Half the length of a longest factor which is its own reverse-complement of a binary word.

Return the Dyck word as binary word.

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 path obtained by adding an initial up and a final down step.