Prove that, for $n\in \Bbb{Z}^+$, $$\left\lfloor{\frac{n}{2}}\right\rfloor+\left\lfloor\frac{\left\lceil\frac{n}{2}\right\rceil}{2}\right\rfloor+\left\lfloor\frac{\left\lceil\frac{\left\lceil\frac{n}{2}\right\rceil}{2}\right\rceil}{2}\right\rfloor+\cdots = n - 1\,,$$ where there are $\lceil{\log_2n}\rceil$ addends on the left-hand side.
I don't know how I could prove this. Any ideas? There is an intimate relationship here with a binary tree where each addend is the number of nodes on that layer, and $n$ is the number of leaves.