Let $\lambda$ denote the Liouville $\lambda $- function. We know that $\lambda$ is multiplicative if we define it for integers $n$.
It is defined here:
https://math.stackexchange.com/posts/3245975/edit
Is the following function also multiplicative? $$\sum_{d|n} \lambda(n/d) 2^{\omega(d) }?$$
$\omega(n):=$ is the distinct prime divisors of $n$.
If so, why? could anyone explain this for me please?
Maybe I'm missing something, but isn't your function the Dirichlet convolution of two multiplicative functions, and as such multiplicative itself?
Proof Let $f, g$ be two multiplicative functions. Let $m, n$ coprime integers. Then a divisor of $m n$ is of the form $d e$, where $d \mid m$ and $e \mid n$.
\begin{align} \sum_{d \mid m, e \mid n} f(m n / d e) g(d e) &= \sum_{d \mid m, e \mid n} f(m/d) f(n /e) g(d) g( e) \\&= \left(\sum_{d \mid m} f(m/d) g(d)\right) \cdot \left(\sum_{e \mid n} f(n /e)g( e)\right) . \end{align}