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

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  • cino hilliard
    ... 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@...>
      >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

      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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    • Andrew Swallow
      ... 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 2 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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