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Old 2007-01-06, 07:55   #23
ltd
 
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Change my reservation i will sieve all noted k for base 17 (riesel and sierpinski)


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Old 2007-01-06, 08:40   #24
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Default Base 6

Did a little work this afternoon of base 6. Using the covering set [7,43,37,31,13] repeating every 24n provides a Sierpinski number 243417.

I will try to do the Riesel later.

Note that the alternative set [7,43,37,31,97] repeating every 24n could also provide a lower Sierpinski value. However 243417 is at 0.73% of the product of this set's cover primes, and there are only 24 values to check, if I was any good at statistics I could tell you what the probability is, but I am not!!

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Robert Smith
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Old 2007-01-06, 08:53   #25
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Candidates for 10

804*10^n+1
1024*10^n+1
2157*10^n+1
2311*10^n+1
2607*10^n+1
2661*10^n+1
2683*10^n+1
3301*10^n+1
3312*10^n+1
3345*10^n+1
3981*10^n+1
4069*10^n+1
4863*10^n+1
5028*10^n+1
5125*10^n+1
5512*10^n+1
5556*10^n+1
5565*10^n+1
6172*10^n+1
6687*10^n+1
6841*10^n+1
7404*10^n+1
7459*10^n+1
7534*10^n+1
7666*10^n+1
7809*10^n+1
7866*10^n+1
8194*10^n+1
8425*10^n+1
8454*10^n+1
8667*10^n+1
8724*10^n+1
8889*10^n+1
8922*10^n+1
8953*10^n+1
9021*10^n+1
9043*10^n+1
9175*10^n+1
9351*10^n+1

1343*10^n-1
1506*10^n-1
1803*10^n-1
1935*10^n-1
2111*10^n-1
2276*10^n-1
2333*10^n-1
3015*10^n-1
3332*10^n-1
3356*10^n-1
4016*10^n-1
4421*10^n-1
4478*10^n-1
4577*10^n-1
5499*10^n-1
5897*10^n-1
6588*10^n-1
6633*10^n-1
6665*10^n-1
7019*10^n-1
7602*10^n-1
8174*10^n-1
8579*10^n-1
9461*10^n-1
9701*10^n-1
9824*10^n-1
10176*10^n-1


candidates for 16 riesel (upto 10,000)

450*16^n-1
1343*16^n-1
1803*16^n-1
1935*16^n-1
2333*16^n-1
3015*16^n-1
3332*16^n-1
4478*16^n-1
4500*16^n-1
4577*16^n-1
5499*16^n-1
5897*16^n-1
6588*16^n-1
6633*16^n-1
6665*16^n-1
7019*16^n-1
7602*16^n-1
8174*16^n-1
8579*16^n-1
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Old 2007-01-06, 11:34   #26
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Default Riesel candidate for base 6

After a bit of fiddling about with [7,43,37,31,13] came up with the riesel candidate 213410 for base 6. 133946 is trivial.

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Robert Smith
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Old 2007-01-06, 12:07   #27
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Default Simple to prove

Sierpinskis and Riesels for the following bases are simple to find and should be relatively simple to prove as they have 2 prime factors in b^2-1 which are not in b-1, and therefore have cover from these new prime factors, every 2n.

14 was the first such case, proven with S-4, R-4
20 is next with S-8, R-8, both proven

The others exhibiting this small covering set, less than base =100, and which therefore should be relatively simple to prove are

29, 32, 34, 38, 41, 44, 50, 54, 56, 59, 62, 64, 65, 68, 69, 74, 76, 77, 83, 84, 86, 89, 90, 92, 94, 98

Someone might want to just run these quickly to prove them. Then we might bash on to find Sierpinskis and Riesels for all other bases up to 100. Tomorrow I start work again so time I can spend on this will be limited.

Regards

Robert Smith
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Old 2007-01-06, 12:49   #28
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Quote:
Originally Posted by robert44444uk View Post
The problem is that, for some values, there is no mooted covering set less than 10^16, therefore the suggested method is inefficient for these bases. Tacking 3 and 15 needs another approach, but I am at a loss to think how to do efficiently.
I agree with that. My method works well if the Riesel/Sierpinski number is relatively small. Once you hit numbers that large it could take weeks, months, or even years to find one.
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Old 2007-01-06, 14:14   #29
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Default Alternative approach

Quote:
Originally Posted by rogue View Post
I agree with that. My method works well if the Riesel/Sierpinski number is relatively small. Once you hit numbers that large it could take weeks, months, or even years to find one.
I have been thinking that covering sets should comprise some prime factor(s) which is/are small enough to appear often enough to allow the creation of the cover. I also noticed that there are relationships of modF, for prime factors F of b^n-1 which are predictable. If n=2, then the sierpinski/ riesel are all either 1modF or(F-1)modF. In fact for n=N, then 2/N of the sierpinskis/ riesels are 1ModF or (F-1)modF. This might allow some type of sieving to take place which would allow some concentration of numbers to be tested.

So an approach would be to define, in principle, what properties covering sets must have. For example it is highly unlikely that a covering set would have no P which has a mulitplicative order in base b of less than 5. But can we prove this is the case? If we can, then we can say for certain that the covering set must have at least one of the prime factors of b^2-1, b^3-1 or b^4-1. Then we should be able to sieve out some k from providing a covering set, and only test those which meet the mod criteria.

Just a few musings on an approach.

Regards

Robert Smith
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Old 2007-01-06, 17:03   #30
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First one down:

88*17^4868+1 is prime.

Lars

Edit:

Next one down:

44*17^6488-1 is prime.

Riesel side done.

Last fiddled with by ltd on 2007-01-06 at 17:32
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Old 2007-01-06, 18:16   #31
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Think, I will try base 18

Quote:
Sierpinski value is 398. 4 k candidates seek primes 18,324,122,381
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Old 2007-01-06, 23:31   #32
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For base 16, I found a covering set [17,13,7,241] So S/R must be less than 372827
For base 32 [3,7,13,17,241]

Last fiddled with by Citrix on 2007-01-06 at 23:32
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Old 2007-01-06, 23:54   #33
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I think that it would make sense to either put up a website or make a sticky thread with the current status for each base. It is beginning to be difficult to follow this thread.
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