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Re: Chains of Generalised Fermats

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  • 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 1 of 5 , Jun 6, 2005
      --- 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 2 of 5 , Jun 7, 2005
        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 3 of 5 , Jun 7, 2005
          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 4 of 5 , Jun 7, 2005
            >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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