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For $n\geqslant0$, $m>0$, $s>t\geqslant0$, $n,m,s,t$ - integers we have $$\sum\limits_{k=0}^{m-1}\left\lfloor{n+ks+t\over ms}\right\rfloor=\left\lfloor{n+t\over s}\right\rfloor$$ $$\sum\limits_{k=0}^{m-1}\left\lceil{n+ks+t\over ms}\right\rceil=\left\lceil{n+t\over s}\right\rceil+m-1$$ $$\sum\limits_{k=0}^{m-1}\left[{n+ks+t\over ms}\right]=\left[{n+t+\lceil{s\over 2}\rceil \lfloor{m\over 2}\rfloor+\lceil{s-1\over 2}\rceil \lfloor{m-1\over 2}\rfloor \over s}\right]$$ All of them are self-discovered. I'm looking for nice and simple closed form (and also maybe more simple proof) for the last.

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    $\begingroup$ Thanks for the upvote. I do realize that $n, s, t$ come from some context you work in, but as $n$ is unbounded, in your expression you can redefine $n := n+t $ and redefine $t:=0$. That should not affect the formulas. $\endgroup$ – JAskgaard Jan 2 at 21:37
  • $\begingroup$ @JAskgaard, you are absolutely right, but I sure it is not so significant. $\endgroup$ – user514787 Jan 2 at 21:41

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