Loading ...
Sorry, an error occurred while loading the content.

10633Re: Is phi(p^2-1)/(p^2-1) bounded?

Expand Messages
  • David Broadhurst <d.broadhurst@open.ac.u
    Jan 3, 2003
    • 0 Attachment
      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
      2, 3, 5, 11, 29, 131, 139, 181, 419, 1429, 17291, 23561,
      23869, 188189, 315589, 483209, 614041, 1624349, 1729001,
      8242961, 15431989, 22486309, 27033161, 36058021, 57762431,
      61577671, 117048931, ...

      (117048931^2-1)/4=
      2*3*5*7*11*13*19*23*29*31*41*73*97

      What comes next?
    • Show all 20 messages in this topic