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18001Re: [PrimeNumbers] additive combinations all prime?

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  • Phil Carmody
    May 5, 2006
      --- Mark Underwood <mark.underwood@...> wrote:
      > Note for below: negatives included as primes.
      > Given positive a,b, a<b, we want b+a and b-a both prime. First
      > solution: (a,b): (1,4)
      > Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a all
      > prime. First solution: (a,b,c) :(1,4,8)
      > Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-b+a, d-
      > c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-c+b+a
      > all prime. (11 combinations.) First solution: (a,b,c,d): (3,5,8,13)
      > (No "fibbing"!)

      -d+c+b-a and d-c-b+a are the same prime
      -d+c-b+a and d-c+b-a are the same prime
      -d-c+b+a and d+c-b-a are the same prime

      > Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations to be
      > prime. First solution: (a,b,c,d,e): (3,10,12,15,27)
      > What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42
      > combinations. All prime? Eeek! If it is solved it won't be on my
      > computer....

      Just line up the sums/differences, so that many combinations evaluate to the
      same thing.

      I think your count of combinations is incorrect. There are 2^(n-1) ways to
      chose the signs. See above examples of miscounting.


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