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-   -   Sieving multiple NewPGen files in a row(How?) (https://www.mersenneforum.org/showthread.php?t=5233)

jasong 2005-12-29 07:45

Sieving multiple NewPGen files in a row(How?)
 
I'm trying to find all primes for the main k/n pairs at Riesel Sieve, but, because of something I believe I've heard about finding primality(I'm not going to quote because I may be wrong), I'm trying to find primes for all the remaining k for n*k^2-1 (not a typo).

I've heard(possibly incorrectly) that it is conjectured that if one form(k*2^n-1) is found prime, than the other form(n*2^k-1) also has a prime, and vice-versa.

Which leads me to my question. Because some of these k take longer than others in this form, is it possible to set criteria for sieving multiple ks in NewPGen(one after the other), then just let it go?

axn 2005-12-29 08:30

I think the form you are looking for is 2^n-k. You can do a Google search for "Dual Sierpinski problem" to find more details.

NewPGen supports the sieving of b^n+/-k with fixed k.

As for sieving multiple k's one after the other, I think you should be able to use the "Sieve Until..." Option. Here you can set stop criteria for how far to sieve a single k, and the increment the k and move on.

You might also want to look at PFGW for such things, since you'll also be able to test the candidates for primality.

jasong 2005-12-29 08:40

[QUOTE=axn1]I think the form you are looking for is 2^n-k. You can do a Google search for "Dual Sierpinski problem" to find more details.

NewPGen supports the sieving of b^n+/-k with fixed k.

As for sieving multiple k's one after the other, I think you should be able to use the "Sieve Until..." Option. Here you can set stop criteria for how far to sieve a single k, and the increment the k and move on.

You might also want to look at PFGW for such things, since you'll also be able to test the candidates for primality.[/QUOTE]
I doublechecked the irc logs, unless the person is dyslexic, it's n*2^k-1.

Also, because of the nature of the problem I decided to undertake, I can't simply increment the k-value by an amount, unless I'm satisfied with 2 ks at a time.

I appreciate your attempt to help me, though.

axn 2005-12-29 08:58

:redface:

[QUOTE=jasong]I've heard(possibly incorrectly) that it is conjectured that if one form(k*2^n-1) is found prime, than the other form(n*2^k-1) also has a prime, and vice-versa.[/QUOTE]

Can you clarify this? when you say "the other form(n*2^k-1) also has a prime", do you mean n*2^k-1 _is_ prime or n*2^<some k> -1 is prime?

At any rate, I highly doubt it if the conjucture is true! All you have to do is find a prime of the form x*2^509203-1 for some x. By the conjucture, you will expect to have a prime 509203*2^x-1. Unfortunately, since 509203 is a Riesel number, there won't be any such primes.

Am I getting close? :confused:

jasong 2005-12-29 09:30

[QUOTE=axn1]:redface:



Can you clarify this? when you say "the other form(n*2^k-1) also has a prime", do you mean n*2^k-1 _is_ prime or n*2^<some k> -1 is prime?

At any rate, I highly doubt it if the conjucture is true! All you have to do is find a prime of the form x*2^509203-1 for some x. By the conjucture, you will expect to have a prime 509203*2^x-1. Unfortunately, since 509203 is a Riesel number, there won't be any such primes.

Am I getting close? :confused:[/QUOTE]
I may have been wrong when I said it could go both ways. If I understood B2 correctly, if n*k^n-1 can be found prime for a certain k, supposedly there's an n(possibly a different one) that makes k*2^n-1 prime for that same k.

I'm not saying it's true, I'm just going by what B2(the runner of Riesel Sieve) told me, and he indicated it was just a conjecture.

axn 2005-12-29 09:43

[QUOTE=jasong]...if n*k^n-1 can be found prime for a certain k...[/QUOTE]

Ok. I am _slightly_ confused. There are 3 different forms you've mentioned.

n*k^2-1
n*2^k-1
n*k^n-1

Which is the form that is needed to have a prime?

The last one is called Generalized Woodall primes -- I think it can be sieved by multisieve.

jasong 2005-12-29 09:49

[QUOTE=axn1]Ok. I am _slightly_ confused. There are 3 different forms you've mentioned.

n*k^2-1
n*2^k-1
n*k^n-1

Which is the form that is needed to have a prime?

The last one is called Generalized Woodall primes -- I think it can be sieved by multisieve.[/QUOTE]
I apologize, that was a typo. (n*2^k-1) and (k*2^n-1)

axn 2005-12-29 10:01

[QUOTE=jasong]I apologize, that was a typo. (n*2^k-1) and (k*2^n-1)[/QUOTE]
Aah! In that case, what I said earlier applies. The conjucture can be "trivially" shown to be false. (Of course, to find a prime x*2^509203-1 is not trivial, but it is relatively straightforward).

Is there some conditions for this conjecture, like we should consider only the Riesel k's?

jasong 2005-12-29 10:15

[QUOTE=axn1]Aah! In that case, what I said earlier applies. The conjucture can be "trivially" shown to be false. (Of course, to find a prime x*2^509203-1 is not trivial, but it is relatively straightforward).

Is there some conditions for this conjecture, like we should consider only the Riesel k's?[/QUOTE]
Hmmmmmmmm, my math education was cut short in the tenth grade, so even if you could prove it false, I probably wouldn't understand it. Unless of course you're claiming you can find an n that makes n*2^509203-1 prime, in which case I would agree that the conjecture is false.

Do you believe you can find an n that causes n*2^509203-1 to be prime?

TTn 2005-12-29 12:22

sieving
 
[QUOTE]Which leads me to my question. Because some of these k take longer than others in this form, is it possible to set criteria for sieving multiple ks in NewPGen(one after the other), then just let it go?[/QUOTE]I believe you will be able to find your answer with RMA.NET version 0.83.

Ranges can be set in the WorkToDo.txt.
It uses a five stage loop procedure, with optional sieving stop points.
Any or all of the following:
You can stop at a maximum bound p.
You can also stop at a particular elimination rate.
You can also stop at a certain processor time total.

Later versions will be able to script a custom set of ranges.
For now you must manually enter parameters for a sequence.

axn 2005-12-29 16:25

[QUOTE=jasong]Unless of course you're claiming you can find an n that makes n*2^509203-1 prime, in which case I would agree that the conjecture is false.

Do you believe you can find an n that causes n*2^509203-1 to be prime?[/QUOTE]
Yes and no. Theoretically, yes -- there is nothing that says you can't have a prime of that form. Practically? It would take a lot (months of CPU time) of computation power. A good candidate for a Distributed Computing project :wink:


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