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I Challenge any and all professor/s of math, are you up to the task?

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  • eamonn@nu.fm
    Approximately 5 years ago after six months of on paper mathematics, I found the most beautiful formula for solving, what I now know to be NP Complete problems.
    Message 1 of 7 , May 27 5:41 PM
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      Approximately 5 years ago after six months of on paper mathematics, I
      found the most beautiful formula for solving, what I now know to be NP
      Complete problems. It wasn't my intent, but after running in to a problem
      whilst coding for which I knew the only solution should be mathematics.

      Over time I realized my solution is most likely the answer to NP Complete.

      I am looking for a mentor, to bridge the gap between my non mathematical
      background, though I have aptitude.

      THE CHALLENGE:

      You will be aware that their are only 5 known Fermat Primes. I challenge
      you to find a positive integer sequence, when factored reveals the hidden
      world of Fermat primes. This discovery I made by observing data in my NP
      math work.

      Please do not be shy to ask me for the solution, but only if you have
      suitable rank, noteriaty and are willing to mentor my important work.

      Regards Eamonn Smyth, a soon to be known mathematician.

      OpenQL Project Team (Quick Language Project)

      I am releasing my technology derived from the maths @
      http://openql.com
      http://sourceforge.net/projects/openql
      http://twitter.com/openql
      http://facebook [Search OpenQL once logged in and search OpenQL]
    • David Broadhurst
      ... There are such persons on this list, if you will be modest enough to confess why you request their scrutiny. ... I am indeed so aware. ... This sounds like
      Message 2 of 7 , May 27 8:03 PM
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        --- In primenumbers@yahoogroups.com,
        eamonn@... wrote:

        > I am looking for a mentor, to bridge the gap between my
        > non mathematical background, though I have aptitude.

        There are such persons on this list, if you
        will be modest enough to confess why you
        request their scrutiny.

        > You will be aware that their are only 5 known Fermat Primes.

        I am indeed so aware.

        > I challenge you to find a positive integer sequence,
        > when factored reveals the hidden world of Fermat primes.

        This sounds like what is known, in my neck of the woods,
        as a "pig in a poke".

        > This discovery I made by observing data in my NP math work.

        Please skip the P/NP rhetoric and proceed to your punch-line.

        > Please do not be shy to ask me for the solution

        Not many folk on this list are shy, by nature.

        > but only if you have suitable rank

        Nor do they accord rank.

        > noteriaty

        But some are indeed notorious.

        > and are willing to mentor

        If you are prepared for that in an open forum.

        > my important work

        With an importance yet to be determined?

        > Regards Eamonn Smyth,
        > a soon to be known mathematician

        That is not for you to decide, Eamonn

        Best regards

        David
      • eamonn@nu.fm
        Message 3 of 7 , May 28 8:02 AM
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          <?php

          /*
          This is free software:
          Math observation and code by Eamonn F. Smyth http://nu.fm
          Positive Integer Sequence when factored reveals hidden world of Fermat
          Primes.
          Feel free to modify this code, if you wish to search for the next set,
          then my instinct says the numbers to be factored will be in the million/s
          digits long. Good Luck if you go hunting, don't forget to credit me if you
          find them.

          This sequence shall be known as the 1159 sequence, check the ending
          numbers and you will understand why.
          You may also call it the Smyth sequence both are correct.

          The important numbers to be factored are 21845 & 1431655765
          Yes I am aware of the missing 3 but it is needed and used to create the
          sequence.
          */

          $maximumBW = 30; //increase this number no matter how big the numbers
          become they honour the 1159 sequence and have beyond coincidence Fermat
          Prime Factors. You might like to modify the code to use bcmaths, if you
          want the big numbers.

          $val = $c = 1;
          for (;$c<$maximumBW+1;$c++){
          if ($c % 2 == 0){
          $val = $val * 2;
          $MDR = $val += 3;
          echo "$MDR\n";
          }else{
          $val = $val * 2;
          $MDR = $val -= 1;
          echo "$MDR\n";

          }
          $MDR++;
          }
          ?>

          Thanks David.

          Kind Regards
          Eamonn.
        • David Broadhurst
          ... http://www.research.att.com/~njas/sequences/A097074 shows that the sequence is a(n) = 2*(2^(n+1) + (-1)^n)/3 - 1 print(vector(14,n,a(n))) [1, 5, 9, 21, 41,
          Message 4 of 7 , May 28 9:36 AM
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            --- In primenumbers@yahoogroups.com, eamonn@... wrote:

            > This sequence shall be known as the 1159 sequence

            http://www.research.att.com/~njas/sequences/A097074
            shows that the sequence is
            a(n) = 2*(2^(n+1) + (-1)^n)/3 - 1

            print(vector(14,n,a(n)))
            [1, 5, 9, 21, 41, 85, 169, 341, 681, 1365, 2729, 5461, 10921, 21845]

            > The important numbers to be factored are 21845 & 1431655765

            a(14) = 2*(2^15 + 1)/3 - 1 = (2^16 - 1)/3 = 21845
            a(30) = 2*(2^31 + 1)/3 - 1 = (2^32 - 1)/3 = 1431655765
            are clearly products of Fermat primes, but
            a(62) = (2^64 - 1)/3 = 5*17*257*641*65537*6700417
            is not.

            End of story?

            David
          • David Broadhurst
            ... a(2*k-2) = (4^k - 1)/3 is clearly composite for k 2. a(2*k-1) = (2*4^k - 5)/3 is a unit for k = 1, a prime for k = 3, 6, 9, 15, 25, 27, 42, 55, 159, 186,
            Message 5 of 7 , May 28 1:03 PM
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              --- In primenumbers@yahoogroups.com,
              "David Broadhurst" <d.broadhurst@...> wrote:

              > http://www.research.att.com/~njas/sequences/A097074
              > shows that the sequence is
              > a(n) = 2*(2^(n+1) + (-1)^n)/3 - 1

              a(2*k-2) = (4^k - 1)/3 is clearly composite for k > 2.

              a(2*k-1) = (2*4^k - 5)/3 is a unit for k = 1, a prime for
              k = 3, 6, 9, 15, 25, 27, 42, 55, 159, 186, 252, 405, 450, 471, 558,
              a probable prime for
              k = 1099, 1215, 2529, 4711, 6300, 10147, 11377,
              and composite for all other positive integers k < 15600.

              The sequence of primes of the form N = (2*4^k - 5)/3
              shares a feature with the Wagstaff sequence
              http://primes.utm.edu/top20/page.php?id=67
              since in the present case one or both of
              N - 1 = 8*(4^(k-1) - 1)/3
              N + 1 = 2*(4^k - 1)/3
              may have many prime factors recorded by the
              Cunningham project and its extensions.

              Thus I easily proved primality for k = 6300:

              Calling N+1 BLS with factored part 52.49%
              and helper 0.17% (157.63% proof)
              (2*4^6300-5)/3 is prime! (22.9328s+0.0190s)

              at 3793 decimal digits.

              Can someone prove primality of a larger member of A97074 ?

              David
            • David Broadhurst
              ... PS: The first gigantic PRP has 11638 digits and has ... David
              Message 6 of 7 , May 28 5:19 PM
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                --- In primenumbers@yahoogroups.com,
                "David Broadhurst" <d.broadhurst@...> wrote:

                > a(2*k-1) = (2*4^k - 5)/3 is a unit for k = 1, a prime for
                > k = 3, 6, 9, 15, 25, 27, 42, 55, 159, 186, 252, 405, 450, 471, 558,
                > a probable prime for
                > k = 1099, 1215, 2529, 4711, 6300, 10147, 11377,
                > and composite for all other positive integers k < 15600.

                PS: The first gigantic PRP has 11638 digits and has
                been lodged with Henri:

                > You have submitted the following probable prime(s)
                > to the PRP Top queue : (2^38659-5)/3,11638
                > Comments : Fermat and Lucas PRP

                David
              • David Broadhurst
                ... Almost as easy to prove prime is Calling N-1 BLS with factored part 26.04% and helper 3.49% (81.60% proof) (2*4^11377-5)/3 is Fermat and Lucas PRP!
                Message 7 of 7 , May 28 9:30 PM
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                  --- In primenumbers@yahoogroups.com,
                  "David Broadhurst" <d.broadhurst@...> wrote:

                  > Calling N+1 BLS with factored part 52.49%
                  > and helper 0.17% (157.63% proof)
                  > (2*4^6300-5)/3 is prime! (22.9328s+0.0190s)

                  Almost as easy to prove prime is

                  Calling N-1 BLS with factored part 26.04%
                  and helper 3.49% (81.60% proof)
                  (2*4^11377-5)/3 is Fermat and Lucas PRP! (48.4337s+0.0105s)

                  with a quick proof at 6850 digits then coming from CHG.

                  But the PRPs (2*4^k - 5)/3 with
                  k = 10147, 19329, 28609, 32089
                  seem to be harder to prove prime.

                  David
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