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

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  • cino hilliard
    ... 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@...>
      >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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    • cino hilliard
      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 2 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@...>
        >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?
        >
        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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      • Andy Swallow
        ... 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 3 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
        • 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 4 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 5 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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