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I want to find the sets of the terms of the sequences $\left( n-2 \left[\frac{n}{2}\right]\right), \left( n-3 \left[\frac{n}{3}\right]\right) $ and more general, if $m$ is natural the set of the terms of $\left( n-m \left[\frac{n}{m}\right]\right)$.

We know that $\left( n-m \left[\frac{n}{m}\right]\right)$ is equal to the largest integer $k$ for which $k \leq \frac{n}{m}<k+1$.

But how can we find a general form for the terms of the given sequences and consequently the desired sets?

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    $\begingroup$ Have you tried just writing what $(n - 2 [n/2])$ equals for $n=1,2,3,4,\ldots$? If you just write things out, you might be able to find a pattern and then formalize your guess into a proof. $\endgroup$ – angryavian Nov 24 '19 at 22:17
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    $\begingroup$ Note: $[n/2]=n/2$ when $n$ is even and $(n-1)/2$ when $n$ is odd $\endgroup$ – J. W. Tanner Nov 24 '19 at 22:17
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    $\begingroup$ $n-m\left[\frac nm\right]=n \mod m$ $\endgroup$ – J. W. Tanner Nov 24 '19 at 22:20
  • $\begingroup$ For $n=3k$ it holds that $\left[ \frac{n}{3}\right]=\frac{n}{3}$, for $n=3k+1$ it holds that $\left[ \frac{n}{3}\right]=\frac{n-1}{3}$ and for $n=3k+2$ it holds that $\left[ \frac{n}{3}\right]=\frac{n-2}{3}$, right? So if $n=3k$ then $n-3 \left[ \frac{n}{3}\right]=0$, if $n=3k+1$ then $n-3 \left[ \frac{n}{3}\right]=1$ and for $n=3k+2$ it holds that $n-3 \left[ \frac{n}{3}\right]=2$, right? $\endgroup$ – Mary Star Nov 24 '19 at 23:34
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For natural number $m$ and integer $n$, $n-m \left[\frac{n}{m}\right] \equiv n \bmod m$. The sets are those of the integers modulo $m$, $\mathbb{Z}/m\mathbb{Z}$.

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