amazing 'networker' in the book I recently read called The Music of

the Primes. And that Wolfram site is amazing.

Mike, I see now that even lesser minds can think alike to greater

minds! One wonders how much of this stuff has been hashed through

even hundreds of years go. Oh well the joy is in the (re)discovery.

You are waxing philosophical when you say

"There is no special "reason" preventing more primes from turning up -

it's just the way the cookie crumbles sometimes."

Perhaps, yet maybe *God* knows the 'reason' why cookies crumble

exactly the way they do :-)

Mark

--- In primenumbers@yahoogroups.com, "nick_honolson"

<nick_honolson@y...> wrote:>

MathWorld-

> "Erdos proved that there exist at least one prime of the form 4k+1

> and at least one prime of the form 4k+3 between n and 2n for all n

>

> 6."

> This was stated at:

> Eric W. Weisstein. "Modular Prime Counting Function." From

> -A Wolfram Web Resource.

by

> http://mathworld.wolfram.com/ModularPrimeCountingFunction.html

>

> Does anyone know where is this proof is?

> Second, how does this NOT prove Mark's question?

>

> I mean when can with n > 6 and m^2 >6,

> n < p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 < p2 < 2*n,

> happen without a prime inside p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 <

> p2?

>

> Note n > 6 and m^2 >6:

>

> If m^2 < n < 2*n <(m+1)^2 then a primes exist inside the squares

> Erdos's and by Bertrand's.

2*n

>

> If n < m^2 < 2*n < (m+1)^2 and both Erdos primes are n < p1 < p2 <

> m^2 then a prime exist inside the squares by at least Bertrand's

> Postulate for prime p3.

>

> If n < m^2 < 2*n < (m+1)^2 and a Erdos prime is m^2 < p1 < p2 <

> or p1 < m^2 < p2 < 2*n then a prime exist inside the squares.

2*n

>

> If m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 <

> or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.

2*n

>

> If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and

> then Bertrand's Postulate holds.

bad

>

> If m < (m+1)^2 < n < 2*n or n < 2*n < m < (m+1)^2, then we have

> unrelated sets in this examination, But a prime p1 < m^2 or (m+1)^2

give

> < p2 will be near enough to m^2 or (m+1)^2 so that p3 can be a

> solution by Bertrand's (or by Bertrand's used multiple times).

>

> Note that n*(n+1) can be subtituded in the logic above because of

> the two nearby primes.

>

>

> As for the also part, Mark see the these pages:

>

> http://mathworld.wolfram.com/AndricasConjecture.html

> http://mathworld.wolfram.com/OmegaConstant.html

>

> Look at the generalization of Andrica's conjecture considers the

> equation

>

> p_(n+1)^x - p_(n)^x = 1

>

> I know that the sign has been changed for your case, but it may

> you insight.

tighter

>

> Nick

>

>

> --- In primenumbers@yahoogroups.com, "Mark Underwood"

> <mark.underwood@s...> wrote:

> >

> >

> > It's true that there is at least one prime between n^2 and (n+1)

> ^2,

> > although I can't recall if it has been proven.

> >

> >

> > Now n*(n+1) lies between n^2 and (n+1)^2, and we observe a

> > result, namely that for n>1 there is always a prime between

groups!)

> >

> > n^2 and n*(n+1)

> >

> > and also between

> >

> > n*(n+1) and (n+1)^2

> >

> > (This is yet another way to divide all odd primes into two

> >

c

> >

> > ******************

> >

> >

> > Also, I was toying with a minimum value of c such that there is

> > always a prime between n^c and (n+1)^c.

> >

> > I'm not totally certain but pretty sure that the lowest value of

> is

> > about 1.5683. Some lower values come very close but no cigar.

> >

> >

> > For example, there is one prime between 12^1.5683 and 13^1.5683.

> That

> > is, 53 lies between 49.258 and 55.846.

> >

> >

> > Let s(n) be the number of primes between n^1.5683 and (n+1)

> ^1.5683.

> >

> >

> > s(n) for n=1 to n=14 is 1,2,1,1,1,2,1,2,1,1,2,1,2,1.

> >

> > s(n) = 1 for 18 values of n from 1 to 100.

> >

> > s(n) = 1 for 33 values of n from 1 to 737. After that s(n) is

> always

> > greater than 1.

> >

> >

> > Mark