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)`.
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.