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Re: Two factors of R(13860)

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  • gchil0@pop.uky.edu
    ... OK, I ll delete mine. ... Sure. Delete the files you ve uploaded and I ll upload the updated versions. Greg
    Message 1 of 16 , Oct 2, 2001
      --- In primenumbers@y..., d.broadhurst@o... wrote:
      > Sorry about that. I had uploaded an old file.
      > Have now remedied that, with your 2 gems at the top:-)

      OK, I'll delete mine.

      > > David...could you also update hd1k.zip with the 2 new factors and
      > > Base10.zip with all 3 new factors?
      > Sorry, my file management skills are minimal.
      > It's already a minor miracle that I managed to
      > get the previous version more or less OK.
      > Please Greg, will you take over as Base10 DB manager?

      Sure. Delete the files you've uploaded and I'll upload the updated
      versions.

      Greg
    • d.broadhurst@open.ac.uk
      ... Sure thing, Boss! That folder is now totally Broadhurst free. Thanks for taking over! David
      Message 2 of 16 , Oct 2, 2001
        Greg Childers commanded:
        > Delete the files you've uploaded
        > and I'll upload the updated versions.
        Sure thing, Boss! That folder is now
        totally Broadhurst free.
        Thanks for taking over!
        David
      • Satoshi TOMABECHI
        Thank you, Greg and David. My program shows that 15793939750*R(13860)/R(10)+1 is prime, however I have little confidence in this result:-) I hope that the
        Message 3 of 16 , Oct 2, 2001
          Thank you, Greg and David.

          My program shows that 15793939750*R(13860)/R(10)+1 is prime,
          however I have little confidence in this result:-)
          I hope that the result is as same as David's.

          The coefficients of cubic equation t3X^3 + t2X^2 + t1X + t0 = 0 are

          t3 = 9499179946...9373462290 1400 digits
          t2 = 2983442091...4520303837 2765 digits
          t1 = -5786178299...9234821312 4165 digits
          t0 = -8297037200...6079883289 2764 digits

          The equation has three real roots r1,r2,r3.

          r1=-1.433940810...*10^{-1401}

          integral part of r2 246804372...4038770926 ( 1382 digits)
          fractional part of r2 0.9565089129....

          integral part of r3 -246804372...8946383526 ( 1382 digits)
          fractional part of r2 0.5941852755....

          Nearest integers of r1,r2 and r3 are not root of the equation.

          # It was not easy to compute roots of the cubic equation.

          Best regards.

          Satoshi Tomabechi
        • Satoshi TOMABECHI
          ... Yes, of course. Furthermore I coded modified KP with additional square tests by you. http://groups.yahoo.com/group/primeform/message/2142 But I have not
          Message 4 of 16 , Oct 2, 2001
            David Broadhurst wrote:
            > Have you coded N+1?
            Yes, of course.
            Furthermore I coded modified KP
            with additional square tests by you.
            http://groups.yahoo.com/group/primeform/message/2142

            But I have not coded combined KP.
            http://groups.yahoo.com/group/primenumbers/message/1743

            > Have you coded 10/3*Max(F1,F2)+Min(F1,F2)>1 ?
            Sorry, I don't know such condition.
            Please tell me reference or pointer to reference.

            Satoshi Tomabechi
          • d.broadhurst@open.ac.uk
            ... By F1 and F2 I meant the fractions of N-1 and N+1. When F1 F2 this condition is what I used for Fib(25561), but it also works for F2 F1. Hence the Max and
            Message 5 of 16 , Oct 2, 2001
              Satoshi TOMABECHI wrote:
              > > Have you coded 10/3*Max(F1,F2)+Min(F1,F2)>1 ?
              > Sorry, I don't know such condition.
              By F1 and F2 I meant the fractions of N-1 and N+1.
              When F1>F2 this condition is what I used for Fib(25561),
              but it also works for F2>F1. Hence the Max and Min.
              David
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