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

[Pieter Moree]

Leroy Quet wrote:

>
> Are there any composite m's where
> sum{k|m, k<= sqrt(m)}  mu(k)
> = 1 ??
Below 500: 240,270,300,360,420,450,480

Note that none of these numbers is squarefree.

The claims I made in my earlier mail on this would imply the following.

For squarefree m the above sum equals 1 iff m is a prime.

If m>1 is squarefull (i.e. all exponents >=2), then trivially the sum
equals zero.

> 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.
>

Clearly size matters...

Bests,
Pieter