20060322, 17:30  #12 
"Robert Gerbicz"
Oct 2005
Hungary
5·17·19 Posts 
On Seventeenorbust forum I've made a prediction:
I've computed the expected n, when we'll find a prime by 50% probability, if all numbers tested up to exponent=10M then: Remaining=7 will reach at 1.3*10^7 Remaining=6 will reach at 1.8*10^7 Remaining=5 will reach at 2.7*10^7 Remaining=4 will reach at 4.5*10^7 Remaining=3 will reach at 9.5*10^7 Remaining=2 will reach at 3.1*10^8 Remaining=1 will reach at 3.1*10^9 Remaining=0 will reach at 4.5*10^13 This isn't very far from the prediction what Yves Gallot has been done in 2001 ( when there were 17 remaining cases ): there is 50% probability to solve the poblem up to N=2^43 ( it is about 8*10^12 ) For Riesel problem the case is much worse: We have 50% chance of solving Riesel problem at n=2^73. ( Gallot gave the prediction n=2^70 ) I've also calculated that by 5% probability we'll complete the project before n=2^50 and by 5% probability we won't complete before n=2^127. See: http://www.rieselsieve.com/forum/viewtopic.php?t=650 
20060322, 17:42  #13  
"Bob Silverman"
Nov 2003
North of Boston
1110101010100_{2} Posts 
Quote:
Please say something about HOW you arrived at these numbers. 

20060322, 18:14  #14  
"Robert Gerbicz"
Oct 2005
Hungary
5·17·19 Posts 
Quote:
k*2^n1 is prime by about w(k)/(n*log(2)) probability, here w(k) is the "weight" of k for Riesel problem and there is a table for this. So the probability that k*2^s1 is composite for all A<s<B is about prod(1w(k)/(n*log(2),n=A,B) but 1x is very close to exp(x) for small x values so this product is about c(A,B,k)=prod(exp(w(k)/(n*log(2)),n=A,B)=exp(w(k)/log(2)*sum(1/n,n=A,B)) but sum(1/n,n=A,B) is about log(B/A) so c(A,B,k)=(B/A)^(w(k)/log(2)). Let f(x,A,B)=prod(c(A,B,k)*x+1c(A,B,k), for all remaining k values) the degree of the polynom is 73 for Riesel problem ( because there are 73 remaining cases ) and you can see that the coefficient of x^j will give the probability that there will be only j remaining k vaules if we complete the project up to N=B and if we suppose that all exponents tested up to A for all k values. So the probability that there will be at most h remaining cases is the sum of the coefficients of x^j from j=0 to h. Last fiddled with by R. Gerbicz on 20060322 at 18:17 

20060324, 00:30  #15  
"Phil"
Sep 2002
Tracktown, U.S.A.
2^{5}·5·7 Posts 
Quote:
http://www.seventeenorbust.com/stats/rangeStatsEx.mhtml 

20060327, 21:59  #16 
"Jason Goatcher"
Mar 2005
3507_{10} Posts 
I don't pretend to understand most of the math in this thread, but a while back I came up with what I guess you'd call a hack to attempt to solve the problem.
I figured I needed to assign a value to each test. It didn't matter how big or small the value is, they just had to accurately describe the value of a test relative to other tests. I came up with two values to deal with:the time a test takes, which in my instance I used calculations which only had a proportionate relationship to the length of time a test takes. The Sierpenski and Riesel problem have separate ways of attempting to determine primality, so I had to use separate equations for each problem. For the Sierpenski problem, I used the equation 1/((n/ln n)^2). The time a test takes is proportionate to (n/ln n)^2, but given even odds, a shorter test is "worth more." The equation for Riesel problem is 7.5^(ln n/524288)*100,(the value for a test is 1/(7.5^(ln n/524288)*100) but the results for the Sierpenski and Riesel problem are NOT interchangeable. Now for the second value, I used a program called pr_prob, which figured the odds any n will yield prime after a certain amount of sieving. This is also an inverse proportion. I multiplied the numbers together and I got some extremely small values. For the Sierpenski problem, I calculated that unless you could sieve an average of 30 values for every you PRP you should be PRP ing. (My calcuations for the Riesel problem are old, so not dependable) Here's the thread, for those interested, although I made some rather embarrassing false starts. 
20060328, 14:54  #17  
"Bob Silverman"
Nov 2003
North of Boston
1110101010100_{2} Posts 
Quote:
single Sierpinski or Riesel Candidate are wrong. I suggest that you look up the timecomplexity function for performing a *single* FFTbased multiplication, then multiply by the number of multiplications needed to test a candidate. And what you call "equations" are not equations. 

20060331, 00:32  #18  
"Jason Goatcher"
Mar 2005
3·7·167 Posts 
Quote:
Besides, no one else has ever managed to come up with anything to solve this statistical problem, which makes my method best, simply because it's a lone candidate. You say it's wrong? Tell me why it's wrong. All you've done so far is insult me, which seems par for the course when it comes to R.D. Silverman. 

20060331, 03:14  #19 
"Jason Goatcher"
Mar 2005
3×7×167 Posts 
Btw, the second paragraph is tongue in cheek.

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