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I considered the identity $\sum_{r=0}^{n}\binom{n}{r}2^r = 3^n$.

Hence $\left \lfloor \sum_{r=0}^{n}\binom{n}{r}2^{r-k} \right \rfloor = \left \lfloor \frac{3^n}{2^k} \right \rfloor$. but now i am stuck.

Can we evaluate this value any better?

Thanks

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    $\begingroup$ What any better are you thinking of ? This is a quite simple expression. $\endgroup$ – Yves Daoust Jan 18 at 23:21
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    $\begingroup$ I think it should be $\lfloor \sum_{r=0}^k\binom{n}r 2^{r-k} \rfloor$. I am pretty sure this cannot be simplified. $\endgroup$ – Mike Earnest Jan 18 at 23:45
  • $\begingroup$ I thought it might be possible to find an expression without the floor value function. I guess probably not. $\endgroup$ – oren1 Jan 19 at 10:12
  • $\begingroup$ I edited the LHS (typo) from $\left \lfloor \sum_{r=0}^{k}\binom{k}{r}2^{r-k} \right \rfloor$ to $\left \lfloor \sum_{r=0}^{n}\binom{n}{r}2^{r-k} \right \rfloor$. does it change something? $\endgroup$ – oren1 Jan 19 at 12:11

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