10639Re: Is phi(p^2-1)/(p^2-1) bounded?
- Jan 3, 2003--- In email@example.com, "David Broadhurst
<d.broadhurst@o...>" <d.broadhurst@o...> wrote:
> Let f(p)=phi(p^2-1)/(p^2-1).<snip>
> Say a prime p is "lowest yet" if there is
> no prime q<p with f(q)<f(p).
> The "lowest yet" sequence begins
> 61577671, 117048931, ...I believe it continues as follows
One can obtain fairly low values of f(p) realtively easily. Consider
So the first 33 primes are factors of p^2-1 and I believe this gives
an f(p) around 0.113
This is significantly smaller than the f(p) around 0.148 for the best
of the minimal examples listed.
To break the 0.10 barrier you need primes up to 257.
To break the 0.05 barrier you need primes up to 75029
By this stage, the numbers are getting rather large.
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