Loading ...
Sorry, an error occurred while loading the content.

RE: Primes in the concatenation with the digits of Pi

Expand Messages
  • cino hilliard
    Hi, I have been investigating some old buddies in the realm of prime numbers. If we take for example the first 999 digits of Pi and append that to the
    Message 1 of 7 , Feb 2, 2004
    • 0 Attachment
      Hi,
      I have been investigating some old buddies in the realm of prime numbers. If
      we take for example
      the first 999 digits of Pi and append that to the sequence of numbers below
      we will get a greater
      than 1000 digit pseudoprime. Pari code is at the end to generate these and
      the pprimes thenselves.

      61,1158,2488,2623,2625,3454,4575,5100,8019,8821,...

      Notice 2623,2635. What, a kind of twin prime? Are there more of these?

      Dirichlet proved in 1837 that every arithmetic progression kn + a where
      (k,a)=1, n=1,2,3,..
      contains an infinite number of primes. The Full proof can be found in
      Lectures on Elementary
      Number Theory by Hans Rademacher pp121 - 135.

      If we set k = 10^999 and a = 314159...4209. then the terms in the above
      sequence are infinite.

      If we set n=1,k=1 we have the progression 1+a. So for a=1,2,3,..
      Dirichlet's theorem proves
      the infinity of primes per se.

      My question is this. Why can't we set n = 0 in the Dirichlet theorem once
      proved? Dirichlet used
      zeta(s) = Sum(n=1..infinity,1/n^s) in his proof. This restricts n=0 unless
      of course we let s=0 and
      this would mean zeta(s) = infinity which dosen't tell us much or too much
      depending on how you
      look at it. Nevertheless, the finished product, kn + a will accept n=0
      without restriction so the
      progression of integers 0 + a = 1,2,3... contain an infinite number of
      primes. I don't see a catch 22
      here since "that which is true is true and cannot be changed." But...

      Pari Code
      f(n) =
      e=floor(Pi*10^998);for(x=0,n,z=x*10^999+e;if(ispseudoprime(z),print1(x",")))

      Have fun in the facinating world of numbers,

      Cino
      "Behavior is not for the pursuit of survival but because of it."

      _________________________________________________________________
      Scope out the new MSN Plus Internet Software � optimizes dial-up to the max!
      http://join.msn.com/?pgmarket=en-us&page=byoa/plus&ST=1
    • Andrew Swallow
      ... these? More of what? Primes close together? Well, yes, it just seems a little difficult to prove it at the moment. ... where ... You seem to be
      Message 2 of 7 , Feb 2, 2004
      • 0 Attachment
        > 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,
        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
        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.

        Hope this helps,

        Andy
      • 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 3 of 7 , Feb 2, 2004
        • 0 Attachment
          >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.

          _________________________________________________________________
          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
        • 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 4 of 7 , Feb 2, 2004
          • 0 Attachment
            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

            _________________________________________________________________
            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
          • 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 5 of 7 , Feb 3, 2004
            • 0 Attachment
              > 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 6 of 7 , Feb 3, 2004
              • 0 Attachment
                >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

                _________________________________________________________________
                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
              • 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 7 of 7 , Feb 3, 2004
                • 0 Attachment
                  > 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
                Your message has been successfully submitted and would be delivered to recipients shortly.