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Old 2005-12-29, 07:45   #1
jasong
 
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"Jason Goatcher"
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Default 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?
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Old 2005-12-29, 08:30   #2
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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.
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Old 2005-12-29, 08:40   #3
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Quote:
Originally Posted by 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.
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.
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Old 2005-12-29, 08:58   #4
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Quote:
Originally Posted by 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.
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?

Last fiddled with by axn on 2005-12-29 at 08:58
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Old 2005-12-29, 09:30   #5
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Quote:
Originally Posted by axn1




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?
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.

Last fiddled with by jasong on 2005-12-29 at 09:31
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Old 2005-12-29, 09:43   #6
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Quote:
Originally Posted by jasong
...if n*k^n-1 can be found prime for a certain k...
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.

Last fiddled with by axn on 2005-12-29 at 09:44
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Old 2005-12-29, 09:49   #7
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Quote:
Originally Posted by 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.
I apologize, that was a typo. (n*2^k-1) and (k*2^n-1)
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Old 2005-12-29, 10:01   #8
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Quote:
Originally Posted by jasong
I apologize, that was a typo. (n*2^k-1) and (k*2^n-1)
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?
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Old 2005-12-29, 10:15   #9
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Quote:
Originally Posted by 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?
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?

Last fiddled with by jasong on 2005-12-29 at 10:15
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Old 2005-12-29, 12:22   #10
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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?
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.
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Old 2005-12-29, 16:25   #11
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Quote:
Originally Posted by 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?
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
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