Jordan's totient function

jomlau 04 Oct 2008 00:05

In my stumbling through Wikipedia, I came across this interesting counterpart to Euler's totient function, claimed by a fellow named Camille Jordan. So, similar to how we have

(1)\begin{align} \sigma(n) = \sum_{d|n} d \end{align}

and

(2)\begin{align} \sigma_k(n) = \sum_{d|n} d^k, \end{align}

we also have a similar generalization for $\phi(n)$. Given $n=p_1^{a_1}\cdots p_k^{a_k}$, we can express Euler's totient function as

(3)\begin{align} \phi(n) = n\prod_{i=1}^k \left(1 - \frac{1}{p_i}\right). \end{align}

Then Jordan's totient function is

(4)\begin{align} J_m(n) = n^m\prod_{i=1}^k \left(1 - \frac{1}{p_1^m}\right). \end{align}

Thus, we have that $\phi(n) = J_1(n)$. Incidentally, we had that $\sigma_0(n) = \nu(n)$, which was pretty cool; $J_0(n)$ is unfortunately a bit less interesting:

(5)\begin{align} J_0(n) = \left\{\!\begin{array}{rl} 1 & \mbox{if }n = 1 \\ 0 & \mbox{if }n > 1 \end{array}\right. \end{align}