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 2008-11-06, 17:33 #1 robert44444uk     Jun 2003 Oxford, UK 1,979 Posts 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
2008-11-06, 18:44   #2
Xentar

Sep 2006

11×17 Posts

Quote:
 Originally Posted by robert44444uk 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"?

2008-11-06, 19:03   #3
rogue

"Mark"
Apr 2003
Between here and the

11001001111012 Posts

Quote:
 Originally Posted by Xentar 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.

 2008-11-06, 20:10 #4 Siemelink     Jan 2006 Hungary 1000011002 Posts 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.
 2008-11-06, 20:22 #5 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 3×1,979 Posts http://mathworld.wolfram.com/BrierNumber.html this definition seems to be a little different
 2008-11-07, 04:27 #6 robert44444uk     Jun 2003 Oxford, UK 36738 Posts 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
2008-11-07, 04:31   #7
robert44444uk

Jun 2003
Oxford, UK

1,979 Posts

Quote:
 Originally Posted by Xentar 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

2008-11-07, 07:06   #8
gd_barnes

May 2007
Kansas; USA

7·11·137 Posts

Quote:
 Originally Posted by robert44444uk 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

 2008-11-07, 10:29 #9 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 134618 Posts 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
 2008-11-07, 10:58 #10 masser     Jul 2003 Behind BB 176010 Posts Link 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
2008-11-07, 20:41   #11

Jan 2006
Hungary

22·67 Posts

Quote:
 Originally Posted by masser 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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