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An Algorithm to generate primes ...???

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  • Kermit Rose
    ******* On Sat Nov 17, 2007 12:30 pm ((PST)) alby7e7 alby7e7@gmail.com alby7e7 Posted I suppose... if GCD(n! , n + 1) = 1 if GCD(n! , n + 1 ) != 1 also
    Message 1 of 2 , Nov 18, 2007
      *******

      On Sat Nov 17, 2007 12:30 pm ((PST))

      "alby7e7" alby7e7@... alby7e7

      Posted


      I suppose...
      if GCD(n! , n + 1) = 1

      if GCD(n! , n + 1 ) != 1

      also (n+1)+1

      so I do GCD(n!, n+2) = 1?
      if yes I do the sum n! + n+2 also I do another GCD with n+3 and n!...

      *******

      Technically, this is a prime search, not a prime generation formula.
      Another difficulty. GCD(n!,q) = 1 does not make sure that q is prime.

      ******


      q might have a factor > n.


      ******

      n! + n +1 = P

      ** but if n > 7


      n + 1 > n! so I will do n+1+1+1+1+1+1+1+1+1...+1k

      ****

      Apparently you meant to say

      ** but if n! < 7


      n + 1 > n! so I will do n+1+1+1+1+1+1+1+1+1...+1k


      ******



      while n+k > n! and GCD(n!,n+k ) = 1

      n! + n + k = P

      According to you is it true?

      it can generate a lot of primes?


      ******

      It's true that a lot of primes will be in this sequence,
      but we won't know which ones are prime unless we do additional testing.

      *****

      *** for example : ***
      EX.1
      n=0

      n! = 1 , n+1 = 2 GCD(1,1)= 1
      1+1 = 2

      EX.2

      n=1


      n!=1 n+1 = 2 GCD(1,2)= 1
      1+2=3

      EX.3

      n=3


      n! = 6
      n+1 = 4 GCD(6,4)=2
      GCD(6,4+1)=1
      6+5 = 11


      EX.4

      n=4

      n! = 24

      n+1 = 5

      GCD(24,5)=1

      24+5 = 29



      ***

      Example of when number in this sequence is not prime.

      n = 12
      n! = 479001600
      n! + n + 1 = 479001613

      479001613b = 29 * 2503 * 6599

      Yet it is true that
      GCD(479001600, 479001613) = 1

      ****
    • 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 2 of 2 , Nov 18, 2007
        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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