MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are a number of generalizations of the Möbius function out there, which can be found by Google. But I'd just like to know if anything has been said about this: For $k \geq 2$, $k \in \mathbb{Z}$, let $\zeta$ be a primitive $k$-th root of unity. Let $\mu_k(n) = 0$ if $n$ is not squarefree. For squarefree $n$, let $\mu_k(n) = \zeta^{\omega(n)}$. Thank you very much.

Update: A recent beautiful paper of Zhi-Wei Sun, On a pair of zeta functions, answers the question contained here very nicely!

share|cite|improve this question
Any particular reason you're interested in this function? Certainly it has a Dirichlet series with an Euler product. I don't know what else can be said about it. – Qiaochu Yuan Dec 20 '11 at 2:29
I was kind of thinking that $\sum_{n \leq x}\mu_k(n)$ would be a way to "measure" the frequencies of $\omega(n) \bmod k$. – Timothy Foo Dec 20 '11 at 2:56

The partial sums of $\mu_k(n)$ are estimated in a paper of Addison:

A Note on the Compositeness of Numbers, A. W. Addison, Proceedings of the American Mathematical Society, Vol. 8, No. 1 (Feb., 1957), pp. 151-154,

Article Stable URL:

Most of the paper treats $\zeta^{\Omega(n)}$, but the (slightly different) function you are interested in is discussed in the closing paragraph.

share|cite|improve this answer
Thank you very much! – Timothy Foo Dec 20 '11 at 6:08

If you are really interested in the sums $\sum_{n\leq x} \mu(n)^2e^{2i\pi a\omega(n)/k}$ (where $a$ could be an arbitrary integer), you should probably just look at what the Delange-Selberg method gives you. (A good reference is Tenenbaum's book on analytic and probabilistic number theory; the versions there may take some time to parse, because they are explicit in every possible parameter, but the result is worth the effort, since you will be able to deal with many more multiplicative functions using the same tools...)

share|cite|improve this answer
Thank you very much! – Timothy Foo Dec 20 '11 at 6:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.