20071108, 10:05  #1 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5·7·13^{2} Posts 
Percent chance of being prime
is there a way of working out what percent chance a number has of being prime based on the highest prime below its square root
less than the square of 3 50% of numbers are prime less than the square of 5 37.5% of numbers are prime that is what i would like be able to work out without actually knowing the figures 
20071108, 12:08  #2  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2×3×5^{2}×73 Posts 
Quote:
Now, what was the question you really meant to ask? Paul 

20071108, 13:16  #3 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5·7·13^{2} Posts 
what i meant was:
how can i work out what percentage of numbers are prime between 2 squares of primes 
20071108, 13:31  #4  
Nov 2003
2^{2}×5×373 Posts 
Quote:
p1^2 and p2^2 is trivially [pi(p2^2)  pi(p1^2)]/(p2^2  p1^2). pi(N) is the prime counting function. This is junior high school level math. WTP????? 

20071108, 14:11  #5 
Aug 2003
Snicker, AL
7×137 Posts 
Just for Paul, we are going to invent a special kind of prime. We'll call it a Schroedinger prime. We never know if it is prime or not until we ask it. Then it tells us either "I'm prime" or else "I'm composite". But until we ask it, it is a Schroedinger prime and nobody knows if it is alive or dead.
Just a little fun on a Thursday morning. DarJones 
20071108, 14:39  #6  
I quite division it
"Chris"
Feb 2005
England
31·67 Posts 
Quote:


20071108, 20:01  #7 
Nov 2004
2^{2}×3^{3}×5 Posts 
I respectfully disagree. I think "primeness" is not subject to the spooky whims of quantum mechanics. I believe a number is prime whether it has been looked at (factored) or not, regardless of the speed or acceleration of the observer relative to the prime. While all else in the universe is at the whim Schroedinger's kitty cat (even existence is not determined until an observer observes), primes are absolute. Primosity is more like the speed of light: absolute, and not subject to the philosophical questions of Schroedinger, Heisenberg, or the old "if a tree falls in the forest when no one is around, does it make a sound?"...recently updated to "if a man makes a statement when there is no woman around to hear it, is he still wrong?"
Norm Last fiddled with by Spherical Cow on 20071108 at 20:03 Reason: Had to change the cat's litter box... 
20071108, 22:55  #8 
"Jason Goatcher"
Mar 2005
110110110011_{2} Posts 
Can't we at least agree that statistics are useful up until the time the status of the number is determined? For example, a lot of people like to sieve enough numbers in a range to give a 90% chance of finding a prime. And when the Prime95 program is run, the sieving and P1 are run according to statistical probabilities. Until you actually know the status of the number, it seems idiotic and/or conceited to keep shouting,"It's either 100% prime or 100% composite." Sure, that's true, but how many people in the general public are in a situation where that matters.
Maybe I should start bringing up the idea of primes with factors that have an imaginary component. In my mind, that's just as relevant. 
20071109, 12:59  #9 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·3·5^{2}·73 Posts 

20071109, 20:55  #10 
May 2003
7·13·17 Posts 
henryzz,
As Silverman said, if p and q are primes, with p>q, then the percentage of primes between p^2 and q^2 is ( pi(p^2)  pi(q^2) )/ (p^2  q^2 ) where pi(x) is the prime counting function. If your values for p and q are small, you can just use this formula in a computer to find the exact value. When I read your initial question, I thought you were asking for the percentage of the numbers up to p^2 which are prime. This is pi(p^2)/p^2, which is asymptotic to p^2/ (log(p^2) p^2) = 1/(2 log(p)). This last formula will give you a good rough estimate. 
20071110, 10:04  #11 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5×7×13^{2} Posts 
how without counting the primes can i work this out
i want to use this for hopefully having p and q being 100 digit primes how long do u think ( pi(p^2)  pi(q^2) )/ (p^2  q^2 ) would take if p and q are 100 digit primes is there a quick way 
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