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Goldbach (again)

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  • Didier van der Straten
    Hello you all Recently there was a discussion about this topic which ended up in obscure fashion, but shortly after, there was a discussion about twin primes
    Message 1 of 6 , Nov 17, 2004
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      Hello you all

      Recently there was a discussion about this topic which ended up in obscure
      fashion,
      but shortly after, there was a discussion about twin primes which triggered
      some thinking
      that can be also developped about Goldbach, as I venture here.

      As a starter I propose a method to evaluate the count of Goldbach
      partitions, based on heuristic formulas
      similar to those about twin primes and about pi(x).

      I have created a page and a table about this, which is draft version for the
      time being :
      http://www.geocities.com/dhvanderstraten/gldbevtxt.html
      http://www.geocities.com/dhvanderstraten/gldbevtbl1.html

      I will develop further as your feedbacks and feelings dictate.
      Rgds

      Didier van der Straten
    • Greg@tell-all.com
      In a message dated 11/17/04 9:42:05 AM Eastern Standard Time, ... I didn t understand your table. 2n = 530 and its count is 14. What are you counting? What is
      Message 2 of 6 , Nov 17, 2004
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        In a message dated 11/17/04 9:42:05 AM Eastern Standard Time,
        vdstrat@... writes:
        > As a starter I propose a method to evaluate the count of Goldbach
        > partitions, based on heuristic formulas
        > similar to those about twin primes and about pi(x).
        >
        > I have created a page and a table about this, which is draft version for the
        > time being :
        > http://www.geocities.com/dhvanderstraten/gldbevtxt.html
        > http://www.geocities.com/dhvanderstraten/gldbevtbl1.html

        I didn't understand your table.
        2n = 530 and its count is 14.
        What are you counting? What is "eval."
        Where is the Excel spreadsheet formula?

        A set of 2n elements has 2^2n subsets including the empty set and 2n itself.
        Where r combinations are expressed as (2^2n/r) then
        2^2n = (2^2n/1) + (2^2n/2) + (2^2n/3) + ... + (2^2n/2^2n)
        Eg. 2^5 = 32 = 1 + 5 + 10 + 10 + 5 + 1
        Alternatively, an unrestricted partition of 5 includes these 5 sums:
        4 + 1,
        3 + 2,
        3 + 1 + 1,
        2 + 2 + 1,
        2 + 1 + 1 + 1,
        1 + 1+ 1 + 1 + 1

        Regardless of whatever it is you are counting, I don't see the
        "clue" that links such counts to the Goldbach Conjecture.

        See if my Stepladder approach offers any clue at:
        http://gmazur.freeyellow.com/Goldbach.com
        (At least to keep the thread from fading into obscurity)

        Greg


        [Non-text portions of this message have been removed]
      • Greg@tell-all.com
        In a message dated 11/17/04 2:39:36 PM Eastern Standard Time, ... Of course I meant http://gmazur.freeyellow.com/Goldbach.htm Sorry for the typo [Non-text
        Message 3 of 6 , Nov 17, 2004
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          In a message dated 11/17/04 2:39:36 PM Eastern Standard Time,
          Greg@... writes:
          > See if my Stepladder approach offers any clue at:
          > http://gmazur.freeyellow.com/Goldbach.com
          > (At least to keep the thread from fading into obscurity)
          >
          Of course I meant
          http://gmazur.freeyellow.com/Goldbach.htm

          Sorry for the typo


          [Non-text portions of this message have been removed]
        • Didier van der Straten
          Dear Greg Thank you for your prompt feedbacks. First I ll try to answer to this one made of 4 questions : A Goldbach partition of an even number 2n is a set
          Message 4 of 6 , Nov 18, 2004
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            Dear Greg

            Thank you for your prompt feedbacks.

            First I'll try to answer to this one made of 4 questions :

            A "Goldbach partition" of an even number 2n is a set of two prime numbers
            the sum of which equals 2n. If n is prime, the set (n, n) is also a Goldbach
            partition.
            I used this vocabulary after reading lots of good internet pages about the
            Goldbach conjecture.

            The "count" I show in my table is the number of distinct Goldbach partitions
            for an even number 2n. One can find an applet on internet that gives the
            list
            of Goldbach partitions for "any" proposed even number.

            The "evaluation" is an attempt to evaluate that said count by a predicting
            formula. Observing how "precise" the result is may give a clue as to "solve"
            the Goldbach conjecture.

            For the moment I do not disclose that formula. I might do so in the future
            depending on further feedbacks and feelings.

            Feel free to continue fruitful interaction. Offline or online as you wish.

            I will be glad to visit again the links you suggested and maybe add
            something
            here about them.

            Rgds.
            Didier

            -----Message d'origine-----
            De : Greg@... [mailto:Greg@...]
            Envoye : mercredi 17 novembre 2004 20:39
            A : vdstrat@...; primenumbers@yahoogroups.com
            Objet : Re: [PrimeNumbers] Goldbach (again)



            In a message dated 11/17/04 9:42:05 AM Eastern Standard Time,
            vdstrat@... writes:
            > As a starter I propose a method to evaluate the count of Goldbach
            > partitions, based on heuristic formulas
            > similar to those about twin primes and about pi(x).
            >
            > I have created a page and a table about this, which is draft version for
            the
            > time being :
            > http://www.geocities.com/dhvanderstraten/gldbevtxt.html
            > http://www.geocities.com/dhvanderstraten/gldbevtbl1.html

            I didn't understand your table.
            2n = 530 and its count is 14.
            What are you counting? What is "eval."
            Where is the Excel spreadsheet formula?

            A set of 2n elements has 2^2n subsets including the empty set and 2n itself.
            Where r combinations are expressed as (2^2n/r) then
            2^2n = (2^2n/1) + (2^2n/2) + (2^2n/3) + ... + (2^2n/2^2n)
            Eg. 2^5 = 32 = 1 + 5 + 10 + 10 + 5 + 1
            Alternatively, an unrestricted partition of 5 includes these 5 sums:
            4 + 1,
            3 + 2,
            3 + 1 + 1,
            2 + 2 + 1,
            2 + 1 + 1 + 1,
            1 + 1+ 1 + 1 + 1

            Regardless of whatever it is you are counting, I don't see the
            "clue" that links such counts to the Goldbach Conjecture.

            See if my Stepladder approach offers any clue at:
            http://gmazur.freeyellow.com/Goldbach.com
            (At least to keep the thread from fading into obscurity)

            Greg


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




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          • Andrey Kulsha
            ... http://www.ieeta.pt/~tos/goldbach.html Best, Andrey :) [Non-text portions of this message have been removed]
            Message 5 of 6 , Nov 18, 2004
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              > I will be glad to visit again the links you suggested and maybe add
              > something here about them.

              http://www.ieeta.pt/~tos/goldbach.html

              Best,

              Andrey :)

              [Non-text portions of this message have been removed]
            • Didier van der Straten
              Dear interested primers Have appreciated feedbacks about my initial draft. I start disclosing my Goldbach partition count evaluation method and formula.
              Message 6 of 6 , Dec 2, 2004
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                Dear interested primers

                Have appreciated feedbacks about my initial draft.

                I start disclosing "my" Goldbach partition count evaluation method and
                formula.

                Please go to my updated page :
                http://www.geocities.com/dhvanderstraten/gldbevtxt.html

                Thanks. I look forward to your continued interest.
                Didier
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