The Möbius Function

The Möbius function is denoted by $\mu(n)$, where $\mu$ is the Greek letter mu.

Depending on the factorisation of $n$, the function takes the values $\{-1, 0, 1\}$.

$\mu(n) = \begin{cases} 1, & \text{if}\, n = 1\\ 0, & \text{if}\, p^2 | n\\ (-1)^k, & \text{if}\, n \,\text{is the product of}\, k \,\text{distinct primes} \end{cases}$

The last says that if $k$ is even, then $\mu(n) = 1$ and if $k$ is odd, then $\mu(n) = -1$.

Consider the numbers $6, 7$ and $8$.

$6 = 2\cdot 3$ so $\mu(6) = 1$

$7 =$  prime so $\mu(7) = -1$

$8 = 2^3$ so $\mu(8) = 0$ since $2^2$ divides $8$

Thus the Möbius function is a multiplicative function :

$\mu(ab) = \mu(a)\mu(b)$,

where gcd$(a, b) = 1$.

The sum of the Möbius function over all the positive divisors of $n$ (including $1$ and itself) is zero except when $n = 1$.

$\displaystyle \sum_{d|n} \mu (d) = \begin{cases} 1, & \text{if} \, n = 1\\ 0, & \text{if} \, n \gt 1 \end{cases}$

For example, the divisors of $6$ are $1, 2, 3$ and $6$ so $1 -1 -1 + 1 = 0$.

The Möbius function is connected to Euler’s totient function by Möbius inversion :

Let $f$ and $g$ be arithmetic (multiplicative) functions such that

$\displaystyle f(n) = \sum_{d|n} g(d)$ for all $n$, then

$\displaystyle g(n) = \sum_{d|n} \mu(d) \, \, f\left(\frac nd\right)$

From the Gaussian identity of Euler’s totient function we have :

$\displaystyle f(n) = \sum_{d|n}\varphi(d)$

When $n = 6, \, \, f(6) = 1 + 1 + 2 + 2 = 6$.

So, by the inversion theorem :

$\displaystyle g(n) =  \varphi(n) = \sum_{d|n} \mu(d) \frac nd$

When $n = 6$ then

$\displaystyle \varphi(6)  = \sum_{d|6} \mu(d) \frac 6d$

Thus

$\displaystyle \varphi(6) = \frac 61 \, – \frac 62 \, – \frac 63 + \frac 66 = 6 \, – 3 \, – 2 + 1 = 2$

Euler’s Totient Function
 

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