When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.

When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately appear, as a function of $x$?

It seems to be around ~ $c \sqrt x$.
Example: $x = 2500$:
2500, 1250, 833, 625, 500, 416, 357, 312, 277, 250, 227, 208, 192, 178, 166, 156, 147, 138, 131, 125, 119, 113, 108, 104, 100, 96, 92, 89, 86, 83, 80, 78, 75, 73, 71, 69, 67, 65, 64, 62, 60, 59, 58, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 43, 42, 41, 40, 40, 39, 39, 38, 37, 37, 36, 36, 35, 35, ...

It cannot occur between term $n$ and term $n+1$ if $\frac{x}{n} - \frac{x}{n+1} \ge 1$, equivalently $x \ge n^2 + n$, equivalently $n \le -\frac{1}{2} + \frac{\sqrt{1+4x}}{2}$.
It must occur, either between term $n$ and $n+1$, or between term $n+1$ and $n+2$, if $\frac{x}{n} - \frac{x}{n+1} \le \frac{1}{2}$, equivalently $x \le \frac{1}{2} n^2 + \frac{1}{2} n$, equivalently $n \ge -\frac{1}{2} + \frac{\sqrt{1+8x}}{2}$.
So the first place it appears is somewhere between the two extremes of $-\frac{1}{2} + \frac{\sqrt{1+4x}}{2}$ and $-\frac{1}{2} + \frac{\sqrt{1+8x}}{2} + 1 = \frac{1}{2} + \frac{\sqrt{1+8x}}{2}$.
