Prove that $J_r(n) = n^r\prod_{p|n}\left(1− \frac{1}{p^r}\right)$.

Question

For $r ≥ 1$, let $J_r(n)$ be Jordan's totient function, i.e., the number of $r$-tuples of integers $a_j$ with $1 ≤ a_j ≤ n$ for $j = 1,\ldots,r$, and $(a_1,\ldots,a_r,n) = 1$. Prove that $$J_r(n) = n^r\prod_{p|n}\left(1− \frac{1}{p^r}\right).$$

Answer 1

Let $$f(n,d) = \# \{ (a_1,\ldots,a_r), a_j\le n,d| a_j\}$$ Then $$\# \{ (a_1,\ldots,a_r), a_j\le n, \gcd(a_1,\ldots,a_r,n)=1\}=\sum_{d|n}\mu(d)f(n,d)=\sum_{d|n}\mu(d)(n/d)^r$$

Answer 2

*Let us prove that $J_k(n)=n^k \displaystyle\prod_{p}\left(1-\frac{1}{p^k}\right)$. We first prove it for $p^a$ for every $p$ prime and $a\in \mathbb{N}_{>0}$ by induction on k. $\,$ It follows for n by multiplicativity.

If $k=1$
The thesis is $J_1(p^a)=p^{a}\left(1-\frac{1}{p}\right)$
$J_1(p^a)=$ #{$a_1$, such that $a_1\le p^a$ and gcd($a_1,p^a)=1\}=\phi(p^a)=p^{a}\left(1-\frac{1}{p}\right)$
Let us suppose it is true for $k-1$, and let's demonstrate it for $k$:
$J_k(p^a)=$ #{{$a_1$,…,$a_{k}\}$, such that $a_i\le p^a$ and gcd($a_1$,…,$a_{k},p^a)=1\}$.
If gcd($a_1$,…,$a_{k-1},p^a)=1$, then $a_k$ can be every number in $\{1,…,p^a\}$
If gcd($a_1$,…,$a_{k-1},p^a)\not=1$, then $a_k\notin\{1p,2p…,p^{a-1}p\}$, so there are only $p^a-p^{a-1}$ options for $a_k$.
Therefore, $\begin{equation}J_k(p^a)={(p^a)}^{(k-1)}\left(1-\frac{1}{p^{k-1}}\right)p^a+{(p^{a-1})}^{(k-1)}(p^a-p^{a-1})\\ \ ={(p^{ak})}\left(1-\frac{1}{p^{k-1}}\right)+{(p^{a-1})}^{k}(p-1)\\ \ ={(p^{ak})}\left(1-\frac{1}{p^{k-1}}\right)+{(p^{ak})}\left(\frac{p}{p^{k}}-\frac{1}{p^{k}}\right)\\ \ ={(p^{ak})}\left(1-\frac{1}{p^{k-1}}+\frac{1}{p^{k-1}}-\frac{1}{p^{k}}\right)\\ \ = {(p^{ak})}\left(1-\frac{1}{p^{k}}\right)\\ \end{equation}$.