1 Introduction and notations
The distribution of the number of descents has been widely studied on several classes of combinatorial objects such as permutations [14], cycles [7, 8], and words [3, 10]. Many interpretations of this statistic appear in several fields as Coxeter groups [4, 11] or lattice theory [5, 12]. One of the most famous result involves the Foata fundamental transformation [9] to establish a onetoone correspondence between descents and excedances on permutations. This bijection provides a more straightforward proof than those of MacMahon [14] for the equidistribution of these two Eulerian statistics.
In this paper, we present a bijection à la Foata on the symmetric group that exchanges pure excedances with special kind of descents defined as a mesh pattern [6] (see below for the definitions of these patterns). Then, we deduce that the popularities but not the distributions of pure descents [2] and pure excedances are the same. They are given by the generalized Stirling number (see A001705 in [15]) where is the th harmonic number. Finally, we conjecture the existence of a bijection on the symmetric group that exchanges pure excedances and while preserving the number of cycles.
Let be the set of permutations of length , i.e., all bijections from into itself. The oneline representation of a permutation is where , . For , the product is the permutation . A cycle is a length permutation satisfying and for . For , we denote by the set of all length permutations admitting a decomposition in a product of disjoint cycles. The set is counted by the signless Stirling numbers of the first kind defined by
where if or , except (see [16, 17]). These numbers also enumerate length permutations having lefttoright maxima, i.e., values such that for (see [16]), and permutations with pure descents, i.e., descents where there is no such that (see [2]). Note that a pure descent can be viewed as an occurrence of the mesh pattern where . Indeed, for a length permutation and a subset , an occurrence of the mesh pattern in a permutation is an occurrence of in with the additional restriction that no element of lies inside the shaded regions defined by , where means the square having bottom left corner in the graphical representation of . For instance, an occurrence of the mesh pattern in Figure 1 corresponds to an occurrence of a pure descent. See [6] for a more detailed definition of mesh patterns.
Regarding this interpretation of pure descents in terms of mesh patterns, we define other kinds of descents by the mesh patterns , with and for . Modulo the fundamental symmetries on permutations (reverse and complement), it is straightforward to see that , , and are respectively in the same distribution class as , and . Then, we deal with only mesh patterns , . We refer to Figure 1 for a graphical illustration. On the other hand, we define a pure excedance as an occurrence of an excedance, i.e. , with the additional restriction that there is no point such that with . Although such a pattern (called ) is not a mesh pattern, we can represent it graphically as shown in Figure 1.
A statistic is an integervalued function from a set of length permutations (we use the boldface to denote statistics). For a pattern , we define the pattern statistic where the image of by is the number of occurrences of in . The popularity of in is the total number of occurrences of over all objects of , that is (see [5] for instance). Below, we present statistics that we use throughout the paper:
We organize the paper as follows. In Section 2, we focus on patterns , . We prove that the statistics and are equidistributed by giving algebraic and bijective proofs. Next, we provide the bivariate exponential generating function for the distribution of , and we deduce that has the same popularity as and , without having the same distribution. In Section 3, we present a bijection on that transports pure excedances into patterns . Notice that the Foata’s first transformation is not a candidate for such a bijection. As a consequence, pure descents and pure excedances are equipopular on , but they do not have the same distribution. Combining all these results, we deduce that patterns , , and are equipopular on the symmetric group . Finally we present two conjectures about the equidistribution of and , and that of and .
2 The statistics ,
For , let be the set of length permutations having occurrences of , and denote by its cardinality. Let be the bivariate exponential generating function . In [2, 13], it is proved that equals the signless Stirling numbers of the first kind (see A132393 in [15]). Indeed, a permutation can be uniquely obtained from an length permutation by one of the two following constructions:

if , then we increase by one all values of greater than or equal to , and we add at the end;

if , then we increase by one all values of greater than or equal to a given value , and we add at the end.
Then, we deduce the recurrence relation with for , and the bivariate exponential generating function is
which proves that .
Below, we prove that also counts length permutations having occurrences of the pattern .
Theorem 1.
The number of length permutations having occurrences of pattern equals .
Proof.
An length permutation can be uniquely obtained from an length permutation by one of the two following constructions:

if , then we increase by one all values of and we add at the end;

if , then we increase by one all values of greater than or equal to given value , , and we add at the end.
We deduce the recurrence relation with the initial condition , and then . ∎
Now, we focus on the distribution of the pattern . Table 1 provides exact values for small sizes.
Theorem 2.
Proof.
Let denote a permutation of length having occurrences of pattern . Let (resp. ) be the number of such permutations satisfying (resp. ). Obviously, we have
A permutation with can be uniquely constructed from an length permutation as . No new occurrences of are created, and we obtain
A permutation satisfying can be uniquely obtained from an length permutation by adding a value on the right side of its oneline notation, after increasing by one all the values greater than or equal to . This construction creates a new pattern if and only if ends with . Thus, we deduce
Combining the equations, we obtain for and
which implies the following differential equation
A simple calculation provides the claimed closed form for the generating function . ∎
Corollary 1.
Proof.
The generating function of the popularity is directly deduced from the bivariate generating function of pattern distribution by calculating
∎
1  2  3  4  5  6  7  8  
0  1  1  1  1  1  1  1  1 
1  1  5  20  84  409  2365  16064  
2  3  35  295  2359  19670  
3  15  315  4480  
4  105  
1  2  6  24  120  720  5040  40320  

The statistic has a different distribution from and , but the three patterns have the same popularity. Below we present a bijection on that transports the statistic to the statistics .
Theorem 3.
There is a onetoone correspondence on such that for any , we have
Proof.
Let be a permutation of length having occurrences of . We decompose
where
 are the tops of the occurrences of , i.e. values such that there does not exist such that ,
 is a maximal sequence such that all its values are lower than ,
 for , is an increasing sequence such that and .
Now we construct an length permutation with pure cycles as follows:
For instance, if then . The map is clearly a bijection on such that equals the number of pure cycles in . ∎
Note that is closely related to the Foata fundamental transformation.
3 The statistic of pure excedances
In order to prove the equidistribution of and , regarding Theorem 3, it suffices to construct a bijection on that transports pure excedances to pure cycles. Here, we first exhibit a bijection on the set of length derangements (permutations without fixed points), then we extend it to the set of all permutations .
Any permutation is uniquely decomposed as a product of transpositions of the following form:
where are integers such that . The transposition array of is defined by , which induces a bijection from to the product set . By Lemma 1 from [1], the number of cycles of a permutation is given by the number of fixed points in . Moreover, it is straightforward to check the two following properties:
 if , then if and only if there is no number such that ;
 if and , then is the minimal element of a cycle of length at least two in .
So, we deduce the following lemma.
Lemma 1.
The transposition array corresponds to a derangement if and only if: there is such that
Given a derangement and its graphical representation . We say that the square is free if all following conditions hold:

Neither nor is a position of a pure excedance;

is not on the first diagonal, i.e. ;

there does not exist such that ;

is not a pure excedance such that and ;

there does not exist , with such that all values of the interval appear on the right of in .
Whenever at least one of the statements above is not satisfied, we say that the square is unfree. Notice that if and are not the positions of a pure excedance, then the square is always free. So, for a column of the graphical representation of such that and are not the positions of a pure excedance, we label by the th free square from the bottom to the top. We refer to Figure 2 for an example of this labelling.
Now we define the map from to the set of transposition arrays of length satisfying the property of Lemma 1.
For a permutation , we label its graphical representation as defined above, and is obtained as follows:

if is a pure excedance in , then we set and ;

otherwise, is the sum of the label of the free square with the number of pure excedances such that .
For instance, if then we obtain (see Figure 2).
Let us consider , . If is a pure excedance of , then we fix and . Otherwise, the square is unfree, and all squares , , are unfree, which implies that the number of free squares in the th column is less than or equal to . This means that lies in . Note that, by construction, all labeled squares do not correspond to any pure excedance. Now let us prove that the square cannot be labeled . Indeed, if then the label of is necessarily at most ; otherwise, if then the fact that is not a pure excedance implies that there is with . Let us choose the lowest with this property. Using (v), the square is unfree, which implies that the label of is less than or equal to minus the minimal number of unfree squares in column , that is . Moreover, the transposition array has exactly fixed points, and for any fixed point there necessarily exists such that . This implies that .
Theorem 4.
The map from to is a bijection such that
Proof.
Since the cardinality of equals that of , and the image of by is contained in , it suffices to prove the injectivity.
Let and , , two derangements in . If and do not have the same pure excedances, then, by construction, and do not have the same fixed points, and thus .
Now, let us assume that and have the same pure excedances. If there is a pure excedance such that then the definition implies . Otherwise the two permutations have the same pure excedances , and for each of them we have . Let be the greatest integer such that (without loss of generality, we assume ). In this case, is not a pure excedance for the two permutations. Thus, (resp. ) is the sum of the label of (resp. ) with the number of pure excedances such that (resp. ). Since we have , the label of is less than the label of , and the number of pure excedances such that is less than or equal to the number of pure excedances such that . Then we have . Then is an injective map, and thus a bijection. ∎
Theorem 5.
There is a onetoone correspondence on the set of length derangements such that for any ,
Proof.
Theorem 6.
The two bistatistics and are equidistribiuted on .
Proof.
Considering Theorem 5, we define the map on . Let be the permutation obtained from by deleting all fixed points and after normalization as a permutation. Let be the set of fixed points of . Then, we set . So, is obtained from by inserting fixed points after a shift of all other entries in order to produce a permutation in . By construction, we have and which completes the proof. ∎
A byproduct of this theorem is
Corollary 2.
The statistics and are equidistributed on .
Theorem 7.
The two statistics and are equidistributed on .
Notice that the Foata’s first transformation is not a candidate for proving the equidistribution of and , while it transports to .
Finally, we present two conjectures for future works.
Conjecture 1.
The two bistatistics and are equidistributed on .
Conjecture 2.
The two bistatistics and are equidistributed on .
Acknowledgements
We would like to greatly thank Vincent Vajnovszki for having offered us Conjecture 2.
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