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• [Andy Swallow] This shows that any arithmetic progression contains infinitely many prime numbers. [Matt Insall] This is a strange statement. When I first saw
Message 1 of 5 , Jul 24, 2003
[Andy Swallow]
This
shows that any arithmetic progression contains infinitely many prime
numbers.

[Matt Insall]
This is a strange statement. When I first saw it, I did not believe it,
so I looked up ``arithmetic progression''. On MathWorld, I found

http://mathworld.wolfram.com/PrimeArithmeticProgression.html

in which the term being defined is not ``arithmetic progression'' but
``prime arithmetic progression''. In this definition, a ``prime arithmetic
progression'' is an arithmetic sequence of primes. Thus I looked up
``arithmetic sequence'', to see if what I recalled was correct. The
url for the MathWorld definition is

http://mathworld.wolfram.com/ArithmeticSequence.html

Now, correct me if I am wrong, but don't the even numbers form an arithmetic
sequence? If one calls arithmetic sequences by the alternate name
``arithmetic
progressions'', then your statement, Andy, seems incorrect, because the
sequence
of even numbers has only one prime in it. On the other hand, if the
definition
of ``arithmetic progression'' requires that the numbers in the
``progression''
be primes, then your statement seems a bit odd also. For there are known
finite
prime arithmetic progressions (see again the entry on MathWorld). That is,
these are prime arithmetic progressions with only finitely many primes in
them.
Thus, your statement seems to need revision, but I do not know what revision
does the trick. Clearly some arithmetic progressions have very few primes
in them
(e.g. the sequence of even numbers), and the prime arithmetic progressions
that are known are typically quite short.
• Ah, but there is another condition : {n, n+m, n+2m, etc} contains infinitely many primes IF gcd(n,m)=1. The even numbers do not follow this pattern, as
Message 2 of 5 , Jul 24, 2003
Ah, but there is another condition :

{n, n+m, n+2m, etc} contains infinitely many primes IF gcd(n,m)=1.

The even numbers do not follow this pattern, as gcd(2,2)=2 <>1.

Look up Dirichlet's theorem on primes in arithmetic progressions, say here:
http://primes.utm.edu/glossary/page.php?prev=divides

Regards,

Paul.

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• Oh, and one more thing - there is a difference between arithmetic progressions containing ONLY primes, of which the longest known has 22 members; and
Message 3 of 5 , Jul 24, 2003
Oh, and one more thing - there is a difference between arithmetic progressions
containing ONLY primes, of which the longest known has 22 members; and
*infinite* arithmetic progressions that contain an infinite number of primes
(and an infinite number of composites).

> -----Original Message-----
> From:
> Sent: 24 July 2003 14:47
> Subject: RE: [PrimeNumbers] Arithmetic Progression?
>
>
> Ah, but there is another condition :
>
> {n, n+m, n+2m, etc} contains infinitely many primes IF gcd(n,m)=1.
>
> The even numbers do not follow this pattern, as gcd(2,2)=2 <>1.
>
> Look up Dirichlet's theorem on primes in arithmetic
> progressions, say here:
> http://primes.utm.edu/glossary/page.php?prev=divides
>
>
> Regards,
>
> Paul.
>

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• Really, dear and highly respectful NP gurus, you may confuse each other as much as you wish, but think about poor laypeople... Please anybody explain in plain
Message 4 of 5 , Jul 24, 2003
Really,
dear and highly respectful NP gurus,
you may confuse each other
as much as you wish,

Please anybody explain in plain language,
what a great truth is in the assertion
that in a*m+b with gcd(a,b)=1
there are infinitely more primes;
(I guess that squares -
which are infinetly more rarer than primes -
are still infinetly many in AP -
or not?)
is it OK, that still primes are infinitely less than
composite terms of AP.
I mean is there any AP with primes which are not
infinitely small part of all terms?
Sorry for silly Qs,
Zak
I'm in no case aiming to be sarcastic
believe or not

Are you sure you want to send this message?
- sure not

<Paul.Jobling@W...> wrote:
> Oh, and one more thing - there is a difference between arithmetic
progressions
> containing ONLY primes, of which the longest known has 22 members;
and
> *infinite* arithmetic progressions that contain an infinite number
of primes
> (and an infinite number of composites).
>
> > -----Original Message-----
> > From:
> > Sent: 24 July 2003 14:47
> > To: 'Matt Insall'; 'primenumbers@yahoogroups.com'
> > Subject: RE: [PrimeNumbers] Arithmetic Progression?
> >
> >
> > Ah, but there is another condition :
> >
> > {n, n+m, n+2m, etc} contains infinitely many primes IF gcd(n,m)=1.
> >
> > The even numbers do not follow this pattern, as gcd(2,2)=2 <>1.
> >
> > Look up Dirichlet's theorem on primes in arithmetic
> > progressions, say here:
> > http://primes.utm.edu/glossary/page.php?prev=divides
> >
> > (Phew - did Google index your pages backwards, Chris??? prev=???)
> >
> > Regards,
> >
> > Paul.
> >
>
>
> __________________________________________________
> Virus checked by MessageLabs Virus Control Centre.
• Umm well sorry Matt, I missed out the important bit! But, prime arithmetic progression? Arithmetic sequence? The most commonly used term is arithmetic
Message 5 of 5 , Jul 24, 2003
Umm well sorry Matt, I missed out the important bit! But, prime
arithmetic progression? Arithmetic sequence? The most commonly used
term is 'arithmetic progression', whatever MathWorld may say. And this
usually refers to the infinite sequence, not a finite one.

Anyway, let a and q be any two integers, which we may assume are
positive. Then the set {qn+a} where n runs from 0 to infinity is
called an 'arithmetic progression', and can also be called an
arithmetic sequence if you so wish. Define pi(x,a,q) as the number of
primes p such that p<=x and p=a mod q. This function is clearly only
of interest when q and a have no common factor, so we'll assume that
(q,a)=1. Then we have the prime number theorem for arithmetic
progressions:

pi(x,a,q) ~ x/(phi(q)log x)

where \$A\$ may be taken as large as we like, for sufficiently large x.
The ~ symbol denotes an asymptotic relationship, but it is possible to
be more explicit about the accuracy of the approximation. This is a
little stronger than Dirichlet's theorem, of course.

The original question was concerned with how much variation there is