Where the thing that looks like a floor function is the floor function. This is an interesting result which I hoped to prove by induction, but ran into trouble applying the inductive hypothesis. The base case is trivial. Here's the progress I made:
Inductive hypothesis: $\sum_{k=0}^m 2^k \binom{m}{k} \binom{m-k}{\lfloor{(m-k)/2}\rfloor}=\binom{2m+1}{m}$ for some $m>0$. Then \begin{align*} &\sum_{k=0}^{m+1} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\ =&\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}+2^{m+1} \binom{m+1}{m+1} \binom{m+1-m+1}{\lfloor{(m+1-(m+1))/2}\rfloor}\\ =&2^{m+1}+\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor} \end{align*} Let $A$ be the set of all $k\in[0,m]$ such that $m+1-k$ is even. Let $B$ be the set of $k\in[0,m]$ such that $m+1-k$ is odd. Observe that for each $k\in[0,m]$, $k$ is in exactly one of $A$ or $B$. It follows that \begin{align*} &\sum_{k=0}^m 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\ =&\sum_{k\in A} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}+\sum_{k\in B} 2^k \binom{m+1}{k} \binom{m+1-k}{\lfloor{(m+1-k)/2}\rfloor}\\ =&\sum_{k\in A} 2^k \binom{m+1}{k} \binom{m+1-k}{(m+1-k)/2}+\sum_{k\in B} 2^k \binom{m+1}{k} \binom{m+1-k}{(m-k)/2} \end{align*}
Splitting the sum into $A$ and $B$ was a last-ditch attempt to form the expression into something to which I could apply the inductive hypothesis. Did I make a mistake somewhere, is this the wrong approach, or am I just missing the next step?