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

[Pieter Moree]

Leroy Quet wrote:

>
> If mu(k) is the Mobius(Moebius) function,
>
> then
>
> sum{k|m, k<= sqrt(m)}  mu(k)

Let us put h(m)=\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??
>
I think this is an interesting problem.

Sequence A086596 is concerned with an invariant that occurs as an
example in
D. Gijswijt and P. Moree, A set-theoretic invariant,
(vide the ArXiv or http://staff.science.uva.nl/~moree/preprints.html)

This invariant, g(m), is an integer and can be written
as
g(t)=(-1)^t/2\sum_{d|p_1...p_t, d\le \sqrt{p_1...p_t}mu(d),

where p_1...p_t is the product of the first t primes.

we thus have g(t)=(-1)^t*h(p_1...p_t)/2.

The beginning of the above sequence
{h(1),h(2),h(3),...}
suggests that the numbers stay
small, but they certainly can get quite large as Tony Noe's
computation of g(t) shows.

I would not be surprised if there is no number C such that
|h(m)|<=C for every integer m.

If m is squarefree and has an odd number of prime factors >=3, then
I suspect that h(m) is even.

If m is squarefree and has an even number of prime factors, then
I suspect that h(m)=0.

The last two assertions follow
(I think) by considering the example in the
quoted paper and replacing p_1,...,p_t by the distinct prime divisors
of m. Quite likely, there is a more direct proof.
I presently have too little time to think about this myself.

Cheers,
Pieter