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Re: [PrimeNumbers] prime gap proof

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  • Jens Kruse Andersen
    ... I think you swapped the two tuples. As others have pointed out, the second (2,4,2) occurs. But the first (2,0,2) does not occur. The 12 which do not occur:
    Message 1 of 4 , Sep 2, 2011
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      pbtoau wrote:
      > If you take the gaps between consecutive odd primes modulo 6, they are
      > either 0, 2 or 4. For 4 consecutive primes, out of 27 permutations,
      > only 15 appear. Can someone point me to an elementary proof of why this
      > is so? Some of the missing ones are obvious (2,2,2). I am particularly
      > interested in (2,0,2) and (2,4,2). The first occurs and the second does
      > not.

      I think you swapped the two tuples.

      As others have pointed out, the second (2,4,2) occurs.
      But the first (2,0,2) does not occur.
      The 12 which do not occur:
      (0, 2, 2), (0, 4, 4), (2, 0, 2), (2, 2, 0), (2, 2, 2), (2, 2, 4),
      (2, 4, 4), (4, 0, 4), (4, 2, 2), (4, 4, 0), (4, 4, 2), (4, 4, 4).

      They are all inadmissible modulo 3, meaning that 3 will divide at least one
      of four numbers with those gaps.
      It may be easier to see what happens if you for each of the four numbers
      write the difference from the first number modulo 6.
      The difference of the first number from itself is 0.
      Gaps (2, 0, 2) corresponds to these differences to the first number modulo 6:
      [0, 2, 2+0, 2+0+2] = [0, 2, 2, 4]
      This as inadmissible modulo 3 because all values modulo 3 occur.

      See more at http://primes.utm.edu/glossary/xpage/ktuple.html
      You can check whether a tuple written on this form is admissible here:
      http://primes.utm.edu/glossary/includes/ktuple.php
      For [0, 2, 2, 4] it says: "fails at prime 3 (term 4)"
      This means the tuple becomes inadmissible modulo 3 when you get to the
      4th term.

      --
      Jens Kruse Andersen
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