floors of positive multiples of an irrational number If $x>0$ is irrational, then the set of fractional parts of $nx$, where $n\in\mathbb{N}$, is dense in $(0,1)$.
These are numbers $nx\bmod{1}$.
Similarily $nx\bmod{N}$ are dense in $(0,N)$ for any $N\in\mathbb{N}$.
Now, what about integer parts of $nx$ modulo something?
I have no idea how to verify the following (my own) conjectures:


*

*for any integer $N\ge 2$ the sequence $(\lfloor x \rfloor\bmod{N}, \lfloor 2x \rfloor\bmod{N}, \lfloor 3x \rfloor\bmod{N}, \dots)$ takes each possible value (i.e. $0,\dots,N-1$) infinitely often.

*for any integer $N\ge 2$ the sequence $(\lfloor x \rfloor\bmod{N}, \lfloor 2x \rfloor\bmod{N}, \lfloor 3x \rfloor\bmod{N}, \dots)$ doesn't become periodic.
 A: For your first conjecture, you can use what you stated earlier, i.e., that

Similarily $nx\bmod{N}$ are dense in $(0,N)$ for any $N\in\mathbb{N}$.

Thus, for every integer $0 \le m \le N - 1$, there are infinitely many $n$ such that
$$m \lt (nx \bmod{N}) \lt m + 1 \tag{1}\label{eq1A}$$
For those infinite # of $n$, you have that $\lfloor nx \rfloor\bmod{N} = m$.
Regarding your second conjecture, use proof by contradiction by first assuming that for some positive integer $n_0$ you have that, for all $n \ge n_0$, the sequence becomes periodic with some period $p \ge 1$. You would thus have all such $n$ that
$$\lfloor nx \rfloor\bmod{N} = \lfloor (n+p)x \rfloor\bmod{N} \tag{2}\label{eq2A}$$
Separate $x$ into its integer & fractional parts, i.e.,
$$x = q + r, \; q \in \mathbb{N}, \; 0 \lt r \lt 1 \tag{3}\label{eq3A}$$
Thus, \eqref{eq2A} becomes
$$\begin{equation}\begin{aligned}
\lfloor n(q + r) \rfloor\bmod{N} & = \lfloor (n+p)(q + r) \rfloor\bmod{N} \\
nq + \lfloor nr \rfloor\bmod{N} & = (n+p)q + \lfloor nr + pr \rfloor\bmod{N}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Next, separate $pr$ into its integer & fractional parts, i.e.,
$$pr = s + t, \; s \in \mathbb{N}, \; 0 \lt t \lt 1 \tag{5}\label{eq5A}$$
Substitute this into \eqref{eq4A} to get
$$\begin{equation}\begin{aligned}
nq + \lfloor nr \rfloor\bmod{N} & = (n+p)q + \lfloor nr + (s + t) \rfloor\bmod{N} \\
nq + \lfloor nr \rfloor\bmod{N} & = (n+p)q + s + \lfloor nr + t \rfloor\bmod{N}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
As you stated earlier,

the set of fractional parts of $nx$, where $n\in\mathbb{N}$, is dense in $(0,1)$.

Thus, there are infinitely many $n$, including above $n_0$, such that you have (where the $\{y\}$ refers to the fractional part of $y$) within the specified non-empty region that
$$\max(0, 1 - r - t) \lt \{nr\} \lt 1 - \max(r, t) \tag{7}\label{eq7A}$$
Choose any such $n = n_1$. Note that, in particular, $\{n_1 r\} \lt 1 - r$. Also, $1 - r \lt \{n_1 r + t\} \lt 1$. However, when $n$ is incremented by $1$, so $n = n_1 + 1$, you have $\{(n_1+1)r\} \lt 1$, so its integral part doesn't change, but $1 \lt \{(n_1+1)r + t\}$, so its integral part increases by $1$. However, this means that if \eqref{eq6A} was true for $n = n_1$, it's not true when $n = n_1 + 1$, so \eqref{eq6A} can't be true for all $n \ge n_0$. This contradicts the original assumption, showing that the sequence is never eventually periodic.
