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Re: the factorization of 2^3780-1

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  • djbroadhurst
    ... Yes, Andrey: you failed to understand that Will expects you to do the remaining Aurifeuillian factorization for yourself. Here, I ask Pari-GP to do it for
    Message 1 of 2 , Apr 18, 2010
      --- In primenumbers@yahoogroups.com,
      Andrey Kulsha <Andrey_601@...> wrote:

      > Will Edgington reports in
      > http://www.garlic.com/~wedgingt/factoredM.txt
      > that this number is fully factored

      > Maybe I misunderstand something there..?

      Yes, Andrey: you failed to understand that Will expects
      you to do the remaining Aurifeuillian factorization
      for yourself.

      Here, I ask Pari-GP to do it for you:

      {wills=[7561,457381,2143261,43887888187572165151544641,
      53557082595126165043094157045628806518641,
      596526861195233662750200340726624395569101];}

      N=subst(polcyclo(3780),x,2);
      will=N/prod(k=1,#wills,wills[k]);

      \\ http://homes.cerias.purdue.edu/~ssw/cun/pmain1209
      \\ 2^2h+1=L.M, L=2^h-2^k+1, M=2^h+2^k+1, h=2k-1.

      k=473; h=2*k-1; L=2^h-2^k+1; M=2^h+2^k+1;

      print(gcd(will,[L,M]))

      [2670383929348266924341948931151336592642126516143095249535829086413119821,
      4432174718656154579820150556638898270574244648997765405952796721]

      David
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