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Old 2020-03-04, 15:59   #34
storm5510
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Quote:
Originally Posted by Happy5214 View Post
Since you're only working with one k, you should use sr1sieve instead of sr2sieve. sr1sieve is significantly faster when working with one or two k values (run separately for two k's), while sr2sieve is better for three or more k values. sr1sieve also has the advantage of outputting (in-place) to an LLR-format file, so there's no need to work with srfile.
I have a second k reserved and I have ran just enough to get it into the Wiki database.

I will make the switch in my batch file to sr1sieve. Thank you!
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Old 2020-04-18, 16:59   #35
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Default 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?
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Old 2020-04-18, 17:10   #36
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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. :)
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Old 2020-04-18, 17:42   #37
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Quote:
Originally Posted by VBCurtis View Post
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. :)
@storm5510:
...Or simply type the function in factorDB: http://factordb.com/index.php?query=10079*2%5En-1 and carefully observe
  • does the function value for every odd n have a factor? What is it? Is it simply a coincidence or can you prove that it will go on for all odd n?
  • does the function value for every other even n have a factor?
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Old 2020-04-18, 17:52   #38
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Quote:
Originally Posted by VBCurtis View Post
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. :)
I found something. 10079 is a natural prime. I started the sieve again. Still no odd n's. I tried it with NewPGen. No odd's there either.

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.
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Old 2020-04-18, 20:26   #39
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Quote:
Originally Posted by storm5510 View Post
A learning experience is right. Natural prime k's will produce either all even n's or all odd n's. Not both.
Not really. Looking at my rieselprime pages (like for Riesel 300-2000) you will spot the marked n-values for twins (red) or Sophie-Germain primes (bold).
(Note: Sophie Germain for a Riesel k-value is, if k*2^n-1 _and_ k*2^(n+1)-1 are prime.)

Keep in mind for Riesel base 2 here:
- k-value ≡ 0 mod 3 -> both even or odd n-values can build a prime
- k-value ≡ 1 mod 3 -> only odd n-values possible (even n-values -> k*2^n-1 are divisible by 3)
- k-value ≡ 2 mod 3 -> only even n-values possible (odd n-values -> k*2^n-1 are divisible by 3)

See also there under "Related" -> "First Twin k" or "First SG" or "Riesel Twin/SG".

The Liskovets-Gallot 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 2020-04-18 at 20:36
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Old 2020-04-19, 14:27   #40
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Originally Posted by kar_bon View Post
Not really. Looking at my rieselprime pages (like for Riesel 300-2000) you will spot the marked n-values for twins (red) or Sophie-Germain primes (bold).
(Note: Sophie Germain for a Riesel k-value is, if k*2^n-1 _and_ k*2^(n+1)-1 are prime.)

Keep in mind for Riesel base 2 here:
- k-value ≡ 0 mod 3 -> both even or odd n-values can build a prime
- k-value ≡ 1 mod 3 -> only odd n-values possible (even n-values -> k*2^n-1 are divisible by 3)
- k-value ≡ 2 mod 3 -> only even n-values possible (odd n-values -> k*2^n-1 are divisible by 3)

See also there under "Related" -> "First Twin k" or "First SG" or "Riesel Twin/SG".

The Liskovets-Gallot Conjecture (see my page) is also here in the forum in search for primes and only 9 are open to prove.
Mathematics was never my strong suit. I understand most of what is above. I am better suited for running, and writing, programs and batch scripts.

The Sophie Germain example above looks a little odd: k*2^n-1 -and- k*2^(n+1)-1. If I were to put this in a program, I would write it (k*2^n-1 -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.
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Old 2020-04-19, 21:00   #41
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First writing and then running a program implies to understand the mathmatics behind to code you use. Your code "(k*2^n-1 -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^n-1 is 2p+1=2*(k*2^n-1)+1=2*k^n-2+1=k*2^(n+1)-1.

In the above example 11 = 3*2^2-1 and 23 = 3*2^3-1, so 3*2^n-1 is prime for n=2 and 3

Hint using PFGW:
put this in a file called test.txt
Code:
ABC2 27*2^$a-1 & 27*2^($a+1)-1
a: from 100 to 150
Run pfgw with "-l test.txt" and you will find a pfgw.log with
Code:
27*2^121-1 
27*2^(121+1)-1
  - Complete Set -
27*2^122-1
The testfile checks all 27*2^n-1 for 100<=n<=150 and find a complete set (n and n+1 build a prime) for n=121.
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Old 2020-04-19, 22:56   #42
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Quote:
Originally Posted by kar_bon View Post
First writing and then running a program implies to understand the mathmatics behind to code you use. Your code "(k*2^n-1 -and- k*2^(n+1)) - 1" is wrong here...
I realized it was wrong while wandering the local grocery with my burglar mask on. I ran it as it as below and received two different answers. The extra spaces are for my bad eyes.

Code:
  405 * 2 ^ 10 - 1 and 405 * 2 ^ (10 + 1) - 1 = 263,167
   (405 * 2 ^ 10 - 1 and 405 * 2 ^ (10 + 1)) - 1 = 262,143
I used hard-coded values for k and n. The second example is not correct. The -1 is applied to the entire sequence and not the right side only. I did not consider that everything inside the parentheses in the original would be subject to an exponential calculation.

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.
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Old 2020-04-20, 08:27   #43
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Quote:
Originally Posted by storm5510 View Post
Code:
  405 * 2 ^ 10 - 1 and 405 * 2 ^ (10 + 1) - 1 = 263,167
What does this mean?
263167 is 257*2^10-1 and nothing of the above on the left side. What are you trying here?

Suggestion for your own code:
- take a Riesel k-value (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^n-1 and the start value n=120?

Last fiddled with by kar_bon on 2020-04-20 at 08:29
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Old 2020-04-20, 14:45   #44
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Quote:
Originally Posted by kar_bon View Post
Note: Sophie Germain for a Riesel k-value is, if k*2^n-1 _and_ k*2^(n+1)-1 are prime.
405*2^10-1 And 405*2^(10+1)-1

Is this not the same form? This is for example only! I never intended, or intend in the future, to run any SG's.
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