Coincidences between average Catalan tableaux

Question

There are Catalan number $C_n$ of standard Young tableaux of shape $(n,n)$, which we view as $2\times n$ matrices. Denote by $P_n$ the average of these matrices: $$ P_n \, := \, \frac{1}{C_n} \, \sum_{A \, \in \, \text{SYT}(n,n)} A $$ Note that $P_n=(p_{ij})$ is a rational matrix with entries monotone increasing in rows and columns. Also, by the 180$^\circ$ rotational symmetry, $p_{ij} + p_{3-i,n+1-j} = 2n+1$. For example, $p_{11}=1$, $p_{2n}=2n$.

The asymptotic density of $P_n$ as $n\to \infty$ is easy to obtain by a direct calculation or via the Brownian excursion (see e.g. here or there), but my question is different.

Question. Let $\beta(n):= \min_{(ij)\ne (kl)} |p_{ij}-p_{kl}|$. Is it true that $\beta(n) = o(1)$?

It would be cool if there was an easy way to see this. I really want a generalization of this result to all large partitions, but at the moment even this is confounding.

UPDATE (May 18, 2019):
Let me explain the motivation behind the question. Recall the 1/3-2/3 conjecture that every poset $\mathcal P=(X,\prec)$ that is not a linear order contains two elements $x,y\in X$ such that $$\frac13 \le P(x\prec y) \le \frac23 $$ For width 2 posets this was shown by Linial in this paper, but I thought that for shapes $(n,n)$ one can improve $1/3$ to perhaps $(1/2-\varepsilon)$, since we know so much about Catalan numbers (including the average of Catalan objects). Linial's proof cannot be easily improved, unfortunately. Now, a beautiful Kahn-Linial proof of the weaker $1/2e$ bound starts with the average LE of $\mathcal P$. If $\beta(n)=o(1)$, their argument plus the (earlier) Grünbaum Theorem implies the $(1/e-\varepsilon)$ bound, already a nice result.

Now, Richard's calculaitons give $p_{17} \to 9949/1024 \approx 9.7158$, $p_{23} \to 9.75$. This means that taking $x=(1,7)$ and $y=(2,3)$ in $(n,n)$ gives $$0.3553 < \frac1e \left(1-\frac{35}{1024}\right) < P(x\prec y) < \frac2e \left(1+\frac{35}{1024}\right) < 0.6447 $$ for $n$ large enough (unless I miscalculated). This improvement over $1/3$ bound is good to know, but surely one can do better.

UPDATE (May 27, 2020):
In our most recent paper we found an upper bound $O\bigl(n^{-5/4}\bigr)$ for the sorting probability for these Catalan posets. This problem was stated in the previous update. Note that we obtained only the upper, but not the lower bounds!

Swee Hong Chan, Igor Pak, Greta Panova,
Sorting probability of Catalan posets, preprint, 2020.

Answer 1

This is not a solution, but rather a long comment. Let $f^{a,b}$ denote the number of standard Young tableaux (SYT) of shape $(a,b)$. The number of SYT $T$ of shape $(n,n)$ with $T_{1d}=k$ is $f^{d-1,k-d}f^{n-k+d,n-d}$. Hence $$ p_{1d} = \frac{1}{C_n}\sum_{k=d}^{2d-1} kf^{d-1,k-d}f^{n-k+d,n-d}. $$ There is a similar formula for $p_{2,d}$, though the number of terms in the sum increases as $n\to\infty$. In particular, \begin{eqnarray*} p_{12} & = & \frac{1}{C_n}\left( 2f^{n,n-2}+3f^{n-1,n-2}\right)\\ & = & \frac{1}{C_n}\left( \frac{2\cdot 3(2n-2)!}{(n+1)!(n-2)!} +\frac{3\cdot (2n-2)!}{n!(n-1)!}\right)\\ & = & \frac{3(3n-1)}{2(2n-1)}. \end{eqnarray*} Write $\bar{p}_{ij}=\lim_{n\to\infty}p_{ij}$ (assuming this limit exists, which I believe is always the case). Thus $\bar{p}_{12}=\frac 94$.

Similarly, $$ p_{21} = \frac{1}{C_n}\left( \sum_{k=2}^{n+1} kf^{n-1,n-k+1}\right). $$ Now \begin{eqnarray*} \frac{f^{n-1,n-k+1}}{C_n} & = & \frac{(2n-k)!(k-1)n!(n+1)!}{n!(n-k+1)!(2n)!}\\ & \to & \frac{k-1}{2^k}. \end{eqnarray*} Thus (assuming we can interchange a limit and an infinite sum) $$ \bar{p}_{21} = \sum_{k\geq 2}\frac{k(k-1)}{2^{k-1}} = 4 $$ (modulo computational error). In general, $\bar{p}_{1d}$ will be given by a finite sum, and $\bar{p}_{2d}$ by an infinite series.

Addendum. I worked out $\bar{p}_{1d}$ in general, namely, \begin{eqnarray*} \bar{p}_{1d} & = & \sum_{k=d}^{2d-1} k(2d-k+1)f^{d-1,k-d}2^{-k}\\ & = & 2^{-2d+1}(d+1)\left(4^d-{2d+1\choose d}\right). \end{eqnarray*} Beginning with $d=2$, the numbers are $$ \frac 94,\ \frac{29}{8},\ \frac{325}{64},\ \frac{843}{128},\ \frac{4165}{512},\ \frac{9949}{1024},\ \frac{185517}{16384},\dots. $$ We can also write $$ \bar{p}_{1,d-1} =2d-\frac{d{2d\choose d}}{4^{d-1}}. $$

Addendum #2. I worked out $\bar{p}_{2d}$. If my computation is correct, then $$ \bar{p}_{2d} = 2d+\frac{d{2d\choose d}}{4^{d-1}}. $$ Compare with the formula for $\bar{p}_{1,d-1}$ above.

Is there a less computational reason for such simple formulas? Do they extend to shapes other than $n(1,1)$, e.g., $n(1,1,1)$ or $n(2,1)$?