sequences from inequalities

[Jud McCranie]

At 05:53 PM 8/30/2004, N. J. A. Sloane wrote:

 >There is a minimum for that ratio.  There is another theorem that says that
> >x/phi(x) <= sqrt( 2x), so the ratio must be unbounded.

%N A097604 Floor( phi(n)*sqrt*2n) ) - n.
...
>%C A097604 This is probably known to be always >= 0 - reference?

"Elementary Number Theory", 4th edition, by David Burton, problem 7a in 
section 7.2 has the equivalent of n/phi(n) <= 2*sqrt(n) - not 
sqrt(2n).  That's probably where I got that inequality, but wrote it down 
incorrectly.