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

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  • 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 1 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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