13 problems

Four-pattern involution equidistribution conjecture. The patterns

A mesh pattern of length $2$ is a pair $(\tau,R)$, with $\tau$ a permutation of length $2$ and $R\subseteq[0,2]\times[0,2]$ the set of shaded boxes. Two patterns are equivalent whe…

Let $p_1$ and $p_2$ be mesh patterns of length $2$. They are equivalent when $s_{n,k}(p_1)=s_{n,k}(p_2)$ for all $n,k\geq0$, where $s_{n,k}(p)$ counts $n$-permutations containing e…

Let $\mathfrak{S}_n$ be the set of permutations of $[n]$, and for a mesh pattern $\mathrm{P}$ let $\mathrm{P}(\sigma)$ denote the number of occurrences of $\mathrm{P}$ in $\sigma$.…

Let a mesh pattern be specified by a set of marked points in a $3\times3$ grid together with a set of shaded cells; let $\sim_{jd}$ denote equality of joint distributions for the c…

Joint distribution equivalence of S21 and S22. The first pair of mesh patterns, with points at $(1,1),(2,2),(3,3)$ and at $(1,3),(2,2),(3,1)$, satisfies $\sim_{jd}$, and the corres…

Let a permutation be sortable by two stacks in series under the restriction that the permutation on the first stack avoids the classical pattern $1324$. Let $\operatorname{Av}(P)$…

Under the Robinson–Schensted–Knuth correspondence, a permutation is mapped to a pair of Young tableaux of the same shape. Let $A$ be the set of permutations whose tableaux avoid th…

Let $p$ be the mesh pattern represented by the diagram encoded in the source, and let $|\mathcal{S}_n(p)|$ denote the number of length-$n$ objects avoiding $p$. Rational generating…

Experimental coincidence classification conjecture. The mesh patterns of size $n$ can be coincidence classified by experimental classification with permutations of size

Distribution conjecture for mesh pattern 3. The distribution of mesh pattern Nr. 3 is the triangle, read by rows, given by

Let $m$ be a mesh pattern with at least one shaded box, and let a random mesh pattern of length $n$ be chosen uniformly. Vanishing containment probability conjecture. The probabili…

Equality conjecture. For all $k\geq 1$, we have