19087Re: [PrimeNumbers] probability of (2*p1*p2) + 1 being prime
- Sep 9 6:53 AMPhil Carmody wrote:
> --- jtrjtrjtr2001 <jtrjtrjtr2001@...> wrote:Phil meant y is odd, but another factor must also be considered.
>> Let p2 be a large prime. We find another large random prime p1, such that
>> y = (2*p1*p2) + 1.
>> Is there any way, one could quantify the probability of y being a prime?
> Well, it's just like any arbitrary number of the same size except that
> it's even, it's not divisible by p1, and it's not divisible by p2.
> Therefore there's a prime density boost of (2/1) * (p1/(p1-1)) *
> Those final two factors are effectively 1.
If q is a random odd prime other than p1 and p2, then q does
not divide y-1. q must divide one of the q-1 numbers from
y to y+q-2, and the chance that q divides y increases from
1/q to 1/(q-1).
So the chance that q does not divide changes from
(q-1)/q to (q-2)/(q-1), which corresponds to multiplying the
chance by ((q-2)/(q-1)) / ((q-1)/q) = q*(q-2)/(q-1)^2.
The situation is just like twin primes:
If p is a large prime then p+2 is odd. If q is a random odd
prime other than p then q does not divide p, so the chance
of q dividing p+2 increases from 1/q to 1/(q-1).
The twin prime constant C_2 = 0.66016... is the product
over all odd primes q of q*(q-2)/(q-1)^2
See for example http://mathworld.wolfram.com/TwinPrimesConstant.html
A random large number n has chance 1/log(n) of being prime.
Our y is odd and also taking C_2 into consideration changes
the chance to 2*C_2/log(y).
Jens Kruse Andersen
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