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First, some background: I will assume the anglophone conventions for Young tableaux in what follows. Given a standard Young tableau $T$ of shape $\lambda$, we can define the cocharge tableau $C(T)$ as follows. Let $n(i)$ be a labeling of the boxes of a $\lambda$-diagram such that $n(i)$ is the box of $T$ containing entry $i$. Let the $n(1)$ box (ie the top-left box) of $C(T)$ be $0$. If entry $n(i)$ of $C(T)$ is $c$, we obtain the next entry as follows: if the $n(i+1)$-th entry of $T$ is below the $n(i)$-th entry, fill box $n(i+1)$ of $C(T)$ with $c+1$. If not, fill it with entry $c$. In other words, every time we have an inversion in $T$, we increase the number we are filling $C(T)$ with. Let $\sum C(T)$ denote the sum of the entries of $C(T)$.

It is clear that the minimum value for $\sum C(T)$ is obtained from the Young tableaux $T$ with first row $1,2,\dots, \lambda_1$, second row $\lambda_1+1, \dots$, and so on. (Filling first left to right, then down.) The maximum value is obtained from the Young tableaux with first column $1,2,\dots, \lambda_1$, and so on. (Filling first top to bottom, then leftwards.) Let the minimum value be $m$ and maximum value be $M$.

My question: I believe that for any $0\leq i\leq M-m$, the number of standard Young tableaux $T$ with $\sum C(T)=m+i$ is exactly equal to the number of standard Young tableaux with $\sum C(T) = M-i$. I've checked this for all partitions of $n$ up to 7 or 8. Does anyone know if there is a reference that has a proof of this, or does anyone have a good explanation for why this should be true?

EDIT: I'll give an example to make clear what I'm asking. Consider, for example, the partition $(3,1,1)$ of 5. There are 6 standard Young tableaux of this shape. Consider the two below $$ T_1 = \begin{pmatrix} 1 & 2 & 3 \\ 4 \\ 5 \end{pmatrix} \hspace{20pt} T_2 = \begin{pmatrix} 1 & 4 & 5 \\ 2 \\ 3 \end{pmatrix} $$ These lead to cocharge tableaux $$ C(T_1) = \begin{pmatrix} 0 & 0 & 0 \\ 1 \\ 2 \end{pmatrix} \hspace{20pt} C(T_2) = \begin{pmatrix} 0 & 2 & 2 \\ 1 \\ 2 \end{pmatrix} $$ with $\sum C(T_1)=3$ and $\sum C(T_2) = 7$. You can show that each of these is the unique tableaux leading to the sum. What I claim, and what is in fact true in this case, is that there is in fact the same number of tableaux with cocharge summing to 4 as there is with cocharge summing to 6. (In this case, this is trivial to verify.)

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Just a sketch of a proof; sorry, this is all I can do with the time I have right now.

If $m\in\mathbb N$ is arbitrary, then an $m$-reciprocal polynomial means a polynomial $p \in \mathbb Z\left[q\right]$ whose coefficient before $q^i$ equals its coefficient before $q^{m-i}$ for every $i\in\mathbb Z$ (this implies that the degree of $p$ is $\leq m$).

Let $\operatorname*{SYT}\left(\lambda\right)$ denote the set of all standard tableaux of shape $\lambda$. You want to prove that the generating function $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\sum C(T)} \in \mathbb Z\left[q\right]$ is an $m$-reciprocal polynomial, where $m = \dbinom{n}{2} + \sum\limits_i\dbinom{\lambda_i}{2} - \sum\limits_i\dbinom{\lambda^{\prime}_i}{2}$ (where we use standard notations: $\lambda^{\prime}$ is the conjugate partition of $\lambda$, and $\mu_i$ denotes the $i$-th entry of $\mu$). Sorry that my $m$ is not your $m$ !

If $T$ is a standard Young tableau of size $n$, we denote by $D\left(T\right)$ the set of all descents of $T$, that is, the set of all $i\in\left\{1,2,...,n-1\right\}$ such that the entry $i+1$ appears in $T$ in a lower row than the entry $i$ appears in. The major index of the standard Young tableau $T$ is defined as $\sum\limits_{i\in D\left(T\right)} i$, and denoted by $\operatorname*{maj}T$. The comajor index of the standard Young tableau $T$ of size $n$ is defined as $\sum\limits_{i\in D\left(T\right)} \left(n-i\right)$, and denoted by $\operatorname*{comaj}T$. In the proof of Proposition 7.19.11 in Richard Stanley's Enumerative Combinatorics, volume 2, it is shown that $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{comaj} T} = \sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$ (actually, it is shown that this holds for skew shapes as well). But it is easily seen that every standard Young tableau $T$ satisfies $\sum C(T) = \operatorname*{comaj} T$, and so this becomes $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\sum C(T)} = \sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$. Hence, it remains to prove that the polynomial $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$ is $m$-reciprocal for $m = \dbinom{n}{2} + \sum\limits_i\dbinom{\lambda_i}{2} - \sum\limits_i\dbinom{\lambda^{\prime}_i}{2}$.

But this should follow easily from plugging $q^{-1}$ instead of $q$ into the formula for $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$ obtained by using Proposition 7.19.11 in Stanley's above-mentioned book and then simplifying the Schur function using (7.104) (again, in Stanley's book).

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