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RE: [PrimeNumbers] A question?

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  • Chris Caldwell
    From: Bob Gilson ... why? Call them n-1, n, and n+1 instead of a, b and c. Then a*c-b is n^2 - n - 1. This is never divisible by 2, so produces twice the
    Message 1 of 8 , Jun 5, 2008
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      From: Bob Gilson
      > Let a,b,c be consecutive natural numbers, where a >1
      > (a*c)-b appears to offer a greater density of primes than the norm;
      why?

      Call them n-1, n, and n+1 instead of a, b and c. Then a*c-b is
      n^2 - n - 1. This is never divisible by 2, so produces twice the
      primes "as usual." It is never divisible by 3, so that gives a
      factor of 3/2. Nor by 7, that gives 7/6. So with these alone
      we expect 3.5 times the usual number.

      CC
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