## 10639Re: Is phi(p^2-1)/(p^2-1) bounded?

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• Jan 3, 2003
> Let f(p)=phi(p^2-1)/(p^2-1).
> Say a prime p is "lowest yet" if there is
> no prime q<p with f(q)<f(p).
> The "lowest yet" sequence begins

<snip>

> 61577671, 117048931, ...

I believe it continues as follows
181333151
267190769
331413809
376754951
636510601
1737265531
3019962791

One can obtain fairly low values of f(p) realtively easily. Consider
for example
p=58531393985146662592474024598667898081212671 prime
p-1=2.5.7.13.19.43.53.67.71.73.43520821168673.98287085283258329
p+1=2^8.3^3.11.17.23.29.31.37.41.47.59.61.79.83.89.97.101.103.107.109.
113.127.131.661

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.

Richard Heylen
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