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Old 2013-08-15, 08:04   #12
Citrix
 
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I think you got the point. Since no one is planning on testing the k beyond 10000-- we can just test the low weight subsequences.
Generally the k that survive up to base=2^5760 are extremely high weight sequences to start with and hence it might be faster just to test them.

What I don't know is, if these subsequences will produces primes as frequently as the other previous low weight sequences have.

Last fiddled with by Citrix on 2013-08-15 at 08:14
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Old 2013-08-15, 09:06   #13
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What I don't know is, if these subsequences will produce primes as frequently as the other previous low weight sequences have.
Well, this needs to be tested...

On the other hand, one could imagine that some of the sub-sequences are fully covered (and therefore eliminated) by some larger (yet unknown) covering set(s) -- making them "Riesel" candidates. The chances for this might be actually larger (due to the much larger base) than for the ordinary k. But I may be wrong with this assumption...
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Old 2013-08-15, 09:27   #14
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Well, this needs to be tested...
One goal would be to bring all the 818 k's with total nash weight of about 200-- to n=2 million to see if any primes can be found.

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Old 2013-08-15, 09:30   #15
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On the other hand, one could imagine that some of the sub-sequences are fully covered (and therefore eliminated) by some larger (yet unknown) covering set(s) -- making them "Riesel" candidates. The chances for this might be actually larger (due to the much larger base) than for the ordinary k. But I may be wrong with this assumption...
This is possible but less likely (almost impossible) given the small size of the k's (k<2^18) and looking at the factorization of 2^(5760*i)-1. There aren't enough factors less than 2^18 to make this possible.

Does this mean that all sub-sequences will eventually produce a prime?
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Old 2013-08-15, 09:50   #16
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This is possible but less likely (almost impossible) given the small size of the k's (k<2^18) and looking at the factorization of 2^(5760*i)-1. There aren't enough factors less than 2^18 to make this possible.
Well, I don't think that the latter always holds. In principle a sequence may also be eliminated by factors larger than k. While the period of a cycle (cycle length) is determined by the product of the covering set, and therefore the "average" size of the k's generated by this cycle, the smallest k's of the cycle may be actually smaller than the largest factor(s) of the set.

IIRC, Joe McLean gave some examples on his pages: http://irvinemclean.com/maths/robintro.htm
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Old 2013-08-15, 09:58   #17
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Well, I don't think that the latter always holds. In principle a sequence may also be eliminated by factors larger than k. While the period of a cycle (cycle length) is determined by the product of the covering set, and therefore the "average" size of the k's generated by this cycle, the smallest k's of the cycle may be actually smaller than the largest factor(s) of the set.

IIRC, Joe McLean gave some examples on his pages: http://irvinemclean.com/maths/robintro.htm
I understand this (that is why I wrote almost impossible and not completely impossible). The larger the numbers of the covering set, the larger the k has to be... but not always true.

It is "almost impossible" to restrict the k to less than 2^18 and use numbers of 40 digits etc (large factors of 2^5760-1) to form the covering set.

The odds of this happening would be in 1 in 10^30 for a 40 digit factor.
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Old 2013-08-15, 11:09   #18
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Well, the point here is that by switching to base 2^5760 you also introduce the "2^i" multipliers to your k, e.g. the transformation k --> k*2^i (i=0...5759). Therefore most of your k's are actually much larger than just 2^18.

This, in turn, also gives rise to much larger (yet unknown) covering sets -- the components of the set are no longer restricted to factors of 2^5760-1.

Last fiddled with by Thomas11 on 2013-08-15 at 11:12
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Old 2013-08-15, 15:41   #19
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Well, the point here is that by switching to base 2^5760 you also introduce the "2^i" multipliers to your k, e.g. the transformation k --> k*2^i (i=0...5759). Therefore most of your k's are actually much larger than just 2^18.

This, in turn, also gives rise to much larger (yet unknown) covering sets -- the components of the set are no longer restricted to factors of 2^5760-1.
The size of the k might have increased, but the number of k's that survive (ie not Riesel number) using the covering set of 2^5760-1 are not that many. Most numbers ended up being Riesel numbers when I tested this.
I estimate around 40,000. So the chance of there existing a larger covering set is very low.
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Old 2013-08-15, 18:27   #20
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The size of the k might have increased, but the number of k's that survive (ie not Riesel number) using the covering set of 2^5760-1 are not that many. Most numbers ended up being Riesel numbers when I tested this.
I estimate around 40,000. So the chance of there existing a larger covering set is very low.
Quote:
Originally Posted by Thomas11 View Post
Well, the point here is that by switching to base 2^5760 you also introduce the "2^i" multipliers to your k, e.g. the transformation k --> k*2^i (i=0...5759). Therefore most of your k's are actually much larger than just 2^18.

This, in turn, also gives rise to much larger (yet unknown) covering sets -- the components of the set are no longer restricted to factors of 2^5760-1.

Thomas11, I agree with what you are saying... but the odds of there existing a larger covering set for these k seems very small and hence unlikely.

It is like asking for 2^p-1 being prime for 2 consecutive primes (p> 1 million). It is possible they exist, but unlikely.
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Old 2013-08-15, 18:39   #21
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Thomas11, I agree with what you are saying... but the odds of there existing a larger covering set for these k seems very small and hence unlikely.
Well, the phrase "covering set" may be a bit missleding here. (Also since we are actually looking for "incomplete" covering sets - otherwise we would yield entirely Riesel numbers).

What I mean is that there might be congruences for larger primes not being members of your original covering set (2^5760-1), or combinations of them (so called "virtual primes") which have just the matching orders to eliminate some (or even just a few) of your remaining sequences. Since they are already "fanned out" to 5760 bits, in my opinion, the chances are higher for such matching congruences.

Anyway, one can just start the search on a bunch of your Ks and see what comes out.
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Old 2013-08-15, 19:34   #22
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Well, the phrase "covering set" may be a bit missleding here. (Also since we are actually looking for "incomplete" covering sets - otherwise we would yield entirely Riesel numbers).

What I mean is that there might be congruences for larger primes not being members of your original covering set (2^5760-1), or combinations of them (so called "virtual primes") which have just the matching orders to eliminate some (or even just a few) of your remaining sequences. Since they are already "fanned out" to 5760 bits, in my opinion, the chances are higher for such matching congruences.

Anyway, one can just start the search on a bunch of your Ks and see what comes out.
I think I get your point. There will for sure be more congruences for larger primes that will reduce the number of candidates left-- but I don't think they will end up being Riesel numbers (no candidates left).
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