6
$\begingroup$

In a joint project, we are currently working on an online combinatorial statistic finder in which (beside other things) want to gather information about combinatorial collections and statistic, see the website http://www.findstat.org.

In this context, I am reading through Dominique Foata and Doron Zeilberger's paper

''Denert's permutation statistic is indeed Euler-Mahonian''

from 1990. In there, they prove that the bistatistic given by the number of exceedences and the Denert index is Euler-Mahonian. Moreover, they say:

"The most natural proof of this result would be in terms of a bijection from $\mathcal{S}_n$ to itself that sends the pair (des,maj) simultaneously to the pair (exc,den). Although it is rather easy to find a bijetion that sends maj to den ..., and it is now trivial ... to find a bijection that sends exc to des, we are unable, at present, to find a bijection that does both at the same time. ... We really hope that such a bijective proof of Denert's conjecture will be found ... ."

So my question is:

Is there an explicit bijection on permutations known that sends thnumber of descents to the number of exceedences, and at the same time the major index to the Denert index?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Yes. G.-N. Han, Distribution Euler-mahonienne : une correspondance, C. R. Acad. Sci. Paris, 310, Série I, 1990, pp. 311-314.

G.-N. Han, Une nouvelle bijection pour la statistique de Denert, C. R. Acad. Sci. Paris, 310, Série I, 1990, pp. 493-496.

G.-N. Han, Une transformation fondamentale sur les réarrangements de mots, Adv. in Math., 105(1), 1994, pp. 26-41.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks Guoniu -- I saw the papers in the meantime but totally forgot that I had been asking the question here... $\endgroup$
    – Christian Stump
    Commented Jan 7, 2014 at 16:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .