Percent chance of being prime

[henryzz]

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

[xilman]

Originally Posted by henryzz

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

Yes, the chance is either 100% or 0%. A number is either prime or not.

Now, what was the question you really meant to ask?

Paul

[henryzz]

what i meant was:
how can i work out what percentage of numbers are prime between 2 squares of primes

[R.D. Silverman]

Originally Posted by henryzz

what i meant was:
how can i work out what percentage of numbers are prime between 2 squares of primes

Let p1 < p2 be your two primes. The percentage of primes between
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?????

[Fusion_power]

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

[Flatlander]

Originally Posted by Fusion_power

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

Yes, this is how things work. The wave function collapses when LLR reaches 100%.

[Spherical Cow]

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

[jasong]

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 P-1 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.

[xilman]

Originally Posted by jasong

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.

Fine by me.

Anyone who doesn't know what we're talking about might find "Gaussian primes" a good starting point.

Paul

[Zeta-Flux]

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.

[henryzz]

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