General Eulerian Numbers and Eulerian Polynomials

Abstract

We will generalize the definitions of Eulerian numbers and Eulerian polynomials to general arithmetic progressions. Under the new definitions, we have been successful in extending several well-known properties of traditional Eulerian numbers and polynomials to the general Eulerian polynomials and numbers.

1. Introduction

Bernoulli [1, pages 95–97] had introduced his famous Bernoulli numbers, denoted by 𝐵_2𝑟 (𝐵_2𝑟+1=0 for 𝑟≥1) to evaluate the sum of the 𝑛th power of the first 𝑚 integers. He then proved the following summation formula: ^𝑚_𝑖=1𝑖^𝑛=𝑚^𝑛+1+𝑚𝑛+12^𝑛+1𝑛+1^⌊𝑟/2⌋_𝑟=1𝑚𝑛+12𝑟^{𝑛−2𝑟+1}(−1)^𝑟+1𝐵_2𝑟, (1) when 𝑛,𝑚≥1.

Two decades later, Euler [2] studied the alternating sum ∑^𝑚_𝑖=1(−1)^𝑖𝑖^𝑛. He ended up with giving the following general result [3, (2.8), page 259]: ^𝑚_𝑖=1𝑖^𝑛𝑡^𝑖=^𝑛_𝑙=1(−1)^𝑛+𝑙𝑛𝑙𝑡^𝑚+1𝑚^𝑙(𝑡−1)^{𝑛−𝑙+1}𝐴_𝑛−𝑙(𝑡)+(−1)^𝑛𝑡𝑡^𝑚−1(𝑡−1)^𝑛+1𝐴_𝑛(𝑡). (2) Another simplified form of ∑^𝑚_𝑖=1𝑖^𝑛𝑡^𝑖 is the following [3, (3.3), page 263]: ^𝑚_𝑖=1𝑖^𝑛𝑡^𝑖=−𝑡^𝑛𝑚+1_𝑘=0𝑛𝑘𝑚^𝑛−𝑘(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)+𝑡(1−𝑡)^𝑛+1𝐴_𝑛(𝑡), (3) where 𝐴_𝑛(𝑡), (𝑛=0,1,2,…) are called Eulerian polynomials and are recursively defined by [3, (2.7), page 264]: 𝐴₀(𝑡)𝐴∶=1,_𝑛(𝑡)=^𝑛−1_𝑘=0𝑛𝑘𝐴_𝑘(𝑡)(𝑡−1)^{𝑛−1−𝑘}. (4)

Note that for ∑^𝑚_𝑖=1𝑖^𝑛𝑡^𝑖, if we put 𝑡=−1, then ∑^𝑚_𝑖=1𝑖^𝑛𝑡^𝑖 becomes the alternating sum of the 𝑛th power of the first 𝑚 positive integers. Furthermore, as 𝑚→∞, we have [3, (3.2), page 263]: 𝐴_𝑛(𝑡)(1−𝑡)^𝑛+1=^∞_𝑗=0𝑡^𝑗𝑗+1^𝑛(𝑛≥0). (5)

Each Eulerian polynomial can be presented as a generating function of Eulerian numbers 𝐴_𝑛,𝑘 [4], also introduced by Euler, as 𝐴_𝑛(𝑡)=^𝑛−1_𝑘=0𝐴_𝑛,𝑘𝑡^𝑘. (6)

Furthermore, the corresponding exponential generating function [3, (3.1), page 262] is 𝐴(𝑡,𝑢)=_𝑛≥0𝐴_𝑛(𝑡)𝑢^𝑛=𝑛!𝑡−1(𝑢(.𝑡−exp𝑡−1)) (7)

The following combinatorial definition of Eulerian numbers was discovered by Riordan in the 1950s.

Definition 1. A Eulerian number 𝐴_𝑛,𝑘 is the number of permutations 𝑝₁𝑝₂𝑝₃…𝑝_𝑛 of the first 𝑛 numbers {1,2,…,𝑛} that have 𝑘 ascents (or descents), that is, 𝑘 places where 𝑝_𝑗<𝑝_𝑗+1 (or 𝑝_𝑗>𝑝_𝑗+1), 1≤𝑗≤𝑛−1.

Example 2. When 𝑛=3,𝑘=1, and 𝐴_3,1=4, because there are four permutations with only one ascent: 132,213,231,312, (8) then 𝐴_𝑛,𝑘 satisfies the recurrence: 𝐴_𝑛,0=1,   (𝑛≥0),𝐴_𝑛,𝑘=0(𝑘≥𝑛) and 𝐴_𝑛,𝑘=(𝑘+1)𝐴_{𝑛−1,𝑘}+(𝑛−𝑘)𝐴_{𝑛−1,𝑘−1}(1≤𝑘≤𝑛−1). (9)

It is well known that Eulerian numbers have the following symmetric property:

Proposition 3. Given a positive integer 𝑛, and 0≤𝑘≤𝑛−1, 𝐴_𝑛,𝑘=𝐴_{𝑛,𝑛−1−𝑘}.

Proof. It is obvious because if a permutation has 𝑘 ascents then its reverse has 𝑛−1−𝑘 ascents.

Furthermore, the values of 𝐴_𝑛,𝑘 can be expressed in a form of a triangular array as shown in Table 1.

Table 1 

First few values of Eulerian numbers 𝐴_𝑛,𝑘.

Besides the recursive formula (9), 𝐴_𝑛,𝑘 can be calculated directly by the following analytic formula [3, (3.5), page 264]: 𝐴_𝑛,𝑘=^𝑘_𝑖=0(−1)^𝑖(𝑘−𝑖+1)^𝑛𝑖().𝑛+10≤𝑘≤𝑛−1 (10)

Since the 1950s, Carlitz ([5, 6]) and his successors have generalized Euler’s results to 𝑞-sequences {1,𝑞,𝑞²,𝑞³,…}. Under Carlitz’s definition, the 𝑞-Eulerian numbers 𝐴_𝑛,𝑘(𝑞) are given by [𝑥]^𝑛=^𝑛−1_𝑘=0𝐴_𝑛,𝑘𝑞𝑛()𝑥+𝑘−1𝑛≥1, (11) where [𝑥]=1−𝑞^𝑥,𝑥𝑛=1−𝑞1−𝑞^𝑥1−𝑞^𝑥−1⋯1−𝑞^{𝑥−𝑛+1}1−𝑞1−𝑞²⋯1−𝑞^𝑛. (12) Then the 𝑞-Eulerian polynomials 𝐴_𝑛(𝑡,𝑞) are defined as 𝐴_𝑛𝑡,𝑞=^𝑛−1_𝑘=0𝐴_𝑛,𝑘𝑡^𝑘(𝑛≥1). (13) Like the traditional Eulerian numbers, the 𝑞-Eulerian numbers 𝐴_𝑛,𝑘(𝑞) have the following recursive formula: 𝐴_1,0𝑞=1,𝐴_𝑛,𝑘𝑞(),𝐴=0if𝑘≤0or𝑘≥𝑛_𝑛,𝑘𝑞=𝑞^{𝑛−1−𝑘}[𝑘+1]𝐴_{𝑛−1,𝑘}𝑞+[𝑛−𝑘]𝐴_{𝑛−1,𝑘−1}𝑞(𝑛≥1). (14)

Similarly to the traditional Eulerian numbers, we can also construct a triangular array for 𝑞-Eulerian numbers as in Table 2.

Table 2 

First values of 𝑞− Eulerian numbers 𝐴_𝑛,𝑘(𝑞).

Definition 4. Given a permutation 𝜋=𝑝₁𝑝₂𝑝₃…𝑝_𝑛 of the first 𝑛 numbers {1,2,3,,…,𝑛}, define functions maj𝜋=_{𝑝𝑗>𝑝𝑗+1}𝑎().𝑗,𝑛,𝑘,𝑖=#𝜋∣maj𝜋=𝑖and𝜋has𝑘ascents (15)

In 1974, Carlitz [6] completed his study of his 𝑞-Eulerian numbers by giving a combinatorial meaning to his 𝑞-Eulerian numbers: 𝐴_𝑛,𝑘𝑞=𝑞^{((𝑚−𝑘+1)(𝑚−𝑘))/2𝑘(𝑛−𝑘−1)}_𝑖=0𝑎(𝑛,𝑛−𝑘,𝑖)𝑞^𝑖, (16) where functions 𝑎(𝑛,𝑘,𝑖) are as defined in Definition 4. Interested readers can find more details about the history of Eulerian numbers, Eulerian polynomials, and the corresponding concepts in 𝑞-environment from [3].

In this paper, instead of studying 𝑞-sequences, we will generalize Euler’s work on Eulerian numbers and Eulerian polynomials to any general arithmetic progression: {𝑎,𝑎+𝑑,𝑎+2𝑑,𝑎+3𝑑,…}. (17)

In Section 2, we will give a new definition of general Eulerian numbers based on a given arithmetic progression as defined in (17). Under the new definition, some well-known combinatorial properties of traditional Eulerian numbers become special cases of our more general results. In Section 3, we will define general Eulerian polynomials. Then, (2), (3), (4), (5), (6), (7), (9), and (10) become special cases of our more general results.

2. General Eulerian Numbers

The traditional Eulerian numbers 𝐴_𝑛,𝑘, play an important role in the well-known Worpitzky’s identity [7]: 𝑥^𝑛=^𝑛−1_𝑘=0𝑛𝐴𝑥+𝑘_𝑛,𝑘,𝑛≥1. (18)

Before we give a general definition of Eulerian numbers based on a given arithmetic progression (17), we shall mention a property associated with the traditional Eulerian numbers 𝐴_𝑛,𝑘.

Proposition 5. Let 𝐴_𝑛,𝑘 be the traditional Eulerian numbers as defined in Definition 1, then ^𝑚_𝑖=1𝑖^𝑛=^𝑛−1_𝑘=0𝐴_𝑛,𝑘𝑚+𝑘+1𝑛+1. (19)

Proof. See [8, (4), page 348].

Given an arithmetic progression (17), we want to define general Eulerian numbers 𝐴_𝑛,𝑘(𝑎,𝑑) so that the important properties of traditional Eulerian numbers such as the recursive formula (9), the triangular array (Table 1), Worpitzky’s Identity (18), and Proposition 5, and so forth, become special cases of more general results under the new definition.

Definition 6. General Eulerian numbers 𝐴_𝑛,𝑘(𝑎,𝑑) associated with an arithmetic progression as in (17) are defined as 𝐴_0,−1=1,𝐴_𝑛,𝑘=0(𝑘≥𝑛or𝑘≤−2) and 𝐴_𝑛,𝑘(𝑎,𝑑)=(−𝑎+(𝑘+2)𝑑)𝐴_{𝑛−1,𝑘}()+(()𝑑)𝐴𝑎,𝑑𝑎+𝑛−𝑘−1_{𝑛−1,𝑘−1}()().𝑎,𝑑0≤𝑘≤𝑛−1 (20)

Like the traditional Eulerian numbers and 𝑞-Eulerian numbers, the first general Eulerian numbers can be presented in the form of a triangular array as in Table 3.

Table 3 

First values of general Eulerian numbers 𝐴_𝑛,𝑘(𝑎,𝑑).

We intentionally choose 𝑘 values to start with −1, because by doing so Table 1 becomes the special case of Table 3 when 𝑎=𝑑=1 in the arithmetic progression: {𝑎,𝑎+𝑑,𝑎+2𝑑,𝑎+3𝑑,…}. (21) In other words, Table 1 corresponds to the sequence of natural numbers: {1,2,3,4,…}. (22) Therefore when 𝑎=𝑑=1, the entries in the first column of Table 3 become zeroes except initially defined 𝐴_0,−1(𝑎,𝑑)=1.

With the new Definition 6, we are able to prove the following two properties. Again note that if 𝑎=𝑑=1, then the following identity is just the conventional Worpitzky’s identity.

Lemma 7 (General Worpitzky’s Identity). Given an arithmetic progression as in (17), (𝑎+(𝑖−1)𝑑)^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗()𝑛𝑎,𝑑𝑖+𝑗,𝑖≥1. (23)

Proof. We will prove Lemma 7 by induction on 𝑛.
(i) When 𝑛=1, using the values in Table 3, 𝑎+(𝑖−1)𝑑=𝐴_1,−1()1𝑎,𝑑𝑖−1+𝐴_1,0()𝑖1𝑎,𝑑. (24)
(ii) Now suppose (𝑎+(𝑖−1)𝑑)^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗()𝑛(()𝑑)𝑎,𝑑𝑖+𝑗then,𝑎+𝑖−1^𝑛+1=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗()𝑛(()𝑑)=𝑎,𝑑𝑖+𝑗𝑎+𝑖−1_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗()𝑎𝑛−𝑑𝑛+𝑑𝑛=𝑎,𝑑𝑖+𝑗𝑗+2𝑖+𝑗𝑖+𝑗+1𝑖+𝑗_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗()𝑎+𝑑+𝑎,𝑑𝑖+𝑗+1𝑛+1−𝑎𝑖+𝑗𝑛+1𝑛−𝑗−1𝑖+𝑗+1𝑛+1_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗()𝑑=𝑎,𝑑𝑗+2𝑖+𝑗𝑛+1_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗()𝑑+𝑎,𝑑𝑎+𝑛−𝑗−1𝑖+𝑗+1𝑛+1_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗()𝑑=𝑎,𝑑−𝑎+𝑗+2𝑖+𝑗𝑛+1_{𝑛𝑗=−1}𝐴_{𝑛,𝑗−1}(𝑎,𝑑)𝑎+𝑛−𝑗𝑑+𝐴_𝑛,𝑗()×𝑑=𝑎,𝑑−𝑎+𝑗+2𝑖+𝑗𝑛+1_{𝑛𝑗=−1}𝐴_𝑛+1,𝑗(𝑎,𝑑).𝑖+𝑗𝑛+1 (25)

By (20) With Lemma 7, we can prove the following Lemma which is a generalization of Proposition 5.

Lemma 8. Given an arithmetic progression as in (17), ^𝑚_𝑖=1(𝑎+(𝑖−1)𝑑)^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗(𝑎,𝑑).𝑚+𝑗+1𝑛+1 (26)

Proof. We will prove Lemma 8 by induction on 𝑚.(i) When 𝑚=1, by Lemma 7, (𝑎+(1−1)𝑑)^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗()𝑛=𝑎,𝑑1+𝑗_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗()𝑛+𝑎,𝑑1+𝑗_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗(𝑎,𝑑),1+𝑗𝑛+1 (27) since ^1+𝑗_𝑛+1=0, for 𝑗=−1,0,1,…,𝑛−1. Therefore, [𝑎+(1−1)𝑑]^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗(𝑎,𝑑)1+𝑗+1𝑛+1, (28) which finishes the base case 𝑚=1.(ii) Now suppose ^𝑚_𝑖=1(𝑎+(𝑖−1)𝑑)^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗(𝑎,𝑑)𝑚+𝑗+1𝑛+1. (29)(iii) Then from Table 3, Definition 6, and Lemma 7, ^𝑚+1_𝑖=1(𝑎+(𝑖−1)𝑑)^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗(𝑎,𝑑)+()𝑚+𝑗+1𝑛+1𝑎+𝑚𝑑^𝑛=^𝑛−1_𝑗=−1𝐴_𝑛,𝑗()+𝑛=𝑎,𝑑𝑚+𝑗+1𝑛+1𝑚+𝑗+1_{𝑛−1𝑗=−1}𝐴_𝑛,𝑗(𝑎,𝑑).𝑚+1+𝑗+1𝑛+1 (30)

The general Eulerian numbers 𝐴_𝑛,𝑘(𝑎,𝑑) can be calculated directly from the following formula, which is a generalization of (10).

Lemma 9. For a given arithmetic progression {𝑎,𝑎+𝑑,𝑎+2𝑑,𝑎+3𝑑,…}, the general Eulerian numbers satisfy 𝐴_𝑛,𝑘(𝑎,𝑑)=^𝑘+1_𝑖=0(−1)^𝑖[(𝑘+2−𝑖)𝑑−𝑎]^𝑛𝑖𝑛+1. (31)

Proof. The following proof is given by inductions on both 𝑛 and 𝑘.
For 𝑛=0,𝑘=−1, 𝐴_0,−1(𝑎,𝑑)=(−1)⁰(𝑑−𝑎)⁰¹₀=1.
For 𝑛=1,𝑘=−1, 𝐴_1,−1(𝑎,𝑑)=(−1)⁰(𝑑−𝑎)¹²₀=𝑑−𝑎.
For 𝑛=1,𝑘=0, 𝐴_1,0(𝑎,𝑑)=(−1)⁰(2𝑑−𝑎)¹²₀+(−1)¹(𝑑−𝑎)¹²₁=𝑎.
Now suppose 𝐴_{𝑛−1,𝑘}(𝑎,𝑑)=∑^𝑘+1_𝑖=0(−1)^𝑖[(𝑘+2−𝑖)𝑑−𝑎]^𝑛−1(^𝑛_𝑖). Then from the recursive formula (20), 𝐴_𝑛,𝑘()=[()𝑑]𝐴𝑎,𝑑−𝑎+𝑘+2_{𝑛−1,𝑘}()+[()𝑑]𝐴𝑎,𝑑𝑎+𝑛−𝑘−1_{𝑛−1,𝑘−1}()=[()𝑑]×𝑎,𝑑−𝑎+𝑘+2_𝑘+1𝑖=0(−1)^𝑖[(𝑘+2−𝑖)𝑑−𝑎]^𝑛−1𝑛𝑖+[𝑎+(𝑛−𝑘−1)𝑑]^𝑘+1_𝑖=1(−1)^𝑖−1×[(𝑘+2−𝑖)𝑑−𝑎]^𝑛−1𝑛=)]𝑖−1byinduction[(𝑘+2𝑑−𝑎^𝑛𝑛0+^𝑘+1_𝑖=1(−1)^𝑖[(𝑘+2−𝑖)𝑑−𝑎]^𝑛−1×()𝑑𝑛𝑖−()𝑑𝑛𝑛𝑖𝑛;𝑘+2𝑛−𝑘−1𝑖−1−𝑎−𝑎𝑖−1 (32) Note that ()𝑑𝑛𝑖−()𝑑𝑛𝑛𝑖𝑛=()𝑑𝑛𝑖−()𝑑𝑛𝑖=()𝑑𝑛𝑖+()𝑑𝑛−()𝑑𝑛𝑖=()𝑑𝑖𝑖𝑖=)]𝑖.𝑘+2𝑛−𝑘−1𝑖−1−𝑎−𝑎𝑖−1𝑘+2𝑛−𝑘−1𝑖−1−𝑎𝑛+1𝑘+2𝑘+2𝑖−1𝑛+1𝑖−1−𝑎𝑛+1𝑘+2𝑛+1−𝑖𝑑𝑛+1−𝑎𝑛+1[(𝑘+2−𝑖𝑑−𝑎𝑛+1 (33) Combine the results above, we have 𝐴_𝑛,𝑘(𝑎,𝑑)=^𝑘+1_𝑖=0(−1)^𝑖[(𝑘+2−𝑖)𝑑−𝑎]^𝑛𝑖𝑛+1. (34)

3. General Eulerian Polynomials

Definition 10. We define the general Eulerian polynomials associated to an arithmetic progression as in (17) as 𝑇_𝑛(𝑡,𝑎,𝑑)=^𝑛−1_𝑘=−1𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1. (35)

Definition 10 is a generalization of the traditional Eulerian polynomials as in (6). The following lemma gives the relation between the general Eulerian polynomials and the traditional Eulerian polynomials.

Lemma 11. Let 𝑇_𝑛(𝑡,𝑎,𝑑) be the general Eulerian polynomials as in Definition 10, 𝑇₀(𝑡,𝑎,𝑑)=1. Then 𝑇_𝑛(𝑡,𝑎,𝑑)=^𝑛−1_𝑘=−1𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝐴_𝑗(𝑡)(𝑎𝑡−𝑎)^𝑛−𝑗, (36) where 𝐴_𝑗(𝑡),𝑗=0,1,2,…,𝑛 are traditional Eulerian polynomials as defined in (4) and (6).

Proof. (i) When 𝑛=0, 𝑇₀(𝑡,𝑎,𝑑)=⁰₀𝑑⁰𝐴₀(𝑡)(𝑎𝑡−𝑎)⁰=1.
(ii) Now suppose 𝑇_𝑛(𝑡,𝑎,𝑑)=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝐴_𝑗(𝑡)(𝑎𝑡−𝑎)^𝑛−𝑗𝑇.Then,_𝑛+1(𝑡,𝑎,𝑑)=^𝑛_𝑘=−1𝐴_𝑛+1,𝑘(𝑎,𝑑)𝑡^𝑘+1=bydefinition_{𝑛𝑘=−1}(𝑎+(𝑛−𝑘)𝑑)𝐴_{𝑛,𝑘−1}(𝑎,𝑑)𝑡^𝑘+1+^𝑛_𝑘=−1(−𝑎+(𝑘+2)𝑑)𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1=^𝑛−1_𝑘=−1(𝑎+(𝑛−𝑘−1)𝑑)𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+2+^𝑛−1_𝑘=−1(−𝑎+(𝑘+2)𝑑)𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1=^𝑛−1_𝑘=−1(𝑎𝑡−𝑎)𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1+^𝑛−1_𝑘=−1𝑑(𝑛𝑡−𝑡+2)𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1+^𝑛−1_𝑘=−1𝑑(1−𝑡)𝑘𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1=(𝑎𝑡−𝑎)𝑇_𝑛()+()𝑇𝑡,𝑎,𝑑𝑑𝑛𝑡−𝑑𝑡+2𝑑_𝑛()+𝑡,𝑎,𝑑_{𝑛−1𝑘=−1}𝑑(1−𝑡)𝑘𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1. (37)(38)
Note that by definition, 𝑇_𝑛(𝑡,𝑎,𝑑)=∑^𝑛−1_𝑘=−1𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1. So 𝑇^′_𝑛(𝑡,𝑎,𝑑)=∑^𝑛−1_𝑘=−1(𝑘+1)𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘, which implies 𝑡𝑇^′_𝑛(𝑡,𝑎,𝑑)=𝑇_𝑛(𝑡,𝑎,𝑑)+^𝑛−1_𝑘=−1𝑘𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1. (39) On the other hand, from [3, (3.4), page 263], we have 𝐴_𝑛(𝑡)=[1+(𝑛−1)𝑡]𝐴_𝑛−1(𝑡)+𝑡(1−𝑡)𝐴^′_𝑛−1(𝑡). (40) With results as in (39) and (40), we have (𝑑𝑛𝑡−𝑑𝑡+2𝑑)𝑇_𝑛(𝑡,𝑎,𝑑)+^𝑛−1_𝑘=−1𝑑(1−𝑡)𝑘𝐴_𝑛,𝑘(𝑎,𝑑)𝑡^𝑘+1=(𝑑𝑛𝑡+𝑑)𝑇_𝑛(𝑡,𝑎,𝑑)+𝑑(1−𝑡)𝑡𝑇^′_𝑛()=()𝑇𝑡,𝑎,𝑑𝑑𝑛𝑡+𝑑_𝑛(𝑡,𝑎,𝑑)+𝑑(1−𝑡)𝑡^𝑛_𝑗=0𝑑^𝑗𝑛𝑗×𝐴^′_𝑗(𝑡)(𝑎𝑡−𝑎)^𝑛−𝑗+𝑎𝑛−𝑗𝐴_𝑗(𝑡)(𝑎𝑡−𝑎)^{𝑛−𝑗−1}=()𝑇byinduction𝑑𝑛𝑡+𝑑_𝑛(𝑡,𝑎,𝑑)−𝑑𝑡𝑛−𝑗𝑇_𝑛()()𝑡𝑡,𝑎,𝑑+𝑑1−𝑡^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝐴_𝑗^{′𝑛−𝑗}=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗+1(𝑎𝑡−𝑎)^𝑛−𝑗×1+𝑗𝑡𝐴_𝑗(𝑡)+𝑡(1−𝑡)𝐴^′_𝑗(𝑡=_𝑛𝑗=0)𝑛𝑗𝑑^𝑗+1(𝑎𝑡−𝑎)^𝑛−𝑗𝐴_𝑗+1(𝑡). (41) Therefore, expression (37) becomes 𝑇_𝑛+1(𝑡,𝑎,𝑑)=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗+1(𝑎𝑡−𝑎)^𝑛−𝑗𝐴_𝑗+1(𝑡)+^𝑛_𝑗=0𝑛𝑗𝑑^𝑗(𝑎𝑡−𝑎)^{𝑛−𝑗+1}𝐴_𝑗(𝑡)=^𝑛+1_𝑗=1𝑛𝑑𝑗−1^𝑗(𝑎𝑡−𝑎)^{𝑛+1−𝑗}𝐴_𝑗(𝑡)+^𝑛_𝑗=0𝑛𝑗𝑑^𝑗(𝑎𝑡−𝑎)^{𝑛−𝑗+1}𝐴_𝑗(𝑡)=^𝑛+1_𝑗=0𝑗𝑑𝑛+1_𝑗𝐴_𝑗(𝑡)(𝑎𝑡−𝑎)^{𝑛+1−𝑗}. (42)

The following result is a generalization of (7).

Lemma 12. Let 𝑇_𝑛(𝑡,𝑎,𝑑) be as defined in Lemma 11. Then 𝑇_𝑛(𝑡,𝑢,𝑎,𝑑)=_𝑛≥0𝑇_𝑛()𝑢𝑡,𝑎,𝑑^𝑛=()((𝑛!𝑡−1exp𝑎𝑢𝑡−1))((.𝑡−exp𝑑𝑢𝑡−1)) (43)

Proof. From Lemma 11𝑇_𝑛(𝑡,𝑢,𝑎,𝑑)=_𝑛≥0𝑇_𝑛()𝑢𝑡,𝑎,𝑑^𝑛𝑛!=_𝑛≥0^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝐴_𝑗(𝑡)(𝑎𝑡−𝑎)^𝑛−𝑗𝑢^𝑛𝑛!=_{𝑛≥0𝑘+𝑙=𝑛0≤𝑘≤𝑛}𝑛!𝑑𝑘!𝑙!^𝑘𝐴_𝑘(𝑡)(𝑎𝑡−𝑎)^𝑙𝑢^𝑘⋅𝑢^𝑙=𝑛!_𝑘≥0𝐴_𝑘(𝑡)(𝑑𝑢)^𝑘𝑘!_𝑙≥0(𝑢𝑎(𝑡−1))^𝑙=()((𝑙!𝑡−1exp𝑎𝑢𝑡−1))((.𝑡−exp𝑑𝑢𝑡−1)) (44)

Using the results from Lemma 12, we can derive the following lemma, which is a general version of (5).

Proposition 13. Given an arithmetic progression as in (17), let 𝑇_𝑛(𝑡,𝑎−𝑑,−𝑑) be the general Eulerian Polynomials associated to the arithmetic progression {𝑎−𝑑,𝑎−2𝑑,𝑎−3𝑑,…}. Then −𝑇_𝑛(𝑡,𝑎−𝑑,−𝑑)(𝑡−1)^𝑛+1=^∞_𝑗=0𝑡^𝑗𝑎+𝑗𝑑^𝑛(𝑛≥0). (45)

Proof. 𝑒^𝑎𝑢1−𝑡𝑒^𝑑𝑢=𝑒^∞𝑎𝑢_𝑗=0𝑡𝑒^𝑑𝑢𝑗=^∞_𝑗=0𝑡^𝑗𝑒^{𝑢(𝑎+𝑗𝑑)}=^∞_𝑗=0𝑡^𝑗∞_𝑛=0𝑢^𝑛𝑎+𝑗𝑑^𝑛=𝑛!_∞𝑛=0𝑢^𝑛𝑛!^∞_𝑗=0𝑡^𝑗𝑎+𝑗𝑑^𝑛. (46)
On the other hand, by Lemma 12, ^∞_𝑛=0−𝑇_𝑛(𝑡,𝑎−𝑑,−𝑑)(𝑡−1)^𝑛+1𝑢^𝑛=1𝑛!𝑡−1^∞_𝑛=0−𝑇_𝑛(𝑡,𝑎−𝑑,−𝑑)(𝑡−1)^𝑛𝑢^𝑛1𝑛!=−𝑡−1^∞_𝑛=0𝑇_𝑛(𝑡,𝑎−𝑑,−𝑑)𝑢𝑛!𝑡−1^𝑛1=−⋅()𝑒𝑡−1𝑡−1^{(𝑎−𝑑)𝑢}𝑡−𝑒^−𝑑𝑢𝑒=−^{(𝑎−𝑑)𝑢}𝑡−𝑒^−𝑑𝑢=𝑒^𝑎𝑢1−𝑡𝑒^𝑑𝑢. (47) By comparing the coefficients of 𝑢^𝑛/𝑛!, we have −𝑇_𝑛(𝑡,𝑎−𝑑,−𝑑)(𝑡−1)^𝑛+1=^∞_𝑗=0𝑡^𝑗𝑎+𝑗𝑑^𝑛(𝑛≥0). (48)

For the finite summation ∑^𝑚_𝑖=1𝑡^𝑖[𝑎+(𝑖−1)𝑑]^𝑛, we have the following property which is a generalization of (2) and (3).

Lemma 14. Let 𝑇_𝑛 be the general Eulerian polynomials as defined in Definition 10: ^𝑚_𝑖=1𝑡^𝑖[𝑎+(𝑖−1)𝑑]^𝑛=^𝑛_𝑙=0𝑛𝑙𝑡^𝑚+1(𝑑𝑚−𝑑)^𝑙(𝑡−1)^{𝑛−𝑙+1}𝑇_𝑛−𝑙()−𝑡𝑡,𝑎,−𝑑₂(𝑡−1)^𝑛+1𝑇_𝑛(𝑡,𝑎,−𝑑_𝑚𝑖=1),𝑡^𝑖[𝑎+(𝑖−1)𝑑]^𝑛=𝑡^𝑚+1(𝑡−1)^𝑛+1𝑇_𝑛(())−𝑡𝑡,𝑎+𝑑𝑚−1,−𝑑₂(𝑡−1)^𝑛+1𝑇_𝑛(𝑡,𝑎,−𝑑). (49)(50)

Proof. We will use (2) and (3) to prove expressions (49) and (50), respectively. ^𝑚_𝑖=1𝑡^𝑖[𝑎+(𝑖−1)𝑑]^𝑛=^𝑚_𝑖=1𝑡^𝑖𝑛_𝑗=0𝑛𝑗𝑎^𝑛−𝑗(𝑖−1)^𝑗𝑑^𝑗=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^{𝑛−𝑗𝑚}_𝑖=1𝑡^𝑖(𝑖−1)^𝑗=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡^𝑚−1_𝑖=1𝑖^𝑗𝑡^𝑖. (51) For (49), if we use (2) to evaluate ∑^𝑚−1_𝑖=1𝑖^𝑗𝑡^𝑖, we have ^𝑚_𝑖=1𝑡^𝑖[𝑎+(𝑖−1)𝑑]^𝑛=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡×^𝑗_𝑙=1(−1)^𝑗+𝑙𝑗𝑙𝑡^𝑚(𝑚−1)^𝑙(𝑡−1)^{𝑗−𝑙+1}𝐴_𝑗−𝑙(𝑡)+(−1)^𝑗𝑡^𝑚−𝑡(𝑡−1)^𝑗+1𝐴_𝑗(𝑡)=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡×^𝑗_𝑙=0(−1)^𝑗+𝑙𝑗𝑙𝑡^𝑚(𝑚−1)^𝑙(𝑡−1)^{𝑗−𝑙+1}𝐴_𝑗−𝑙(𝑡)+^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡(−1)^𝑗−𝑡(𝑡−1)^𝑗+1𝐴_𝑗(𝑡)=𝐼+𝐼𝐼;𝐼𝐼=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡(−1)^𝑗−𝑡(𝑡−1)^𝑗+1𝐴_𝑗(𝑡)=−𝑡²(𝑡−1)^𝑛𝑛+1_𝑗=0𝑛𝑗(−𝑑)^𝑗(𝑎(𝑡−1))^𝑛−𝑗𝐴_𝑗(𝑡)𝑡=−²(𝑡−1)^𝑛+1𝑇_𝑛(𝑡,𝑎,−𝑑). (52) This gives the second term in (49). 𝐼=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡×^𝑗_𝑙=0(−1)^𝑗+𝑙𝑗𝑙𝑡^𝑚(𝑚−1)^𝑙(𝑡−1)^{𝑗−𝑙+1}𝐴_𝑗−𝑙(𝑡)=^𝑛𝑛_{𝑙=0𝑗=𝑙}𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡(−1)^𝑗+𝑙𝑗𝑙×𝑡^𝑚(𝑚−1)^𝑙(𝑡−1)^{𝑗−𝑙+1}𝐴_𝑗−𝑙(𝑡)=^𝑛𝑛_{𝑙=0𝑗=𝑙}𝑛!!𝑑𝑗!𝑛−𝑗^𝑗𝑎^𝑛−𝑗𝑡(−1)^𝑗+𝑙×𝑗!!𝑡𝑙!𝑗−𝑙^𝑚(𝑚−1)^𝑙(𝑡−1)^{𝑗−𝑙+1}𝐴_𝑗−𝑙(𝑡)=^𝑛_𝑙=0^𝑛−𝑙_𝑘=0𝑛!()!𝑑𝑛−𝑘−𝑙^𝑘+𝑙𝑎^{𝑛−𝑘−𝑙}𝑡(−1)^𝑘×1𝑡𝑙!𝑘!^𝑚(𝑚−1)^𝑙(𝑡−1)^𝑘+1𝐴_𝑘(𝑡)=𝑘=𝑗−𝑙_𝑛𝑙=0(𝑚−1)^𝑙𝑛!𝑡^𝑚+1𝑑^𝑙()!×𝑙!𝑛−𝑙_{𝑛−𝑙𝑘=0}(−𝑑)^𝑘𝑎^{𝑛−𝑘−𝑙}(𝑛−𝑙)!𝑘!(𝑛−𝑘−𝑙)!(𝑡−1)^𝑘+1𝐴_𝑘(𝑡)=^𝑛_𝑙=0𝑛𝑙(𝑚−1)^𝑙𝑡^𝑚+1𝑑^𝑙(𝑡−1)^{𝑛−𝑙+1}×^𝑛−𝑙_𝑘=0𝑘()𝑛−𝑙−𝑑^𝑘(𝑎𝑡−𝑎)^{𝑛−𝑘−𝑙}𝐴_𝑘(𝑡)=^𝑛_𝑙=0𝑛𝑙𝑡^𝑚+1(𝑑𝑚−𝑑)^𝑙(𝑡−1)^{𝑛−𝑙+1}𝑇_𝑛−𝑙(𝑡,𝑎,−𝑑), (53) which gives the first term of (49). So we have proved (49) by using expression (2).
For (50), if we use (3) to evaluate ∑^𝑚−1_𝑖=1𝑖^𝑗𝑡^𝑖, we have ^𝑚_𝑖=1𝑡^𝑖[𝑎+(𝑖−1)𝑑]^𝑛=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗𝑡×−𝑡^𝑚𝑗_𝑘=0𝑗𝑘(𝑚−1)^𝑗−𝑘(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)+𝑡(1−𝑡)^𝑗+1𝐴_𝑗(𝑡)=−𝑡^𝑛𝑚+1_𝑗=0𝑛𝑗𝑑^𝑗𝑎^{𝑗𝑛−𝑗}_𝑘=0𝑗𝑘(𝑚−1)^𝑗−𝑘(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)+^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗(1−𝑡)^−𝑗𝑡²𝐴1−𝑡^𝑗(𝑡)=𝐼𝐼𝐼+𝐼𝑉;𝐼𝑉=^𝑛_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗(1−𝑡)^−𝑗𝑡²𝐴1−𝑡^𝑗(𝑡)=^𝑛_𝑗=0𝑛𝑗(−𝑑)^𝑗𝑎^𝑛−𝑗(𝑡−1)^𝑛−𝑗𝑡²(1−𝑡)(𝑡−1)^𝑛𝐴_𝑗(𝑡)𝑡=−²(𝑡−1)^𝑛+1𝑇_𝑛(𝑡,𝑎,−𝑑), (54) by Lemma 11. Thus, we have obtained the second part of (50). 𝐼𝐼𝐼=−𝑡^𝑛𝑚+1_𝑗=0𝑛𝑗𝑑^𝑗𝑎^𝑛−𝑗×^𝑗_𝑘=0𝑗𝑘(𝑚−1)^𝑗−𝑘(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)=−𝑡^{𝑛𝑚+1𝑗}_𝑗=0𝑘=0𝑛!!⋅𝑗!𝑛−𝑗𝑗!𝑘!𝑗−𝑘!×𝑑^𝑗𝑎^𝑛−𝑗(𝑚−1)^𝑗−𝑘(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)=−𝑡^𝑛𝑚+1_𝑘=0𝑛!𝑘!(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)×^𝑛_𝑗=𝑘1!!()𝑛−𝑗𝑗−𝑘𝑚−1^𝑗−𝑘𝑑^𝑗𝑎^𝑛−𝑗=−𝑡^𝑛𝑚+1_𝑘=0𝑛!𝑘!(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)×^𝑛−𝑘_𝑙=01()!()𝑙!𝑛−𝑘−𝑙𝑚−1^𝑙𝑑^𝑙+𝑘𝑎^{𝑛−𝑙−𝑘}𝑙=𝑗−𝑘=−𝑡^𝑛𝑚+1_𝑘=0𝑛𝑘𝑑^𝑘(1−𝑡)^𝑘+1𝐴_𝑘(𝑡)(𝑎+𝑑(𝑚−1))^𝑛−𝑘=𝑡^𝑚+1(𝑡−1)^𝑛𝑛+1_𝑘=0𝑛𝑘(−𝑑)^𝑘(𝑡−1)^𝑛−𝑘×𝐴_𝑘(𝑡)(𝑎+𝑑(𝑚−1))^𝑛−𝑘=𝑡^𝑚+1(𝑡−1)^𝑛+1𝑇_𝑛(𝑡,𝑎+𝑑(𝑚−1),−𝑑), (55) which gives the first term of (50).

Acknowledgments

The authors want to express their deep appreciation to Professor Dominique Foata and the referee for their kind help.