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## Prime between n^2 and (n+1)^2

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• 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,
Message 1 of 5 , Feb 28, 2005
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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 tighter
result, namely that for n>1 there is always a prime between

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 groups!)

******************

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 c 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
• ... No, it hasn t. ... If this result is proved, then the above follows immediately. So it hasn t been proven either. ... By the PNT, the n-th prime is ~n log
Message 2 of 5 , Feb 28, 2005
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On Monday 28 February 2005 17:29, you 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.

No, it hasn't.

>
> Now n*(n+1) lies between n^2 and (n+1)^2, and we observe a tighter
> result, namely that for n>1 there is always a prime between
>
> 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 groups!)

If this result is proved, then the above follows immediately. So it hasn't
been proven either.

>
> ******************
>
>
> 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 c 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.

By the PNT, the n-th prime is ~n log n, so it follows that all c's are
acceptable at infinity, since any power > 1 of n grows faster than n log n.
Whatever value of c that ends up being the smallest possible is just a quirk
of the start of the sequence. If you disregard the `first' primes, then any c
= 1 + e will do (for a suitable definition of `first' that's a function of
e).

> 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.

See the PNT argument above.

Décio

[Non-text portions of this message have been removed]
• 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:
Message 3 of 5 , Mar 2, 2005
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"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 MathWorld-
-A Wolfram Web Resource.
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 by
Erdos's and by Bertrand's.

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

If m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 < 2*n
or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.

If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and 2*n
then Bertrand's Postulate holds.

If m < (m+1)^2 < n < 2*n or n < 2*n < m < (m+1)^2, then we have bad
unrelated sets in this examination, But a prime p1 < m^2 or (m+1)^2
< 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 give
you insight.

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 tighter
> result, namely that for n>1 there is always a prime between
>
> 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 groups!)
>
>
> ******************
>
>
> 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 c
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
• The Erdos proof is available at Ask Dr. Math - not that I understand it! Bob nick_honolson wrote: Erdos proved that there exist at
Message 4 of 5 , Mar 2, 2005
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The Erdos proof is available at Ask Dr. Math - not that I understand it!

Bob

nick_honolson <nick_honolson@...> wrote:

"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 MathWorld-
-A Wolfram Web Resource.
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 by
Erdos's and by Bertrand's.

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

If m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 < 2*n
or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.

If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and 2*n
then Bertrand's Postulate holds.

If m < (m+1)^2 < n < 2*n or n < 2*n < m < (m+1)^2, then we have bad
unrelated sets in this examination, But a prime p1 < m^2 or (m+1)^2
< 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 give
you insight.

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 tighter
> result, namely that for n>1 there is always a prime between
>
> 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 groups!)
>
>
> ******************
>
>
> 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 c
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

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• Thank you Nick, that was helpful! Erdos figured prominently as an amazing networker in the book I recently read called The Music of the Primes. And that
Message 5 of 5 , Mar 2, 2005
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Thank you Nick, that was helpful! Erdos figured prominently as an
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:
>
> "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
MathWorld-
> -A Wolfram Web Resource.
> 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
by
> Erdos's and by Bertrand's.
>
> 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 <
2*n
> or p1 < m^2 < p2 < 2*n then a prime exist inside the squares.
>
> If m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 <
2*n
> or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.
>
> If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and
2*n
> then Bertrand's Postulate holds.
>
> 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
> < 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
give
> you insight.
>
> 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
tighter
> > result, namely that for n>1 there is always a prime between
> >
> > 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
groups!)
> >
> >
> > ******************
> >
> >
> > 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
c
> 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
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