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Old 2020-03-27, 09:35   #23
KEP
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Quote:
Originally Posted by LaurV View Post
Sorry for waking up this old thread, but I didn't want to create a new one, and the subject of the current one seems suitable for my silly question:


Why 81 was chosen as the conjectured k for Riesel base 1024?
The covering set for k=81 is (5 and 41) and there is a few notes on this site: http://www.noprimeleftbehind.net/cru...es-powers2.htm wich may mean something to you. But I recon that the remaining single k does not have a covering set and therefor does not meet the demand for being a conjectured k
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Old 2020-03-27, 09:48   #24
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Quote:
Originally Posted by LaurV View Post
Sorry for waking up this old thread, but I didn't want to create a new one, and the subject of the current one seems suitable for my silly question:


Why 81 was chosen as the conjectured k for Riesel base 1024?

Because it is the lowest k that has a known covering set of numeric factors for R1024. A factor of 5 covers even-n and a factor of 41 covers odd-n.

Smaller k's k==(1 mod 3, 1 mod 11, & 1 mod 31) have a single trivial factor that eliminates them or others (k=9 & 36) need algebraic factors to eliminate them from consideration.

k=29 does not have a known covering set and is not eliminated by one of the conditions above so we believe it will eventually have a prime. But like many other bases that is just a conjecture at this point. :-)

Last fiddled with by gd_barnes on 2020-03-27 at 09:51
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Old 2020-03-27, 14:48   #25
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An example is Sierp base 311, all k's that between a multiple of 3 and a multiple of 13 have the covering set {3,13}, the first such k's are 14, 25, 53, 64, 92, 103, 131, 142, ..., however all such k's < 142 are also trivial (i.e. gcd(k+1,311-1) is not 1), thus the conjectured k for Sierp base 311 is 142. (this conjecture is proven, all k's except 10, 46, and 76 have an easy prime)
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Old 2020-03-27, 14:54   #26
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Then why did you eliminate 9 and 36?

(the tree I am barking at, is the fact that 81 is a square)
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Old 2020-03-27, 15:14   #27
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Quote:
Originally Posted by LaurV View Post
Then why did you eliminate 9 and 36?

(the tree I am barking at, is the fact that 81 is a square)
gd_barnes is right, to extend his post:
to get a riesel/sierpinski cover you have to use d_i divisors in each remainder class in the covering set, you can't use algebraic factors, you can't use even in those cases where you could use, like in your example for k=81 we could cover the n==0 mod 2 case, because 81*1024^n-1 has an algebraic factor. For this reason we can rule out say k=9, it is sure that this sequence contains no prime, but it has no covering set. The proof could be hard/impossible(?), but just see the prime factorization for n=54: http://factordb.com/index.php?id=1100000000033081963 .

ps. you could also cover the n==0 mod 1 so every integer for k=81 with an algebraic factor.

Last fiddled with by R. Gerbicz on 2020-03-27 at 15:17
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Old 2020-03-27, 15:43   #28
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Quote:
Originally Posted by LaurV View Post
Then why did you eliminate 9 and 36?

(the tree I am barking at, is the fact that 81 is a square)
This is like the base 125 case, the CK for both sides of base 125 are 8, and both 125 and 8 are cubes, thus 8*125^n+-1 both have algebra factors.
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Old 2020-03-27, 15:45   #29
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Quote:
Originally Posted by R. Gerbicz View Post
gd_barnes is right, to extend his post:
to get a riesel/sierpinski cover you have to use d_i divisors in each remainder class in the covering set, you can't use algebraic factors, you can't use even in those cases where you could use, like in your example for k=81 we could cover the n==0 mod 2 case, because 81*1024^n-1 has an algebraic factor. For this reason we can rule out say k=9, it is sure that this sequence contains no prime, but it has no covering set. The proof could be hard/impossible(?), but just see the prime factorization for n=54: http://factordb.com/index.php?id=1100000000033081963 .

ps. you could also cover the n==0 mod 1 so every integer for k=81 with an algebraic factor.
9*1024^n-1 is unlikely to have a covering set, like the simple cases: 3*2^n+-1, 5*2^n+-1, 7*2^n+-1, 9*2^n+-1, 11*2^n+-1, etc. they are unlikely to have a covering set, but also no proof.
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Old 2020-03-27, 15:51   #30
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Quote:
Originally Posted by sweety439 View Post
9*1024^n-1 is unlikely to have a covering set, like the simple cases: 3*2^n+-1, 5*2^n+-1, 7*2^n+-1, 9*2^n+-1, 11*2^n+-1, etc. they are unlikely to have a covering set, but also no proof.
Well... since all prime factors of k*2^n+-1 are odd, thus if there exist an n such that k*2^n+-1 or its dual (|2^n+-k|) is power of 2 (including 1), then k*2^n+-1 cannot have a covering set.

Thus these k*2^n+-1 cannot have a covering set:

1*2^n+1 (for n=0, the value is 2)
3*2^n+1 (for n=0, the value is 4)
7*2^n+1 (for n=0, the value is 8)
15*2^n+1 (for n=0, the value is 16)
31*2^n+1 (for n=0, the values is 32)
1*2^n-1 (the dual form |2^n-1|, for n=1, the value is 1)
3*2^n-1 (for n=0, the value is 2, also the dual form |2^n-3|, for n=2, the value is 1)
5*2^n-1 (for n=0, the value is 4)
7*2^n-1 (the dual form |2^n-7|, for n=3, the value is 1)
9*2^n-1 (for n=0, the value is 8)
15*2^n-1 (the dual form |2^n-15|, for n=4, the value is 1)
17*2^n-1 (for n=0, the value is 16)
31*2^n-1 (the dual form |2^n-31|, for n=5, the value is 1)
33*2^n-1 (for n=0, the value is 32)
etc.

However, there is no proof that 5*2^n+1, 9*2^n+1, 11*2^n+-1, 13*2^n+-1, 17*2^n+1, ... have no covering set (i.e. the sequence of the smallest prime factor of k*2^n+-1 is unbounded above).

Last fiddled with by sweety439 on 2020-03-27 at 15:54
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Old 2020-03-27, 19:59   #31
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Put more simply for R1024, for k=81 all n's have a numeric covering set of factors [5, 41]. For k=9 and 36 that is not the case. They require algebraic factors for the even n to eliminate them.
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Old 2020-03-27, 20:31   #32
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Quote:
Originally Posted by gd_barnes View Post
Put more simply for R1024, for k=81 all n's have a numeric covering set of factors [5, 41]. For k=9 and 36 that is not the case. They require algebraic factors for the even n to eliminate them.
You need less, for example for k=9 you need the algebraic factor only for n==0 mod 6, because the 9*1024^n-1 sequence for
n==1 mod 2 is divisible by 5
n==2 mod 6 is divisible by 7
n==4 mod 6 is divisible by 13

Ofcourse you can go further so where you need the algebraic factor for less than 1/6 part of the cases. But you will need in at least one remainder class the algebraic factor (without a proof).
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Old 2020-03-28, 03:21   #33
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Ok. Thanks all three (yep, including sweety, in spite of the fact that most of his/her posts are clutter, he/she gave the example with 125 )

I have learned something new today. I was thinking that stuff with algebraic factors are excluded completely (because they are always composite, so Mr. Riesel respective Sierpinsky, would have nothing to "prove" in this case. It is clear that, if you have algebraic factorization, then k*b^n-1 is always composite. Nothing to prove).

The "covering set" should be applied to what is left after getting rid of the sequences with algebraic factors.

It seems I was wrong (which I still refuse to accept, but that is not your problem )

Last fiddled with by LaurV on 2020-03-28 at 03:21
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