Kluyver's Formula for Ramanujan's Sum

Theorem

Let $q \in \N_{>0}$.

Let $n \in \N$.

Let $\map {c_q} n$ be Ramanujan's sum.

Let $\mu$ denote the Möbius function.

Then:

$\ds \map {c_q} n = \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}$

where $\divides$ denotes divisibility.

Proof

Let $\alpha \in \R$.

Let $e: \R \to \C$ be the mapping defined as:

$\map e \alpha := \map \exp {2 \pi i \alpha}$

Let $\zeta_q$ be a primitive $q$th root of unity.

Let:

$\ds \map {\eta_q} n := \sum_{1 \mathop \le a \mathop \le q} \map e {\frac {a n} q}$

By Complex Roots of Unity in Exponential Form this is the sum of $n$th powers of all $q$th roots of unity.

Therefore:

$\ds \map {\eta_q} n = \sum_{d \mathop \divides q} \map {c_d} n$

By the Möbius Inversion Formula, this gives:

$\ds \map {c_q} n = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}$

Now by Sum of Powers of Primitive Complex Roots of Unity, we have:

$\ds \map {\eta_d} n = \begin{cases} d & : d \divides n \\ 0 & : d \nmid n \end{cases}$

Therefore:

\(\ds \map {c_q} n\)

\(=\)

\(\ds \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}\)

\(\ds \)

\(=\)

\(\ds \sum_{\substack {d \mathop \divides q \\ d \mathop \nmid n} } \map {\eta_d} n \map \mu {\frac q d} + \sum_{\substack {d \mathop \divides q \\ d \mathop \divides n} } \map {\eta_d} n \map \mu {\frac q d}\)

\(\ds \)

\(=\)

\(\ds 0 + \sum_{\substack {d \mathop \divides q \\ d \mathop \divides n} } d \map \mu {\frac q d}\)

\(\ds \)

\(=\)

\(\ds \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}\)

Common Divisor Divides GCD

as required.

$\blacksquare$

Source of Name

This entry was named for Jan Cornelis Kluyver.