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

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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.
Message 1 of 4 , Jan 2, 2004
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
• 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
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

If you go to the message list for Oct, 2001, you can pick up on the

-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
• 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
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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