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RE: [PrimeNumbers] RE: Anyone like to prove primality of a Mersenne cofactor?

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  • Jon Perry
    How s the factorization of 2^(2^n)-1 coming along? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/
    Message 1 of 7 , Mar 31, 2003
      How's the factorization of 2^(2^n)-1 coming along?

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths/
      http://www.users.globalnet.co.uk/~perry/DIVMenu/
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
    • Jon Perry
      Voodoo De Ja!!! Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP
      Message 2 of 7 , Mar 31, 2003
      • Paul Leyland
        Will Edgington and I try to keep our tables of Mersenne factorizations up to date by synchronizng every week or so. The latest indicated that another
        Message 3 of 7 , Apr 1, 2003
          Will Edgington and I try to keep our tables of Mersenne factorizations up to date by synchronizng every week or so. The latest indicated that another Mersenne number had been completely factored. Neither Will nor I have the resources at the moment to prove the 2193-digit cofactor of M(7417) is prime. It's certainly a strong pseudoprime.

          If anyone would care to complete the proof, please let Will and me know the result. The known prime factors of M(7417) are 118673, 16269026327 and 3888241452787718190543521.

          Apologies for not crediting the person who found the largest of the three factors given above. I don't know who he or she is.

          Paul
        • Michael Bell
          Hi, Well I would do, but Primo doesn t seem to like wine. Has anyone had any success with that, or does anyone know of a Linux ECPP tool with comparable
          Message 4 of 7 , Apr 1, 2003
            Hi,

            Well I would do, but Primo doesn't seem to like wine. Has anyone had any
            success with that, or does anyone know of a Linux ECPP tool with comparable
            speed?

            Mike.

            Paul Leyland wrote:
            > Will Edgington and I try to keep our tables of Mersenne factorizations
            > up to date by synchronizng every week or so. The latest indicated that
            > another Mersenne number had been completely factored. Neither Will nor
            > I have the resources at the moment to prove the 2193-digit cofactor of
            > M(7417) is prime. It's certainly a strong pseudoprime.
            >
            > If anyone would care to complete the proof, please let Will and me know
            > the result. The known prime factors of M(7417) are 118673, 16269026327
            > and 3888241452787718190543521.
            >
            > Apologies for not crediting the person who found the largest of the
            > three factors given above. I don't know who he or she is.
            >
            > Paul
            >
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          • David Broadhurst
            (2^7417-1)/(1930694161304071*3888241452787718190543521) 2193 c8 2002 Mersenne cofactor, ECPP
            Message 5 of 7 , Apr 1, 2003
              (2^7417-1)/(1930694161304071*3888241452787718190543521) 2193 c8 2002
              Mersenne cofactor, ECPP
            • David Broadhurst
              These are the 6 smallest unproven probably prime Mersenne cofactors known to me: (2^14561-1)/8074991336582835391 (2^17029-1)/418879343
              Message 6 of 7 , Apr 1, 2003
                These are the 6 smallest unproven probably prime
                Mersenne cofactors known to me:

                (2^14561-1)/8074991336582835391

                (2^17029-1)/418879343

                (2^20887-1)/(694257144641*3156563122511*28533972487913*\
                1893804442513836092687)

                (2^28759-1)/226160777

                (2^28771-1)/104726441

                (2^32531-1)/(65063*25225122959)

                Updates welcomed!

                David Broadhurst
              • jbrennen
                ... About as well as the factorization of 2^(2^n)+1 (the Fermat numbers). ... 2^(2^n)-1 == prod(i=0,n-1,2^(2^i)+1) So, completely factored up to 2^(2^12)-1.
                Message 7 of 7 , Apr 1, 2003
                  --- In primenumbers@yahoogroups.com, "Jon Perry" <perry@g...> wrote:
                  > How's the factorization of 2^(2^n)-1 coming along?

                  About as well as the factorization of 2^(2^n)+1 (the Fermat numbers).

                  :)


                  2^(2^n)-1 == prod(i=0,n-1,2^(2^i)+1)


                  So, completely factored up to 2^(2^12)-1.

                  2^(2^13)-1, not yet factored.
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