# Question on Divisor Sum over the Liouville Function $\lambda(d)=(-1)^{\omega(d)}$

This question assumes the following:

• $$\nu(n)$$ is the number of distinct primes in the factorization of $$n$$,
• $$\omega(n)$$ is the number of prime factors counting multiplicities in the factorization of $$n$$,
• $$\lambda(n)$$ is defined to be $$(-1)^{\omega(n)}$$, and
• $$\delta_{i,j}=\cases{0,& i\ne j \\ 1,& i=j}\quad$$ is the Kronecker delta function.

Question: Has the following formula been proven (or disproven) and if not, can it be?

(1) $$\quad\sum\limits_{d|n} 2^{\nu(d)+\nu\left(\frac{n}{d}\right)}\lambda(d)=\delta_{n,1}=\cases{0,& n\ne 1 \\ 1,& n=1}$$

The case $$n=1$$ is trivial, so my interest is a proof for the case $$n>1$$.

The Liouville function $$\lambda(n)$$ is closely related to the Möbius function $$\mu(n)$$, and formula (1) above seems somewhat analogous to formula (2) below.

(2) $$\quad\sum_\limits{d|n}\mu(d)=\delta_{n,1}$$

• Never saw it but perhaps using the fact tha $2^{\nu(n)}=\sum\limits_{d|n} \lambda_{-1}(d)$. Note that $2^{\nu(n)}$ is also tied to $\sigma(n^2)$ math.stackexchange.com/questions/2068054/… – Collag3n Jun 29 '19 at 4:50
• So you define $\omega$ but then use $\lambda$ in the summation? – Robert Soupe Jun 29 '19 at 12:25
• @RobertSoupe I defined $\omega$ (\omega) because I intended to use it to define $\lambda$ in the title, but there was an error in the title which originally used $\Omega$ (\Omega) which I corrected after noticing your comment. – Steven Clark Jun 29 '19 at 13:25
• @Collag3n The question/answer at math.stackexchange.com/questions/2073242/… also perhaps provides some insight. – Steven Clark Jun 29 '19 at 13:31

$$2^{\nu(n)} \lambda(n),2^{\nu(n)}$$ are multiplicative functions so $$f(n)=\sum_{d | n}\lambda(d) 2^{\nu(d)}2^{\nu(n/d)}$$ is multiplicative ie. $$\begin{split} f(n) &= \prod_{p^k \| n} f(p^k)=\prod_{p^k \| n} \left(\sum_{d | p^k}\lambda(d) 2^{\nu(d)}2^{\nu(p^k/d)}\right)\\ &=\prod_{p^k \| n} \left(\sum_{l=0}^k \lambda\left(p^l\right) 2^{\nu(p^l)}2^{\nu(p^k/p^l)}\right)\\ &= \prod_{p^k \| n} \left[2+\left(\sum_{l=1}^{k-1} (-1)^l 4\right) + 2(-1)^k \right] \end{split}$$