20200304, 15:59  #34  
Random Account
Aug 2009
U.S.A.
3104_{8} Posts 
Quote:
I will make the switch in my batch file to sr1sieve. Thank you! 

20200418, 16:59  #35 
Random Account
Aug 2009
U.S.A.
3104_{8} Posts 
No odd n's.
I have been running k = 10079 for a while now and I have nearly got it to n = 1,000,000. Something seemed strange about the process. I saw it a short time ago. There are no odd values of n anywhere. I kept the results for all that I have ran. No odd n's to be seen. I checked what is left of the original sieve. No odd n's in it either.
I used srsieve/sr1sieve to sieve to p = 15e11. Something went really wrong here. I believe the best thing to do is to scrap everything and run the sieve again, and watch it closely. Thoughts anyone? 
20200418, 17:10  #36 
"Curtis"
Feb 2005
Riverside, CA
4368_{10} Posts 
Sounds like a learning opportunity! I suggest using excel (or some code, if you have such skills I do not) to calculate exponents from n = 1 to n = 30 (or more), and then trial divide and see which primes take out the odd exponents.
The experience is likely to improve your trust in the sieve programs. :) 
20200418, 17:42  #37  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,041 Posts 
Quote:
...Or simply type the function in factorDB: http://factordb.com/index.php?query=10079*2%5En1 and carefully observe


20200418, 17:52  #38  
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Aug 2009
U.S.A.
2^{2}×401 Posts 
Quote:
Back in the winter, I wrote a short Perl script to generate a list of natural primes. I picked a different natural, 13399. Sieving it produced all odd n's. No evens. I tried a third natural, 180077. It produced all even n's. A learning experience is right. Natural prime k's will produce either all even n's or all odd n's. Not both. 

20200418, 20:26  #39  
Mar 2006
Germany
101100100010_{2} Posts 
Quote:
(Note: Sophie Germain for a Riesel kvalue is, if k*2^n1 _and_ k*2^(n+1)1 are prime.) Keep in mind for Riesel base 2 here:  kvalue ≡ 0 mod 3 > both even or odd nvalues can build a prime  kvalue ≡ 1 mod 3 > only odd nvalues possible (even nvalues > k*2^n1 are divisible by 3)  kvalue ≡ 2 mod 3 > only even nvalues possible (odd nvalues > k*2^n1 are divisible by 3) See also there under "Related" > "First Twin k" or "First SG" or "Riesel Twin/SG". The LiskovetsGallot Conjecture (see my page) is also here in the forum in search for primes and only 9 are open to prove. Last fiddled with by kar_bon on 20200418 at 20:36 

20200419, 14:27  #40  
Random Account
Aug 2009
U.S.A.
644_{16} Posts 
Quote:
The Sophie Germain example above looks a little odd: k*2^n1 and k*2^(n+1)1. If I were to put this in a program, I would write it (k*2^n1 and k*2^(n+1))  1. The "1" would apply to the entire sequence and not just the right side. If it is only the right side, then it seems this part could be reduced to k*2^n by removing the "+1" and "1." Adding the extra set of parenthesis clarifies the order of of execution in the first case. Without them, a language compiler may not apply the "1" to the entire sequence because of the "and" in between. I would have to write something short and simple to see if there is a difference. 

20200419, 21:00  #41 
Mar 2006
Germany
2·3·5^{2}·19 Posts 
First writing and then running a program implies to understand the mathmatics behind to code you use. Your code "(k*2^n1 and k*2^(n+1))  1" is wrong here.
Please read an article about Sophie Germain, like on Wikipedia. If p is prime and 2p+1 is also a prime, p is called a Sophie Germain prime. Example: p=11, so 2p+1=23, both prime, so 11 is a Sophie Germain. For p=k*2^n1 is 2p+1=2*(k*2^n1)+1=2*k^n2+1=k*2^(n+1)1. In the above example 11 = 3*2^21 and 23 = 3*2^31, so 3*2^n1 is prime for n=2 and 3 Hint using PFGW: put this in a file called test.txt Code:
ABC2 27*2^$a1 & 27*2^($a+1)1 a: from 100 to 150 Code:
27*2^1211 27*2^(121+1)1  Complete Set  27*2^1221 
20200419, 22:56  #42  
Random Account
Aug 2009
U.S.A.
644_{16} Posts 
Quote:
Code:
405 * 2 ^ 10  1 and 405 * 2 ^ (10 + 1)  1 = 263,167 I became familiar with PFGW when I first began running Riesel searches before switching to LLR. I experimentsd with the ABC2 form but I never used an "and" operator in one. 

20200420, 08:27  #43 
Mar 2006
Germany
B22_{16} Posts 
What does this mean?
263167 is 257*2^101 and nothing of the above on the left side. What are you trying here? Suggestion for your own code:  take a Riesel kvalue (preferred k divisible by 3)  calculate the value p = k * 2^n  1  if p is prime, calculate q = 2*p + 1  if q is prime, p is SG Have you tried Batalovs suggestion, say for 27*2^n1 and the start value n=120? Last fiddled with by kar_bon on 20200420 at 08:29 
20200420, 14:45  #44 
Random Account
Aug 2009
U.S.A.
2^{2}×401 Posts 

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