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2009-09-04, 22:59   #23
Zuzu

Sep 2009

11 Posts

Quote:
 Originally Posted by philmoore Well, Zuzu, I hope that you're right and I'm wrong, as your figures certainly look more optimistic! I will take another look at it this weekend, when I'm not doing school work. That said, I admit that statistics is not my strong suit, although I felt like I had a good "seat of the pants" feel for it when I was doing graduate work in physics. I estimated the 1.32153 expected number as follows: each candidate $N$has probability of being prime equal to ${e^\gamma\ln F}\over{\ln N}$, where $F$ is the factoring depth. There are approximately 328,300 candidates left in the sieve for all exponents n between 0 and 50 million, so I assumed that the candidates are equally distributed and replaced the sum by an integral for n from 4.65 million to 50 million. Next, I assumed that since the probabilities for each candidate are independent, that if $p$ is the probability of at least one candidate being prime, then $p^2$ is the probability of at least two, $p^3$ is the probability of at least three, and so forth. Summing these probabilities gives the expected number of primes as $p\over{1-p}$, so if we set this equal to the expected number 1.32, we get p as about 1.32/2.32, or about 56.9%. Note that this gives a 32.3% chance of at least two primes if we search all the candidates, but note that there is only about a 50% chance that the two primes will then belong to different k values. I also translated this into the language of Poisson processes, where $e^{-w\log_2(n_f/n_i)}$ was the probability of finding no primes for all exponents between $n_i$ and $n_f$. My weights for $w$ were .17700 for 41693 and .17566 for 40291. (Divide these weights by 2 and you will see that they are a little smaller than those on Payam Samidoost's webpage, but those weights were computed by Brennen's Proth weight applet with a fairly low cutoff.) I will look up my notes from the last discovery and see if I can resolve this. Any comments are welcome, as I may be making some wrong assumptions here!
Phil,

Thank you for having provided the relevant information: the number of surviving candidates for a given factoring depth (sieving depth ?). Then the probability of an individual candidate being prime is approximately independent of the Proth weight. I'll make the calculations and confirm or infirm my previous considerations...

Last fiddled with by Zuzu on 2009-09-04 at 23:01 Reason: misleading formulation, needing precision

2009-09-05, 18:13   #24
Jeff Gilchrist

Jun 2003

7·167 Posts

Quote:
 Originally Posted by paleseptember As suggested, I went and bought a lottery ticket. Strangely enough, no joy.
Did you make sure all the numbers you picked on the ticket were PRP as well? If not...

Jeff.

 2009-09-06, 00:44 #25 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 2·13·43 Posts We are now listed on Henri and Renaud Lifchitz' page PRP Records with first, second, and fourth places! Our "production score" climbed from 45.70 to 47.76, and the record probable prime I found a year ago is now in seventh place. Go team!
 2009-09-09, 06:48 #26 geoff     Mar 2003 New Zealand 100100001012 Posts So that is why they call Australia the lucky country. Not just gold mines and sunshine, they have all the big PRPs too :-) Well done again Ben!

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