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

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  • richard_heylen <richard_heylen@yahoo.co.
    Jan 3, 2003
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      --- In primenumbers@yahoogroups.com, "David Broadhurst
      <d.broadhurst@o...>" <d.broadhurst@o...> wrote:
      > 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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