**The Question in its original form has been answered, but there is a follow-up, see the end of the post.**

A border-strip is a skew Young diagram that does not contain a $2 \times 2$-box. A border-strip with $n$ boxes can be encoded as a binary string of length $n-1$ by starting from the lower left corner and writing $0$ for every right step and $1$ for every up step, see the example below.

If $\epsilon=\epsilon_1 \epsilon_2 \dotsb \epsilon_{n-1} \in \{ 0,1\}^{n-1}$ is a binary string, then we denote the set of all standard Young tableau of shape $\epsilon$ as $\textrm{SYT}(\epsilon)$. One such standard Young tableau is seen below.

On any Standard Young tableau $T$, one can define the major index $\textrm{maj}(T)$ as follows. A descent of $T$ is an entry $i$ such that $i+1$ appears strictly below $i$ in $T$. Define $\textrm{maj}(T)$ as the sum of all descents of $T$. For example, the major index of the standard Young tableau above is $3+5+8=16$. The major index generating polynomial of $\epsilon$ is then defined as

$$f^\epsilon(q)=\sum_{T \in \textrm{SYT}(\epsilon)}q^{\textrm{maj}(T)}.$$

Is anything known about major index generating polynomial of standard border-strip tableaux? If $\epsilon$ is a hook, that is $\epsilon_1 = \dotsb = \epsilon_k =1$ and $\epsilon_{k+1} = \dotsb = \epsilon_{n-1}=0$, then the answer is known by using the $q$-hook formula as

$$f^\epsilon(q)=\begin{bmatrix} n-1 \\ k \end{bmatrix}_q$$

However, I could not find any known theory about the general case. The particular problem I am interested in is the following: Let $1 \leq i <n/2$. If $\epsilon_i=1$ and $\epsilon_{n-i-1}=0$ then put $\epsilon'=\epsilon_1\epsilon_2 \dotsb \epsilon_{i-1} 0 \epsilon_{i+1}\dotsb \epsilon_{n-i-2} 1 \epsilon_{n-i}\dotsb \epsilon_{n-1}$. I conjecture that if $\omega=e^{2\pi i/n}$ is a primitive $n$:th root of unity, then $$f^\epsilon(\omega)f^{\epsilon'}(\omega)=1.$$

I have verified this conjecture for $n \leq 4$ and have verified several special cases for bigger $n$ as well.

**UPDATE:** My supervisor was able to prove an even stronger assertion. Suppose that we have a skew diagram $\lambda/\mu$ of size $n$, that is, $|\lambda/\mu|=n$. Then

$$f^{\lambda/\mu}(\omega)=\begin{cases}(-1)^{\textrm{ht}(\lambda/\mu)} & \text{if } \lambda/\mu \text{ is a border-strip}, \\ 0 & \text{otherwise}.\end{cases}$$

Here, the height $\textrm{ht}(\lambda/\mu)$ is the number of rows that $\lambda/\mu$ touches minus one. So the height of the border-strip in the Figure is $3$. He used the suggested Proposition 7.19.11 in Richard Stanley's Enumerative Combinatorics Volume 2 and the Murnaghan–Nakayama rule. It is not hard to see that this implies the original Conjecture (note that border-strips given by $\epsilon$ and $\epsilon'$ have the same height). However, I think it would be interesting to know if the equality above could be proved by a more straightforward, combinatorial argument.

Enumerative Combinatorics Volume 2? It reduces your question to computing $s_\epsilon\left(1,q,q^2,\ldots\right)$. This is an evaluation of a skew Schur function, but it has the shape of a ribbon so it may be easier to deal with than a usual Schur. $\endgroup$ – darij grinberg Mar 19 '19 at 14:58