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Re: [PrimeNumbers] An Algorithm to generate primes ...???

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  • Steve Palmer
    ... I don t think that was the intended claim. I think the claim was probably meant to be that [GCD (n!, n+1 ) = 1] implies that [n! + n + 1] is prime. It
    Message 1 of 2 , Nov 18, 2007
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      On Sun, 18 Nov 2007, Kermit Rose wrote:
      > Another difficulty. GCD(n!,q) = 1 does not make sure that q is prime.

      I don't think that was the intended claim. I think the claim was probably
      meant to be that [GCD (n!, n+1 ) = 1] implies that [n! + n + 1] is prime.

      It looks to me like maybe it's [finding/generating] series A073309,
      http://www.research.att.com/~njas/sequences/A073309. (Primes of the form
      n! + n + 1)

      Although I'm not sure I can explain it clearly, it makes intuitive sense
      to me for factors from 2 through n+1.

      A.) n+1 is not a factor of n!, so it can't be a factor of [n! + (n+1)].
      B.) Thinking of a number line, GCD=(n!,n+1) = 1 implies that the (n+1)
      portion of [n! + (n + 1)] can't be evenly split into chunks which also fit
      evenly into the (n!) part, so factors {2..n}, from n!, are not factors of
      [n! + n + 1].

      So I think factors from 2 through (n+1) can be ruled out. I'm not clear
      about factors bigger than n+1 though. Maybe they can't be factors of [n!
      + n + 1], but I haven't noticed a reason why and I don't have more time to
      think about it right now.

      I have to admit I didn't follow all the 1+1+1+1 stuff. I skipped straight
      to the examples, so maybe I'm misunderstanding the original post. I hope
      not.

      Steve
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