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

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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
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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