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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.

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    $\begingroup$ Another trivial observation that might be useful is that $p_{i,j} \geq p_{i-1,j} + 1$ and $p_{i,j} \geq p_{i,j-1} + 1$. $\endgroup$ – Sam Hopkins May 15 at 15:48
  • $\begingroup$ Observation 1: for $n \leq 12$: the minimum is obtained at a unique pair $(i,j) \neq (k,l)$ and its 180°-rotation counterpart. $\endgroup$ – Christian Stump May 15 at 16:07
  • $\begingroup$ Observation 2: $|p_{14} - p_{22}| \leq 1$ and $|p_{12}-p_{21}| \leq 2$ for all $4 \leq n \leq 13$, so either of these pairs might be a candidate for a witness for an affirmative answer to your question. $\endgroup$ – Christian Stump May 15 at 16:08
  • $\begingroup$ @ChristianStump: I seriously doubt $(p_{12},p_{21})$ could work. $\endgroup$ – Sam Hopkins May 15 at 16:20
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    $\begingroup$ @ChristianStump: $a_{12}$ will be greater than $a_{21}$ (by only 1) about $1/4$ of the time, because there are $C_{n-1}$ tableaux where the first column is $1,2$. But $a_{21}$ will be greater than $a_{12}$ by at least 1, and in fact sometimes much more, about $3/4$ of the time. So $p_{21}-p_{12}$ is gonna be at least $1/2$, not $o(1)$. $\endgroup$ – Sam Hopkins May 15 at 16:39
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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 $n(2,1)$?

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    $\begingroup$ Richard, nope - I had a completely different thought. It is quite likely that for some number theoretic reason you don't have a coincidence of the kind you are describing. Rather, when looking at $(p_{1j} - 2j)$ and scaling everything to $[0,1]$ interval you get density given by the semicircle projection. Same for $(p_{2j} - 2j)$ but the semicircle is concave, not convex. Instead of looking at the beginning $j=O(1)$ as you suggest, look at the middle range. There is no obvious relation between two sets of values, so one would guess that $\beta(n) = O(1/\log n)$. $\endgroup$ – Igor Pak May 16 at 6:02
  • $\begingroup$ One more thing - if you make denominators $4^n$, the numerator sequence is known and computed rather far: oeis.org/A018218 If the other sequence is also known, more values can be checked. $\endgroup$ – Igor Pak May 16 at 6:39
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    $\begingroup$ Echoing what Igor is saying, it seems conceivable that $\beta(n) = o(1)$ without having $\overline{p}_{ij}=\overline{p}_{kl}$ for any $ij, kl$ because there could be e.g. a sequence $(a_n,b_n)$ of indices depending on $n$ with $|p_{1a_{n}} - p_{2b_n}| = o(1)$. $\endgroup$ – Sam Hopkins May 16 at 15:48
  • $\begingroup$ Asymptotically, Richard's formulas give $2d\pm c\sqrt{d}$ formula which look exactly right from the semicircle POV. $\endgroup$ – Igor Pak yesterday
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    $\begingroup$ @SamHopkins: Indeed, let $n\to \infty$. Let $a(d) := \overline p_{1d}$, $b(d):=\overline p_{2,d}$. Then $a(d) = 2d-c\sqrt{d} - c' + O(1/\sqrt{d})$, $b(d) = 2d+c/\sqrt{d} + c' + O(1/\sqrt{d})$, for some explicit $c,c'$. Then one should be looking at the fractional part of $a(d+\sqrt{d})-b(d) = 2c\sqrt{d} +c'' + O(1/\sqrt{d})$. It seems, $\sqrt{n}$ mod 1 is ergodic and rates of convergence are relatively well understood arxiv.org/abs/1311.6387 Thus, if one is careful one can possibly derive $\beta(n) = o(1)$ from here. $\endgroup$ – Igor Pak yesterday

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