20081106, 17:33  #1 
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 
20081106, 18:44  #2 
Sep 2006
11×17 Posts 

20081106, 19:03  #3 
"Mark"
Apr 2003
Between here and the
1100100111101_{2} Posts 
I presume he is referring to k where both k*2^n+1 and k*2^n1 are composite for all n.

20081106, 20:10  #4 
Jan 2006
Hungary
100001100_{2} 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. 
20081106, 20:22  #5 
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 
20081107, 04:27  #6 
Jun 2003
Oxford, UK
3673_{8} 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 20081107 at 04:46 
20081107, 04:31  #7 
Jun 2003
Oxford, UK
1,979 Posts 

20081107, 07:06  #8  
May 2007
Kansas; USA
7·11·137 Posts 
Quote:
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 

20081107, 10:29  #9 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
13461_{8} 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^n1 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^%d1;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 
20081107, 10:58  #10 
Jul 2003
Behind BB
1760_{10} Posts 
Link
Last fiddled with by masser on 20081107 at 10:58 
20081107, 20:41  #11 
Jan 2006
Hungary
2^{2}·67 Posts 
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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