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Re: [PrimeNumbers] Conjecture: There is always more than 3 primes between two consecutive non 0 cubes

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  • Andy Swallow
    Rathe weak conjecture, considering that the number of primes seems to be always increasing in the sequence you ve submitted. Let x=n^3, for some integer n.
    Message 1 of 4 , Jan 2, 2004
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      Rathe weak conjecture, considering that the number of primes seems to be
      always increasing in the sequence you've submitted.

      Let x=n^3, for some integer n. Then the distance between n^3 and (n+1)^3
      is of order n^2, i.e. x^(2/3). It is in fact known that the prime number
      theorem holds in such intervals (see Ivic's book on the zeta function,
      for instance). So the number of primes between n^3 and (n+1)^3 is of
      order n^2/log n, roughly.

      Similarly for any higher power, before anyone goes rushing off to play
      with their computers. So for instance, the number of primes between any
      consecutive 4th or 5th powers is as would be expected by the PNT.

      The real challenge is the problem originally posted, that of finding
      primes between consecutive squares, or equivalently, showing the
      existence of primes in intervals (x, x+x^(1/2)). The best that can be
      done is replacing the 1/2 by 21/40+e, for any small e>0.

      Andy
    • richard042@yahoo.com
      Related, Chris Caldwell sent a post on 10/18/01 which seemed to state that the version of this conjecture for a single prime is proven assuming the Riemman
      Message 2 of 4 , Jan 2, 2004
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        Related, Chris Caldwell sent a post on 10/18/01 which seemed to state
        that the version of this conjecture for a single prime is proven
        assuming the Riemman Hypothesis. I looked for his original message
        which seems to have been removed? Regardless, it's entirety is
        included in the first response from Phil

        http://groups.yahoo.com/group/primenumbers/message/3419

        If you go to the message list for Oct, 2001, you can pick up on the
        thread for more info.

        -Dick

        --- In primenumbers@yahoogroups.com, Andy Swallow <umistphd2003@y...>
        wrote:
        >
        > Rathe weak conjecture, considering that the number of primes seems
        to be
        > always increasing in the sequence you've submitted.
        >
        > Let x=n^3, for some integer n. Then the distance between n^3 and
        (n+1)^3
        > is of order n^2, i.e. x^(2/3). It is in fact known that the prime
        number
        > theorem holds in such intervals (see Ivic's book on the zeta
        function,
        > for instance). So the number of primes between n^3 and (n+1)^3 is of
        > order n^2/log n, roughly.
        >
        > Similarly for any higher power, before anyone goes rushing off to
        play
        > with their computers. So for instance, the number of primes between
        any
        > consecutive 4th or 5th powers is as would be expected by the PNT.
        >
        > The real challenge is the problem originally posted, that of finding
        > primes between consecutive squares, or equivalently, showing the
        > existence of primes in intervals (x, x+x^(1/2)). The best that can
        be
        > done is replacing the 1/2 by 21/40+e, for any small e>0.
        >
        > Andy
      • Andrew Swallow
        I guess we re coming at it from two slightly different angles. When I say the conjecture is proved, I mean for all sufficiently large cubes. There is of course
        Message 3 of 4 , Jan 2, 2004
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          I guess we're coming at it from two slightly different angles.

          When I say the conjecture is proved, I mean for all sufficiently large
          cubes. There is of course a cutoff point, which message 3419 talks
          about, below which it is not proved, but which may be too high for
          computer methods to reach.

          But the Riemann Hypothesis certainly shouldn't matter too much.

          For the record, the following is true *without* needing the Riemann
          Hypothesis:

          "Let r>=3 be an integer. There exists a prime between n^r and (n+1)^r,
          for all sufficiently large n"

          It's actually easier to prove it for the higher values of r. The proof
          is a tad long though, so I'm not going to write it out...

          It's the r=2 case that is the real problem. Of course, if you allow
          for non-integral values of r, then it's true for r>40/19.

          As for the smaller values of n, I don't know much about that. I find
          the methods involved in the 'sufficiently large' proofs more
          interesting.

          Andy
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