2
$\begingroup$

Let $\omega$ be a positive real number, and define: $$\mathbf{1}_{\omega}\left(n\right)\overset{\textrm{def}}{=}\left(-1\right)^{n}\binom{-\omega}{n}=\binom{\omega-1+n}{n}$$ for all positive integers $n$, and let:$$\zeta_{\omega}\left(s\right)\overset{\textrm{def}}{=}\sum_{n=1}^{\infty}\frac{\mathbf{1}_{\omega}\left(n\right)}{n^{s}}$$ Since $\mathbf{1}_{\omega}\left(1\right)=\omega>0$, $\mathbf{1}_{\omega}$ possesses an inverse with respect to Dirichlet convolution, which I denote by $\mu_{\omega}$, with: $$\sum_{n=1}^{\infty}\frac{\mu_{\omega}\left(n\right)}{n^{s}}=\frac{1}{\zeta_{\omega}\left(s\right)}$$ When $\omega=1$, $\mathbf{1}_{\omega}$ becomes the constant function $1$ and $\mu_{\omega}$ becomes the Möbius function from number theory.

To be clear, this is a reference request: I am looking to see if there has been any work done with these functions. In particular, I am interested in:

• If there is any "official name" for these functions, what is it?

• Closed-form expressions for $\mu_{\omega}\left(n\right)$ (analogous to how the möbius function can be computed in terms of the prime factors of its inputs)

• Asymptotics/formulas for $\mu_{\omega}\left(n\right)$, $\sum_{k=1}^{n}\mu_{\omega}\left(k\right)$ and $\sum_{d\mid n}\mu_{\omega}\left(d\right)$ as $n\rightarrow\infty$.

I'm making tedious progress brute-forcing the formulas by hand, but since this is only tangential to what I'm actually working on, it would be of much help if it turned out that someone had already done that drudgery for me. ;)

$\endgroup$
4
  • 3
    $\begingroup$ It is considered good form to cross-link to other questions you ask on the same topic. $\endgroup$ Commented Dec 8, 2020 at 0:51
  • $\begingroup$ Yes, but that linked question is also mine. ;) $\endgroup$
    – MCS
    Commented Dec 9, 2020 at 1:43
  • 2
    $\begingroup$ I know, the point is that if someone answers that question and not this one, someone could come across this (assumed to be unanswered) question and duplicate effort on it. $\endgroup$ Commented Dec 9, 2020 at 1:50
  • $\begingroup$ Gotcha. I'll keep that in mind for future questions. $\endgroup$
    – MCS
    Commented Dec 9, 2020 at 2:15

0

You must log in to answer this question.