The number of descents of a binary word.

Return the Dyck path corresponding to the parallelogram polyomino obtained by applying Delest-Viennot's bijection.
Let $D$ be a Dyck path of semilength $n$. The parallelogram polyomino $\gamma(D)$ is defined as follows: let $\tilde D = d_0 d_1 \dots d_{2n+1}$ be the Dyck path obtained by prepending an up step and appending a down step to $D$. Then, the upper path of $\gamma(D)$ corresponds to the sequence of steps of $\tilde D$ with even indices, and the lower path of $\gamma(D)$ corresponds to the sequence of steps of $\tilde D$ with odd indices.
The Delest-Viennot bijection $\beta$ returns the parallelogram polyomino, whose column heights are the heights of the peaks of the Dyck path, and the intersection heights between columns are the heights of the valleys of the Dyck path.
This map returns the Dyck path $(\gamma^{(-1)}\circ\beta)(D)$.

The Foata bijection $\phi$ is a bijection on the set of words of given content (by a slight generalization of Section 2 in [1]).
Given a word $w_1 w_2 ... w_n$, compute the image inductively by starting with $\phi(w_1) = w_1$. At the $i$-th step, if $\phi(w_1 w_2 ... w_i) = v_1 v_2 ... v_i$, define $\phi(w_1 w_2 ... w_i w_{i+1})$ by placing $w_{i+1}$ on the end of the word $v_1 v_2 ... v_i$ and breaking the word up into blocks as follows.

• If $w_{i+1} \geq v_i$, place a vertical line to the right of each $v_k$ for which $w_{i+1} \geq v_k$.
• If $w_{i+1} < v_i$, place a vertical line to the right of each $v_k$ for which $w_{i+1} < v_k$.

In either case, place a vertical line at the start of the word as well. Now, within each block between vertical lines, cyclically shift the entries one place to the right.
For instance, to compute $\phi(4154223)$, the sequence of words is

• 4,
• |4|1 -- > 41,
• |4|1|5 -- > 415,
• |415|4 -- > 5414,
• |5|4|14|2 -- > 54412,
• |5441|2|2 -- > 154422,
• |1|5442|2|3 -- > 1254423.

So $\phi(4154223) = 1254423$.

Return the Dyck word as binary word.