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The Kostka number $K_{\lambda \mu}$ gives the number of SSYT of shape $\lambda$ with content $\mu$. Are there any results regarding Kostka numbers for "partial contents" (I do not know if there is a term for this already used in the literature)? By this, I mean something like "How many SSYT exist of shape $\lambda$ with exactly $a_1$ 1 entries, $a_2$ 2 entries, and any number of entries from $\{3, \ldots, n\}$?" This would be a sum of Kostka numbers, something like $$\tilde{K}_{\lambda (a_1, a_2)} = \sum_{\substack{x_3, \ldots, x_n \\ \mu = (a_1, a_2, x_3, \ldots, x_n) \\ a_1 + a_2 + x_3 + \dots + x_n = |\lambda|}} K_{\lambda \mu}\,.$$

Any results in this direction would be very helpful for computing certain coefficients that appear in the anomaly cancellation equations for 6D supergravity theories.

Edit: As mentioned in the comment by Jair Taylor, these could also be expressed in terms of skew Kostka numbers, something like $$\tilde{K}_{\lambda (a_1, a_2)} = \sum_{\substack{\mu \\ |\mu| = a_1 + a_2}} K_{\mu (a_1, a_2)} \sum_{\substack{\gamma \\ |\gamma| = |\lambda| - a_1 - a_2}} K_{\nu / \mu, \gamma}\,.$$ To me, this seems somewhat more involved than the original sum, but it could be simplified if there is some analogue of the hook-content formula for skew tableaux.

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    $\begingroup$ The entries containing $3, \ldots, n$ forms a skew SSYT here, so you can get a sum involving skew Kostka numbers. $\endgroup$ – Jair Taylor Apr 5 '18 at 21:56
  • $\begingroup$ I had thought of that, but that sum seems less useful even than the one given above, because I would have to sum over all possible shapes of the tableau containing 1s and 2s, and then for each term multiply the associated Kostka number by the sum over all contents of skew Kostka numbers for the corresponding skew diagram. Is there a simplification I'm missing? $\endgroup$ – Andrew Patrick Turner Apr 6 '18 at 2:29
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    $\begingroup$ Since you're summing all the skew Kostka numbers, this is the same as evaluating the Schur function at $1,1, \ldots, 1$. I've posted a more detailed answer below. $\endgroup$ – Jair Taylor Apr 6 '18 at 3:14
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Let $\mu'$ be the partition $1^{n_1}2^{n_2}\cdots l^{n_l}$ with $|\mu'| = k$. Any SSYT of shape $\lambda$ with $n_1$ $1$'s, $n_2$ $2$'s, etc. and any number of $l+1$'s, $l+2$'s, ... is given by

1) an SSYT of some shape $\lambda'$ and content $\mu'$, and

2) an skew SSYT of shape $\lambda / \lambda'$.

Hence the number of $SSYT$'s satisfying these conditions is

$$\sum_{\lambda' \vdash k} K_{\lambda', \mu'} \sum_{\nu} K_{\lambda / \lambda', \nu}$$

The sum $$\sum_{\nu} K_{\lambda / \lambda', \nu}$$ is the skew Schur function $s_{\lambda/\lambda'}$ evaluated at $x_1 = 0, \ldots, x_l = 0, x_{l+1} = 1, x_{l+2} =1, \ldots, x_{n} = 1$; or equivalently, $x_1 = x_2 = \ldots = x_{n-l} = 1$, $x_{n-l+1} = \cdots = 0$. Hence your the your count is

$$\sum_{\lambda' \vdash k} K_{\lambda', \mu'} s_{\lambda/\lambda'}(1, 1, \ldots, 1).$$

This should be somewhat faster to compute than your count, since there are various ways of evaluating Schur function such as the Jacobi-Trudi identity. It may be possible to simplify this more - I am not sure.

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  • $\begingroup$ Ah, that's very clever. For numerically finding these values, my current approach is to compute the Schur function $s_\lambda(x_1, \ldots, x_l, 1, \ldots, 1)$, and extract the coefficient for the term I want. I wonder if this is similarly efficient. It would be nice to have something closed form, in order to get closed forms for the coefficients I am computing, but I know that is probably asking too much for anything associated with Kostka numbers, since computing them is #P-complete. $\endgroup$ – Andrew Patrick Turner Apr 6 '18 at 4:54
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    $\begingroup$ Ah, I see. That method would also work. Note that you can extract the coefficient of $x_1^{n_1} \cdots x_l^{n_l} = x^{\mu'}$ in a symmetric function $X$ by taking the Hall inner product $\langle X, h_{\mu'} \rangle$ where $h_\mu$ is the complete homogeneous symmetric function. I'm not sure if this is faster, but it's easy to do in, e.g., Sage. $\endgroup$ – Jair Taylor Apr 6 '18 at 5:16

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