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Chains of Generalised Fermats

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  • garychaffey2
    I have been looking at the following chains of GFN. And would be interested if anybody else has lloked at these and if a longer chain is known or could be
    Message 1 of 5 , Jun 6, 2005
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      I have been looking at the following chains of GFN. And would be
      interested if anybody else has lloked at these and if a longer chain is
      known or could be found.
      Let b be an even number
      p(1)=b^2+1
      p(2)=(2*p(1))^2+1
      p(3)=(2*p(2))^2+1
      .
      .
      .
      p(n)=(2*p(n-1))^2+1
      I have found a few b such that the first three terms in this sequence
      are prime but does a longer chain exist?
      Any comments welcome
      Gary Chaffey
    • Mark Underwood
      ... ... chain is ... sequence ... Hi Gary, I ve checked prime factors up to 23, and 13 is the only one which gets in the way of a long
      Message 2 of 5 , Jun 6, 2005
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        --- In primenumbers@yahoogroups.com, "garychaffey2"
        <garychaffey2@y...>
        wrote:
        > I have been looking at the following chains of GFN. And would be
        > interested if anybody else has lloked at these and if a longer
        chain
        is
        > known or could be found.
        > Let b be an even number
        > p(1)=b^2+1
        > p(2)=(2*p(1))^2+1
        > p(3)=(2*p(2))^2+1
        > .
        > .
        > .
        > p(n)=(2*p(n-1))^2+1
        > I have found a few b such that the first three terms in this
        sequence
        > are prime but does a longer chain exist?



        Hi Gary,

        I've checked prime factors up to 23, and 13 is the only one which
        gets in the way of a long string of primes. As it turns out, any
        starting residue n of 13 when iterated through

        (2*n)^2 +1

        will eventually yield a zero residue . The maximum number of primes
        which could be obtained is 5. (Based only on prime factors 23 and
        less.)

        It's interesting: The chain will never produce a factor of 3,7,11,19
        or 23. It may or may not produce a factor of 5 or 17, depending on the
        starting residue mod 5 or mod 17. But it will always produce a
        factor of 13. Talk about an unlucky number :)

        Mark
      • mikeoakes2@aol.com
        ... You might like to look at my Quadratic-map prime chains NMBRTHRY post of last year:
        Message 3 of 5 , Jun 7, 2005
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          In an email dated 6/6/2005 5:43:37 pm GMT Daylight time, "garychaffey2" <garychaffey2@...> writes:

          >I have been looking at the following chains of GFN. And would be
          >interested if anybody else has lloked at these and if a longer chain is
          >known or could be found.
          >Let b be an even number
          >p(1)=b^2+1
          >p(2)=(2*p(1))^2+1
          >p(3)=(2*p(2))^2+1
          >.
          >.
          >.
          >p(n)=(2*p(n-1))^2+1
          >I have found a few b such that the first three terms in this sequence
          >are prime but does a longer chain exist?
          >Any comments welcome

          You might like to look at my "Quadratic-map prime chains" NMBRTHRY post of last year:
          http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0402&L=nmbrthry&F=&S=&P=922

          -Mike Oakes
        • Jens Kruse Andersen
          ... Then also look at http://www.primepuzzles.net/puzzles/puzz_137.htm It says Yves Gallot found the smallest chain of 6 before Mike: 7072833120^1+1
          Message 4 of 5 , Jun 7, 2005
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            Mike Oakes wrote:

            > You might like to look at my "Quadratic-map prime chains" NMBRTHRY post
            > of last year:
            > http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0402&L=nmbrthry&F=&S=&P=922

            Then also look at http://www.primepuzzles.net/puzzles/puzz_137.htm
            It says Yves Gallot found the smallest chain of 6 before Mike:
            7072833120^1+1
            7072833120^2+1
            7072833120^4+1
            7072833120^8+1
            7072833120^16+1
            7072833120^32+1

            I have long considered going for 7 and maybe 8, but been occupied with other
            projects.

            A tip which has often been useful to me:
            If you are interested in something related to a specific large number then
            search the number at http://www.google.com
            Searching Mike's 7072833120 quickly reveals Gallot was first.

            --
            Jens Kruse Andersen
          • mikeoakes2@aol.com
            ... Thx for tip, Jens. I was unaware of Yves s work, but it seems google is indeed getting quite intelligent these days, if it can respond to a single
            Message 5 of 5 , Jun 7, 2005
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              >A tip which has often been useful to me:
              >If you are interested in something related to a specific large number then
              >search the number at http://www.google.com

              Thx for tip, Jens.
              I was unaware of Yves's work, but it seems google is indeed getting quite "intelligent" these days, if it can respond to a single integer like that :-)
              Must remember in future...

              Mike
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