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Re: [PrimeNumbers] proving the Riemann hypothesis

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  • Peter Kosinar
    ... The original poster might have intended to say that there is no -constant- upper bound on the gap between two consecutive primes -- which is obviously
    Message 1 of 40 , Feb 1, 2007
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      >> If the RH is false, the gap between prime P1 and the
      >> next prime P2 could be anywhere from 1 to infinitely
      >> large.
      >
      > False. Your attempt at a proof ends here.

      The original poster might have intended to say that there is no -constant-
      upper bound on the gap between two consecutive primes -- which is
      obviously true.

      Regardless of that, this statement:

      >> I can prove that the gap must be smaller than P1xP1-P1. If P1 is 5,
      >> this means that the gap between 5 and its next prime must be smaller
      >> than 5x5 - 5 = 20

      is a consequence of Bertrand's postulate (proved by Chebyshev, or
      Tchebycheff, or some other proper transcription :-) ), which says that the
      gap after prime P is at most of size P.

      Peter

      --
      [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
    • Werner D. Sand
      For example 2 adjacent gaps cannot be equal if they aren t multiple of 6. For example the gap between 2 pairs of twins is at least 4. For example each prime
      Message 40 of 40 , Feb 7, 2007
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        For example 2 adjacent gaps cannot be equal if they aren't multiple of
        6. For example the gap between 2 pairs of twins is at least 4. For
        example each prime number has the form 2n+/-1, 3n+/-1, 4n+/-1, 6n+/-1.
        Each pair of twins has the form 12n+-1, there are approximate formulas
        for the nth prime and the number of primes < x and so on. You cannot
        call all this random ore unpredictable. Of course the prime numbers are
        distributed as regularly as possible, that's a tautology. In
        mathematics everything is as regular as possible. Is pi random? Build
        P=2,357111317192329…, and you have the same case as pi. Consider the
        primes to be an irrational number, and there are no problems. If you
        mean there is no formula f(n) which produces primes for each n, then
        you are right. In this sense primes are random. (I am not quite sure –
        there is a formula p=[k^n^3] (H.W.Mills) which is said to produce only
        prime numbers). If you define "formula" as an algorithm, as a
        calculation instruction such as the sieve of Eratosthenes, then the
        primes are not random but simply what they are. Perhaps the compound
        numbers are random? Or are they only non-transparently complicated?

        Werner
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