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For a labeled tree $T$ on $\{1,2,...,n\}$ an inversion of $T$ is a pair $1 < i < j \leq n$ such that $j$ belongs to the unique path from $1$ to $i$ (we think of $T$ as being rooted at $1$). Let $\mathrm{inv}(T)$ denote the number of inversions of $T$.

Define the generating function $f(q) := \sum_{T} q^{\mathrm{inv}(T)}$ where the sum is over all labeled trees on $\{1,2,...,n\}$.

Then it is known that $f(-1)$ is the number of alternating permutations in $\mathfrak{S}_n$ (i.e., the so-called "Euler number"). See e.g. Goulden-Jackson 3.3.49(d).

Question: Is there a simple proof of this result via a sign-reversing involution?

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J.-J. Pansiot, Nombres d'Euler et Inversions dans les Arbres, Europ. J. Combin. 3 (1982), 259–262, uses a sign-reversing involution to show that $f(-1)$ is the number of increasing trees on $[n]$ in which every vertex other than the root has an even number of children. In Pansiot's paper he gives a reference to a paper of Viennot (that I have not looked at) that shows that these trees are counted by the Euler numbers.

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  • $\begingroup$ Perfect! Thanks! $\endgroup$ Oct 17 '19 at 9:28

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