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#1 |
"Jason Goatcher"
Mar 2005
3·7·167 Posts |
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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^n-1, 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^n-1 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^n-1 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 2007-03-28 at 02:51 |
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#2 |
Aug 2005
24 Posts |
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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 On-Line 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? |
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#3 | |
Cranksta Rap Ayatollah
Jul 2003
12018 Posts |
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#4 | |
Nov 2003
164448 Posts |
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#5 | |
Tribal Bullet
Oct 2004
5·709 Posts |
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jasonp |
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#6 | |
Jun 2003
Oxford, UK
7F716 Posts |
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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 pseudo-cover 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^3-1, which gives you a form of k*2^3-1, k*2^6-1, k^2^9-1...(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 2007-05-29 at 13:45 |
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