The first entry of the permutation.
This can be described as 1 plus the number of occurrences of the vincular pattern ([2,1], {(0,0),(0,1),(0,2)}), i.e., the first column is shaded, see [1].
This statistic is related to the number of deficiencies St000703The number of deficiencies of a permutation. as follows: consider the arc diagram of a permutation $\pi$ of $n$, together with its rotations, obtained by conjugating with the long cycle $(1,\dots,n)$. Drawing the labels $1$ to $n$ in this order on a circle, and the arcs $(i, \pi(i))$ as straight lines, the rotation of $\pi$ is obtained by replacing each number $i$ by $(i\bmod n) +1$. Then, $\pi(1)-1$ is the number of rotations of $\pi$ where the arc $(1, \pi(1))$ is a deficiency. In particular, if $O(\pi)$ is the orbit of rotations of $\pi$, then the number of deficiencies of $\pi$ equals
$$ \frac{1}{|O(\pi)|}\sum_{\sigma\in O(\pi)} (\sigma(1)-1). $$

Return the permutation obtained by reading the entries of the tableau row by row, starting with the bottom-most row in English notation.

The promotion of a standard Young tableau.
This map replaces the largest entry of the tableau with a zero, uses the jeu de taquin to move it to the origin, and finally increases all entries by one.

Sends a permutation to its image under the Foata bijection.
The Foata bijection $\phi$ is a bijection on the set of words with no two equal letters. It can be defined by induction on the size of the word:
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.
To compute $\phi([1,4,2,5,3])$, the sequence of words is

• $1$
• $|1|4 \to 14$
• $|14|2 \to 412$
• $|4|1|2|5 \to 4125$
• $|4|125|3 \to 45123.$

In total, this gives $\phi([1,4,2,5,3]) = [4,5,1,2,3]$.
This bijection sends the major index (St000004The major index of a permutation.) to the number of inversions (St000018The number of inversions of a permutation.).