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?


1 Answer 1


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!

  • 2
    $\begingroup$ Indeed, it's closely related to what Tian An Wong, Snehal Shekatkar, and I studied in this paper (again using Selberg-Delange): doi.org/10.5802/jtnb.1066 $\endgroup$ Commented Sep 10, 2020 at 22:52

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