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10656RE: [PrimeNumbers] Re: Is phi(p^2-1)/(p^2-1) bounded?

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  • Phil Carmody
    Jan 4, 2003
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      --- Jon Perry <perry@...> wrote:

      Quoting me:

      > 'I expect it to drift downards so it's not well-defined.
      > (or maybe it is, maybe it's zero. On average numbers have 1/eps distinct
      > divisors, i.e. a divergent number. That's got to take a toll on the phi
      > value. Any sample up to 300000# is puny compared with the sizes of almost
      > all integers...)'

      > 'Therefore the highest value you will find will be
      > 1/2*2/3 = 1/3 from p=3,5,17
      > and 1/3-eps from numbers with a few prime factors larger than 2 or 3 in p+1
      > and p-1.'

      Where the 3 was already pointed out as a typo.

      > Cough, cough. You make these up, or do they come naturally?

      OK John. Which of the two statements do you think is wrong?
      And why?

      Come on, show us the flaws, I yearn to be enlightened by your razer-sharp
      mathematical quill. I'll even fill in the ellipses, if you like, as have a
      feeling you're getting confused by my elision.

      �10 to Oxfam for each statement you persuade me to retract. I trust you'll


      The answer to life's mystery is simple and direct:
      Sex and death. -- Ian 'Lemmy' Kilminster

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