20050102, 07:54  #1 
Jun 2003
1,553 Posts 
Why Base 5 and Weights
Just curious, why base 5 and not 3 or 7?
Also, could you provide the weights for the remaining k's? Use : b5 for http://pages.prodigy.net/chris_nash/psieve.html Citrix Last fiddled with by Citrix on 20050102 at 08:05 
20050103, 02:54  #2  
Mar 2003
New Zealand
13·89 Posts 
Quote:
Code:
$ ./psieve3.exe 10918 b5 ***************************************** * PSIEVE 3.21 Chris Nash, Paul Jobling * * Thanks to Joe McLean for suggestions! * ***************************************** 10918 n=1 mod 2  factor 3 Best %  50.00 for modulus 2 n=1 mod 3  factor 31 n=4 mod 5  factor 11 n=2 mod 6  factor 7 Best %  83.33 for modulus 6 n=3 mod 9  factor 19 n=12 mod 16  factor 17 n=14 mod 17  factor 409 Best %  88.89 for modulus 18 n=6 mod 19  factor 191 n=0 mod 30  factor 61 Best %  90.00 for modulus 30 n=36 mod 42  factor 127 n=18 mod 42  factor 43 n=45 mod 69  factor 139 n=62 mod 82  factor 83 n=0 mod 89  factor 179 Best %  93.33 for modulus 90 n=60 mod 94  factor 2069 n=20 mod 152  factor 457 n=91 mod 155  factor 311 n=120 mod 173  factor 3461 n=10 mod 188  factor 12409 n=168 mod 196  factor 197 n=5 mod 209  factor 419 n=185 mod 215  factor 431 n=172 mod 226  factor 227 n=176 mod 232  factor 33409 n=138 mod 232  factor 233 n=33 mod 239  factor 479 n=17 mod 245  factor 491 n=176 mod 254  factor 509 255:254 

20050103, 14:23  #3 
Jun 2003
1,553 Posts 
You also need to use the e option. it will in the end say that this many candidates are left, which will be the weight.
Citrix Last fiddled with by Citrix on 20050103 at 14:23 
20050106, 19:53  #4 
Jun 2003
Oxford, UK
3·5^{4} Posts 
Why base 5?
In answer to Citrix's point, I had been looking at Sierpinski/Riesels of the form k.(2^x+1))^n+/1, where x=1,2,3.... because these are the only base forms which give non trivial solutions for k. See http://groups.yahoo.com/group/primeform/message/4773
and David Broadhurst's elegant reply. x=0 is the classic series, subject to extensive literature and the the SoB search x=1 is already being extensively researched and looks horribly difficult because the lowest mooted k is in the 10 million range both Sierpinski and Riesel the next x=2 is the focus of this search and gives a sensible number of candidates up to the lowest proven values of k both Sierpinski and Riesel, and the Sierpinski is easier because there are less candidates Hope this answers your point. Regards Robert Smith 
20050118, 22:40  #5 
Jun 2003
1,553 Posts 
could you also provide the average and total weight for the remaining k's.

20050118, 23:14  #6  
Jun 2003
1553_{10} Posts 
Quote:
Citrix 

20050119, 14:51  #7  
Mar 2003
New Zealand
13×89 Posts 
Quote:


20050119, 15:46  #8 
Jun 2003
1,553 Posts 
just do w1+w2+w3+...Wn to get total
then for average w1+w2+w3+....Wn/n for n k's left. I hope this post is more clear. Citrix 
20050119, 16:04  #9 
Mar 2003
New Zealand
1157_{10} Posts 
OK, I see what you mean now.

20050119, 16:24  #10  
Jun 2003
Oxford, UK
3×5^{4} Posts 
Quote:
See http://groups.yahoo.com/group/primeform/message/4388 and resulting long string of replies Regards Robert Smith 

20050120, 17:50  #11  
Jun 2003
11000010001_{2} Posts 
Quote:
Robert, It would be intresting to see if k's that are multiple of 3 or 5 or both can ever generate a sierpinski or riesel number for base 2? I will try to work on this later this week or as soon as I get some time and see what I can come up with. Base 5 takes too long as most of the optimizations for base 2 that make them super fast don't work for base 5. (I'm not sure if base 4 counts, but the smallest sierpinski number is k=5 for that base and I can prove that ) Citrix 

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