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Re: F = 20z^2 - n and (5r^2 -F)/n

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  • djbroadhurst
    ... Kevin sent me a correct answer, in decimal format. But for that he needed 454 characters. The solution can be written in less than 20 characters, using a
    Message 1 of 7 , Mar 2, 2010
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > Puzzle: (p^2 + 11)/20 is a square for the primes
      > p = 3, 13, 67, 4253, 21587, 6950947 ...
      > Find the 17th prime in this sequence.

      Kevin sent me a correct answer, in decimal format. But for
      that he needed 454 characters. The solution can be written
      in less than 20 characters, using a form that OpenPFGW
      readily understands. So the puzzle is still open, for a
      compact answer.

      Meanwhile, congratulations to Kevin, for a valid result

      David
    • djbroadhurst
      ... Kevin has now found a reasonably compact answer as a linear combination of 3 Fibonacci numbers. In fact the answer can be written in terms of only 2:
      Message 2 of 7 , Mar 3, 2010
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@...> wrote:

        > Puzzle: (p^2 + 11)/20 is a square for the primes
        > p = 3, 13, 67, 4253, 21587, 6950947 ...
        > Find the 17th prime in this sequence.

        Kevin has now found a reasonably compact answer
        as a linear combination of 3 Fibonacci numbers.
        In fact the answer can be written in terms of only 2:

        Solution [in 19 characters]:

        F(2166)/2+3*F(2168)

        Proof [by exhaustion, with primality proving]:

        P(n,s) = fibonacci(6*n)/2 + 3*fibonacci(6*n+2*s);
        {c=0;for(n=0,361,forstep(s=-1,1,2,p=P(n,s);
        if(issquare((p^2+11)/20)&&isprime(p),c++;
        print1(s*n" "))));print([c]);}

        0 -1 1 -3 3 5 -11 15 -45 -47 -65 75 105 253 -303 -359 361 [17]

        Comments: Dario Alpern's http://www.alpertron.com.ar/QUAD.HTM
        may be used to prove that my Fibonacci method is exhaustive.
        A probable continuation of the sequence of primes is as follows:

        F(3966)/2+3*F(3964), 829
        F(5334)/2+3*F(5336), 1116
        F(8682)/2+3*F(8684), 1816
        F(15066)/2+3*F(15068), 3150
        F(15438)/2+3*F(15436), 3227
        F(41166)/2+3*F(41168), 8604
        F(50874)/2+3*F(50876), 10633
        F(114702)/2+3*F(114704), 23972
        F(117978)/2+3*F(117980), 24657

        with the number of digits of each PRP given, after the comma.
        The last 3 have been consigned to Henri's repository of PRPs.

        David
      • Kevin Acres
        ... My solution, in 29 characters, is: 3/2*F(2169)+5*F(2164)+F(2161) David s sequence, including the non-prime values, starting with the 13,67 pair can be
        Message 3 of 7 , Mar 4, 2010
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          At 03:15 PM 4/03/2010, djbroadhurst wrote:

          > > Puzzle: (p^2 + 11)/20 is a square for the primes
          > > p = 3, 13, 67, 4253, 21587, 6950947 ...
          > > Find the 17th prime in this sequence.
          >
          >Kevin has now found a reasonably compact answer
          >as a linear combination of 3 Fibonacci numbers.
          >In fact the answer can be written in terms of only 2:
          >
          >Solution [in 19 characters]:
          >
          >F(2166)/2+3*F(2168)

          My solution, in 29 characters, is:

          3/2*F(2169)+5*F(2164)+F(2161)

          David's sequence, including the non-prime values, starting with the
          13,67 pair can be reproduced by the following:

          Let f1=F(12*n+9), f2=F(12*n+4), f3=F(12*n+1), a=f1+2*f2 and o=1/2*f1+3*f2+f3

          Then the lower of each pair is a-o and the higher is a+o or:

          1/2*f1-f2-f3 and 3/2*f1+5*f2+f3

          In the case of the p454 just substitute n for 180 in the above assignments.

          The pari/gp script to generate the sequence follows:

          for(i=0,10,
          f1=fibonacci(12*i+9);
          f2=fibonacci(12*i+4);
          f3=fibonacci(12*i+1);
          o=1/2*f1+3*f2+f3;
          a=f1+2*f2;
          print([a-o,a+o]);
          );

          I strayed a little from the puzzle by solving for 180z^2-p^2-11
          before trying to re-create the sequence from 13/67 onwards.

          Once again, Dario's page at: http://www.alpertron.com.ar/QUAD.HTM was
          of great help.


          Kevin.
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