20140527, 14:28  #1 
"W. Byerly"
Aug 2013
1423*2^21790231
142_{8} Posts 
probability a number is prime with a weighted k.
given a riesal prime (k*2^N1 ), a nash weight of the k, and a range of n, calculate the expected number of primes in the range.
I have done my research and it look like the average nash weight is 1751.542, so would am I correct in saying that this nash weight would be considered the same as the average prime number density? As an example, say the k value is 51 and the n range is 1M 2M. The nash weight is 1550, divide it by 1751.542 and get ~0.885 times the average prime density. Then take the midpoint of the n ranges > 1.5M (I know you will have to check each individual one, but for simplicities sake) and then plug this into the probability of finding a prime number formula: 0.885* (1/ln(51*2^1,500,000)) which comes out to .00000085 and then multiply this by 1000000 candidates to test to get .85, so we can say we will find an average of 0.85 primes in this range, or 85% chance of finding 1 prime right? Last fiddled with by Trilo on 20140527 at 14:29 
20140527, 15:29  #2 
Aug 2006
2·2,969 Posts 
If the expected number of new primes is 0.85, this means the chance of a prime is about

20140527, 20:15  #3 
"W. Byerly"
Aug 2013
1423*2^21790231
2×7^{2} Posts 

20140528, 05:59  #4  
"Curtis"
Feb 2005
Riverside, CA
2×7×11×29 Posts 
Quote:
The Nash weight is a measure of how many candidates survive a small sieve. To take sieving (and thus weight) into account, see: http://www.mersenneforum.org/showpos...&postcount=157 Curtis 

20140528, 13:58  #5  
"W. Byerly"
Aug 2013
1423*2^21790231
2×7^{2} Posts 
Quote:


20140601, 22:55  #6 
Dec 2011
After milion nines:)
2^{3}·13^{2} Posts 
If you use formula from that thread you will get
(1.781*ln(389651833000000000))/(3000000*ln(2)) = 0,0000346908898293 0,0000346908898293 is constant since all sieve we get from Primegrid is sieved on same depth ( I use sieve from 3M  4M) So you just need to multiply number of candidates in sieve with 0,0000346908898293 and you will get expected number of primes in range 3M  4M regardless of K. 25193 ( number of candidates in sieve of K 51) * 0,0000346908898293= 0.87 primes in range 34M for k= 31 it is 34986*0,0000346908898293= 1,21 primes in that range. So you need number with nash value around 2050 , and that number will have around 1 primes in that range. Last fiddled with by pepi37 on 20140601 at 23:00 Reason: add more text 
20140602, 08:21  #7  
"Curtis"
Feb 2005
Riverside, CA
1000101110010_{2} Posts 
Quote:


20140606, 15:55  #8  
"W. Byerly"
Aug 2013
1423*2^21790231
98_{10} Posts 
Quote:
so in other word, using the example in that post with 2 primes, 1 (1 1/14997)^{(1/2)*18283}= 45.6% chance to find 2 primes in that range 

20140606, 16:34  #9  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Quote:
1.21^k * e^(1.21) / k! For k=3: 1.21^3 * e^(1.21) / 3! ~= 0.088 Note that for k=0, the formula simplifies to e^(1.21), which should look familiar:You may also want to calculate the probability of >=3 primes, which should be (1  (probability of 0)  (probability of 1)  (probability of 2)) Last fiddled with by MiniGeek on 20140606 at 16:36 

20140606, 17:45  #10  
"W. Byerly"
Aug 2013
1423*2^21790231
2×7^{2} Posts 
Quote:


20140606, 18:21  #11  
"W. Byerly"
Aug 2013
1423*2^21790231
2·7^{2} Posts 
Quote:
with expected number at = 0.076 and k= 1: (.076^1)*(e^.076)/1 comes out at 7.04% but 1.e^.076 comes out to 7.31% Last fiddled with by Trilo on 20140606 at 18:23 

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