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Science

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!"
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Seventeen or Bust Nixes Three Sierpinski Candidates

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  • by jonnyfish ( 224288 ) <jonathan.m.fisherNO@SPAMgmail.com> on Saturday December 07, 2002 @11:04AM (#4832698) Journal
    Way back when I was first learning DirectDraw (shudder), I wrote a program to generate pixel colors based on x, y, and time (ticks) values. I played with trig functions a little bit, trying to generate some plasma effects. Eventually I got bored of that and tried some operations using the bitwise operators and, amazingly, I got it to generate the Sierpinski triangle using something like 2 or 3 bitwise ands per pixel. I say amazingly because all I know about Sierpinski triangles is the name, so I generated it completely by accident. Needless to say, I was very surprised the first time I saw it.
    • What you made was a simple two state cellular automaton.... here's a link: http://www.wolframscience.com/preview/nks_pages/?N KS0263.gif No No Not Wolfram! (don't forget to remove the space)
      • Anyways... Oddly enough I found an extra jab at Wolfram at the bottom of this story's link: 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.
      • I still don't know how I did it :P
    • It's a really easy effect to create - take the X coordinate, then AND with the Y coordinate. Use one colour for 0 and another for everything else, and you've got yourseif a Sierpinski triangle! Try it :)
  • by cyberlemoor ( 624985 ) on Saturday December 07, 2002 @11:18AM (#4832768)
    ..can be found here [prothsearch.net] (a slightly more detailed explanation than the one at the link the author gives).
  • It was the magical "collapse sections" button.

    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)

    by Scarblac ( 122480 ) <slashdot@gerlich.nl> on Saturday December 07, 2002 @03:57PM (#4834161) Homepage

    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 k is not a Sierpinski number by finding an n so that k*2^n+1 is 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 :)

    • by iltzu ( 22928 )
      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.

      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 to how, ask the guy who proved it...

      • For people who have not done lots of number theory ( I have done some and this is mostly over my head), A very rough analogy: you know any linear algebra? what a vector space is? a bunch of (simple) objects (the basis set) can describe any object (eg a line) in the (vector) space. Partly what the Sierpinski test is about is finding a bunch of primes that can be used to describe all numbers of the forms k.2^n+1. I could be wrong - if so please correct me. These suckers fascinate me (as do all prime number problems)
    • by Omkar ( 618823 )
      I don't know (I'm a high school student), but you can use a technique (math induction) to prove the divisibility of numbers. For example:


      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.
  • Quote for the article:

    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...

Dennis Ritchie is twice as bright as Steve Jobs, and only half wrong. -- Jim Gettys

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