The number of right ropes of a permutation. Let $\pi$ be a permutation of length $n$. A raft of $\pi$ is a non-empty maximal sequence of consecutive small ascents, [[St000441]], and a right rope is a large ascent after a raft of $\pi$. See Definition 3.10 and Example 3.11 in [1].

The genus of a permutation. The genus $g(\pi)$ of a permutation $\pi\in\mathfrak S_n$ is defined via the relation $$ n+1-2g(\pi) = z(\pi) + z(\pi^{-1} \zeta ), $$ where $\zeta = (1,2,\dots,n)$ is the long cycle and $z(\cdot)$ is the number of cycles in the permutation.

The number of characters of the symmetric group whose value on the partition is zero. The maximal value for any given size is recorded in [2].

The multiplicity of the standard representation in the Kronecker square corresponding to a partition. The Kronecker coefficient is the multiplicity $g_{\mu,\nu}^\lambda$ of the Specht module $S^\lambda$ in $S^\mu\otimes S^\nu$: $$ S^\mu\otimes S^\nu = \bigoplus_\lambda g_{\mu,\nu}^\lambda S^\lambda $$ This statistic records the Kronecker coefficient $g_{\lambda,\lambda}^{(n-1)1}$, for $\lambda\vdash n > 1$. For $n\leq1$ the statistic is undefined. It follows from [3, Prop.4.1] (or, slightly easier from [3, Thm.4.2]) that this is one less than [[St000159]], the number of distinct parts of the partition.

Half of the largest even part of an integer partition. The largest even part is recorded by [[St000995]].

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 $4+3+1 = 8$.