Ramanujan sum

For positive integers $s$ and $n$, the complex number

is referred to as a Ramanujan sum, or a Ramanujan trigonometric sum.Since $e^{2\pi i}=1$, an equivalent definition is

$$c_s(n)=\sum _{k\in r(n)}{e}^{2\pi iks/n}$$

where $r(n)$ is some reduced residue system mod $n$, meaning anysubset of $\mathbb{Z}$ containing exactly one element of eachinvertible residue class mod $n$.

Using a symmetry argument about roots of unity, one can show

Applying Möbius inversion, we get

$$c_s(n)=\sum _{\begin{array}{c}d|n\\ d|s\end{array}}\mu (n/d)d=\sum _{d|(n,s)}\mu (n/d)d$$

which shows that $c_s(n)$ is a real number, and indeed an integer.In particular $c_s(1)=\mu (s)$. More generally,

$$c_{st}(mn)=c_s(m)c_t(n)\text{ if }(m,t)=(n,s)=1.$$

Using the Chinese remainder theorem, it is not hard to show thatfor any fixed $n$, the function $s\mapsto c_s(n)$ is multiplicative:

$$c_s(n)c_t(n)=c_{st}(n)\text{ if }(s,t)=1.$$

If $m$ is invertible mod $n$, then the mapping $k\mapsto km$is a permutation of the invertible residue classes mod $n$. Therefore

$$c_s(mn)=c_s(n)\text{ if }(m,s)=1.$$

Remarks: Trigonometric sums often makeconvenient apparatus in number theory, since anyfunction on a quotient ring of $\mathbb{Z}$ definesa periodic function on $\mathbb{Z}$ itself, and conversely. Foranother example, see Landsberg-Schaar relation.

Some writers use different notation from ours, reversing the rolesof $s$ and $n$ in the expression $c_s(n)$.

The name “Ramanujan sum” was introduced by Hardy.