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Re: (p^2 + 11)/20 is a square

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  • 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 1 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 2 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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