20070106, 02:13  #12 
"Jason Goatcher"
Mar 2005
5×701 Posts 

20070106, 03:56  #13  
Jun 2003
Oxford, UK
2^{4}×7×17 Posts 
Quote:
Regards Robert Smith 

20070106, 04:07  #14  
Jul 2003
wear a mask
13×113 Posts 
Quote:
jasong, IIRC, NewPGen allows increasing k, too. But in the menu for "sieve until" it only says n. If you pick the fixed k sieves, and then select updating n under that menu heading, NewPGen actually updates the k values. I think this confusion is mentioned in the README file that comes with NewPGen. I'll check this out and get back to you... have fun, masser Edit: I just checked the latest Windows version of NewPGen and it seems to work for only base 2. Sorry, jasong  that's a dead end for finding Riesel/Sierpinski numbers....I would follow Rogue's suggestion above and use srsieve or psieve, located here: http://pages.prodigy.net/chris_nash/psieve.html along with instructions on using it. Last fiddled with by masser on 20070106 at 04:42 

20070106, 05:26  #15 
Jun 2003
3^{2}×5^{2}×7 Posts 
Could someone find the covering set for base 8 k=14 on the riesel side. Jasong seems to be correct, it seems like base 6 riesel number.
For base=9 on the riesel side a perfect square can never provide a prime. For example 36*9^n1 can be factorized trivially into 6*3^n+1 and 6*3^n1. I think this proves 74 as the smallest riesel for base 9. same goes for base=16 and k=9 on the riesel side. edit: reserving base=16 k=1to 10000 on sierpinski side. 114 numbers left at n=128. not using trivial solutions like 4,8 etc.. or numbers such that k=2 (mod 3) or k=4 (mod 5). Also removing 4*a^2 as this leads to gaussian factorization. edit4: 120 does not appear to be a riesel number for base 16. I really think there might be something wrong with the calculations. (120*16^71 is 3PRP! (0.0030s+0.0004s)) Last fiddled with by Citrix on 20070106 at 06:23 
20070106, 05:28  #16  
Jun 2003
Oxford, UK
2^{4}·7·17 Posts 
Quote:
Regards Robert Smith 

20070106, 06:36  #17  
Jun 2003
Oxford, UK
2^{4}×7×17 Posts 
How it is done
Quote:
The method I use, which needs programming to run efficiently, (I am doing all of the caculations in Excel or even manually in some instances), involves (for the Sierpinski case) (i) factoring b^n1 for n (=A) which are small or highly composite (especially 6,8,12,16,20,24,30,32,36,40,48,60,64...). (ii) calculating the multiplicative order base b of all of the prime factors (iii) summing the reciprocals of the multiplicative orders for all non trivial prime factors of a given A (iv) if the sum in (iii) is greater than 1, then there is probably a covering set of a combination of the prime factors of that b^A1 (v) discover possible combinations of those prime factors for a given A that give rise to covering sets (vi) construct a table with values 1..A and P(1), P(2)..P(x) where P's are the prime factors of b^A1, and discover all combinations of P's which cover all values 1..A (this is not easy to do, unless you are good mathematician or programmer)  I tend to use a greedy algorithm to discover the sets and their positions (vii) mark the first instance of value 1..A where the prime factor occurs in the table (viii) for a particular P(x) calculate a K value which divides K.b^n+1, n from 1 to P(x), express as KmodP(x) (ix) use the Chinese Remainder Theorem, using K(1)..K(x)modP(1)..P(x) to discover the k, for all combinations in (vi) (x) discard all k which are also trivial (k*b^n1 is divisible by any of the factors of b1) (xi) the lowest k discovered by CRM is a low Sierpinski number for that covering set. (xii) repeat for other covering sets, although the likely smallest case is the simplest covering set For example, in the Sierpinski 13 case: (i) 13^41 = 28560=2^4*3*5*7*17 (ii) 2,3 multiplicative order =1 (trivial), 7 = M(2), 5 and 17 are M(4) (iii) 1/2+1/4+1/4=1 (iv) therefore possible covering set (v) only one possible set of factors as sum of reciprocals is =1 (vi) covering sets given by (a) 7 in positions 1 and 3, 5 in 2 and 17 in 4, (b) 7 in 1 and 3, 5in 4 and 17 in 2 (c) 7 in positions 2,4, 5 in 1, and 17 in 3, (d) 7 in 2,4, 5 in 3 and 17 in 1. (vii) (a) 7,1..5,2..17,4 etc (viii) (a) 1mod7, 1mod5, 16mod17 (b).... (ix) (a) 526, (b) 239, (c) 293 (d) 132 (x) and (xi) trivial k are provided by 1mod2 and 1mod3 , and 132 is neither (xii) it is quicker to test for primes of k.b^n+1, with k<132 than to look at other covering sets in this instance. First primes are generally low values of n and in this instance, only k=48 and 120 created a problem, however 120*13^1552+1 is prime, as is 48*13^6267+1. Anyone up to program? Regards Robert Smith 

20070106, 06:42  #18  
Jun 2003
Oxford, UK
770_{16} Posts 
Base 16 error
Quote:
Regards Robert Smith 

20070106, 07:36  #19  
Jun 2003
Oxford, UK
2^{4}×7×17 Posts 
Base 16 corrected
Quote:
Sierpinski 66741 Riesel 33965 Still possible lower values from the covering set [7,13,19,37,73] repeating every 9n. Regards Robert Smith 

20070106, 07:36  #20 
Jun 2003
627_{16} Posts 
Base 16
16 numbers left under 10,000 tested to n=4000 (or 16,000 in base 2). Stopping now.
186*16^n+1 2158*16^n+1 2857*16^n+1 2908*16^n+1 3061*16^n+1 4885*16^n+1 5886*16^n+1 6348*16^n+1 6663*16^n+1 6712*16^n+1 7212*16^n+1 7258*16^n+1 7615*16^n+1 7651*16^n+1 7773*16^n+1 8025*16^n+1 
20070106, 07:41  #21 
Jun 2003
Oxford, UK
770_{16} Posts 
Base 8 reconfirmed

20070106, 07:48  #22  
Jun 2003
Oxford, UK
2^{4}·7·17 Posts 
Square trivials
Quote:
Are there other trivial solutions we should look out for? Aurifeullians? Regards Robert Smith 

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