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19087Re: [PrimeNumbers] probability of (2*p1*p2) + 1 being prime

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  • Jens Kruse Andersen
    Sep 9 6:53 AM
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      Phil Carmody wrote:
      > --- jtrjtrjtr2001 <jtrjtrjtr2001@...> wrote:
      >> 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)) *
      > (p2/(p2-1))
      > Those final two factors are effectively 1.

      Phil meant y is odd, but another factor must also be considered.
      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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