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I have a "very" indirect proof that the following fact is true for every Young diagram $\lambda \vdash n$ and every $r \in \{0,\dotsc,n\}$:

\begin{equation} d_\lambda = \sum_{a \in \mathrm{AC}(\lambda)} | \{T \in \mathrm{SYT}(\lambda \cup a) \, | \, \mathrm{maj}_{n+1}(T) = r \} |, \end{equation} where $d_\lambda$ is the dimension of the symmetric group irrep $\lambda$ or, in other words, $d_\lambda = |\mathrm{SYT}(\lambda)|$; $\mathrm{SYT}(\lambda)$ is the set of standard Young tableaux of shape $\lambda$; $\mathrm{AC}(\lambda)$ is the set of addable cells of the Young diagram $\lambda$; $\lambda \cup a$ is the Young diagram obtained from $\lambda$ by adding the cell $a$; $\mathrm{maj}_{n+1}(T)$ is the major index of the standard Young tableaux $T$ modulo $n+1$.

I assume this fact must be proven somewhere in combinatorics literature, but I am unable to find the relevant paper/book. A reference would be highly appreciated.

Moreover, I am interested in understanding the following number for every $\lambda \vdash n$, every $r \in \{0,\dotsc,n\}$ and every $d \leq \ell(\lambda)$: \begin{equation} f_{\lambda,d}(r) = \sum_{a \in \mathrm{AC}_d(\lambda)} | \{T \in \mathrm{SYT}(\lambda \cup a) \, | \, \mathrm{maj}_{n+1}(T) = r \} |, \end{equation} where $\ell(\lambda)$ is the number of rows of the Young diagram $\lambda$ and $\mathrm{AC}_d(\lambda)$ is the set of addable cells of the Young diagram $\lambda$ with row coordinate at most $d$. Clearly, $f_{\lambda,\ell(\lambda)+1}(r) = d_\lambda$ for every $r$, but what can we say about other values of $d$?

In particular, the number $f_{\lambda,\ell(\lambda)}(r)$ is especially interesting to me. Note, that understanding $f_{\lambda,\ell(\lambda)}(r)$ is equivalent to understanding \begin{equation} d_\lambda - f_{\lambda,\ell(\lambda)}(r) = | \{T \in \mathrm{SYT}(\lambda \cup (\ell(\lambda)+1,1)) \, | \, \mathrm{maj}_{n+1}(T) = r \} |, \end{equation} where $(\ell(\lambda)+1,1)$ is an addable cell at row $\ell(\lambda)+1$ and column $1$.

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    $\begingroup$ Joshua Swanson, jpswanson.org should be able to answer, or give a reference. $\endgroup$
    – Per Alexandersson
    Nov 19 at 10:04
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    $\begingroup$ One observation, which you might already know, is that the number of SYT in your union over $a$ in $AC(\lambda)$ is $(n+1)d_{\lambda}$. This is because they index a basis for the induced representation from $S_n$ to $S_{n+1}$. So your claim is equivalent to knowing that $\mathrm{maj} \mod (n+1)$ is equidistributed. $\endgroup$
    – Christian Gaetz
    Nov 19 at 11:57
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    $\begingroup$ Using standard results on major indices (Proposition 7.19.11 in Stanley's EC2) and the Pieri rule, your claim is equivalent to the following: For any partition $\lambda$ of $n$, the formal power series $\left(1-q^2\right)\left(1-q^3\right)\cdots\left(1-q^{n+1}\right) s_\lambda\left(1,q,q^2,q^3,\ldots\right)$ is a polynomial in $q$ that is divisible by $1 + q + q^2 + \cdots + q^n$. But this is easy, since the $\left(1-q^{n+1}\right)$ factor on the left hand side can be rewritten as $\left(1-q\right)\left(1 + q + q^2 + \cdots + q^n\right)$, and the ... $\endgroup$
    – darij grinberg
    Nov 19 at 17:42
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    $\begingroup$ ... power series $\left(1-q^2\right)\left(1-q^3\right)\cdots\left(1-q^n\right) \left(1-q\right) s_\lambda\left(1,q,q^2,q^3,\ldots\right)$ is just the generating function for the major index on SYTs of shape $\lambda$ (again Proposition 7.19.11), thus a polynomial. $\endgroup$
    – darij grinberg
    Nov 19 at 17:44
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    $\begingroup$ Quick remark on how I lost the $1-q$ factor at the beginning: Summing over all $a \in \operatorname{AC}\left(\lambda\right)$ is equivalent to summing over all partitions $\mu$ of the form "$\lambda$ plus one box". But on the level of Schur functions, this is equivalent to multiplying by $h_1 = x_1 + x_2 + x_3 + \cdots$ (by the Pieri rule). Upon principal specialization, this means multiplying by $h_1\left(1, q, q^2, q^3, \ldots\right) = 1 / \left(1-q\right)$. Thus, the $1-q$ factor in Proposition 7.19.11 gets cancelled. $\endgroup$
    – darij grinberg
    Nov 19 at 17:46

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