Question

What's known about this generalization of the Moebius function?

Answer 1

\[\Large \mu_k(n) = (e^{i 2\pi / k})^n\] it has the same property that for squarefree numbers it takes on this value and 0 for all others. k=2 is the original moebius function.

Answer 2

@ganeshie8

Answer 3

It seems to be obviously multiplicative still and contains more information than the original mobius function since we can always relate it back to the original one like this: \[\Large \mu _4(n)=i^n \\ \Large \mu_2(n)=(-1)^n = \mu(n) \text{ original Moebius function}\] \[\Large \mu_2(n)=(-1)^n =( i^2)^n = (i^n)^2= [\mu_4(n)]^2\] So we can always just square this version to get the original one back.

I'm just playing around here, I came up with this on my own so if there's like some official or better way of generalizing the moebius function I would be interested to know