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Factorization of 4 consecutive 10673-digit numbers

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  • tjw99
    Let n = 245363571*2^35426-3 n = 3*349*111602773267*953666301013*43440278284896679*prp10630 n+1 = 245363571*2^35426-2 = 2*(245363571*2^35425-1) (2*prime) n+2 =
    Message 1 of 2 , Apr 27, 2010
      Let n = 245363571*2^35426-3

      n = 3*349*111602773267*953666301013*43440278284896679*prp10630
      n+1 = 245363571*2^35426-2 = 2*(245363571*2^35425-1) (2*prime)
      n+2 = 245363571*2^35426-1 (prime)
      n+3 = 245363571*2^35426 = 2^35426*3^2*1117*24407

      My recent search for a gigantic CC3 yielded about 260 gigantic Sophie Germain prime pairs. For each SG prime pair p, 2p+1, I attempted to factor 2p-1 and 2p+3, first by trial division, and then by ECM. Trial division and ECM up to B1=2000 yielded nothing but composite cofactors, but ECM at B1=11000 finally produced a 10630-digit PrP cofactor.

      n-1 = 245363571*2^35426-4 = 2^2*5*720418081*209358848887*C10652
      n+4 = 245363571*2^35426+1 = 5*7*17*67*137*C10666

      n-1 and n+4 have been tested with 74 ECM curves at B1=11000.

      Tom
    • Jens Kruse Andersen
      ... Congratulations! http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm is updated. -- Jens Kruse Andersen
      Message 2 of 2 , Apr 28, 2010
        Tom wrote:
        > Let n = 245363571*2^35426-3
        >
        > n = 3*349*111602773267*953666301013*43440278284896679*prp10630
        > n+1 = 245363571*2^35426-2 = 2*(245363571*2^35425-1) (2*prime)
        > n+2 = 245363571*2^35426-1 (prime)
        > n+3 = 245363571*2^35426 = 2^35426*3^2*1117*24407

        Congratulations!
        http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm
        is updated.

        --
        Jens Kruse Andersen
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