- 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:>

to be

> Rathe weak conjecture, considering that the number of primes seems

> always increasing in the sequence you've submitted.

(n+1)^3

>

> Let x=n^3, for some integer n. Then the distance between n^3 and

> 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

play

> order n^2/log n, roughly.

>

> Similarly for any higher power, before anyone goes rushing off to

> 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.

be

>

> 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

> 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 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