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Re: Equation for consecutive prime numbers

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  • djbroadhurst
    ... Yes, it seems to a very weak claim that the equation is satisfied for every pair of consecutive primes. Yet there is no proof known to humankind! We would
    Message 1 of 4 , Jan 19, 2011
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      --- In primenumbers@yahoogroups.com,
      Maximilian Hasler <maximilian.hasler@...> wrote:

      > Letting p=q+d ...
      > This is true whenever d^2/2 < p+q = 2q+d,
      > for d=1 or any even value of d.
      > No other condition on parity (let alone primality)
      > of p,q seems required.

      Yes, it seems to a very weak claim that the equation
      is satisfied for every pair of consecutive primes.

      Yet there is no proof known to humankind!

      We would like the gap d = p - q to be O(p^theta),
      with theta < 1/2. That is not proven.
      Not by Cramer, not by Hoheisel, not by
      Montgomery, not by Heath-Brown.

      Please see pp 253-254 of Ribenboim's book.

      Even the assumption of the Riemann hypothesis
      is not enough to prove the author's claim,
      since that leads only to d = O(log(p)*sqrt(p)),
      as far as any human knows.

      Of course we strongly believe, like Cramer, that
      d = O(log(p)^2). But belief ain't proof.

      David
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