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Factorization of 6 consecutive 2139-digit integers

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  • djbroadhurst
    Application of a Prouhet-Tarry-Escott solution: factorization of 6 consecutive 2139-digit integers Let b = 5026578700; m = 13416739015680*b; x =
    Message 1 of 4 , Dec 19, 2009
      Application of a Prouhet-Tarry-Escott solution:
      factorization of 6 consecutive 2139-digit integers

      Let b = 5026578700; m = 13416739015680*b;
      x = (2^67+28683395)*b; p25 = 6805713408228112745527499;
      y = (8*x^6+24*x^5+50*x^4+54*x^3+41*x^2+12*x-148)^2;
      N = (y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-3;
      then
      p2088 = N/(3*5*115523*2364592753*16987001959*p25);
      p2131 = (N+1)/(2^2*7^2*462899);
      p2132 = (N+4)/(13*652739);
      p2116 = (N+5)/(2^5*5^2*37*241*28843*794200385809);
      are prime and the factorizations of (N+2) and (N+3) in
      http://physics.open.ac.uk/~dbroadhu/cert/ifacpte6.zip
      contain no prime with more than 172 digits.

      Comments: This sets a record at http://tinyurl.com/ycxabks
      for both 5 and 6 consecutive factorizations. The algebraic
      factorization of (N+2)*(N+3) comes from a sextic
      substitution in Prouhet-Tarry-Escott solution 10.12.151 of
      http://www.math.uni.wroc.pl/~jwr/PTE resulting in 4 linear
      factors, 5 quadratics, 12 cubics, a quartic and 15 sextics.
      OpenPFGW and GMP-ECM found 44 triplets of probable primes,
      each promising a run of 5 factorizations after NFS effort on
      sextics. One was extended to a run of 6, by extracting the
      prime p25. After using ECM and Msieve's MPQS, 9 composite
      factors of (N+2)*(N+3) remained for GGNFS. Two composites
      had GNFS difficulty less than 122, leaving 7 sextics with
      SNFS difficulty 180. Pari-GP and Primo proved primality of
      factors below 180 digits and above 2000 digits,
      respectively. Neither of the composite numbers
      c2117 = (N - 1)/(2*6911*18959*675413*81134281)
      c2118 = (N + 6)/(3*167858252466989944843)
      is likely to have a prime divisor with less than 30 digits.

      David Broadhurst, 19 December, 2009
    • Jens Kruse Andersen
      ... Congratulations to both! http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm is now updated. -- Jens Kruse Andersen
      Message 2 of 4 , Dec 21, 2009
        David Broadhurst wrote:
        > Application of a Prouhet-Tarry-Escott solution:
        > factorization of 6 consecutive 2139-digit integers
        >
        > http://physics.open.ac.uk/~dbroadhu/cert/ifacpte6.zip

        Tom wrote:
        > Factorization of 4 consecutive 5257-digit numbers
        > Let n = 297079965*2^17434-1
        >
        > n = 4363*22567639*p5246
        > n+1 = 297079965*2^17434 = 2^17434*3^2*5*7*13*72547
        > n+2 = 297079965*2^17434+1 (prime)
        > n+3 = 297079965*2^17434+2 = 2*(297079965*2^17433+1) (2*prime)

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

        --
        Jens Kruse Andersen
      • djbroadhurst
        Application of a multigrade equality found by Tarry: factorization of 8 consecutive 850-digit integers Let x = 429*(2^50 + 9096237); y = (16*x^6 - 72*x^4 +
        Message 3 of 4 , Dec 22, 2009
          Application of a multigrade equality found by Tarry:
          factorization of 8 consecutive 850-digit integers

          Let x = 429*(2^50 + 9096237);
          y = (16*x^6 - 72*x^4 + 81*x^2 - 25)^2;
          N = 72*(y - 2^2)*(y - 2^8)*(y - 5^4)*(y - 21^2)/14! - 3;
          then
          p849 = N/3;
          p812 = (N + 1)/(2*9277*9862129*3520556383*110169881898372133);
          p850a = (N + 2);
          p850b = (N + 5)/2;
          p845 = (N + 6)/(3*19*1009);
          p844 = (N + 7)/(2^2*208577);
          are prime and the factorizations of (N + 3) and (N + 4) in
          http://physics.open.ac.uk/~dbroadhu/cert/ifacgast.zip
          contain no prime with more than 108 digits.

          Comments: Algebraic factorization of (N + 3)*(N + 4) comes from
          a sextic substitution in a multigrade equality discovered in
          1913 by Gaston Tarry, amateur extraordinaire:
          http://www-history.mcs.st-and.ac.uk/Biographies/Tarry.html
          OpenPFGW, GMP-ECM, Pari-GP and Primo achieved the rest.
          Remarkably, p849, p850a and p850b have BLS proofs, with
          helpers of less than 16 digits from the factorization of
          N + 3 = 3*(p849 + 1) = p850a + 1 = 2*(p850b - 1).
          Neither of the composites
          c837 = (N - 1)/(2^2*95819*27609121);
          c824 = (N + 8)/(5*311137*114685080442874011127);
          is likely to have a prime divisor with less than 25 digits.

          David Broadhurst, 22 December, 2009
        • Jens Kruse Andersen
          ... Congratulations! The record page is updated again. -- Jens Kruse Andersen
          Message 4 of 4 , Dec 22, 2009
            David wrote:
            > Application of a multigrade equality found by Tarry:
            > factorization of 8 consecutive 850-digit integers

            Congratulations!
            The record page is updated again.

            --
            Jens Kruse Andersen
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