>From: "Andrew Swallow" <umistphd2003@...>

From the prime page

>To: primenumbers@yahoogroups.com

>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,

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

Little result?

>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

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

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

>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.

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?

>

Cino

>Hope this helps,

>

>Andy

>

Sometime I get so mad I could eat fried chicken.

_________________________________________________________________

There are now three new levels of MSN Hotmail Extra Storage! Learn more.

http://join.msn.com/?pgmarket=en-us&page=hotmail/es2&ST=1- 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@...>

I make my case even easier.

>To: primenumbers@yahoogroups.com

>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@...>

> >To: primenumbers@yahoogroups.com

> >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?

>

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

_________________________________________________________________

What are the 5 hot job markets for 2004? Click here to find out.

http://msn.careerbuilder.com/Custom/MSN/CareerAdvice/WPI_WhereWillWeFindJobsIn2004.htm?siteid=CBMSN3006&sc_extcmp=JS_wi08_dec03_hotmail1 > I find this to be a huge result way bigger than say Wiles proof of FLT

Well I see your point, the proof of FLT is a little complex for most

> 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.

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.

Well no, it was just unclear what you wanted to vary when. Fixing *any*

>

> Is there only one k and a for which the statement is true?

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

Well do you think this sequence contains infinitely many primes? Of

> 1,1,1,1,1,1,1,1,1,...

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>From: Andy Swallow <umistphd2003@...>

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

>To: primenumbers@yahoogroups.com

>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

>

Does this mean you now agree that the infinitude of primes can be

> > 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*

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

Well, while 6n+4 does not satisfy the Dirichlet criteria, I can combine it

>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

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.

True. However, this does not preclude me to combine the progression 0n+a

>Dirichlet's proof depended on the characters modulo k. If k=0, there are

>no characters, and therefore the proof fails.

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

_________________________________________________________________

Let the new MSN Premium Internet Software make the most of your high-speed

experience. http://join.msn.com/?pgmarket=en-us&page=byoa/prem&ST=1> My statement still stands:

Dirichlet's theorem proves that there are an infinite number of primes

>

> You can prove that there is an infinite number of primes using

> Dirichlet's theorem.

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