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Factorization of 7 consecutive 1404-digit integers

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  • David Broadhurst
    Let m=67440294559676054016000 y=(m*(10^96+10624986)+22)^2 N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4 then N+k is factorized for k in [0,6].
    Message 1 of 1 , Aug 27, 2007
      Let

      m=67440294559676054016000
      y=(m*(10^96+10624986)+22)^2
      N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4

      then N+k is factorized for k in [0,6].

      Proof:
      http://physics.open.ac.uk/~dbroadhu/cert/ifac7.zip

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

      Neighbouring composites:
      c1360=(N-1)/(2^2*37*113*988979*2759209353591763*2338830543576644077)
      c1362=(N+7)/(2^2*480023*1408111*32195489*5342864083947636432359)
      subjected to 400 ECM curves at B1=250000

      David Broadhurst
      27 August 2007
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