10} Posts 
I am not sure your formulas are right. I got 1/23 per candidate instead of 1/25 from your formula for the numbers I am testing currently. The odds of a number of the form k*b^n+c being prime(c small) is 1 in (n*ln(b)+ln(k))
The odds of a random candidate surviving sieving upto s is 1 in (1.781*ln(s)). Thus the odds of each candidate surviving sieving is (1.781*ln(s))/(n*ln(b)+ln(k)). 1.781 is an approximation of exp(gamma) http://www.wolframalpha.com/input/?i...7s+constant%29 I have attached a spreadsheet including these formulas. It closely matches the results I have been getting so far in my search for 7tuples. I discovered just after sieving upto 46P that once k reaches a certain level pfgw is forced to resort to more general methods of fft which slows it down by a factor of 2.5. I need to sieve deeper in future. Last fiddled with by henryzz on 20120430 at 20:12 Reason: explanation of 1.781 
20120430, 20:50  #72 
Mar 2004
575_{8} Posts 
Our Approaches are just different. I did not use a constant, because i just use the quotient between lieve Limit and candidate. So even the base of the log has no influence anymore.
I see one main difference: I use P(prime) = log(Sievedepth)/(log(Candidate)/2) You use P(prime) = log(Sievedepth)/log(Candidate) I am careful with low sieving levels because tuplets are not independant. After sieving to some limit, I think my approximation are more or less accurate. About the probabilities of 6 and 7 Tuple: Here we should maybe take care about rounding errors: 1(13*10^10))^150000000 Here there are just a few significant digits raised to a huge power. Could you tell me some information about your search? How do you count 150000000 candidates? k*2^1000+c or ((210*k)+q)*2^1000+c ? How many candidates were left after sieving how deep (6 tuple and 7 tuple)? 
20120430, 23:01  #73 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
13461_{8} Posts 
I sieved upto only 10M for most of the range using gsieve. 150M is the number of candidates remaining after the sieving for the range 046P. The sieving only took a couple of days on a single quad. My n is exactly 1000.
Each candidate is a k*2^n1 that needs to be tested(and if prime pfgw will test +1 +5 etc). It seems that you effectively use 2 instead of 1.781 as your constant. This isn't bad but it does build up when you get to 7tuples. 2^7=128 while 1.781^7=~57 
20120503, 06:13  #74 
Jan 2007
Germany
2×11×19 Posts 
Hello members !
Can everybody of you writing a sieve like APSIEVE (without PRPing) with features:  sparse sieve like NewPGen  ktuplets up to k=10  sieving up to 10^13...?  form k*p#+a1,a2,... ? best, Norman Last fiddled with by Cybertronic on 20120503 at 06:15 
20120505, 12:22  #75 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
3·1,979 Posts 
I believe it should be possible to modify gsieve to do that. It should pretty much just need the function initprimes modifying. Unfortunately I don't understand that function so I can't do it.

20120721, 21:07  #76 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
13461_{8} Posts 
I have tested n=1000 for 7tuples upto 34P.
I have found 324 quads and 15 quins so far. I have sieved until 81P so testing should go fast for a bit. I hope to have found a run or two of six primes by then. Note this isn't a 6tuple as a 6tuple isn't contained within a 7tuple. 
20130622, 19:32  #77 
Jun 2009
2^{2}×173 Posts 
It's me again...
I'm trying to adapt the codes in this thread to a new brainfart of mine, but it seems to be beyond my capabilities. I'd like to search for triplets of the form k*x 1, +1, +5 with x being a precomputed constant of several thousands of digits. I know k has to be a multiple of 6. I also know that for the x I'm using to test this, a must end in 2 or 8. That leaves a = 12, 18, 42, 48, 72, 78 etc. which makes two strings of numbers a=30*n+12 a=30*n+18 with n = [0...chosenlimit] I thought about doing two runs, one for the 30*n+12 version and one for 30*n+18. I have a PFGW script to precompute a table (text file) which contains lines of p, x MOD p for all primes p below a chosen sieve limit, then use this information to go through an array of booleans and set those to 0 where (a*(x MOD p)) equals +1, p1 or p5. And this is where the fun starts. I seem to be unable to determine which a values to cancel out. I can find them by hand by simply trying from the low end upwards to the first hit, then go up by steps of the size p, but I don't seem to be able to create the formula for finding the first one. Plus this would probably be much more efficient on a bit array rather than booleans. I know the codes here in this thread contain all the stuff I need. I keep looking at the init_indexes function from Sex_2.pas in post #34, but what is happening there is beyond my grasp. So any hints or help would be greatly appreciated. In case you're wondering about my motivation for this: I'd like to try and go for a triple too large to prove with PRIMO. So I use a number x that makes a N+1 or N1 proof possible for all three members of the triple. I tried this for a small example (a few dozen digits) to make sure it works by using PFGW and brute force. For larger numbers I need a good sieve though. I have a working PFGW "sieve" script, but it's based on the FACTORIZE function which makes it too slow for large numbers. 
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