## primes between squares calcs

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• Aah! It s good to see my favorite math problem, the number of primes between squares, reappear on primenumbers, and while it is up, I would like to mention
Message 1 of 5 , Dec 8, 2010
Aah! It's good to see my favorite math problem, the number of primes between squares, reappear on primenumbers, and while it is up, I would like to mention Capelle's wondeful conjecture A, posted here on primenumbers Sept 4, 2005, message number 16999. It is an extension of Legendre's conjecture, which he uses as a basis for his expansive formula.
Capelle conjecture A is:
Pi((m+1)^n) - Pi(m^n) >= m^(n-2) for n>=2, m>=1 .
Legendre's conjecture claims there is at least one prime between the squares of consecutive integers, and is Capelle's case with n=2 and m = 1.

I have some ideas how Capelle's conjecture can be made even more powerful, and I have done several handfulls of calculations to check the viability of my approach, but I of course need to do a lot more calculations. Can anyone direct me to a database that tells me the number of primes between two positive integers, or perhaps has someone checked out Capelle's conjecture A to any significant degree? Any leads or previous work would lighten my task.

Thanks, Bill Oscarson
Message 2 of 5 , Dec 9, 2010
--- In primenumbers@yahoogroups.com, "Bill" <bill2math@...> wrote:
>
> Aah! It's good to see my favorite math problem, the number of primes between squares, reappear on primenumbers, and while it is up, I would like to mention Capelle's wondeful conjecture A, posted here on primenumbers Sept 4, 2005, message number 16999. It is an extension of Legendre's conjecture, which he uses as a basis for his expansive formula.
> Capelle conjecture A is:
> Pi((m+1)^n) - Pi(m^n) >= m^(n-2) for n>=2, m>=1 .
> Legendre's conjecture claims there is at least one prime between the squares of consecutive integers, and is Capelle's case with n=2 and m = 1.
>
> I have some ideas how Capelle's conjecture can be made even more powerful, and I have done several handfulls of calculations to check the viability of my approach, but I of course need to do a lot more calculations. Can anyone direct me to a database that tells me the number of primes between two positive integers, or perhaps has someone checked out Capelle's conjecture A to any significant degree? Any leads or previous work would lighten my task.
>
> Thanks, Bill Oscarson
>

Patrick Capelle
• OOPs! I left out a in my definition of Legendre s conjecture. Legendre s conjecture is of course for n=2 and m = 1. My apologies Bill
Message 3 of 5 , Dec 9, 2010
OOPs! I left out a ">" in my definition of Legendre's conjecture.
Legendre's conjecture is of course for n=2 and m >= 1.
My apologies Bill
• ... Intuitively Legendre s conjecture would appear to be true: the rate of growth of the gap between squares looks like it would far exceed the rate of growth
Message 4 of 5 , Dec 12, 2010
--- In primenumbers@yahoogroups.com, "Bill" <bill2math@...> wrote:
>

> Legendre's conjecture claims there is at least one prime between the squares of consecutive integers, and is Capelle's case with n=2 and m = 1.
>

Intuitively Legendre's conjecture would appear to be true: the
rate of growth of the gap between squares looks like it would
far exceed the rate of growth in the average gap between primes.
However the pattern of these gaps can be very uneven and I believe that
it HAS been proven that the gap between primes can be arbitrarily
large. This does not disprove Legendre's conjecture - probably
the only way to do that would be to find a counterexample. Good
with that project!

a.
• ... The proof is trivial. For prime p, all integers n s.t. p!+1
Message 5 of 5 , Dec 12, 2010
On Sun, 2010-12-12 at 10:13 +0000, Aldrich wrote:
> However the pattern of these gaps can be very uneven and I believe
> that
> it HAS been proven that the gap between primes can be arbitrarily
> large. This does not disprove Legendre's conjecture - probably

The proof is trivial. For prime p, all integers n s.t.
p!+1 < n <= p!+p are divisible by at least one prime <= p and p can be
taken arbitrarily large.

Paul
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