20091124, 01:48  #12  
A Sunny Moo
Aug 2007
USA (GMT5)
3×2,083 Posts 
Quote:
Secondly, I was thinking the same thing as henryzz, that possibly there's something we don't yet know that's causing our predictions to be way too high. I wonder if there's something specific to the Dual Sierpinski conjecture that makes it more "primedense" than others? 

20091124, 01:57  #13  
May 2007
11^{2} Posts 
Quote:
e 

20091124, 02:04  #14  
May 2007
1111001_{2} Posts 
Quote:
On the other hand, to up our chances for the last sequence, maybe we need for mister prime of FOB to have him help us with the last one. Last one is the charm to paraphrase a common saying, yea right.... 

20091124, 03:31  #15 
"Phil"
Sep 2002
Tracktown, U.S.A.
2135_{8} Posts 
MiniGeek calls it the butterfly effect: because he took a single work file 4.80M, that caused this latest prime to be discovered by Engracio rather than Jeff. Jeff, you have my permission to find the next one, but let me say in advance, I will be pleased no matter who finds it!
These primes certainly seem to be consistently showing up early  I just hope this trend continues a little longer. Maybe mdettweiler and henryzz are right, that there is something we don't know that is making the problem a little easier than predicted. 
20091124, 04:13  #16 
May 2007
11^{2} Posts 
Sounds like someone with a little bit of time and lots of know how, they can bite into and be busy for a while.

20091125, 02:51  #17 
Mar 2003
New Zealand
13×89 Posts 
Congratulations engracio! What a breakthrough for this project.

20091125, 03:13  #18 
Feb 2008
13_{8} Posts 
Congrats engracio!

20091127, 12:13  #19 
Sep 2009
11_{10} Posts 
Congrats to Engracio and all the group, we're lucky indeed !!!
Regarding the stats for the last prime: The 63M figure for 50% prob looks correct, but I don't understand why "We shouldn't expect 1 prime until 191.63M" according to MiniGeek... BTW much congrats and luck, I didn't expect a PRP so early, the 50% prob was around 17M indeed... (sorry for being spartan in the form but I'm abroad, having not much time...) Last fiddled with by Zuzu on 20091127 at 12:19 Reason: Pbs with emoticons 
20091127, 12:46  #20  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10253_{8} Posts 
Quote:
e^(x)=y is an equation meaning that when you expect x random occurrences, the chances of it not happening are y. We're looking for the chances of it happening (i.e. finding a prime), so we really want 1e^(x). (note that I left that out up there because for y=0.5, a 50% chance, they're the same) For the one I showed there, I wanted to know the expected number of primes before we had a 50% chance of finding one. It's about 0.693. The 5.15M*2^(X/.19166) part lets me smooth out calculating between doublings, where I know we expect .19166 primes. This means I can see the size we should expect to get to for any expected number of primes. By the same idea, 191.63M=5.15M*2^(1/.19166), and 1e^(1)=0.632... You can see this is probably about right by noting that .19166*5 is a little under 1, and if you double 5.15M five times you get 164.8M, which is a little smaller than the number I calculated. Last fiddled with by MiniGeek on 20091127 at 12:48 

20091128, 15:32  #21 
"Phil"
Sep 2002
Tracktown, U.S.A.
1,117 Posts 

20091202, 09:08  #22 
Sep 2009
11 Posts 
OK I understand your calculation: 191M is the exponent n for which the probability of finding no PRP between 1 and n is equal to exp(1) = 0.368, provided that no PRP has been found between 1 and 5.15M (thus neglecting the possibility of finding a PRP by doublechecking).
IMO the reference probability is more based on 50% but as you have written, for each p in ]0,1[ one can similarily calculate the exponent n(p) for which the probability of finding no PRP between 1 and n is equal to p: n(p)=5.15M*2^(ln(p)/0.19166)) For low values of p, n(p) increases dramatically: 7.54M for 0.9; 14.58M for 0.75; 63.17M for 0.5 but 775M for 0.25; 21G for 0.1; 261G for 0.05 and 88T for 0.01. There is a reasonable probability of finding the last PRP before say 50M (even before 10M if lucky as we were) but not finding any before say 1G is not to be ruled out. OTOH for the SoB project, the mean number of primes per doubling being 1.04, the range of outcomes is less extended for the next prime. Provided that no prime has been found before 17.2M and neglecting doublecheck, n(p) would be equal to 18.5M for 0.9; 20.8M for 0.75; 27.3M for 0.5; 43.4M for 0.25; 80.1M for 0.1; 127M for 0.05 and 373M for 0.01. Last fiddled with by Zuzu on 20091202 at 09:21 Reason: Error corrected 
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