Max value of x/phi(x)
Benoit Cloitre
abcloitre at wanadoo.fr
Tue Aug 31 13:32:15 CEST 2004
> Since phi(x) = x prod{ 1 - 1/p_i } for all distinct primes p_i that
> divide x,
> this is the same as asking for a lower bound on prod{ 1 - 1/p_i }.
>
> Clearly there is no absolute lower bound, and any new minimum on that
> product will first be reached when x is the product of the first k
> primes
> for some k; I guess you can use prime density theorems to get a rough
> idea of what k needs to be for a given minimum, but I'm not sure
> exactly
> how you'd go about that.
And a Mertens theorem said : limit n-->infty log(p_n)*prod{i=1,n, 1 -
1/p_i }=exp(Eulergamma)
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