## Seventeen or Bust Nixes Three Sierpinski Candidates 19

Craigj0 writes

*"In just 8 days Seventeen or bust has removed three Sierpinski candidates after people have been trying for years. Seventeen or bust is a distributed attack on the Sierpinski problem. You can find the first two press releases here(1) and here(2), the third is still to come. More information about Sierpinski numbers can be found here. Finally they could always use some more people so join!"*
## Speaking of Sierpinski... (Score:4, Interesting)

## Re:Speaking of Sierpinski... (Score:1)

## It's Illegal to reply to your own messages (Score:1)

The math links on this site go to Eric Weisstein's World of Mathematics, hosted by Wolfram Research, makers of Mathematica. (As a sidenote, Wolfram's headquarters are located a stone's throw from the University of Illinois campus.) Eric makes us wonder why the rest of us are allowed to waste his oxygen; but in any case, we thank him.## Re:Speaking of Sierpinski... (Score:1)

## Re:Speaking of Sierpinski... (Score:2)

## More information on the Sierpinksi problem (Score:5, Informative)

## Re:Can someone explain why this article is filtere (Score:2, Informative)

## Sussed it out. (Score:1)

WTF that does I'll never know, but it needed to be checked. Maybe that's why there are almost no replies to this thread...

## How to prove this? (Score:5, Interesting)

I've read around a bit now, I've even installed their client (wasn't currently doing any other distributed stuff, so why not), but I still don't understand the math well. I understand you can prove a number

kisnota Sierpinski number by finding annso thatk*2^n+1is prime. The lowest known Sierpinski number is 78557. There are now only 14 lower numbers left for which there's no fitting n found yet, and they're searching for them.Now what I don't understand is how Sierpinski-ness can be proven, how they know there's not some huge n that makes 78557*2^n+1 prime after all; and I can't find the info. There's a class of numbers that are Sierpinski by construction (apparently) but they are much higher than this one. I guess there's no quick easy answer, I just have to read the literature, and I'm not going to... There are too many contrived number properties out there, and too much other stuff to do :)

## Re:How to prove this? (Score:3, Informative)

Here's some info [astate.edu], though the exact construction of the proof isn't give. Apparently, it's possible to prove that for any

n, 78557*2^n+1 is divisible by one of a finite (and quite small) number of primes. As tohow, ask the guy who proved it...## As I understand it ... a very rough analogy (Score:1)

## Re:How to prove this? (Score:3, Insightful)

8^n - 1 = 7m where for any n m,n are +ve integers

Proof:

the statement is true for n=1 (trivial)

assume it holds for n = some +ve int. k.

8^k - 1 = 7s

Consider the next case:

8^(k+1) - 1

= 8(8^k) - 1

= 8(7s +1) -1

= 56s + 7

= 7(8s+1) clearly divisible by 7.

=>The assertion holds for n=k+1 if it holds for n=k So since n = 1 holds, the assertion holds for all +ve int. n. I'm sure that the techniques these guys use are far more complicated and sophisticated, but it is possible to prove things like that.

## Heh (Score:2)

So if you've been thinking that you can't find a prime without multiple computers that are always on, this just goes to show you obviously can.Ya! Way to rub it in their faces! I mean, what kind of idiots didn't already understand that you can find prime numbers on multiple computers...especially computers that are on!

That'll teach 'em...