Expanding on my comment, $Q_k(n) = \left\lfloor \frac{n}{k} \right\rfloor = \frac{n - (n \bmod k)}{k} = \frac{n-a_k(n)}{k}$, where $a_k(n) = n \bmod k$ can be thought of as a recurrence defined by $a_k(n) = n \, \mid_{\, n=0, 1, \dots, k-1}$ and $a_k(n) = a_k(n-k)$ for $n \ge k$. The characteristic polynomial of the recurrence is $z^k-1$, so $a_k(n) = \sum_{i=0}^{k-1} c_i \omega_i^n$ where $\omega_i$ are the $k^{th}$ roots of unity and $c_i$ are constants that can be determined from the initial conditions.
However, calculating the actual expression for $a_k(n)$ is rather laborious. For a shortcut, we can use the following, proved for example here using either complex numbers or trig identities:
$$ \delta_{k,i}(n) = \frac{1}{k} \sum_{j=0}^{k-1} \cos \frac{j\,2(n-i)\pi}{k} = \begin{cases} 1 & n \equiv i \pmod{k}\\ 0 & \text{otherwise} \end{cases} $$Then:
$$ a_k(n) = \sum_{i=1}^{k-1} i\,\delta_{k,i}(n)=\frac{1}{k}\,\sum_{i=1}^{k-1}\left(i\,\sum_{j=0}^{k-1} \cos \frac{j\,2(n-i)\pi}{k}\right) \tag{1} $$For example:
$$ \begin{align} a_2(n) &= \frac{1}{2}\left(1 - \cos \,n\pi\right)\\ &= 0, 1, 0, 1, 0, 1, \dots \end{align} $$ $$ \begin{align} a_3(n) &= \frac{1}{3}\left(1 + \cos\frac{2(n - 1)\pi}{3} + \cos\frac{4(n - 1)\pi}{3} + 2 \left(1 + \cos \frac{2(n - 2) \pi }{3} + \cos\frac{4(n - 2)\pi}{3}\right) \right)\\ &= 0, 1, 2, 0, 1, 2, \dots \end{align} $$Substituting $a_k(n)$ from $(1)$ back into $Q_k(n) = \frac{n-a_k(n)}{k}$ gives an explicit form for $Q_k(n)$ in terms of cosines of multiples of $\frac{2\pi}{k}$.