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Factorization of 4 consecutive 5638-digit numbers

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  • tjw99
    Let n = 25390425*2^18703-1 n = 7*15569*150611*1338793*P5622 n+1 = 25390425*2^18703 = 2^18703*3*5^2*43*7873 n+2 = 25390425*2^18703+1 (prime) n+3 =
    Message 1 of 2 , Jan 23, 2010
      Let n = 25390425*2^18703-1

      n = 7*15569*150611*1338793*P5622
      n+1 = 25390425*2^18703 = 2^18703*3*5^2*43*7873
      n+2 = 25390425*2^18703+1 (prime)
      n+3 = 25390425*2^18703+2 = 2*(25390425*2^18702+1) (2*prime)

      n-1: 25390425*2^18703-2 = 2*4657349734574297*C5622
      n+4: 25390425*2^18703+3 = 3^2*13*47*10528525069*3236067162807210923*C5606

      The 5622-digit cofactor of n has been proven prime by Primo, with a certificate at http://xenon.stanford.edu/~tjw/pp/p5622.zip

      The two primes 25390425*2^18702+1 and 25390425*2^18703+1 are a CC2 (2nd kind) byproduct of a recent CC3 search.

      n-1 and n+4 have been tested with 221 ECM curves at B1=50000.

      Tom
    • Jens Kruse Andersen
      ... Congratulations on improving your month old 5257-digit record in http://tech.groups.yahoo.com/group/primeform/message/9948 That also used a CC2 from a CC3
      Message 2 of 2 , Jan 25, 2010
        Tom wrote:
        > Let n = 25390425*2^18703-1
        >
        > n = 7*15569*150611*1338793*P5622
        > n+1 = 25390425*2^18703 = 2^18703*3*5^2*43*7873
        > n+2 = 25390425*2^18703+1 (prime)
        > n+3 = 25390425*2^18703+2 = 2*(25390425*2^18702+1) (2*prime)
        >
        > The 5622-digit cofactor of n has been proven prime by Primo, with a
        > certificate at
        > http://xenon.stanford.edu/~tjw/pp/p5622.zip
        >
        > The two primes 25390425*2^18702+1 and 25390425*2^18703+1
        > are a CC2 (2nd kind) byproduct of a recent CC3 search.

        Congratulations on improving your month old 5257-digit record in
        http://tech.groups.yahoo.com/group/primeform/message/9948
        That also used a CC2 from a CC3 search.
        I have verified the certificate and updated
        http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm

        --
        Jens Kruse Andersen
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