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

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• 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
reciprocate?

Phil

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

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