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It is a well-known exercise that $C_n = \chi_{(n,n)}(1)=\chi_{(n,n)}^{1^n}$ where $C_n$ is the $n$th Catalan number and $\chi_{(n,n)}^{1^n}$ is the character of the irrep $(n,n)$ on conjugacy class $1^n = (1,1,\cdots,1)$.

A related identity that I stumbled across for fixed-point-free involutions is that $\chi_{(n,n)}^{2^n} = \pm \binom{n}{\lfloor n/2 \rfloor}$.

I haven't been able to find this stated as an exercise in a textbook or online manuscript, and it seems too obvious to have been overlooked. If you have seen this identity before and recall the reference, I would be happy to know.

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    $\begingroup$ Representing $(n,n)$ on a $2$-runner abacus, this follows very quickly from 2.7.27 in James and Kerber: we have to make $n$ upward bead moves, one bead moving up $\lfloor n/2 \rfloor$ times, and the other $\lceil n/2 \rceil$ times. (Posted as a comment since I'm sure you know this already, and its only a short proof, not the identity.) $\endgroup$
    – Mark Wildon
    Commented Jul 14, 2016 at 9:47
  • $\begingroup$ Feels like it is related to number of SYTs of two-row shape of equal lengths, which are easy to see are in bijection with say Dyck paths... (see Stanleys book on Catalan objects) $\endgroup$
    – Per Alexandersson
    Commented Jul 14, 2016 at 12:21

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This is a special case of Corollary 2.2 in S. Fomin and N. Lulov, On the number of rim-hook tableaux, taking $r=2$ and $\lambda = (n,n)$. It's probably easier to use the formula in the proof than the corollary itself.

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  • $\begingroup$ Exactly what I was looking for Mark. Thanks! I guess the the theory of rim-hook tableaux is a bit too specialized for any of its consequences to appear as exercises. $\endgroup$
    – Nathan Lindzey
    Commented Jul 14, 2016 at 19:07

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