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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
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      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/
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    • 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
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      • David Broadhurst
        (2^7417-1)/(1930694161304071*3888241452787718190543521) 2193 c8 2002 Mersenne cofactor, ECPP
        Message 3 of 7 , Apr 1, 2003
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          (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 4 of 7 , Apr 1, 2003
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            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 5 of 7 , Apr 1, 2003
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              --- 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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