>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?

>

>Hope this helps,

>

>Andy

>

Cino

Sometime I get so mad I could eat fried chicken.

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