The spin of an integer partition. The Ferrers shape of an integer partition $\lambda$ can be decomposed into border strips. The spin is then defined to be the total number of crossings of border strips of $\lambda$ with the vertical lines in the Ferrers shape. The following example is taken from Appendix B in [1]: Let $\lambda = (5,5,4,4,2,1)$. Removing the border strips successively yields the sequence of partitions $$(5,5,4,4,2,1), (4,3,3,1), (2,2), (1), ().$$ The first strip $(5,5,4,4,2,1) \setminus (4,3,3,1)$ crosses $4$ times, the second strip $(4,3,3,1) \setminus (2,2)$ crosses $3$ times, the strip $(2,2) \setminus (1)$ crosses $1$ time, and the remaining strip $(1) \setminus ()$ does not cross. This yields the spin of $(5,5,4,4,2,1)$ to be $4+3+1 = 8$.

The dinv adjustment of an integer partition. The Ferrers shape of an integer partition $\lambda = (\lambda_1,\ldots,\lambda_k)$ can be decomposed into border strips. For $0 \leq j < \lambda_1$ let $n_j$ be the length of the border strip starting at $(\lambda_1-j,0)$. The dinv adjustment is then defined by $$\sum_{j:n_j > 0}(\lambda_1-1-j).$$ The following example is taken from Appendix B in [2]: Let $\lambda=(5,5,4,4,2,1)$. Removing the border strips successively yields the sequence of partitions $$(5,5,4,4,2,1),(4,3,3,1),(2,2),(1),(),$$ and we obtain $(n_0,\ldots,n_4) = (10,7,0,3,1)$. The dinv adjustment is thus $4+3+1+0 = 8$.

The order of a permutation. $\operatorname{ord}(\pi)$ is given by the minimial $k$ for which $\pi^k$ is the identity permutation.

The length of the longest cycle of a permutation.

The number of bases of the positroid corresponding to the permutation, with all fixed points counterclockwise.

The number of circled entries of the shifted recording tableau of a permutation. The diagram of a strict partition $\lambda_1 < \lambda_2 < \dots < \lambda_\ell$ of $n$ is a tableau with $\ell$ rows, the $i$-th row being indented by $i$ cells. A shifted standard Young tableau is a filling of such a diagram, where entries in rows and columns are strictly increasing. The shifted Robinson-Schensted algorithm [1] associates to a permutation a pair $(P, Q)$ of standard shifted Young tableaux of the same shape, where off-diagonal entries in $Q$ may be circled. This statistic records the number of circled entries in $Q$.

The depth of the binary word interpreted as a path. This is the maximal value of the number of zeros minus the number of ones occurring in a prefix of the binary word, see [1, sec.9.1.2]. The number of binary words of length $n$ with depth $k$ is $\binom{n}{\lfloor\frac{(n+1) - (-1)^{n-k}(k+1)}{2}\rfloor}$, see [2].

The number of visible inversions of a permutation. A visible inversion of a permutation $\pi$ is a pair $i < j$ such that $\pi(j) \leq \min(i, \pi(i))$.

The cardinality of the support of a permutation. A permutation $\sigma$ may be written as a product $\sigma = s_{i_1}\dots s_{i_k}$ with $k$ minimal, where $s_i = (i,i+1)$ denotes the simple transposition swapping the entries in positions $i$ and $i+1$. The set of indices $\{i_1,\dots,i_k\}$ is the '''support''' of $\sigma$ and independent of the chosen way to write $\sigma$ as such a product. See [2], Definition 1 and Proposition 10. The '''connectivity set''' of $\sigma$ of length $n$ is the set of indices $1 \leq i < n$ such that $\sigma(k) < i$ for all $k < i$. Thus, the connectivity set is the complement of the support.

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 [[St000703]] 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 $$\frac{1}{|O(\pi)|}\sum_{\sigma\in O(\pi)} (\sigma(1)-1).$$