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Consecutive numbers all factored

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  • jbrennen
    Just a quick thought I had -- I was wondering if anyone has heard of any results in this area... For a given n =2, what is the largest n-tuple of consecutive
    Message 1 of 4 , Nov 21, 2002
      Just a quick thought I had -- I was wondering if anyone has
      heard of any results in this area...

      For a given n>=2, what is the largest n-tuple of consecutive integers
      for which the prime factorization of each member is known?

      Is there any method for constructing arbitrarily large
      n-tuples of this form? I'm guessing that there isn't, and
      that all of the "large" such n-tuples known today depend on
      very large primes.

      For instance, for n==2, an obvious answer is:
      (2^13466917-1,2^13466917).

      For n==3, I can easily construct:

      (33218925*2^169690-1,33218925*2^169690,33218925*2^169690+1)

      For n==5, go find the largest known BiTwins and we get:

      (928223645*2503#*2^2-2,
      928223645*2503#*2^2-1,
      928223645*2503#*2^2,
      928223645*2503#*2^2+1,
      928223645*2503#*2^2+2)

      So I'm curious, what is the longest titanic factorized n-tuple?
      The above-mentioned 5-tuple is titanic. Can someone give me
      6 consecutive titanic numbers along with their prime factorizations?
    • Paul Jobling
      Hi Jack, ... The only way that occurs to me right now is about as hard as finding a CC5, for which the record is ~500 digits. This would be by finding a highly
      Message 2 of 4 , Nov 22, 2002
        Hi Jack,

        > So I'm curious, what is the longest titanic factorized n-tuple?
        > The above-mentioned 5-tuple is titanic. Can someone give me
        > 6 consecutive titanic numbers along with their prime factorizations?

        The only way that occurs to me right now is about as hard as finding a CC5,
        for which the record is ~500 digits. This would be by finding a highly
        composite N (eg k.2^n or p#) such that the following 5 numbers are all prime:

        3N-1, 6N+1, 6N-1, 3N+1, 2N+1.

        Then the factorisations of the 6 numbers

        6N-2, 6N-1, 6N, 6N+1, 6N+2, 6N+3

        could all be found.

        Any technique that relies on just 4 primes fails because then there would have
        to be a number that was divisible by 3 but for which there was no trivial
        factorisation, I think.

        Regards,

        Paul.


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      • Paul Jobling
        ... And to prove the concept, I proved the following numbers prime: 98497038*300#-1 98497038*300#+1 98497038*300#/2-1 98497038*300#/2+1 98497038*300#/3+1
        Message 3 of 4 , Nov 22, 2002
          > 3N-1, 6N+1, 6N-1, 3N+1, 2N+1.
          >
          > Then the factorisations of the 6 numbers
          >
          > 6N-2, 6N-1, 6N, 6N+1, 6N+2, 6N+3
          >
          > could all be found.

          And to prove the concept, I proved the following numbers prime:

          98497038*300#-1
          98497038*300#+1
          98497038*300#/2-1
          98497038*300#/2+1
          98497038*300#/3+1

          Giving complete factorisations for the 6 numbers

          98497038*300#+k k=-2...+3

          Regards,

          Paul.

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        • Andrey Kulsha
          If we find such 33-digit n that (6*n)^6/2-1, (6*n)^6/3-1, (6*n)^6-5, (6*n)^6/6-1, (6*n)^6-7 are all prime, then 11 numbers of about 200 digits from (6*n)^6-9
          Message 4 of 4 , Nov 22, 2002
            If we find such 33-digit n that (6*n)^6/2-1, (6*n)^6/3-1, (6*n)^6-5, (6*n)^6/6-1, (6*n)^6-7 are all prime, then 11 numbers of about 200 digits from (6*n)^6-9 through (6*n)^6+1 might be easily factored with a good chance.

            Best wishes,

            Andrey


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