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Re: [PrimeNumbers] p-1 isn't a random integer

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  • Décio Luiz Gazzoni Filho
    As expected, my idea proposed below is not new. David Broadhurst and Phil Carmody (both missed members of this list, unlike the Goldbach-proving spammers)
    Message 1 of 3 , Feb 4, 2005
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      As expected, my idea proposed below is not new. David Broadhurst and Phil
      Carmody (both missed members of this list, unlike the Goldbach-proving
      spammers) pointed me towards the twin-prime constant C2 = 0.660162, which is
      the ratio between Mertens' product and `my' product below. One can use
      Cohen's methods from http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi to
      compute C2 to high-precision, as pointed out by David Broadhurst. Mike Oakes
      also pointed out (off list) an interesting site regarding the twin-prime
      constant, the page by Xavier Gourdon at
      http://numbers.computation.free.fr/Constants/Primes/twin.html. If this isn't
      enough, the owners of Crandall & Pomerance's book can have a look at chapter
      1, who has a derivation of the twin prime conjecture.

      Décio

      On Thursday 03 February 2005 23:49, you wrote:
      > For p prime, the integer p-1 is an important quantity in certain analyses
      > (particularly if they involve the group (Z/pZ)*), but as I recall, this
      > integer is usually considered as `a random integer in the vicinity of p' in
      > such analyses (except perhaps by considering that it is even).
      >
      > However, a neat idea occurred to me: just as the probability that 2 divides
      > this integer is skewed in comparison to a random integer (it's 1 in this
      > case but 1/2 for a random integer), so is the probability that it is
      > divisible by any other prime. Consider for instance the residue class of p
      > mod 3; unless p itself is 3, then p == 1 or 2 mod 3, with equal probability
      > for each residue class. So p-1 == 0 or 1 mod 3 also with equal probability.
      > Thus, p-1 is divisible by 3 with probability 1/2, not 1/3 as expected.
      > Using the same argument, p-1 is divisible by a given prime q with
      > probability 1/(q-1), not 1/q.
      >
      > The first implication that occurred to me is that one shouldn't use
      > Mertens' theorem in any analyses involving p-1 where p is prime. For
      > instance, I computed the probability that primes from 3 to 31337 divide a
      > given integer using Mertens' theorem; the result is 0.1085. Using PARI/GP
      > to evaluate the product (1 - 1/(p-1)) for primes p from 3 to 31337, I
      > obtained 0.07157. The difference is quite real.
      >
      > I'll try to work out an estimate, similar to Mertens' theorem, but for the
      > product (1 - 1/(p-1)) instead. However, I'm pretty sure David Broadhurst or
      > Mike Oakes will beat me to it (:
      >
      > I don't know how useful this is (probably not useful at all), but I thought
      > it was neat and worth mentioning.


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