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Re: Hardy Littlewood Conjecture

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  • djbroadhurst
    ... Let s try to parse that carefully. (I failed to do so, previously, sorry.) Proposition: Let x be a positive integer for which there exists at least one
    Message 1 of 3 , Jan 6, 2011
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      --- In primenumbers@yahoogroups.com,
      "ALBERTO" <arfzelaya@...> wrote:

      > Let p any prime of the form 6k+-1
      > and n any integer number
      ...
      > if an integer x is not equal to pn+-k
      > it won't be equal to 6ab+-a+-b

      Let's try to parse that carefully.
      (I failed to do so, previously, sorry.)

      Proposition: Let x be a positive integer for which there exists
      at least one triplet (a,b,s) with integers a >= b > 0
      and a sign s = +/-1 such that x = 6*a*b + s*(a+b).
      Then there exists at least one such triplet for which
      at least one of 6*a + s and 6*b + s is prime.

      Proof: 6*x + 1 = (6*a + s)*(6*b + s) has at least one prime divisor.

      Example 1: With x = 1627604, there are 5 triplets,
      yielding only 1 prime, since 6*x + 1 = 5^10.

      Example 2: With x = 6197024, there are 63 triplets,
      yielding 7 primes, since 6*x + 1 = 23#/6.

      Question for Alberto: What has this to do with H+L ?

      David (with apologies for earlier stupidity)
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