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

[Leroy Quet]

I wrote:
>[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

 
For the terms of this sequence given on the EIS, the (positive) 1's seem 
to only occur at prime indexes m (and m =1) (if I counted the terms 
by-hand correctly).

Are there any composite m's where
sum{k|m, k<= sqrt(m)}  mu(k)
= 1 ??

I know that the exponents raising the primes in any m does not *alone* 
determine the value of this sum,
because, if m = the squarefree product of 3 distinct primes, for example,
 sum{k|m, k<= sqrt(m)}  mu(k) 
= 0 if m/p < p (as when m = 42), and
= -2 if m/p > p (as when m = 30),

where p is the largest of the 3 primes dividing m.

thanks,
Leroy Quet