20070328, 02:45  #1 
"Jason Goatcher"
Mar 2005
3×7×167 Posts 
Could a Distributed Computing approach help find the smallest Brier number?
I ask this question in total ignorance of the math involved. I throw this out there just to see if people think DCing could help this problem.
For those of you who don't know what a Brier number is, I'll explain. When you're searching for large prime numbers, there's a basic formula, k*b^n+c. For the most part, most people concentrate where b equals 2, and c is plus or minus 1. For k*2^n1, there are k's where all n from 1 to infinity will give the equation a composite number, and it's the same thing for k*2^n+1, but usually with different numbers. The smallest number for k*2^n1 is believed to be 509203, for +1, I think it's 78557(not sure). A Brier number is a number, k, where both k*2^n1 and k*2^n+1 will yield composite numbers for all n from 1 to infinity. My question is,"Is it possible to use distributed computing to help solve the lowest Brier number problem in a timely fashion?" For the record, I know searching for primes below the smallest known Brier number is kind of idiotic since it's so unbelievably huge. I'm wondering if there is a possibility to do some sort of covering set research, in an attempt to find a smaller number. Last fiddled with by jasong on 20070328 at 02:51 
20070328, 03:30  #2 
Aug 2005
20_{8} Posts 
The mathworld.com description for Brier Number currently says
"The first few are 878503122374924101526292469, 3872639446526560168555701047, 623506356601958507977841221247, ... " The comment field of Sequence A076335 in the The OnLine Encyclopedia of Integer Sequences states "These are just the smallest examples known  there may be smaller ones." Don't one of these two statements need to be corrected or is there no rigor in the field of mathematics? 
20070328, 07:41  #3  
Cranksta Rap Ayatollah
Jul 2003
1010000001_{2} Posts 
Quote:


20070328, 16:22  #4  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}·3·7·89 Posts 
Quote:


20070328, 17:05  #5  
Tribal Bullet
Oct 2004
3,547 Posts 
Quote:
jasonp 

20070529, 13:30  #6  
Jun 2003
Suva, Fiji
2^{3}·3·5·17 Posts 
Quote:
http://www.primepuzzles.net/problems/prob_029.htm I have been corresponding with Prof. Caldwell on techniques for Sierpinski/Riesel any base and he suggests there are alternatives to the traditional covering set approach (using primes modulo a certain base) which, in particular, look at known algebraic factoring methods to allow for a pseudocover to replace one or more of the primes in the cover, or will allow for a smaller covering set. For example x^3  a^3 = (x  a)(x^2 + ax + a^2) When you set a=1, then this becomes x^31, which gives you a form of k*2^31, k*2^61, k^2^91...(pseudo order 3 base 2) as k*2^(3*integer) is still a cube if k is a cube. Other useful factoring tools might be used to give pseudo cover, for both + and  cases. Aurifeuillian for example. This type of strategy has not been investigated as far as I am aware for the Brier case. But it would take a lot of brain power, which I am in limited supply, to investigate all the possibilities. But if it produced a smaller Brier, then it would be a breakthrough. Last fiddled with by robert44444uk on 20070529 at 13:45 

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