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• ## RE: [PrimeNumbers] Re: Primes in the concatenation with the digits of Pi

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• ... From the prime page An arithmetic sequence (or arithmetic progression) is a sequence (finite or infinite list) of real numbers for which each term is the
Message 1 of 7 , Feb 2, 2004
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>From: "Andrew Swallow" <umistphd2003@...>
>Subject: [PrimeNumbers] Re: Primes in the concatenation with the digits of
>Pi
>Date: Mon, 02 Feb 2004 22:11:57 -0000
>
>
> > 61,1158,2488,2623,2625,3454,4575,5100,8019,8821,...
> >
> > Notice 2623,2635. What, a kind of twin prime? Are there more of
>these?
>
>More of what? Primes close together? Well, yes, it just seems a little
>difficult to prove it at the moment.
>
> > Dirichlet proved in 1837 that every arithmetic progression kn + a
>where
> > (k,a)=1, n=1,2,3,..
> > contains an infinite number of primes.
>
>You seem to be misunderstanding what an arithmetic progression is,

From the prime page
An arithmetic sequence (or arithmetic progression) is a sequence (finite or
infinite list) of real numbers for which each term is the previous term plus
a constant (called the common difference). For example, starting with 1 and
using a common difference of 4 we get the finite arithmetic sequence: 1, 5,
9, 13, 17, 21; and also the inifinite sequence
1, 5, 9, 13, 17, 21, 25, 29, . . ., 4n+1, . . .
Obviously here we have n varying 0,1,2,3,4,...

I say starting with 1 and using a comon difference of 1 we get the finite
arithmetic sequence
1,2,3,4 and the infinite sequence of 1,2,3,4,5,6,...,1n+1

I also say starting with 1 and using a comon difference of 0 we get the
finite arithmetic sequence
1,1,1,1 and the infinite sequence of 1,1,1,1,1,1,...,0n+1

>what varies, and what doesn't. Let k and a be any integers. Then the
>sequence a,a+k,a+2k,a+3k,a+4k,etc is an arithmetic progression.
>Dirichlet's little result shows that, provided (k,a)=1, that sequence

Little result?

This was a fantastic result. Iit used Euler's idea of a proof of an an
infinity of primes with
zeta(s) = sum(n=1..infinity,1/n^s) and his product formula for zeta(s) =
Prod(p,1/(1-(1/p^s)) for
p ranging over all primes.

For s = 1+e>0 the summation formula for zeta(s) converges. Now zeta(1+e) =
Sum(n=1..infinity,1/(n^(1+e))) -> inf as e -> 0. This implies if there were
only a finite number
of primes the product formula for zeta(1+e) wouild have a finite number of
terms in the product
formula contradicting zeta(s) = Prod(p,1/(1-(1/p^s)) for p ranging over all
primes.

I find this to be a huge result way bigger than say Wiles proof of FLT
because that proof will never
be understood by more than a few people. Besides, it was based on another
conjecture on modular
forms which under certain conditions implied the truth of FLT and Wiles
proved that little result.

You misunderstood my question. I was just asking if I could set n=0 after
the proof forbid
such a thing in the first place.

>contains infinitely many primes. Your question seemed just confused
>over what varied. Only n varies, you can't fix n and vary k and/or a,
>and deduce an infinity of primes in other types of set.

I am not meaning to fix n and vary a and k. I want to vary all three a,k,n.

Is there only one k and a for which the statement is true?
for k =0 and a =1 we have for n=0,1,2,...,0n+1
1,1,1,1,1,1,1,1,1,...
for k =1 and a =1 we have for n=0,1,2,...,1n+1
1,2,3,4,5,6,7,8,9...
for k =2 and a =1 we have for n=0,1,2,...,2n+1
1,3,5,7,9,11,13,...
for k =3 and a =1 we have for n=0,1,2,...,3n+1
1,4,7,11,14,17,....
for k =4 and a =1 we have for n=0,1,2,...,4n+1
1,5,9,13,17,21,...
1,6,11,16,21,26,...5n+1
1,7,13,19,25,31,...6n+1
....
....
Notice the second term if this compound set is the set of positive integers
and the set of positive
integers have an infinity of primes in them by Dirichlets theorem.

What are you talking about only n varies?

>
>Hope this helps,
>
>Andy
>

Cino
Sometime I get so mad I could eat fried chicken.

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• I make my case even easier. Keep in mind I am not talking about prime progressions. Every odd number not divisible by 3 can be expressed in distinctly one of
Message 1 of 7 , Feb 2, 2004
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I make my case even easier. Keep in mind I am not talking about prime
progressions.

Every odd number not divisible by 3 can be expressed in distinctly one of
two forms:
6n+1 or 6n+5. By distinctly I mean no number can be of both forms. Then,
by Dirichlet's Theorem, the arithmetic progression 6n+1 has an infinite
number of primes and
by Dirichlet's Theorem, the arithmetic progression 6n+5 has an infinite
number of primes
for n=1,2,..

Since every prime > 3 is an odd number, the infinite number of primes in
6n+1 and 6n+5
constitute the infinity of primes > 3. Throw in 2 and 3 and we have them
all. :-)

We could use similar arguments to prove the same for numbers of the form
4n+1,4n+3
2n+1.
etc

Cino

>From: "cino hilliard" <hillcino368@...>
>Subject: RE: [PrimeNumbers] Re: Primes in the concatenation with the digits
>of Pi
>Date: Tue, 03 Feb 2004 00:44:43 +0000
>
>
> >From: "Andrew Swallow" <umistphd2003@...>
> >Subject: [PrimeNumbers] Re: Primes in the concatenation with the digits
>of
> >Pi
> >Date: Mon, 02 Feb 2004 22:11:57 -0000
> >
> >
> > > 61,1158,2488,2623,2625,3454,4575,5100,8019,8821,...
> > >
> > > Notice 2623,2635. What, a kind of twin prime? Are there more of
> >these?
> >
> >More of what? Primes close together? Well, yes, it just seems a little
> >difficult to prove it at the moment.
> >
> > > Dirichlet proved in 1837 that every arithmetic progression kn + a
> >where
> > > (k,a)=1, n=1,2,3,..
> > > contains an infinite number of primes.
> >
> >You seem to be misunderstanding what an arithmetic progression is,
>
>From the prime page
>An arithmetic sequence (or arithmetic progression) is a sequence (finite or
>infinite list) of real numbers for which each term is the previous term
>plus
>a constant (called the common difference). For example, starting with 1 and
>using a common difference of 4 we get the finite arithmetic sequence: 1, 5,
>9, 13, 17, 21; and also the inifinite sequence
>1, 5, 9, 13, 17, 21, 25, 29, . . ., 4n+1, . . .
>Obviously here we have n varying 0,1,2,3,4,...
>
>I say starting with 1 and using a comon difference of 1 we get the finite
>arithmetic sequence
>1,2,3,4 and the infinite sequence of 1,2,3,4,5,6,...,1n+1
>
>I also say starting with 1 and using a comon difference of 0 we get the
>finite arithmetic sequence
>1,1,1,1 and the infinite sequence of 1,1,1,1,1,1,...,0n+1
>
>
> >what varies, and what doesn't. Let k and a be any integers. Then the
> >sequence a,a+k,a+2k,a+3k,a+4k,etc is an arithmetic progression.
> >Dirichlet's little result shows that, provided (k,a)=1, that sequence
>
>Little result?
>
>This was a fantastic result. Iit used Euler's idea of a proof of an an
>infinity of primes with
>zeta(s) = sum(n=1..infinity,1/n^s) and his product formula for zeta(s) =
>Prod(p,1/(1-(1/p^s)) for
>p ranging over all primes.
>
>For s = 1+e>0 the summation formula for zeta(s) converges. Now zeta(1+e) =
>Sum(n=1..infinity,1/(n^(1+e))) -> inf as e -> 0. This implies if there were
>only a finite number
>of primes the product formula for zeta(1+e) wouild have a finite number of
>terms in the product
>formula contradicting zeta(s) = Prod(p,1/(1-(1/p^s)) for p ranging over all
>primes.
>
>I find this to be a huge result way bigger than say Wiles proof of FLT
>because that proof will never
>be understood by more than a few people. Besides, it was based on another
>conjecture on modular
>forms which under certain conditions implied the truth of FLT and Wiles
>proved that little result.
>
>
>You misunderstood my question. I was just asking if I could set n=0 after
>the proof forbid
>such a thing in the first place.
>
> >contains infinitely many primes. Your question seemed just confused
> >over what varied. Only n varies, you can't fix n and vary k and/or a,
> >and deduce an infinity of primes in other types of set.
>
>I am not meaning to fix n and vary a and k. I want to vary all three a,k,n.
>
>Is there only one k and a for which the statement is true?
>for k =0 and a =1 we have for n=0,1,2,...,0n+1
>1,1,1,1,1,1,1,1,1,...
>for k =1 and a =1 we have for n=0,1,2,...,1n+1
>1,2,3,4,5,6,7,8,9...
>for k =2 and a =1 we have for n=0,1,2,...,2n+1
>1,3,5,7,9,11,13,...
>for k =3 and a =1 we have for n=0,1,2,...,3n+1
>1,4,7,11,14,17,....
>for k =4 and a =1 we have for n=0,1,2,...,4n+1
>1,5,9,13,17,21,...
>1,6,11,16,21,26,...5n+1
>1,7,13,19,25,31,...6n+1
>....
>....
>Notice the second term if this compound set is the set of positive integers
>and the set of positive
>integers have an infinity of primes in them by Dirichlets theorem.
>
>What are you talking about only n varies?
>
I make my case even easier.
Every odd number not divisible by 3 can be expressed in distinctly one of
two forms:
6n+1 or 6n+5.
By Dirichlet's Theorem, the arithmetic progression 6n+1 has an infinite
number of primes.
By Dirichlet's Theorem, the arithmetic progression 6n+5 has an infinite
number of primes.
Since every prime > 3 is an odd number, the infinite number of primes in
6n+1 and 6n+5
constitute the infinite number of primes. We throw in 2 and 3 just for
completeness sake.

Cino

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• ... Well I see your point, the proof of FLT is a little complex for most people. But that point of view could also be said to de-value the achievments of some
Message 1 of 7 , Feb 3, 2004
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> I find this to be a huge result way bigger than say Wiles proof of FLT
> because that proof will never
> be understood by more than a few people. Besides, it was based on another
> conjecture on modular
> forms which under certain conditions implied the truth of FLT and Wiles
> proved that little result.

Well I see your point, the proof of FLT is a little complex for most
people. But that point of view could also be said to de-value the
achievments of some of the more "detailed" results on primes in A.P.'s

> I am not meaning to fix n and vary a and k. I want to vary all three a,k,n.
>
> Is there only one k and a for which the statement is true?

Well no, it was just unclear what you wanted to vary when. Fixing *any*
k and a with (k,a)=1, and varying only n, you get infinitely many
primes.

> for k =0 and a =1 we have for n=0,1,2,...,0n+1
> 1,1,1,1,1,1,1,1,1,...

Well do you think this sequence contains infinitely many primes? Of
course not. Every integer divides zero. Or perhaps no integer divides
zero. Is zero an integer? Anyway, you can't say that (0,1)=1.
Dirichlet's proof depended on the characters modulo k. If k=0, there are
no characters, and therefore the proof fails.

Andy
• ... It is as easy to de-value as it is to criticize. ... Does this mean you now agree that the infinitude of primes can be demonstrated by Dirlichlet s
Message 1 of 7 , Feb 3, 2004
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>From: Andy Swallow <umistphd2003@...>
>Subject: Re: [PrimeNumbers] Re: Primes in the concatenation with the digits
>of Pi
>Date: Tue, 3 Feb 2004 10:21:21 +0000
>
>
> > I find this to be a huge result way bigger than say Wiles proof of FLT
> > because that proof will never
> > be understood by more than a few people. Besides, it was based on
>another
> > conjecture on modular
> > forms which under certain conditions implied the truth of FLT and Wiles
> > proved that little result.
>
>Well I see your point, the proof of FLT is a little complex for most
>people. But that point of view could also be said to de-value the
>achievments of some of the more "detailed" results on primes in A.P.'s

It is as easy to de-value as it is to criticize.

>
> > I am not meaning to fix n and vary a and k. I want to vary all three
>a,k,n.
> >
> > Is there only one k and a for which the statement is true?
>
>Well no, it was just unclear what you wanted to vary when. Fixing *any*

Does this mean you now agree that the infinitude of primes can be
demonstrated by Dirlichlet's
theorem? How about the 6n+1 and 6n+5 argument which you do not address?

>k and a with (k,a)=1, and varying only n, you get infinitely many
>primes.
>
> > for k =0 and a =1 we have for n=0,1,2,...,0n+1
> > 1,1,1,1,1,1,1,1,1,...
>
>Well do you think this sequence contains infinitely many primes? Of
>course not. Every integer divides zero. Or perhaps no integer divides

Well, while 6n+4 does not satisfy the Dirichlet criteria, I can combine it
with all the
(k,a) = 1 sets if I so desire as I did with 0n+1 just to complete the
integers formed by the
second term. This was redundancy. But redundancy is allowed as in the two
sets 2n+1 and 3n+1
where there are repeated numbers. It could well have been left out since 1
is not prime anyway
and the infinite dirichlet produced set of second terms of the rows formed
by
1n+1 -> 2
2n+1 -> 3
3n+1 -> 4
...
...
was sufficient to prove my point. Well, had I done that some of your thunder
would have been
calmed.:-)

>zero. Is zero an integer? Anyway, you can't say that (0,1)=1.
>Dirichlet's proof depended on the characters modulo k. If k=0, there are
>no characters, and therefore the proof fails.
True. However, this does not preclude me to combine the progression 0n+a
with the
Dirichlet allowed ones. 0n+7 would produce infinitly many primes albiet not
necessarily different.

My statement still stands:

You can prove that there is an infinite number of primes using Dirichlet's
theorem. Moreover,
you can prove the infinitude in a infinite number of ways. well, if you have
time and space.

Have fun,
Cino

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• ... Dirichlet s theorem proves that there are an infinite number of primes in certain infinite subsets of the integers. So the fact that the total number of
Message 1 of 7 , Feb 3, 2004
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> My statement still stands:
>
> You can prove that there is an infinite number of primes using
> Dirichlet's theorem.

Dirichlet's theorem proves that there are an infinite number of primes
in certain infinite subsets of the integers. So the fact that the
total number of primes is infinite follows immediately...

Apologies if I misunderstood your original message and its'
intentions.

Andy
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