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Factorization of 6 consecutive 1404-digit integers

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  • David Broadhurst
    Let m = 67440294559676054016000 a = (y-22^2)*(y-61^2)*(y-86^2)*(y-127^2)*(y-140^2)*(y-151^2)/(6*m) b =
    Message 1 of 3 , Aug 20, 2007
      Let

      m = 67440294559676054016000
      a = (y-22^2)*(y-61^2)*(y-86^2)*(y-127^2)*(y-140^2)*(y-151^2)/(6*m)
      b = (y-35^2)*(y-47^2)*(y-94^2)*(y-121^2)*(y-146^2)*(y-148^2)/m

      Then b = 6*a+1. Moreover, with

      y = (m*(10^96+9581328)+22)^2

      each element of

      [3*a-1, 6*a-1, (3*a+1)/(23*24943*13071241), 2*a+1]

      is prime.

      By factorizing all 24 of the 119-digit algebraic cofactors
      in a and b, we obtain complete factorizations of the 6
      successive 1404-digit integers 6*a+k-2 with k in [0,5].

      Proof:
      http://physics.open.ac.uk/~dbroadhu/cert/ifac6a.zip

      Software used:
      GGNFS
      GMP-ECM
      Msieve
      OpenPFGW
      Pari-GP
      Primo

      Neighbouring composites:
      c1376=(2*a-1)/(59*104882036089*123707274359671);
      c1403=(3*a+2)/2;
      subjected to 320 ECM curves at B1=250000

      David Broadhurst
      20 August 2007
    • Jens Kruse Andersen
      ... Congratulations! http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated. The algebraic cofactors have been upgraded from relatively
      Message 2 of 3 , Aug 20, 2007
        David Broadhurst wrote:
        > By factorizing all 24 of the 119-digit algebraic cofactors
        > in a and b, we obtain complete factorizations of the 6
        > successive 1404-digit integers 6*a+k-2 with k in [0,5].

        Congratulations!
        http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated.
        The algebraic cofactors have been upgraded from
        "relatively easy to factor" to "possible to factor"
        since your 1104-digit record for 6 numbers.

        Donovan Johnson has set a 2135-digit record for 5 numbers.
        It contains a twin prime but it and others were computed to
        set this record, so it doesn't count as a record with an "origin".

        --
        Jens Kruse Andersen
      • David Broadhurst
        ... updated. I liked the peculiarity that 3/4 of the 1400-digit primes were provable by BLS. Thay was not my intent; it just happened that way: the
        Message 3 of 3 , Aug 21, 2007
          --- In primeform@yahoogroups.com, "Jens Kruse Andersen"
          <jens.k.a@...> wrote:


          > Congratulations!
          > http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is
          updated.

          I liked the peculiarity that 3/4 of the 1400-digit primes
          were provable by BLS. Thay was not my intent;
          it just happened that way: the 3-constellation
          [3*a-1, 6*a-1, 2*a+1] fell out of pfgw -f -d,
          unbidden.

          In the long term, the k=5 record will need
          a 4-constellation, to avoid ECPP. But I leave
          that to the master sievers of linear forms.

          I prefer shallow sieving of 12th (resp. 8th)
          order polynomials for k = 6,7 (resp. 8,9,10),
          followed by serious ECM in a minority of
          promising cases.

          I chose 1400 digits for k=6 so that any
          significant later improvement would entail SNFS,
          rather than GNFS, which works fine for me
          below 120 digits.

          David
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