k is coprime to m, 1 <= k <= m/r

From: Leroy Quet (qqquet_at_mindspring.com)
Date: 09/09/04 

Date: 9 Sep 2004 06:53:35 -0700

In general, if phi(m) = number of positive integers which are <= m and
are coprime to m,

for r = positive integer, q >= 0,

limit{m->oo}

(1/(m^q phi(m))) sum{1<=k<=m/r, GCD(k,m)=1} k^q

=

1/(r^(q+1) (q+1)).

I started wondering, as a result of the above,
how does phi(m/r,m) compare with phi(m)/r,
where phi(n,m) is number of positive integers <= n and coprime to m?

phi(m/r,m) = phi(m)/r - sum{k|m} mu(k) {m/(kr)},

where {x} is the fractional part of x,
and where mu() is the Mobius (Moebius) function.

But I do not know how to approximate the sum.
(It would definitely be bounded by +-2^(b(m)-1), where b(m) is the
number of distinct primes dividing m.)

I am guessing there are some interesting questions related to this
somehow.

thanks,
Leroy Quet

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