20061201, 22:08  #12 
Jan 2006
Hungary
268_{10} Posts 
I am currently sieving with Multisieve Woodall numbers. I use the range from 1.3 million to 2 million. I still have 30,000 candidates left. Do I understand this discussion correctly that Multisieve is not really meant for such a range?
If this is so, how can I sieve quicker? Citrix, would your sieve be amendable to do Woodalls? Or should I stick with Multisieve but switch to short ranges (10,000) containing 500 candidates? Have a nice day, Willem. 
20061201, 23:58  #13  
"Mark"
Apr 2003
Between here and the
3^{2}·719 Posts 
Quote:


20061202, 05:04  #14  
Feb 2005
2^{2}·3^{2}·7 Posts 
Quote:


20061226, 05:42  #15 
Jun 2003
5×317 Posts 
@ rouge.
I was able to modify your program yesterday, to suit my needs, but it did not become as fast as we thought it would have become. I just increased the size of the array that stores all the values of 2^1, 2^2... (mod p). Though I did see some speed up. I am not sure how to speed up the program further. Any thoughts. Also something is wrong with the time function. The time per candidate keeps on jumping between 15 sec and 60 secs. (I did not modify this code, so its not me) 
20061226, 13:39  #16  
"Mark"
Apr 2003
Between here and the
1100101000111_{2} Posts 
Quote:
I'm not aware of any issues with the timing code. If you find something let me know. 

20061226, 19:56  #17  
Jun 2003
1585_{10} Posts 
Quote:
Currently I have only sieved upto 350M. (The sieve is quite slow, about 80M/hr. Not sure if this series will be worth testing.) So, it will be a long time before I cross 2^31, and if the program gets slower after that.... For the timing issue I think it depends on the distribution of factors. Sometimes 10 factors are found in a min and then no factors for 5+ mins. I think this would screw up the timing? 

20070131, 08:26  #18 
Einyen
Dec 2003
Denmark
3,203 Posts 
I sieved cullen numbers from n=1 to n=5,000,000 up to 2.5G (2.5 * 10^{9}). I found 4,847,529 factors so that leaves 152,457 unknowns plus the 14 known cullen primes.
Here is all 5 million n's with factors: cullenfactors.zip (18 Mb). The 14 primes have "PRIME" instead of a factor, and the 152,457 unknowns have "?" instead of a factor. Here is ABC format file for the 109,042 unknowns between n=1,500,000 (current progress according to http://www.prothsearch.net/cullenrange.html ) and n=5,000,000: cullen1.5M5M.zip 
20070131, 19:22  #19 
Aug 2002
Buenos Aires, Argentina
13×107 Posts 
p*2^{p}+1 cannot be prime if p>3 is a prime twin.
First suppose p=1 (mod 3) and let p=2q+1 (because p is odd). Operating modulo 3 we find: p*2^{p}+1 = 1*2^{2q+1}+1 = 2^{2q}*2+1 = (2^{2})^{q}*2+1 = 1^{q}*2+1 = 2+1 = 0 (mod 3) The other possibility is p=2 (mod 3) and p+2 also a prime. Operating modulo p+2 we find: p*2^{p}+1 = (2)*2^{p}+1 = 2^{p+1}+1 = 1+1 = 0 (mod p+2) So in both cases p*2^{p}+1 is composite. 
20070201, 01:17  #20  
Feb 2006
BrasÃlia, Brazil
3·71 Posts 
Dario,
Modular arithmetic still looks like a strange realm to me. So, I'm not sure if I understand what you've written: Quote:
I can't understand the part I've put in bold. Since both sides have a "+1" term, I assume I can change that particular equality to: 2^{p+1} = 1 (mod p+2) I still can't understand it even if I multiply both sides by (1) (can I do that?), giving: 2^{p+1} = 1 (mod p+2) Testing with small primes, I can see it works. But why? Thanks a lot, Bruno 

20070201, 03:59  #21 
Einyen
Dec 2003
Denmark
3,203 Posts 
This is just Fermat's Little Theorem with p+2 = prime (p being twin prime) and a=2.

20070201, 12:14  #22 
Aug 2002
Buenos Aires, Argentina
56F_{16} Posts 
In that case p does not need to be prime. When p=1 (mod 3) the argument shown above demonstrates that p*2^{p}+1 is multiple of 3.
Last fiddled with by alpertron on 20070201 at 12:15 
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