## Problem 4

Posted 04/02/2015 and updated 05/21/2015
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Suppose $T$ is a standard Young tableau of shape $\lambda=n^j$. If we replace the $k$-th largest element in $T$ by $k$ and rotate $T$ 180 degrees, then we get a new tableau, say $P$. If $P=T$, we say $T$ is symmetric. Exercise P is about the number of symmetric Young tableaux of shape $n^2$.

$a(n)$ is the number of symmetric standard Young tableaux of shape $(n^3)$.

For example, $a(1)=1$, $a(2)=3$.

Find $a(n)$.

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Updated 05/21/2015 and contributed by Ran Pan and Dun Qiu

For $n\ge 1$,
$a(2n)=\frac{(3n)!}{n!n!(n+1)!}$ and $a(2n+1)=\frac{(3n-2)!}{(n-1)!n!n!}$.
The sequence $a(n)$ is counted by A215294.

Based on formulas found by Dun, using Zeilberger's algorithm, we could check whether a hypergeometric series is equal to a simpler expression. The details of Zeilberger's algorithm can be found in the book A=B by Petkovsek, Wilf and Zeilberger.

Updated 05/10/2015 and contributed by Dun Qiu

Suppose $\lambda_k=(n+k,n,n-k)$ is a partition of $3n$, $\mu_k=(n+k+1,n,n-k)$ is a partition of $3n+1$ and $f^\lambda$ is the number of standard Young tableaux of shape $\lambda$.

$a(2n)=\sum_{k=0}^n f^{\lambda_k}$ and $a(2n+1)=\sum_{k=0}^n f^{\mu_k}.$

A few initial terms of $a(n)$ for $n\ge 1$ are $1, 3, 6, 30, 70$. $a(n)$ seems to coincide with A215294 on OEIS.