Question: Let $\omega_k$ be the number of distinct prime divisors of k. What is the asymptotic growth of $C_n := \sum_{k=1}^n 2^{\omega_k}$?

Thank you for considering this elementary question. Below I give some motivation for this problem and some of my progress.

**Motivation:**
There are a couple. The first is purely number theoretic. Note that because $2^{\omega_k} = \sum_{d | k} |\mu(d)|$, we have $C_n = \sum_{k=1}^n \sum_{d|k} |\mu(d)|,$ and thus $C_n$ is the asymptotic growth of absolute values of the Möbius function (in other contexts this relates to the Prime Number Theorem, see here).

The second motivation comes from geometric group theory. The *commensurability index* of $A, B \leq G$ (all groups) is $[A : A \cap B][B: A \cap B]$. The *full commensurability growth function* assigned to a pair $A \leq B$ is defined to be
$$
C_n(A,B) = \# \{ \Delta \leq B : c(\Delta, A) \leq n \}.
$$
This is a generalization of the subgroup growth function to pairs and it can be infinite in natural settings. The question above is the case $G(\mathbb{Z}) = \mathbb{Z} \leq \mathbb{R} = G(\mathbb{R})$ of the following

Problem:Compute the full commensurability growth function for the pairs $G(\mathbb{Z}) \leq G(\mathbb{R})$ where $G$ is a unipotent linear algebraic group.

**Some progress:**
Write $f \preceq g$ if there exists $C > 0$ such that $f(n) \leq C g(C n)$.
In Proposition .3, it is shown that $n (\log(n))^{\log(2)} \leq C_n \preceq n(\log(n))$.