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 $k=2$:


$\lambda\_2(1)=1, \hspace{2 mm} \lambda\_2(p\_1)=1, \hspace{2 mm} \lambda\_2(p\_1p\_2)=-1, \hspace{2 mm} \lambda\_2(p\_1p\_2p\_3)=-1 \hspace{2 mm} \text{ and so on...}.$


-Is there anythig known about this function?


-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 $$\sum\_{n<x} \lambda\_2(n) = o(x),$$
smilar to $\lambda\_1$ should we expect square root cancelation on sum of $\lambda\_2$ as well?



Generalization of the The Liouville Lambda function
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!


