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Old 2008-11-06, 17:33   #1
robert44444uk
 
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Default Brier any base

Not wishing to disturb anyone from what they are doing, but maybe it is time to explore Brier numbers any base.

Briers are k such that k*2^n+/-1 are never prime

Without thinking at all deeply, (it is late and the wine is having its wicked way) I have to assume they exist for b>2
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Old 2008-11-06, 18:44   #2
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Quote:
Originally Posted by robert44444uk View Post
Briers are k such that k*2^n+/-1 are never prime
Ehh, maybe I don't understand the question.. but aren't that the numbers, seventeen or bust (for example) is looking for?
SoB is looking for primes - so the remaining k could be "briers"?
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Old 2008-11-06, 19:03   #3
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Quote:
Originally Posted by Xentar View Post
Ehh, maybe I don't understand the question.. but aren't that the numbers, seventeen or bust (for example) is looking for?
SoB is looking for primes - so the remaining k could be "briers"?
I presume he is referring to k where both k*2^n+1 and k*2^n-1 are composite for all n.
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Old 2008-11-06, 20:10   #4
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For base 20:
the lowest Riesel is 8.
the lowest Sierpinski is 8.

If a Brier number for a base is a number that is both a Riesel and a Sierpinski then the lowest Brier number for base 20 is 8.

Willem.
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Old 2008-11-06, 20:22   #5
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http://mathworld.wolfram.com/BrierNumber.html
this definition seems to be a little different
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Old 2008-11-07, 04:27   #6
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Anyway, the concept is clearer now I have slept and shaken off the wine.

So the Mathworld looks odd because they have inverted not only Sierpinski and Riesel but also k and n. Written by someone doubly dyslexic. It is only conjectured that k=878503122374924101526292469 is smallest for n=2.

Siemelink has already spotted that Brier any base can give up results easily where the lowest k for either is the same, then this is the lowest Brier as well. Same with many of those where low k=4 i.e. b=14mod15

Last fiddled with by robert44444uk on 2008-11-07 at 04:46
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Old 2008-11-07, 04:31   #7
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Quote:
Originally Posted by Xentar View Post
Ehh, maybe I don't understand the question.. but aren't that the numbers, seventeen or bust (for example) is looking for?
SoB is looking for primes - so the remaining k could be "briers"?
SoB is looking for numbers k such that k*2^n+1 never prime, not k*2^n+&-1
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Old 2008-11-07, 07:06   #8
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Quote:
Originally Posted by robert44444uk View Post
SoB is looking for numbers k such that k*2^n+1 never prime, not k*2^n+&-1
The "&" sign clears it up. I didn't know what you meant either.

Mathworld needs to correct their k and n. There's even a link on that page to a different page on their own site where the k and n are correct. They need to be consistent with the rest of the math world as well as themselves. lol

This would be an interesting exercise. Willem has already come up with one solution. I think I'll mess around a little with this myself.


Gary
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Old 2008-11-07, 10:29   #9
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am i right in thinking that this modified script will prove brier conjuctures
Code:
SCRIPT
DIM base,  2
DIM min_k, 1
DIM max_k, 100000
DIM max_n, 1000
OPENFILEAPP k_file,pl_remain.txt
OPENFILEAPP p_file,pl_prime.txt
DIMS tmpstr
DIM n
DIM k_step, 1
DIM k
IF (base % 2 == 1) THEN SET k_step, 2
IF (base % 2 == 1) && (min_k % 2 == 1) THEN SET min_k, min_k + 1
SET k,min_k - k_step
 
LABEL next_k
SET k, k + k_step
IF (k > max_k)   THEN GOTO END
SET n, 0
 
LABEL next_n
SET n, n + 1
PRP k*base^n-1
 
IF (ISPRIME)   THEN GOTO Riesel_Prime_found
PRP k*base^n+1
IF (ISPRIME)   THEN GOTO Sierpinski_Prime_found
IF (n < max_n) THEN GOTO next_n
 
SETS tmpstr,%d*%d^n+-1;k;base;
WRITE k_file,tmpstr
GOTO next_k
 
LABEL Riesel_Prime_found
SETS tmpstr,%d*%d^%d-1;k;base;n;
WRITE p_file,tmpstr
GOTO next_k
LABEL Sierpinski_Prime_found
SETS tmpstr,%d*%d^%d+1;k;base;n;
WRITE p_file,tmpstr
GOTO next_k
END
it is a modification of the script posted in the base 101 thread
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Old 2008-11-07, 10:58   #10
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Before this gets out of hand...

http://www.primepuzzles.net/problems/prob_049.htm

GIYF

Last fiddled with by masser on 2008-11-07 at 10:58
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Old 2008-11-07, 20:41   #11
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Quote:
Originally Posted by masser View Post
Before this gets out of hand...
That site is down for the moment, so allow me to dream. I would try to find the briers like this:
1) generate a series of Riesels R1, R2,..
2) generate a series of Sierpinskis S1, S2,
3) for every Rx loop through Sy
4a) if the cover set of Rx equals the cover set of Sy and Rx = Sy it is a Brier
4b) if the cover sets are not equal there must be a Brier where Rx + a*product(Cover of Rx) = Sy + b*product(Cover of Sy)

Cheers, Willem.
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