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Factorization of 4 consecutive 5257-digit numbers

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  • tjw99
    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 =
    Message 1 of 2 , Dec 19, 2009
      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)

      n-1 = 2*4846109*10719641497868929976071*C5228
      n+4 = 3*11*C5256

      The 5246-digit cofactor of n has been proven prime by Primo, with a certificate at http://xenon.stanford.edu/~tjw/pp/p5246.zip

      The two primes 297079965*2^17433+1 and 297079965*2^17434+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
    • djbroadhurst
      ... Well done, Tom. I imagine that it will not be long until you beat the k=5 record without needing PTE. I remark that 2500 digits would be ample to scare off
      Message 2 of 2 , Dec 20, 2009
        --- In primeform@yahoogroups.com,
        "tjw99" <tjw99@...> wrote:

        > The two primes 297079965*2^17433+1 and 297079965*2^17434+1
        > are a CC2 (2nd kind) byproduct of a recent CC3 search.

        Well done, Tom. I imagine that it will not be long
        until you beat the k=5 record without needing PTE.
        I remark that 2500 digits would be ample to scare
        off the SNFS brigade :-)

        Best regards

        David
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