Consider the two-row skew shape $\lambda_n = (2n+1,n)/(1)$. The number of standard Young tableaux of this shape is $\binom{3n}{n}-\binom{3n}{n-2}$ (since one can easily biject this to the set of non-skew shape $(2n+1,n)$, see A026004 in the OEIS.
Let us now consider the $q$-analog $Y_n(q):= \sum_{T \in SYT(\lambda_n)} q^{maj(T)}$. By a result in this paper, we know that there is some cyclic group $C_n$ of order $n$ acting on $SYT(\lambda_n)$ such that $(SYT(\lambda_n),C_n,Y_n(q))$ satisfies the cyclic sieving phenomenon.
Now for the interesting observation: $Y_n(1)$ counts the number of non-crossing forests with exactly two trees, on $n+2$ vertices. This set also satisfies a cyclic sieving phenomenon under rotation, see this paper by S. Kluge. Note that this is a cyclic group action of order $n+2$.
By computer experiments, it seems that the $q$-analog $Y_n(q)$ evaluates to the same values at $n+2$th roots of unity the $q$-analog on non-crossing forests considered by Kluge: $$ NCF_{n+2,2}(q) = \frac{[n+2]}{[2n+2]} \binom{3n+1}{n}_q. $$
Conjecture: if $\xi$ is either an $n$th root of unity OR an $(n+2)$th root of unity, then $Y_n(\xi) = NCF_{n+2,2}(\xi)$. I think this should be straightforward to prove by using results from this preprint, combined with the q-Lucas theorem.
Question: Can one find a bijection showing that the set of non-crossing forests with 2 trees on $n+2$ vertices is equal to the number of SYTs of shape $\lambda_n$? The cyclic group action of order $n+2$ (rotation) on non-crossing forests, what does this correspond to on the SYTs?
Motivation: the previously known cyclic sieving results on SYTs have order $n$, where $n$ usually is a divisor of the number of boxes in the shape. (The group action, if known, is the promotion operator.) Establishing a bijection above, would give a more exotic example, where the fake-degree polynomial $Y_n$ is now evaluated at $n+2$th roots of unity instead.
