Apparently, from time to time, someone will come up with a newly discovered,
very large
prime number.

However, it has been my understanding (or misunderstanding) that, if true, RH
would
transform the PNT from a good estimate to an exact formula for generating the
primes up to
n or for coming up with the nth prime.

I'm guessing that this must not be the case, since if it were, one could simply
assume the
truth of RH (as, apparently, many mathematicians do in their mathematical
papers), amend
the PNT in line with this, and proceed to generate n primes (or any particular
nth prime).

Is something like this being done to generate all those zeros on the critical
line?

Is there any simple(?) way of describing the relationship between the zeroes and
the primes?

Thanks.

You know when you create the sieve of Eratosthenes you use all the
primes, as generators of sub-sieves, whose value is less than the
square root of n? Well what about if you assume, purely in the
abstract, you can break a rule and increase the value of n
considerably - double it, treble it, quadruple it etc. while *not*
introducing any new sub-sieves.... Is it within the realm of known
mathematics to be able to predict for this extended region, with an
accuracy similar to that established by way of the Prime Number
Theorem, how many places there are where no sieve components fall?

Cheers,
Tom

--- gulland68 <tmgulland@...> wrote:
> You know when you create the sieve of Eratosthenes you use all the
> primes, as generators of sub-sieves, whose value is less than the
> square root of n? Well what about if you assume, purely in the
> abstract, you can break a rule and increase the value of n
> considerably - double it, treble it, quadruple it etc. while *not*
> introducing any new sub-sieves.... Is it within the realm of known
> mathematics to be able to predict for this extended region, with an
> accuracy similar to that established by way of the Prime Number
> Theorem, how many places there are where no sieve components fall?

Yes. Let's call q=sqrt(n) in the above.
The numbers you're talking about are "numbers with no factors less than q" and
crop up in all kinds of contexts.

Look at the sublinear techniques for prime counting, for example.

There's nothing particularly interesting about them, primality-wise their
behaviour is as described by Dirichlet's theoreom. (in summary -
uninteresting.)

Phil

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Do the margins representing maxima and minima for the predictions as
to the number of places where no sieve components fall, closely
resemble those of the PNT (just at a different gradient)? Are the
predictions as accurate as with the PNT?
Cheers,
Tom

--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
>
> --- gulland68 <tmgulland@...> wrote:
> > You know when you create the sieve of Eratosthenes you use all the
> > primes, as generators of sub-sieves, whose value is less than the
> > square root of n? Well what about if you assume, purely in the
> > abstract, you can break a rule and increase the value of n
> > considerably - double it, treble it, quadruple it etc. while *not*
> > introducing any new sub-sieves.... Is it within the realm of known
> > mathematics to be able to predict for this extended region, with an
> > accuracy similar to that established by way of the Prime Number
> > Theorem, how many places there are where no sieve components fall?
>
> Yes. Let's call q=sqrt(n) in the above.
> The numbers you're talking about are "numbers with no factors less
than q" and
> crop up in all kinds of contexts.
>
> Look at the sublinear techniques for prime counting, for example.
>
> There's nothing particularly interesting about them, primality-wise
their
> behaviour is as described by Dirichlet's theoreom. (in summary -
> uninteresting.)
>
> Phil
>
> ()  ASCII ribbon campaign      ()    Hopeless ribbon campaign
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>
>
>
>
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> with the Yahoo! Search weather shortcut.
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>

Can you please reply in group, so that others can see the discussion.
If you're using the web interface, then select 'reply to all' in the pulldown
above the textarea.

--- gulland68 <tmgulland@...> wrote:
> --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
> > --- gulland68 <tmgulland@...> wrote:
> > > You know when you create the sieve of Eratosthenes you use all the
> > > primes, as generators of sub-sieves, whose value is less than the
> > > square root of n? Well what about if you assume, purely in the
> > > abstract, you can break a rule and increase the value of n
> > > considerably - double it, treble it, quadruple it etc. while *not*
> > > introducing any new sub-sieves.... Is it within the realm of known
> > > mathematics to be able to predict for this extended region, with an
> > > accuracy similar to that established by way of the Prime Number
> > > Theorem, how many places there are where no sieve components fall?
> >
> > Yes. Let's call q=sqrt(n) in the above.
> > The numbers you're talking about are "numbers with no factors less
> than q" and
> > crop up in all kinds of contexts.
> >
> > Look at the sublinear techniques for prime counting, for example.
> >
> > There's nothing particularly interesting about them, primality-wise
> their
> > behaviour is as described by Dirichlet's theoreom. (in summary -
> > uninteresting.)
>
> To me this *is* interesting. You are saying, are you not, that the
> reason we can make predictions as to the abundance of primes is not
> specifically dependent upon the understanding of how the number scale
> really operates, but rather we can make similar kinds of prediction
> irrespective of whether or not the new sieves are brought in in the
> manner that they truly occur. I wonder, are the bounds of the
> predictions more or less fine - i.e. is it more or less accurate -
> than those for the PNT in regard to the number scale itself?

Basically, each prime removes, when considering the whole of the integers, a
fixed proportion, 1/p, of the remaining candidates, independently of all other
primes. Dirichlet goes even further and says that all non-zero residues moduli
are indistinguishable in behaviour.

Things only get interesting when you start looking at sieves of finite ranges
and then there's an error term associated with that 1/p which probably causes
more confusion amongst cranks than any other detail in mathematics. (Or at
least cranks who think they've proved the Goldbach conjecture, which is about
half of them!)

I highly recommend getting hold of (a loan of from your local university
library) Hans Riesel's /Prime Numbers and Computer Methods for Factorisation/
('PNaCMfF' for short), which has an excellent section on the relation between
the limitting behaviour of sieving and the prime number theorem. In particular
- they disagree with each other! The details are subtle, so best not to try and
reproduce a sloppy version.

Phil

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A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet?

Fixing that:
--- gulland68 <tmgulland@...> wrote:
> > --- gulland68 <tmgulland@...> wrote:
> > Yes. Let's call q=sqrt(n) in the above.
> > The numbers you're talking about are "numbers with no factors less
> than q" and
> > crop up in all kinds of contexts.
> >
> > Look at the sublinear techniques for prime counting, for example.
> >
> > There's nothing particularly interesting about them, primality-wise their
> > behaviour is as described by Dirichlet's theoreom. (in summary -
> > uninteresting.)

> Do the margins representing maxima and minima for the predictions as
> to the number of places where no sieve components fall, closely
> resemble those of the PNT (just at a different gradient)? Are the
> predictions as accurate as with the PNT?

Nope, in sieving the error term explodes (relatively). That's why kooks make
pronouncements of proofs of Goldbach, as they've completely forgotten the error
terms which swamp everything they're dealing with.

For more about extremal behaviour, see 'prime gaps' and the two
Hardy-Littlewood conjectures. All documented on http://primepages.org/ .

Phil

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If one could prove that the residue of mod(P) over all larger primes
was equally likely to be 1,2,3...p(n-1), would that in any way prove or
be equivalent to the Riemann hypothesis?  This equal likelihood seems
apparent to me but does it prove anything?

In other words, if, over all primes, the residue obtained after
division by, say, 5 was equally likely to be 1,2,3 or 4 or stayed
within a certain sized error term such as with coin flips staying
within the square root of the number of flips, would that say something
useful about the even-ness of the distribution of the primes?

Regards,

Wes

--- Wes <6bullocks@...> wrote:
> If one could prove that the residue of mod(P) over all larger primes
> was equally likely to be 1,2,3...p(n-1), would that in any way prove or
> be equivalent to the Riemann hypothesis?  This equal likelihood seems
> apparent to me but does it prove anything?

No that's just Dirichlet's theorem. Nothing new there at all.

Phil

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On 4/4/07, Phil Carmody <thefatphil@...> wrote:
> --- Wes <6bullocks@...> wrote:
> > If one could prove that the residue of mod(P) over all larger primes
> > was equally likely to be 1,2,3...p(n-1), would that in any way prove or
> > be equivalent to the Riemann hypothesis?  This equal likelihood seems
> > apparent to me but does it prove anything?
>
> No that's just Dirichlet's theorem. Nothing new there at all.
>

Really?  I thought Dirichlet's theorem just said there were infinitely
many of each, not that they were equally likely.
--Joshua Zucker

--- Joshua Zucker <joshua.zucker@...> wrote:
> On 4/4/07, Phil Carmody <thefatphil@...> wrote:
> > --- Wes <6bullocks@...> wrote:
> > > If one could prove that the residue of mod(P) over all larger primes
> > > was equally likely to be 1,2,3...p(n-1), would that in any way prove or
> > > be equivalent to the Riemann hypothesis?  This equal likelihood seems
> > > apparent to me but does it prove anything?
> >
> > No that's just Dirichlet's theorem. Nothing new there at all.
>
> Really?  I thought Dirichlet's theorem just said there were infinitely
> many of each, not that they were equally likely.

Yes, Dirichlet's original formulation says just that. I can't find
out when it was strengthened to include the density statement, but
anyway, the density statement does hold true. As the proof of the
density statement depends on the very tools Dirichlet himself created
I wouldn't be surprised if he revisited it himself. None of the usual
sources (Primepages, Mathworld, PlanetMath, PNaCP) provide a reference
for the stronger statement. I'll go off and do some research...

Phil

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--- Joshua Zucker <joshua.zucker@...> wrote:
> On 4/4/07, Phil Carmody <thefatphil@...> wrote:
> > --- Wes <6bullocks@...> wrote:
> > > If one could prove that the residue of mod(P) over all larger primes
> > > was equally likely to be 1,2,3...p(n-1), would that in any way prove or
> > > be equivalent to the Riemann hypothesis?  This equal likelihood seems
> > > apparent to me but does it prove anything?
> >
> > No that's just Dirichlet's theorem. Nothing new there at all.
>
> Really?  I thought Dirichlet's theorem just said there were infinitely
> many of each, not that they were equally likely.

This is an interesting hunt. It _appears_ that Dirichlet's theorem does prove
that /if/ the primes along the AP have a natural density, then that natural
density must be 1/phi(n). However, it doesn't actaully prove that the natural
density exists.

More research required, but I'm almost entirely out of resources to refer to
now. Does anyone have a Hardy&Littlewood or a Bach&Shallit, to see if either of
those make mention of the density clause?

Phil

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--- Phil Carmody <thefatphil@...> wrote:
> --- Joshua Zucker <joshua.zucker@...> wrote:
> > On 4/4/07, Phil Carmody <thefatphil@...> wrote:
> > > --- Wes <6bullocks@...> wrote:
> > > > If one could prove that the residue of mod(P) over all larger primes
> > > > was equally likely to be 1,2,3...p(n-1), would that in any way prove or
> > > > be equivalent to the Riemann hypothesis?  This equal likelihood seems
> > > > apparent to me but does it prove anything?
> > >
> > > No that's just Dirichlet's theorem. Nothing new there at all.
> >
> > Really?  I thought Dirichlet's theorem just said there were infinitely
> > many of each, not that they were equally likely.
>
> This is an interesting hunt. It _appears_ that Dirichlet's theorem does prove
> that /if/ the primes along the AP have a natural density, then that natural
> density must be 1/phi(n). However, it doesn't actaully prove that the natural
> density exists.
>
> More research required, but I'm almost entirely out of resources to refer to
> now. Does anyone have a Hardy&Littlewood or a Bach&Shallit, to see if either
> of
> those make mention of the density clause?

On sci.math, Erick Bryce Wong said:

> I believe Dirichlet proved the 1/phi(n) term but for analytic density (also
> known as Dirichlet density).  This statement is strictly weaker than natural
> density: a sequence of primes with natural relative density necessarily has
> Dirichlet density, but the converse does not hold.

Which I think is strong enough support for my original claim.
However, I'm not familiar with what analytic density is, so shall have to do a
bit more background reading. It looks like this might be a place to start:
http://www.math.ubc.ca/~hoek/Teaching/Pnthm.pdf
It's got all the right words in it, the only question is whether they make
sense to me!

Phil

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Hello all:

   I have obtained a limit that related The Riemann Zeta function Zeta(i)
   the prime gaps pi-p(i-1)  where i is integer and the nth prime number pn.

   Limit pn/Sum[Zeta(i)*(pi-p(i-1)),{i,2,n}]=1

   Even more :

   Limit  pn-Sum[Zeta(i)*(pi-p(i-1)),{i,2,n}]=bk

   Where bk is a new mathematical constant   bk=0.544991092223.....

   Questions:

   1) Does someone can prove the firs limit?
   2) Does someone can prove that the second limit converges?
   3) Is bk related to other constants?

   Sincerely

   Sebastián Martín Ruiz
   Avda. de Regla 43 Chipiona 11550 Spain
   phone: 686746131


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[Non-text portions of this message have been removed]

Wow, I get two copies - I'm so lucky!

--- Sebastian Martin <sebi_sebi@...> wrote:
> Hello all:
>
>   I have obtained a limit that related The Riemann Zeta function Zeta(i)
>   the prime gaps pi-p(i-1)  where i is integer and the nth prime number pn.
>
>   Limit pn/Sum[Zeta(i)*(pi-p(i-1)),{i,2,n}]=1

>   1) Does someone can prove the firs limit?

Yes, it's trivial and uninteresting.

>   2) Does someone can prove that the second limit converges?

Yes, it's trivial and uninteresting.

Exactly the same type of relation holds true for any sequence that converges to
1, not just {Zeta(i)}. There are an infinitude of these, your relation doesn't
connect Zeta to the primes in any way.

Take the fine structure constant alpha ~1/137, and replace Zeta(i) by
alpha/CF_i(alpha) where CF_i is the i-th continued fraction convergent.

Congratulations, your expression connects prime gaps to the fine structure
constant.

Not!

Phil

Phil

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> >
> >   Limit pn/Sum[Zeta(i)*(pi-p(i-1)),{i,2,n}]=1
>
> >   1) Does someone can prove the firs limit?
>
> Yes, it's trivial and uninteresting.
>


Speaking of converging to one, I have wondered if there is an exponent
k such that

(1/2)^k + (1/3)^k + (1/5)^k + (1/7)^k + ... (1/p)^k + ...

has a limit  of one.  I just tried  k = sqrt(2), and the first million
primes yields a sum of about .97486

For comparison, I just tried the same thing on the natural numbers
starting at two.

(1/2)^k + (1/3)^k + (1/4)^k + (1/5)^k + (1/6)^k + ...

If k = sqrt(3), the first million terms sums to about .99378


Just some reversion, no need for comment. A reflective Easter, everyone.

Mark



.



> >   2) Does someone can prove that the second limit converges?
>
> Yes, it's trivial and uninteresting.
>
> Exactly the same type of relation holds true for any sequence that
converges to
> 1, not just {Zeta(i)}. There are an infinitude of these, your
relation doesn't
> connect Zeta to the primes in any way.
>
> Take the fine structure constant alpha ~1/137, and replace Zeta(i) by
> alpha/CF_i(alpha) where CF_i is the i-th continued fraction convergent.
>
> Congratulations, your expression connects prime gaps to the fine
structure
> constant.
>
> Not!
>
> Phil
>
> Phil
>
> ()  ASCII ribbon campaign      ()    Hopeless ribbon campaign
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>
>
>
>
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> Download and play hundreds of games for free on Yahoo! Games.
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>

> I have wondered if there is an exponent
> k such that
>
> (1/2)^k + (1/3)^k + (1/5)^k + (1/7)^k + ... (1/p)^k + ...
>
> has a limit  of one.

1/2^k + 1/3^k + ... + 1/p^k + ...

just equals to

Sum[MoebiusMu[n]*Log[Zeta[n*k]]/n, {n, 1, +Infinity}]

which converges rapidly; so it can be used to estimate the waned value of k as
1.3994333287263303...

Best,

Andrey


[Non-text portions of this message have been removed]

> 1.3994333287263303...

1.399433328726330318202807214745644327904727429484383941274765822888062492487247\
800233390475384227...

(if you need more digits)

Best,

Andrey

[Non-text portions of this message have been removed]

As we know, the number of primes up to n is about n/log(n). Given
this, it is easy to show that the number of primes between n^2 and
(n+1)^2 is also about n/log(n).

How much does the actual count of primes between n^2 and (n+1)^2
differ from n/log(n) ? On a very cursory inspection, it seems the
prime count is no more than sqrt(n) removed from n/log(n).

Mark

1 * 1 = 7 * 13 = 11 * 11 = 17 * 23 = 19 * 19 = 29 * 29 = 1 mod 30

For the integer k,
If for every value of n,  none of

m1 = (k-n)/(30*n+1)
m2 = ( k - 7 * n - 3) / (30 * n + 13)
m3 = (k - 11 * n - 4)/(30 * n + 11)
m4 = (k - 13 * n - 3)/(30 * n + 7)
m5 = (k - 17 * n - 13) / (30 * n + 23)
m6 = (k - 19 * n - 12) / (30 * n + 19)
m7 = (k - 23 * n - 13) / (30 * n + 17)
m8 = (k - 29 * n - 28) / (30 * n + 29)

are an integer,
then
(30 * k + 1) is a prime.


Derivation:


(30 m + 1) * (30 n + 1) = (30 k + 1)
30 * 30 * m * n + 30 n + 30 m + 1 = 30 k + 1

k = 30 * m * n + m + n

30 * m * n + m = k - n

m = (k - n) / (30 * n + 1)

Maximum n is found by:

n = (k - n) / (30 * n + 1)

30 * n^2 + n = k - n

30 * n^2 + 2 n - k = 0

max n = - 1 + sqrt(1 + 30 k)

(30 m + 7) * (30 n + 13) = (30 k + 1)

30 * 30 * m * n + 7 * 30 n + 13 * 30 m + 91 = 30 k + 1

k = 30 * m * n + 7 * n + 13 * m + 3

30 * m * n + 13 * m = k - 7 * n - 3

m = ( k - 7 * n - 3) / (30 * n + 13)

max n is found by

n * (30 * n + 13)  = (k - 7 * n - 3)

30 * n^2 + 13 n = k - 7 * n - 3

30 * n^2 + 20 n - k + 3 = 0

max n = -10 + sqrt(100 + 30 * (k-3) )



etc

Hi,

Mike Oakes said:
2^11795+11795 137-PRP by PFGW (PRIMO certification would take c. 5
days)


My computer says:
[PRIMO - Task Report]
Version=2.3.1
WebSite=http://www.ellipsa.net/
Task=Certification
ID=B2E63036D59E4
Created=04-05-2007 03:58:18 pm

[Common]
Path=C:\Christ\Primo 2.3.1\
Selected=1
Processed=1
Certified=1
Candidate #1=Certified, 94h 3mn 48s

[Candidate #1]
Input=primo-B2E5901A2ECA1-001.tmp
Report=primo-B2E5901A2ECA1-001.cr
Output=primo-B2E5901A2ECA1-001.out
Status=Candidate certified prime


So, both numbers have been proven prime.

I'll run Cert_val on the second certificate and publish both after that.

Christ van Willegen

--- In primenumbers@yahoogroups.com, "Mark Underwood"
<mark.underwood@...> wrote:
>
>
> As we know, the number of primes up to n is about n/log(n). Given
> this, it is easy to show that the number of primes between n^2 and
> (n+1)^2 is also about n/log(n).
>
> How much does the actual count of primes between n^2 and (n+1)^2
> differ from n/log(n) ? On a very cursory inspection, it seems the
> prime count is no more than sqrt(n) removed from n/log(n).
>
> Mark
>


Just did some prime counting and so far it holds that
the number of primes between n^2 and (n+1)^2 is within the range
n/log(n) +/- sqrt(n). At least for n up to about 47,000.

There have been some close calls though. Here are the cases where the
difference between the actual prime count and n/log(n) was at least 75
percent of the square root of n.

Format: (n, primecount between n^2 and (n+1)^2, percentage)

(696,81)
(696,85,81)
(1760,204,75)
(2456,268,94)
(2761,390,79)
(3516,486,93)
(3788,508,78)
(5266,675,83)
(9980,1168,84)
(10706,1250,93)
(15646,1718,78)
(23515,2458,79)
(23924,2503,84)
(28678,2923,76)
(28678,2923,76)
(32460,2986,77)
(39590,3904,83)

Mark

http://primes.utm.edu/howmany.shtml#table

http://numbers.computation.free.fr/Constants/Primes/Pix/pixproject.html

http://www.ieeta.pt/~tos/primes.html

http://www.ieeta.pt/~tos/primes/1d22.txt.gz

     Congrats to Prof. Tomas Oliveira e Silva!

     Best,

     Andrey

[Non-text portions of this message have been removed]

This is a very surprising result.  It is over 14 million lower than
the likely value that Gourdon found.  Has it been checked by
computing an offset value and sieving the difference?

Best regards,

David Baugh

--- In primenumbers@yahoogroups.com, "Andrey Kulsha" <Andrey_601@...>
wrote:
>
> http://primes.utm.edu/howmany.shtml#table
>
>
http://numbers.computation.free.fr/Constants/Primes/Pix/pixproject.htm
l
>
> http://www.ieeta.pt/~tos/primes.html
>
> http://www.ieeta.pt/~tos/primes/1d22.txt.gz
>
>     Congrats to Prof. Tomas Oliveira e Silva!
>
>     Best,
>
>     Andrey
>
> [Non-text portions of this message have been removed]
>

Proof:

pi[(n+1)^2]-pi(n^2) ~ (PNT)
[(n+1)^2]/ln[(n+1)^2] - (n^2)/ln(n^2) =
[(n+1)^2]/2ln(n+1) - (n^2)/(2ln n) =
[(n+1)^2]/2[ln n + ln(1+1/n)] - (n^2)/(2ln n) -> (n->inf)
[(n+1)^2]/(2ln n) - (n^2)/(2ln n) =
(2n+1)/(2ln n) =
n/ln n + 1/(2ln n) -> (n->inf)
n/ln n ~
pi(n)

qed

Werner


--- In primenumbers@yahoogroups.com, "Mark Underwood"
<mark.underwood@...> wrote:
>
> --- In primenumbers@yahoogroups.com, "Mark Underwood"
> <mark.underwood@> wrote:
> >
> >
> > As we know, the number of primes up to n is about n/log(n). Given
> > this, it is easy to show that the number of primes between n^2 and
> > (n+1)^2 is also about n/log(n).
> >
> > How much does the actual count of primes between n^2 and (n+1)^2
> > differ from n/log(n) ? On a very cursory inspection, it seems the
> > prime count is no more than sqrt(n) removed from n/log(n).
> >
> > Mark
> >
>
>
> Just did some prime counting and so far it holds that
> the number of primes between n^2 and (n+1)^2 is within the range
> n/log(n) +/- sqrt(n). At least for n up to about 47,000.
>
> There have been some close calls though. Here are the cases where
the
> difference between the actual prime count and n/log(n) was at least
75
> percent of the square root of n.
>
> Format: (n, primecount between n^2 and (n+1)^2, percentage)
>
> (696,81)
> (696,85,81)
> (1760,204,75)
> (2456,268,94)
> (2761,390,79)
> (3516,486,93)
> (3788,508,78)
> (5266,675,83)
> (9980,1168,84)
> (10706,1250,93)
> (15646,1718,78)
> (23515,2458,79)
> (23924,2503,84)
> (28678,2923,76)
> (28678,2923,76)
> (32460,2986,77)
> (39590,3904,83)
>
> Mark
>

In the same way one proves that the number of primes between
cubes amounts to approximately:

N = pi(n+1)^3-pi(n^3) ~
(n/ln(n))*(n+1) ~
pi(n)*(n+1),
better ~ n*pi(n).

WDS

--- In primenumbers@yahoogroups.com, "Werner D. Sand"
<Theo.3.1415@...> wrote:
>
> Proof:
>
> pi[(n+1)^2]-pi(n^2) ~ (PNT)
> [(n+1)^2]/ln[(n+1)^2] - (n^2)/ln(n^2) =
> [(n+1)^2]/2ln(n+1) - (n^2)/(2ln n) =
> [(n+1)^2]/2[ln n + ln(1+1/n)] - (n^2)/(2ln n) -> (n->inf)
> [(n+1)^2]/(2ln n) - (n^2)/(2ln n) =
> (2n+1)/(2ln n) =
> n/ln n + 1/(2ln n) -> (n->inf)
> n/ln n ~
> pi(n)
>
> qed
>
> Werner
>
>
> --- In primenumbers@yahoogroups.com, "Mark Underwood"
> <mark.underwood@> wrote:
> >
> > --- In primenumbers@yahoogroups.com, "Mark Underwood"
> > <mark.underwood@> wrote:
> > >
> > >
> > > As we know, the number of primes up to n is about n/log(n).
Given
> > > this, it is easy to show that the number of primes between n^2
and
> > > (n+1)^2 is also about n/log(n).
> > >
> > > How much does the actual count of primes between n^2 and (n+1)^2
> > > differ from n/log(n) ? On a very cursory inspection, it seems
the
> > > prime count is no more than sqrt(n) removed from n/log(n).
> > >
> > > Mark
> > >
> >
> >
> > Just did some prime counting and so far it holds that
> > the number of primes between n^2 and (n+1)^2 is within the range
> > n/log(n) +/- sqrt(n). At least for n up to about 47,000.
> >
> > There have been some close calls though. Here are the cases where
> the
> > difference between the actual prime count and n/log(n) was at
least
> 75
> > percent of the square root of n.
> >
> > Format: (n, primecount between n^2 and (n+1)^2, percentage)
> >
> > (696,81)
> > (696,85,81)
> > (1760,204,75)
> > (2456,268,94)
> > (2761,390,79)
> > (3516,486,93)
> > (3788,508,78)
> > (5266,675,83)
> > (9980,1168,84)
> > (10706,1250,93)
> > (15646,1718,78)
> > (23515,2458,79)
> > (23924,2503,84)
> > (28678,2923,76)
> > (28678,2923,76)
> > (32460,2986,77)
> > (39590,3904,83)
> >
> > Mark
> >
>

Hi ,

> --- Phil Carmody <thefatphil@...> wrote:
> > --- Joshua Zucker <joshua.zucker@...> wrote:
> > > On 4/4/07, Phil Carmody <thefatphil@...> wrote:
> > > > --- Wes <6bullocks@...> wrote:
> > > > > If one could prove that the residue of mod(P) over all larger
primes
> > > > > was equally likely to be 1,2,3...p(n-1), would that in any way
prove or
> > > > > be equivalent to the Riemann hypothesis?  This equal
likelihood seems
> > > > > apparent to me but does it prove anything?
> > > >
> > > > No that's just Dirichlet's theorem. Nothing new there at all.
> > >
> > > Really?  I thought Dirichlet's theorem just said there were infinitely
> > > many of each, not that they were equally likely.
> >
> > This is an interesting hunt. It _appears_ that Dirichlet's theorem
does prove
> > that /if/ the primes along the AP have a natural density, then that
natural
> > density must be 1/phi(n). However, it doesn't actaully prove that
the natural
> > density exists.

Phil , I apologize for being such a bad correspondent , I owe you so
nuch correspondence , but I recall back in June mentioning that some
related literature is appearing under the rubric "prime race" .
Another term is "Chebyshev Bias" .

HTH .

Cheers ,

Walter

Why can't prime numbers be analyzed in terms of a sequential prime generating
algorithm?

Due to some very basic playing with the numbers, I can immediately see that you
can create a sum that outputs primes in sequence.

Is it simply a matter of big primes and the exponential use of processing power
required to calculate them past a certain point, or is there something more
fundamental preventing primes from being described?

I see a very simple, definite pattern that describes the distribution of prime
numbers. Where is the mystery? It would seem that factorization of large
numbers, and discovering the fundamentals behind quirks like the rule of 9s and
so on would be more practical in cracking primes as a whole, rather than
focussing on prime sequences.

I probably just don't know enough, though... can someone perhaps explain even a
simple aspect of primes that isn't explained by simple arithmetic?

I don't mean to sound cocky, either, I'm just honestly confused by the focus on
prime numbers, as opposed to more general work with numbers.

Thanks very much for any responses :)


---------------------------------
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[Non-text portions of this message have been removed]

> I don't mean to sound cocky, either, I'm just honestly
> confused by the focus on prime numbers, as opposed to more
> general work with numbers.

Most of the properties of integers in general depend on their prime
factorizations, so the study of primes is fundamental to the
study of numbers in general.  To put it another way, to
understand how a building will respond to stress, you need to
know what it is made of.  The multiplicative building blocks
of the integers are the primes.

When you take your Sieve of Eratosthenes and consider values of n well
beyond the square of the highest prime that you have selected to give
a filter, does anybody know what sort value of n it would be when,
among all those places - I guess you could call them component
coincidences - where more than one filter's component falls, the mean
number of filter components will be four?(That's obviously counting
all powers of p as being represented by one sole sieve component i.e.
there's no new filter made especially for squares of 3, for cubes of
3, for squares of 5 etc.)

Cheers,

Tom

This amused me:

http://www.xkcd.com/c247.html

Of course the pedant in me makes me think that the representation is a mixed
radix one, and that 2:53 is in fact (2*6+5)*10+3. This leaves you with smaller
numbers to factor, at the cost of performing the mixed-radix evaluation.

Phil


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