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Strengthening Goldbach

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  • mikeoakes2@aol.com
    Hi people Woke this morning with: every integer = 2 mod 4 can be expressed as the sum of 2 primes each = 3 mod 4 It seems ok (excluding 2). Mike [Non-text
    Message 1 of 6 , May 29, 2002
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      Hi people

      Woke this morning with:

      "every integer = 2 mod 4 can be expressed as the sum of 2 primes each = 3 mod
      4"

      It seems ok (excluding 2).

      Mike


      [Non-text portions of this message have been removed]
    • Jon Perry
      ... mod ... Probably true - Erdos proved between n and 2n there exist a prime of the form 1mod4 and 3mod4, and although not directly related, I imagine the to
      Message 2 of 6 , May 29, 2002
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        >"every integer = 2 mod 4 can be expressed as the sum of 2 primes each = 3
        mod
        >4"

        Probably true - Erdos proved between n and 2n there exist a prime of the
        form 1mod4 and 3mod4, and although not directly related, I imagine the to
        fact are.

        A gap in this would probably disrupt my view of the Prime Number
        Distribution.

        Similar conjectures, obviously 1mod4, also GC is true for primes of a twin
        prime pair except for a certain number of exceptions.

        A Dirichlet version of GC might be: In any AP {a+bx : (a,b)=1} there exists
        N such that all primes >=N are the sum of two primes in AP.

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com


        -----Original Message-----
        From: mikeoakes2@... [mailto:mikeoakes2@...]
        Sent: 29 May 2002 08:35
        To: primenumbers@yahoogroups.com
        Subject: [PrimeNumbers] Strengthening Goldbach
      • mikeoakes2@aol.com
        In a message dated 29/05/02 18:57:25 GMT Daylight Time, perry@globalnet.co.uk ... Jon, [I take it you meant to say ...all even integers = N... ] I m afraid
        Message 3 of 6 , May 29, 2002
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          In a message dated 29/05/02 18:57:25 GMT Daylight Time, perry@...
          writes:


          > A Dirichlet version of GC might be: In any AP {a+bx : (a,b)=1} there exists
          > N such that all primes >=N are the sum of two primes in AP.
          >

          Jon,
          [I take it you meant to say "...all even integers >= N..."]

          I'm afraid your version is trivially false.
          Take e.g. a=1, b=6, and note that no integer <> 2 MOD 6 is so representable.

          Mike



          [Non-text portions of this message have been removed]
        • Jon Perry
          ... representable. So it would seem that the only AP s capable of such a feat are 1mod4 and 3mod4. We could define a covering set of AP s that do satisfy GC.
          Message 4 of 6 , May 29, 2002
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            >I'm afraid your version is trivially false.
            >Take e.g. a=1, b=6, and note that no integer <> 2 MOD 6 is so
            representable.

            So it would seem that the only AP's capable of such a feat are 1mod4 and
            3mod4.

            We could define a covering set of AP's that do satisfy GC.

            Jon Perry
            perry@...
            http://www.users.globalnet.co.uk/~perry/maths
            BrainBench MVP for HTML and JavaScript
            http://www.brainbench.com


            -----Original Message-----
            From: mikeoakes2@... [mailto:mikeoakes2@...]
            Sent: 29 May 2002 20:51
            To: primenumbers@yahoogroups.com
            Subject: Re: [PrimeNumbers] Strengthening Goldbach


            In a message dated 29/05/02 18:57:25 GMT Daylight Time,
            perry@...
            writes:


            > A Dirichlet version of GC might be: In any AP {a+bx : (a,b)=1} there
            exists
            > N such that all primes >=N are the sum of two primes in AP.
            >

            Jon,
            [I take it you meant to say "...all even integers >= N..."]

            I'm afraid your version is trivially false.
            Take e.g. a=1, b=6, and note that no integer <> 2 MOD 6 is so representable.

            Mike



            [Non-text portions of this message have been removed]



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          • Jon Perry
            Talking of GC, has anyone heard a rumour that the prize for the conjecture is going to be re-instated? Jon Perry perry@globalnet.co.uk
            Message 5 of 6 , May 30, 2002
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              Talking of GC, has anyone heard a rumour that the prize for the conjecture
              is going to be re-instated?

              Jon Perry
              perry@...
              http://www.users.globalnet.co.uk/~perry/maths
              BrainBench MVP for HTML and JavaScript
              http://www.brainbench.com


              -----Original Message-----
              From: Jon Perry [mailto:perry@...]
              Sent: 29 May 2002 21:00
              To: primenumbers@yahoogroups.com
              Subject: RE: [PrimeNumbers] Strengthening Goldbach



              >I'm afraid your version is trivially false.
              >Take e.g. a=1, b=6, and note that no integer <> 2 MOD 6 is so
              representable.

              So it would seem that the only AP's capable of such a feat are 1mod4 and
              3mod4.

              We could define a covering set of AP's that do satisfy GC.

              Jon Perry
              perry@...
              http://www.users.globalnet.co.uk/~perry/maths
              BrainBench MVP for HTML and JavaScript
              http://www.brainbench.com


              -----Original Message-----
              From: mikeoakes2@... [mailto:mikeoakes2@...]
              Sent: 29 May 2002 20:51
              To: primenumbers@yahoogroups.com
              Subject: Re: [PrimeNumbers] Strengthening Goldbach


              In a message dated 29/05/02 18:57:25 GMT Daylight Time,
              perry@...
              writes:


              > A Dirichlet version of GC might be: In any AP {a+bx : (a,b)=1} there
              exists
              > N such that all primes >=N are the sum of two primes in AP.
              >

              Jon,
              [I take it you meant to say "...all even integers >= N..."]

              I'm afraid your version is trivially false.
              Take e.g. a=1, b=6, and note that no integer <> 2 MOD 6 is so representable.

              Mike



              [Non-text portions of this message have been removed]



              Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
              The Prime Pages : http://www.primepages.org



              Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/





              Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
              The Prime Pages : http://www.primepages.org



              Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
            • mikeoakes2@aol.com
              ... mod 4 No-one has said it has already been conjectured, so I assume it hasn t. Here s some relevant data to back it up. Given a cutoff n, if you total up
              Message 6 of 6 , Jun 2, 2002
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                In a message dated 29/05/02 08:35:16 GMT Daylight Time, I wrote:
                > every integer = 2 mod 4 can be expressed as the sum of 2 primes each = 3
                mod 4

                No-one has said it has already been conjectured, so I assume it hasn't.
                Here's some relevant data to back it up.
                Given a cutoff n, if you total up the number of ways each of the integers
                less than n which are = 2 mod 4 can be expressed as the sum of 2 primes each
                = 1 mod 4 (total1), and do the same for = 3 mod 4 (total3), you get the
                following:-

                n total1 total3 total3/total1
                10 5 3 0.6
                100 105 107 1.01904762
                1000 4045 4344 1.07391842
                10000 209824 217155 1.03493881
                100000 12602761 12773813 1.01357258
                1000000 834142745 837815535 1.00440307

                So the stronger conjecture, that the ratio => 1 as n => infinity, seems
                eminently reasonable.

                Anyone for a proof? (only joking...)

                Mike


                [Non-text portions of this message have been removed]
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