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p(i)# divides n!

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  • John W. Nicholson
    p(i)# | (n+1)! where p(i) is the largest prime
    Message 1 of 1 , Dec 2, 2001
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      p(i)# | (n+1)!
      where p(i) is the largest prime <= n! and p(i)# is primorial.
      So a gap same size n can be found at (n+1)!/p(i)#.

      This helps to find g to be smaller, but p(i)# is still huge.
      Also, by the same logic as with n! can be applied to p(i)# to have a gap
      after p(i)#+1 of minimum size p(i)+1.

      --
      John W. Nicholson mailto:JohnW.Nicholson@...
      http://web2.airmail.net/johnwn1/
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