How to interpret the next expression of the Euler totient function?

Question

We know that the original definition of the Euler-totient function is as follows: Φ(n). However, the other day I came across the following expression, which is nothing more than the following: Φd(n). And my question would be, how can the above Euler-totient function be expressed by a formula? Forgive me for not being able to describe the expression Φd(n) nicely, but I don’t know how to describe it properly. Thank you for your answers!

Answer 1

The question has more than one possible answer.

$1.$ The expression is $\Phi_d(n)$, where $\Phi_d(X)$ is the $d$-th cyclotomic polynomial. Here $\phi$ was confused with $\Phi$.

$2.$ Starting from $\phi(n)=\sum_{\stackrel{1\le k \le n}{(k,n)=1}} 1$ we could consider $$ \phi_d(n)=\sum_{\stackrel{1\le k \le n}{(k,n)=1}} k^d. $$ Then $\phi_0(n)=\phi(n)$. However, this doesn't give something new for $d=1$, because $\phi_d(n)=\frac{1}{2}n\phi(n)$, see here:

For $n > 2, n \in \mathbb{Z}$, show the sum of integers coprime to $n$ in the range $[1,n-1]$ is equal to $\frac{1}{2}n \phi(n)$

For $d=2$ we obtain $$\bbox[5px,border:2px solid #00A000]{ \frac{1}{6} \varphi(n) \left(2n^2 + (-1)^{|\mathcal{P}|} \prod_{p\in \mathcal{P}} p\right).}$$