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Let $S_n$ be the set of permutations on $\{1,2,\dotsc,n\}$. A peak-value of $\pi$ is some $\pi_i$ such that $\pi_{i-1} < \pi_i > \pi_{i+1}$, where $1<i<n$. Let $PV(\pi)$ denote the set of peak-values.

A run of $\pi \in S_n$ is a maximal subsequence of elements in increasing order. Given $\pi$, we can list the runs increasingly based the first entries. Let us call this operation runsort. A run of $\pi \in S_n$ is a maximal subsequence of elements in increasing order. For example, $$ \mathrm{runsort}(458 \, 23 \, 6 \, 17) = 17\,23\,458\,6. $$

O. Nabawanda and I have with some effort (arxiv preprint) been able to show that $$ \sum_{ \pi \in S_n} t^{|PV(\pi)|} = \sum_{ \pi \in S_n} t^{|PV(\mathrm{runsort}(\pi))|}. $$ We actually show that a multivariate identity (which not only keeps track of the number of peaks, but the actual set of peak-values) holds. We construct a bijection through a recursive process, and the bijection is not canonical.

I wonder if perhaps there is some easy bijective argument which proves the univariate identity above, where the bijection is natural or canonical in some sense.

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  • $\begingroup$ I checked all the statistics on permutations in findstat, whether your set-valued equidistribution could be refined, without success. However, I did not (yet) check statistics composed with maps. Do you have an idea what might make sense checking? $\endgroup$ Commented Apr 30, 2021 at 19:04
  • $\begingroup$ @MartinRubey Ah, It can for sure be refined - one can refine by the set of first values in runs (before sorting). There are a few other statistics one refine with. But that might make it more complicated (perhaps)... $\endgroup$ Commented Apr 30, 2021 at 19:10
  • $\begingroup$ I made a silly mistake. There are indeed some nice refinements. As you say, descent bottoms, and - simultaneoiusly - the number of left to right minima and the number of occurrences of the consecutive pattern 132. I think that this is elegant enough to be worth proving. $\endgroup$ Commented May 2, 2021 at 20:16
  • $\begingroup$ @MartinRubey left-to-right-minima you say, hmm... that is very interesting, I have not thought about that! Consecutive 132-patterns we observed, and I believe it is easy to include a proof of this in the framework we have. $\endgroup$ Commented May 2, 2021 at 20:42
  • $\begingroup$ Very limited data (all permutations of at most 9 elements) suggest that one can drop both occurrences of "number" in my previous comment, stipulating that an occurrence of a consecutive pattern 132 is defined by its middle letter. $\endgroup$ Commented May 3, 2021 at 10:32

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