Sum{k|m, k<= sqrt(m)} mu(k)

[Leroy Quet]

[I just posted this set of questions to sci.math, but I do not expect 
many responses there.]



If mu(k) is the Mobius(Moebius) function,

then

sum{k|m, k<= sqrt(m)}  mu(k)

gives us the sequence:
1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, -1, 1, 0, 0, 0,...

(Sequence A068101 in the EIS:
http://www.research.att.com/projects/OEIS?Anum=A068101 )


What can be said about this sequence, such as its asymptotics,
the expected number of 0's, +-1's, +-2's, etc below any given value,
and (perhaps) even what its closed-form might be??


All I can deduce right now is that, unspectacularly,

sum{m=1 to n}   sum{k|m, k<= sqrt(m)}  mu(k)  =

sum{1<=k<=sqrt(n)} mu(k) (floor(n/k) +1 -k).


Thanks,
Leroy Quet