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Inaugural puzzle

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  • David Broadhurst
    p1 is the smallest titanic prime of the form sqrt(3*x^2-2) with integer x. p2 is the smallest titanic prime of the form sqrt(5*y^2-4) with integer y. p3 is the
    Message 1 of 8 , Jan 15, 2009
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      p1 is the smallest titanic prime of the form
      sqrt(3*x^2-2) with integer x.

      p2 is the smallest titanic prime of the form
      sqrt(5*y^2-4) with integer y.

      p3 is the smallest prime such that
      2*p1*p2*p3-1 is prime.

      Find a metrical connection between p3 and liberty.

      David Broadhurst
      ----------------------------------------------------------------
      The Open University is incorporated by Royal Charter (RC 000391),
      an exempt charity in England and Wales and
      a charity registered in Scotland (SC 038302).
    • Phil Carmody
      From: David Broadhurst ... Is sqrt the exact square root defined over only squares, or merely the integer square root with truncation defined over the
      Message 2 of 8 , Jan 15, 2009
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        From: David Broadhurst
        > p1 is the smallest titanic prime of the form
        > sqrt(3*x^2-2) with integer x.

        Is 'sqrt' the exact square root defined over only squares, or merely the integer square root with truncation defined over the naturals?

        Phil
      • David Broadhurst
        ... It is indeed. David Broadhurst ... The Open University is incorporated by Royal Charter (RC 000391), an exempt charity in England and Wales and a charity
        Message 3 of 8 , Jan 16, 2009
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          --- In primenumbers@yahoogroups.com, Phil Carmody
          <thefatphil@...> wrote:
          >
          > From: David Broadhurst
          > > p1 is the smallest titanic prime of the form
          > > sqrt(3*x^2-2) with integer x.
          >
          > Is 'sqrt' the exact square root defined over only squares

          It is indeed.

          David Broadhurst
          ----------------------------------------------------------------
          The Open University is incorporated by Royal Charter (RC 000391),
          an exempt charity in England and Wales and
          a charity registered in Scotland (SC 038302).
        • Jens Kruse Andersen
          Spoiler warning. ... I suck at diophantine equations so I didn t even try but just bruteforced the first few x values which gave integer square root. A quick
          Message 4 of 8 , Jan 16, 2009
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            Spoiler warning.

            David Broadhurst wrote:
            > p1 is the smallest titanic prime of the form
            > sqrt(3*x^2-2) with integer x.
            >
            > p2 is the smallest titanic prime of the form
            > sqrt(5*y^2-4) with integer y.
            >
            > p3 is the smallest prime such that
            > 2*p1*p2*p3-1 is prime.
            >
            > Find a metrical connection between p3 and liberty.

            I suck at diophantine equations so I didn't even try but just
            bruteforced the first few x values which gave integer square root.
            A quick lookup of the x sequences in OEIS and there they were.
            http://www.research.att.com/~njas/sequences/A001835 has comment:
            Terms are the solutions to: 3x^2-2 is a square.
            http://www.research.att.com/~njas/sequences/A001519 has comment:
            Terms for n>1 are the solutions to : 5x^2-4 is a square.

            A little PARI+PFGW later and I get a 1673-digit p1,
            a 1223-digit p2, and p3 = 12241.
            The metrical connection is left as an exercise for the reader
            (actually, I don't know and didn't look hard).
            I haven't run Primo on p1 and p2. Knowing David there might be
            easy proofs but I'm not looking. OK, I'm lazy.
            Assuming they are prime, PFGW proved 2*p1*p2*p3-1.

            --
            Jens Kruse Andersen
          • David Broadhurst
            ... Well done, Jens! Primo timing: Total=2h 11mn 27s DecimalSize=1673 ... No, that s too large. Try re-reading the OEIS page :-) David Broadhurst ... The Open
            Message 5 of 8 , Jan 16, 2009
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              --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
              <jens.k.a@...> wrote:

              > A little PARI+PFGW later and I get a 1673-digit p1

              Well done, Jens!

              Primo timing:
              Total=2h 11mn 27s
              DecimalSize=1673

              > a 1223-digit p2

              No, that's too large. Try re-reading the OEIS page :-)

              David Broadhurst
              ----------------------------------------------------------------
              The Open University is incorporated by Royal Charter (RC 000391),
              an exempt charity in England and Wales and
              a charity registered in Scotland (SC 038302).
            • Jens Kruse Andersen
              ... Oops. My PARI/GP script to find p2 was correct but it output decimal expansions and didn t stop after the first. I had a bunch of open windows and failed
              Message 6 of 8 , Jan 16, 2009
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                David Broadhurst wrote:
                > --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
                > <jens.k.a@...> wrote:
                >> a 1223-digit p2
                >
                > No, that's too large. Try re-reading the OEIS page :-)

                Oops. My PARI/GP script to find p2 was correct but it output decimal
                expansions and didn't stop after the first. I had a bunch of open windows
                and failed to scroll up to see two earlier prp's when I copied the result
                later. How sloppy. The old PARI window is still open and I just checked the
                output. The first 3 titanic prp's of p2 type has 1001, 1002, 1223 digits.

                I now get 1673-digit p1, 1001-digit p2, and p3=151. Primo is running on p2.
                PFGW has proved p3 where p1 was enough help for an easy proof.
                I still don't know what the metric connection is.

                --
                Jens Kruse Andersen
              • David Broadhurst
                ... Congratulations on completing the maths. ... Solution: 1) p1 = 3*lucasU(4,1,2925) - lucasV(4,1,2925)/2 is the smallest titanic prime in the integer
                Message 7 of 8 , Jan 16, 2009
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                  --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
                  <jens.k.a@...> wrote:

                  > I now get 1673-digit p1, 1001-digit p2, and p3=151.

                  Congratulations on completing the maths.

                  > I still don't know what the metric connection is.

                  Inaugural puzzle:

                  > p1 is the smallest titanic prime of the form
                  > sqrt(3*x^2-2) with integer x.
                  >
                  > p2 is the smallest titanic prime of the form
                  > sqrt(5*y^2-4) with integer y.
                  >
                  > p3 is the smallest prime such that
                  > 2*p1*p2*p3-1 is prime.
                  >
                  > Find a metrical connection between p3 and liberty.

                  Solution:

                  1) p1 = 3*lucasU(4,1,2925) - lucasV(4,1,2925)/2
                  is the smallest titanic prime in the integer sequence
                  http://www.research.att.com/~njas/sequences/A001834
                  with a[1] = 1, a[2] = 5, a[n] = 4*a[n-1] - a[n-2],
                  ensuring that a[n]^2 = 3*x^2 - 2, with integer x.
                  Then a[n] = 3*lucasU(4,1,n) - lucasV(4,1,n)/2 solves
                  the recursion, in terms of generalized Lucas numbers.

                  PFGW test:
                  Calling N-1 BLS with factored part 4.90%
                  and helper 1.80% (16.51% proof)
                  3*lucasU(4,1,2925)-lucasV(4,1,2925)/2
                  is Fermat and Lucas PRP!

                  Primo timing:
                  Total=2h 11mn 27s
                  DecimalSize=1673

                  2) p2 = (5*F(2*2394) - L(2*2394))/2
                  is the smallest titanic prime in the integer sequence
                  http://www.research.att.com/~njas/sequences/A002878
                  with a[1] = 1, a[2] = 4, a[n] = 3*a[n-1] - a[n-2],
                  ensuring that a[n]^2 = 5*y^2 - 4, with integer y.
                  Then a[n] = (5*F(2*n) - L(2*n))/2 solves the recursion,
                  in terms of the Fibonacci and classical Lucas numbers.

                  PFGW test:
                  Calling N+1 BLS with factored part 7.61%
                  and helper 3.79% (26.69% proof)
                  (5*F(2*2394)-L(2*2394))/2
                  is Fermat and Lucas PRP!

                  Primo timing:
                  Total=19mn 41s
                  DecimalSize=1001

                  3) p3 = 151
                  is the smallest prime such that 2*p1*p2*p3-1 is prime.

                  PFGW test with helpers p1 and p2:
                  Calling N+1 BLS with factored part 100.00%
                  and helper 0.65% (300.68% proof)
                  151*(3*lucasU(4,1,2925)-lucasV(4,1,2925)/2)*(5*F(2*2394)-L(2*2394))-1
                  is prime!

                  4) Then by googling
                  > 151 prime
                  we are directed, at very first choice, to Chris Caldwell's page
                  http://primes.utm.edu/curios/page.php/151.html
                  reminding us of the Statue of Liberty.

                  I reflected on the poem inscribed within its pedestal
                  http://www.libertystatepark.com/emma.htm
                  which served to remind me of a time when the
                  United States of America were a better nation
                  than that belittled so wantonly on 20/03/03
                  by the ignominious Bush and his shameless toady, Blair:
                  http://tech.groups.yahoo.com/group/primenumbers/message/11863
                  http://tech.groups.yahoo.com/group/primenumbers/message/11875
                  http://tech.groups.yahoo.com/group/primenumbers/message/11883
                  http://tech.groups.yahoo.com/group/primenumbers/message/12042

                  Let us dare to hope for some improvement:
                  > There is a call to life a little sterner,
                  > And braver for the earner, learner, yearner.

                  David
                • Shane
                  You ve got it backwards. The USA is still a toady of England, and your Royal lowness. My native american heritage, is slightly offended, by it s European
                  Message 8 of 8 , Jan 18, 2009
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                    You've got it backwards.
                    The USA is still a toady of England, and your Royal lowness.
                    My native american heritage, is slightly offended, by it's European
                    guests. I'm just plain ashamed of my english lineage.
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