Determine all positive integers $m$ with the following property:
The sequence $a_{0}, a_{1}, a_{2}, \ldots$ with $a_{0}=\frac{2 m+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$ contains at least one integer.
Hint: $\lfloor x\rfloor$ denotes the greatest integer part (integer function) of $x$.
It holds that $a_{0}=m+\frac{1}{2}$ and $a_{1}=a_{0}\left\lfloor a_{0}\right\rfloor=\left(m+\frac{1}{2}\right) \cdot m=m^{2}+\frac{m}{2}$. This expression is obviously an integer for even $m$, so the desired property of the sequence is present here.
Furthermore, for $m=1$, we have $a_{0}=\frac{3}{2},\left\lfloor a_{0}\right\rfloor=1$, and for $a_{k}=\frac{3}{2}$, it always follows that $a_{k+1}=\frac{3}{2} \cdot 1=\frac{3}{2}$. Thus, there is no integer term in the sequence.
Now let $m \geq 3$ be odd. There exists a unique representation $m=2^{p} \cdot n_{0}+1$ with $p \geq 1, n_{0}$ odd and $p, n_{0} \in \mathbb{N}$. Thus, $a_{0}=2^{p} n_{0}+\frac{3}{2}$, and for $a_{k}=2^{p} n_{k}+\frac{3}{2}$ ( $n_{k}$ odd), it follows that $a_{k+1}=\left(2^{p} n_{k}+\frac{3}{2}\right)\left(2^{p} n_{k}+1\right)=2^{p-1}\left(2^{p+1} n_{k}^{2}+5 n_{k}\right)+\frac{3}{2}$, where the term in parentheses is odd. Therefore, there exists a representation $a_{k+1}=2^{p-1} n_{k+1}+\frac{3}{2}$ with odd $n_{k+1}$. By increasing $k$ by 1, $p$ is simultaneously decreased by 1, so that $a_{k+p}=2^{0} n_{k+p}+\frac{3}{2}$ holds and $\left\lfloor a_{k+p}\right\rfloor$ is even. Then $a_{k+p+1}$ is an integer, and thus the desired property of the sequence is also present for every $m \geq 3$. Therefore, all positive integers except 1 belong to the desired set.

Hint: The occasional practice of calculating with "fractional residue classes" modulo a power of two often led to errors, such as the incorrect answer that no numbers with a remainder of 1 modulo 8 belong to the set.