Generalization of the The Liouville Lambda function

Let $$n=p^{\alpha_1}_1 \cdots p^{\alpha_m}_m,$$ and define $$\lambda_k(n)= (-1)^{ [\frac{\Omega(n)}{k} ]},$$ where $$\Omega(n)= \alpha_1 + \cdots + \alpha_k,$$ and $$[\cdot]$$ is the floor function.

For $$k=1$$, $$\lambda_1$$ is the Liouvilles Lambda function. For $$k=2$$:

$$\lambda_2(1)=1, \hspace{2 mm} \lambda_2(p_1)=1, \hspace{2 mm} \lambda_2(p_1p_2)=-1, \hspace{2 mm} \lambda_2(p_1p_2p_3)=-1 \hspace{2 mm} \text{ and so on...}.$$

-If for $$\Re(s)>1$$ we define $$L(s, \lambda_2):= \sum_{n=1}^{\infty} \frac{\lambda_2(n)}{n^s},$$
Is there any connection between $$L_{\lambda_2}(s)$$ and the zeros of the Riemann zeta function?
-I guess (without knowing how to prove) that $$\sum_{n smilar to $$\lambda_1$$ should we expect square root cancelation on sum of $$\lambda_2$$ as well?
Let's just consider the case $$k=2$$; you can try to generalize this argument for larger $$k$$. For $$k=2$$, $$\sum_{n\le x} \lambda_2(n) = \sum_{\substack{ n\le x \\ \Omega(n) = 0,1 \mod 4}} 1 - \sum_{\substack{ n\le x \\ \Omega(n) = 2,3 \mod 4}} 1.$$ This can be expressed as $$\text{Re} \sum_{n\le x} i^{\Omega(n)} + \text{Im} \sum_{n\le x}i^{\Omega(n)},$$ and now one can use Selberg--Delange. This will show an asymptotic expansion for $$\sum_{n\le x} i^{\Omega(n)}$$, and the leading term in the asymptotic is $$C x (\log x)^{i-1},$$ for a suitable constant $$C$$. Thus you can find an asymptotic for your sum, which will be of the form $$A \cos(\log \log x) \frac{x}{\log x} + B \sin(\log \log x) \frac{x}{\log x},$$ for suitable constants $$A$$ and $$B$$. Weird! But then you're also looking at a weird object!