📂Number Theory

Analytic Number Theory and the Liouville Function

Definition ¹

Let’s consider a prime number $p_{1} , \cdots , p_{k}$ and a natural number $n$ represented as $n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}$. The arithmetic function $\lambda$ defined as follows is called the Liouville function. $$ \lambda (n) = (-1)^{a_{1} + \cdots a_{k}} $$

Basic Properties

• [1] Liouville series: $n$ is a perfect square, then $1$, otherwise $0$. In other words, $$ \sum_{d \mid n} \lambda (d) = \begin{cases} 1 &, n \text{ is a square} \\ 0 & , \text{otherwise}\end{cases} $$
• [2] Completely multiplicative: For all $m,n \in \mathbb{N}$, $\lambda (mn) = \lambda (m) \lambda (n)$

Description

$$ \begin{matrix} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \ \lambda (n) & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 & 1 \\ \sum_{d \mid n} \lambda (d) & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} $$ The Liouville function might not be intuitively understood just by its definition similar to the Möbius function. Interestingly, following the proof of theorem [1] and the Möbius inversion formula, one can confirm that: $$ \lambda^{-1} (n) = \left| \mu (n) \right| , \qquad \forall n \in \mathbb{N} $$

Proof

[2]

$$ n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}} \\ m = q_{1}^{b_{1}} \cdots q_{r}^{b_{r}} $$ Let two natural numbers $m,n$ be represented as mentioned above. Then, according to the definition of the Liouville function, $$ \lambda (nm) = (-1)^{a_{1} + \cdots a_{k} + b_{1} + \cdots + b_{r}} = (-1)^{a_{1} + \cdots a_{k} }(-1)^{ b_{1} + \cdots + b_{r}} = \lambda (n) \lambda (m) $$

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[1]

Properties of multiplicative functions:

• (2): It’s equivalent that $f$ is a multiplicative function and for all prime numbers $p_{1} , \cdots , p_{r}$ and all $a_{1} , \cdots, a_{r} \in \mathbb{N}$, $f \left( p_{1}^{a_{1}} \cdots p_{r}^{a_{r}} \right) = f \left( p_{1}^{a_{1}} \right) \cdots f \left( p_{r}^{a_{r}} \right)$.
• (6): If $f$ is completely multiplicative, then $F(n) := \sum_{d \mid n} f(d)$ is multiplicative.

If $g$ is defined as $g(n) := \sum_{d \mid n} \lambda (n)$, then theorem [2] ensures that $\lambda$ is completely multiplicative, hence by corollary (6), $g$ is multiplicative. If $g$ is a multiplicative function, by corollary (2) we can break down $g$, which means it’s sufficient to calculate only for a fixed prime number $p$ and $a \in \mathbb{N}$, $g \left( p^{a} \right)$. $$ \begin{align*} g \left( p^{a} \right) =& \sum_{d \mid p^{a}} \lambda (d) \\ =& 1 + \lambda \left( p \right) + \cdots \lambda \left( p^{a} \right) \\ =& 1 - 1 + 1 - 1 + \cdots + (-1)^{a} \\ =& \begin{cases} 0 &, a \text{ is odd} \\ 1 &, a \text{ is even} \end{cases} \end{align*} $$ Therefore, $g(n)$ is represented solely by the product of $0$ and $1$, and if $a$ is odd for even one $p$, $0$ is multiplied resulting in $g(n) = 0$.

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