Abstract

Since the 1950s, mathematicians have successfully interpreted the traditional Eulerian numbers and -Eulerian numbers combinatorially. In this paper, the authors give a combinatorial interpretation to the general Eulerian numbers defined on general arithmetic progressions .

1. Introduction

Definition 1. Given a positive integer , define as the set of all permutations of . For a permutation , is called an ascent of if ; is called a weak exceedance of if .

It is well known that a traditional Eulerian number is the number of permutations that have weak exceedances [1, page 215]. And satisfies the recurrence: , , ,

Besides the recursive formula (1), can be calculated directly by the following analytic formula [2, page 8]:

Definition 2. Given a permutation , define functions

Since the 1950s, Carlitz [3, 4] and his successors have generalized Euler’s results to -sequences . Under Carlitz’s definition, the -Eulerian numbers are given by where functions are as defined in Definition 2.

In [5], instead of studying -sequences, the authors have generalized Eulerian numbers to any general arithmetic progression

Under the new definition, and given an arithmetic progression as defined in (5), the general Eulerian numbers can be calculated directly by the following equation [5, Lemma 2.6]: Interested readers can find more results about the general Eulerian numbers and even general Eulerian polynomials in [5].

2. Combinatorial Interpretation of General Eulerian Numbers

The following concepts and properties will be heavily used in this section.

Definition 3. Let be the set of -permutations with weak exceedances. Then . Furthermore, given a permutation , let , where .

Given a permutation , it is known that can be written as a one-line form like , or can be written in a disjoint union of distinct cycles. For written in a cycle form, we can use a standard representation by writing (a) each cycle starting with its largest element and (b) the cycles in increasing order of their largest element. Moreover, given a permutation written in a standard representation cycle form, define a function as to be the permutation obtained from by erasing the parentheses. Then is known as the fundamental bijection from to itself [6, page 30]. Indeed, the inverse map of the fundamental bijection function is also famous in illustrating the relation between the ascents and weak exceedances as follows [2, page 98].

Proposition 4. The function gives a bijection between the set of permutations on with ascents and the set .

Example 5. The standard representation of permutation is , and ; ; has ascents, while has weak excedances because , , , and .

Now suppose we want to construct a sequence consisting of vertical bars and the first positive integers. Then the vertical bars divide these numbers into compartments. In each compartment, there is either no number or all the numbers are listed in a decreasing order. The following definition is analogous to the definition of [2, page 8].

Definition 6. A bar in the above construction is called extraneous if either (a)it is immediately followed by another bar; or(b)each of the rest compartment is either empty or consists of integers in a decreasing order if this bar is removed.

Example 7. Suppose , ; then in the following arrangement the , , and bars are extraneous.

Now we are ready to give combinatorial interpretations to the general Eulerian numbers . First note that (6) implies that is a homogeneous polynomial of degree with respect to and . Indeed, where

The following theorem gives combinatorial interpretations to the coefficients , .

Theorem 8. Let the general Eulerian numbers be written as in (8). Then

Proof. We can check the result in (10) for two special values and quickly. By (2), when , ; when , . Therefore, (10) is true for and .
Generally, for , we write down bars with compartments in between. Place each element of in a compartment. If none of the bars is extraneous, then the arrangement corresponds to a permutation with ascents. Let be the set of arrangements with at most one extraneous bar at the end and none of integers locating in the last compartment. We will show that .
To achieve that goal, we use the Principle of Inclusion and Exclusion. There are ways to put numbers into compartments with elements avoiding the last compartments.
Let be the number of arrangements with the following features:  none of sits in the last compartment;  each arrangement in has at least extraneous bars.  in each arrangement in , any two extraneous bars are not located right next to each other.
Then the Principle of Inclusion and Exclusion shows that
Now we consider the value of , where . Suppose that we have compartments with bars in between. There are ways to insert numbers into these compartments with first integers avoiding the last compartment and list integers in each component in a decreasing order. Then insert separating extraneous bars into positions. So we get Plug formula (12) into (11); we have .
Given an arrangement , if we remove the bars, then we obtain a permutation . So without confusion, we just use the same notation to represent both an arrangement in set and a permutation on . Now for each , either  (case 1) has no extraneous bar and none of locates in the last compartment or (case 2) has only one extraneous bar at the end.
If is in case 1, then has ascents since each bar is non-extraneous. And the last compartment of is nonempty. Therefore the last cycle of has to be . In other words, since none of locates in the last compartment. And by Proposition 4, .
If is in case 2, then has ascents since only the last bar is extraneous. Note that in this case, the arrangement with no elements of in the compartment second to the last or the last nonempty compartment has been removed by the Principle of Inclusion and Exclusion. Equivalently, at least one number of has to be in the compartment second to the last. So the last cycle of has to be , and . Also by Proposition 4, .
Combing all the results above, statement (10) is correct.

The next Theorem describes some interesting properties of the coefficients .

Theorem 9. Let the coefficients be as described in Theorem 8. Then, (1), for any ;(2), for all .

Before we can prove Theorem 9, we need the following lemma which is also interesting by itself.

Lemma 10. Given a positive integer , then for any .

Proof. First of all, given a positive integer , we define a function as follows: For instance, for , . is obviously a bijection of to itself.
Now for some fixed , suppose , and . For any , we write in the standard representation cycle form. So and has ascents by Proposition 4. Now we compose with the bijection function as just defined. Then has ascents, which implies that has weak excedances. So . Note that the last cycle of has to be . Therefore, . Since both and are bijection functions, gives a bijection between and .

Now we are ready to prove Theorem 9.

Proof of Theorem 9. For part , by Theorem 8, For part , also by Theorem 8,

Remark 11. Using the analytic formula of as in (9), part 2 of Theorem 9 implies the following identity: where is a positive integer, and .

3. Another Combinatorial Interpretation of and

In pursuing the combinatorial meanings of the coefficients , the authors have found some other interesting properties about permutations. The results in this section will reveal close connections between the traditional Eulerian numbers and , where or .

One fundamental concept of permutation combinatorics is inversion. A pair is called an inversion of the permutation if and [6, page 36]. The following definition provides the main concepts of this section.

Definition 12. For a fixed positive integer , let and (or is not an inversion) and (or is an inversion).
It is obvious that . The following theorem interprets coefficients and in terms of and .

Theorem 13. Let the coefficients of the general Eulerian numbers be written as in (9). and are as defined in Definition 12. Then (1),(2).

Proof. For part , by Theorem 8, , where , . Given a permutation and , then both and belong to , so one of them has to be in . If and , then , but . Therefore, .
Part can be proved using exactly the same method. So we leave it to the readers as an exercise.

and are interesting combinatorial concepts by themselves. Note that generally speaking, . Indeed, .

Theorem 14. For any positive integer , the sets and are defined in Definition 12. Then for .

Proof. It is an obvious result of part 2 of Theorems 9 and 13.

Our last result of this paper is the following theorem which reveals that both and take exactly the same recursive formula as the traditional Eulerian numbers as shown in (1).

Theorem 15. For a fixed positive integer , let and be as defined in Definition 12; then

Proof. A computational proof can be obtained straightforward by using (9) and Theorem 13. But here we provide a proof in a flavor of combinatorics.
Idea of the Proof. For (18), given a permutation , for each position with , we insert into a certain place of , such that the new permutation is in . There are such positions, so we can get new permutations in . Similarly, if , for each position with , and the position at the end of , we insert into a specific position of and the resulting new permutation is in . There are such positions, so we can get new permutations in . We will show that all the permutations obtained from the above constructions are distinct, and they have exhausted all the permutations in
For any fixed , then . We classify into the following disjoint cases:
Case a. Consider that with . So .  , and ; , and ; , , and ; , , and ; , and .
Case b. Consider that . So for some and :  ; , and ; , and ; .Based on the classifications listed above, we can construct a map by applying the idea of the proof we have illustrated at the beginning of the proof. To save space, the map is demonstrated in Table 1. From Table 1 we can see that in each case, the positions of inserting are all different. So all the images obtained in a certain case are different. Since all the cases are disjoint, all the images are distinct.
Similarly, for each , then . We classify into the following disjoint cases.
Case c. Consider that with . So :  , and ; , and ; , ; , ; ; .
Case d. Consider that . So :  ; .To prove (19), we use a similar idea of proof as shown above. If , for each position with , we insert into a certain place of to get . If , for each position with , and the position where , we insert into a specific position of to obtain . Such a map is illustrated in Table 2. And the distinct images under exhaust all the permutations in .