Higher-order Euler-type polynomials and their applications

Abstract

In this paper, we construct generating functions for higher-order Euler-type polynomials and numbers. By using the generating functions, we obtain functional equations related to a generalized partial Hecke operator and Euler-type polynomials and numbers. A special case of higher-order Euler-type polynomials is eigenfunctions for the generalized partial Hecke operators. Moreover, we give not only some properties, but also applications for these polynomials and numbers.

AMS Subject Classification:08A40, 11F25, 11F60, 11B68, 30D05.

1 Introduction

In this section, we define generalized partial Hecke operators and we give some notation for these operators. Also, we define generalized Euler-type polynomials, Apostol-Bernoulli polynomials and Frobenius-Euler polynomials.

Throughout this paper, we use the following notations:

$\mathbb{N}=\{1,2,\dots \}$, ${\mathbb{N}}_0=\{0,1,2,\dots \}=\mathbb{N}\cup \{0\}$. Also, as usual, ℤ denotes the set of integers, ℝ denotes the set of real numbers and ℂ denotes the set of complex numbers. We assume that $ln(z)$ denotes the principal branch of the multi-valued function $ln(z)$ with an imaginary part $\mathrm{\Im }(ln(z))$ constrained by $-\pi <\mathrm{\Im }(ln(z))\le \pi$. Furthermore, $0^n=1$ if $n=0$, and $0^n=0$ if $n\in \mathbb{N}$.

$$N(M)=(N_1,N_2,\dots ,N_M),$$

where $M\in \mathbb{N}$ and $N_1,N_2,\dots ,N_M\in \mathbb{N}$.

Let $a\in \mathbb{N}$ and ${\chi }_{a,N(M)}$ be a function depending on $a,N_1,N_2,\dots ,N_M$ such that

$${\chi }_{a,N(M)}:{\mathbb{N}}_0\to \mathbb{C}.$$

${\chi }_{a,N(M)}$ is defined by

$${\chi }_{a,N(M)}(k)=\prod _{j=1}^M{\xi }^k(N_j),$$

where $0\le k\le a-1$, $j\in \{1,2,\dots ,M\}$ and

$$\xi (N_j)=e^{\frac{2\pi i}{N_j}}.$$

${\chi }_{a,N(M)}$ satisfies the following properties:

• (i) ${\chi }_{a,N(M)}$ is a periodic function with $N_1{N}_2\cdots N_M$.

• (ii) If we take $N_1\ge 2$ and $N_2=N_3=\cdots =N_M=1$, we have

  $${\chi }_{a,(N_1,1,1,\dots ,1)}(k)={\xi }^k(N_1){\xi }^k(1){\xi }^k(1)\cdots {\xi }^k(1)={\xi }^k(N_1).$$

We note that replacing $N(M)$ by $(N_1,1,1,\dots ,1)$, ${\chi }_{a,N(M)}$ is reduced to ${\xi }^k(N_1)$ (cf. [1]).

Let $\mathbb{C}[x]$ be a ring of polynomials with complex coefficients. By using ${\chi }_{a,N(M)}$, we give the following definition.

Definition 1.1 [2]

Let $P\in \mathbb{C}[x]$. The generalized partial Hecke operator of $T_{{\chi }_{a,N(M)}}$ is defined by

$$T_{{\chi }_{a,N(M)}}(P(x))=\sum _{k=0}^{a-1}{\chi }_{a,N(M)}(k)P\left(\frac{x+k}{a}\right).$$

The operator $T_{{\chi }_{a,N(M)}}$ satisfies the following properties:

• (i) $T_{{\chi }_{a,N(M)}}$ is linear on $\mathbb{C}[x]$ and

  $$T_{{\chi }_{a,N(M)}}:\mathbb{C}[x]\to \mathbb{C}[x].$$

• (ii) $T_{{\chi }_{a,N(M)}}$ preserves the degree of the polynomials on $\mathbb{C}[x]$.

• (iii) If we take $N_1\ge 2$ and $N_2=N_3=\cdots =N_M=1$, we have

  $$T_{{\chi }_{a,N_1}}(P(x))=\sum _{k=0}^{a-1}{\xi }^k(N_1)P\left(\frac{x+k}{a}\right).$$

Remark 1.2 Setting $N(M)=(N_1,1,1,\dots ,1)$, $T_{{\chi }_{a,(N_1,1,1,\dots ,1)}}$ is reduced to $T_{{\chi }_{a,N_1}}$ (cf. [1]).

The generating function of generalized Euler-type numbers $P_{n,N(M)}$ is given by

$${\mathcal{F}}_{N(M)}(t)=\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}\frac{t^n}{n!}=\frac{{\prod }_{j=1}^M\xi (N_j)-1}{-1+e^t{\prod }_{j=1}^M\xi (N_j)}$$

[2].

Now, we give the definition of Euler-type polynomials as follows.

Definition 1.3 [2]

The polynomial $P_{n,N(M)}$ is defined by means of the following generating function:

$${\mathcal{F}}_{N(M)}(t,x)=\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}(x)\frac{t^n}{n!}=\frac{(({\prod }_{j=1}^M\xi (N_j))-1)e^{tx}}{({\prod }_{j=1}^M\xi (N_j))e^t-1},$$

(1)

where

$$|t+\sum _{j=1}^M\frac{2\pi i}{N_j}|<2\pi .$$

The polynomial $P_{n,N(M)}$ satisfies the following properties:

• (i) $P_{n,N(M)}\in \mathbb{C}[x]$.

• (ii) $P_{n,N(M)}$ is a polynomial with degree n and depends on $N_1,N_2,\dots ,N_M$.

• (iii) If we take $N_1\ge 2$ and $N_2=N_3=\cdots =N_M=1$, we have

  $$\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N_1}(x)\frac{t^n}{n!}=\frac{({\xi }_{N_1}-1)e^{tx}}{{\xi }_{N_1}{e}^t-1},$$

where

$$|t+\frac{2\pi i}{N_1}|<2\pi .$$

• (iv) We derive the following functional equation:

  $${\mathcal{F}}_{N(M)}(t,x)={\mathcal{F}}_{N(M)}(t)e^{tx},$$

  (2)

  so that, obviously,

  $$P_{n,N(M)}(0)=P_{n,N(M)}.$$

We now are ready to define Euler-type numbers and polynomials with order k.

Definition 1.4 Euler-type numbers with order k, $P_{n,N(M)}^{(k)}$, are defined by means of the following generating functions:

$${\mathcal{F}}_{N(M)}^{(k)}(t)=\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k)}\frac{t^n}{n!},$$

(3)

where $k\in \mathbb{N}$ and

$$|t+\sum _{j=1}^M\frac{2\pi i}{N_j}|<2\pi .$$

Euler-type polynomials with order k are given by the following functional equation:

$${\mathcal{F}}_{N(M)}^{(k)}(t,x)={\mathcal{F}}_{N(M)}^{(k)}(t)e^{tx}=\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k)}(x)\frac{t^n}{n!}.$$

(4)

We see that

$${\mathcal{F}}_{N(M)}^{(0)}(t,x)=e^{tx}.$$

Thus we obtain

$$P_{n,N(M)}^{(0)}(x)=x^n.$$

Remark 1.5 Substituting $k=1$ into (4), we get (2). Therefore, (3) reduces to (1); that is,

$$P_{n,N(M)}^{(1)}(x)=P_{n,N(M)}(x)$$

so that, obviously,

$$P_{n,N(M)}^{(1)}(0)=P_{n,N(M)}.$$

By using (4) and (3), we obtain

$$\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k)}(x)\frac{t^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k)}\frac{t^n}{n!}\sum _{n=0}^{\mathrm{\infty }}{x}^n\frac{t^n}{n!}.$$

Therefore, we get the following theorem.

Theorem 1.6

$$P_{n,N(M)}^{(k)}(x)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)x^{n-j}{P}_{j,N(M)}^{(k)}.$$

(5)

Hence, we arrive at the following definition.

Definition 1.7 Euler-type polynomials with order k, $P_{n,N(M)}^{(k)}$, are defined by means of the following generating functions:

$${\mathcal{F}}_{N(M)}^{(k)}(t,x)=\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k)}(x)\frac{t^n}{n!},$$

(6)

where

$$|t+\sum _{j=1}^M\frac{2\pi i}{N_j}|<2\pi .$$

Note that there is one generating function for each value of k. These are given explicitly as follows:

$$\begin{array}{rcl}{\mathcal{F}}_{N(M)}^{(k)}(t,x)& =& {\left(\frac{-1+{\prod }_{j=1}^M\xi (N_j)}{-1+e^t{\prod }_{j=1}^M\xi (N_j)}\right)}^k{e}^{tx}\\ =& \sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k)}(x)\frac{t^n}{n!}.\end{array}$$

We derive the following functional equation:

$${\mathcal{F}}_{N(M)}^{(k+l)}(t,x)={\mathcal{F}}_{N(M)}^{(k)}(t,x){\mathcal{F}}_{N(M)}^{(l)}(t).$$

(7)

By using the above functional equation, we arrive at the following theorem.

Theorem 1.8

$$P_{n,N(M)}^{(k+l)}(x)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)P_{j,N(M)}^{(k)}(x)P_{n-j,N(M)}^{(l)}.$$

(8)

Proof By using (3), (6) and (7), we get

$$\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(k+l)}(x)\frac{t^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}(\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)P_{j,N(M)}^{(k)}(x)P_{n-j,N(M)}^{(l)})\frac{t^n}{n!}.$$

By comparing the coefficients of $\frac{t^n}{n!}$ on both sides of the above equation, we get the desired result. □

Substituting $x=0$ into (8), we obtain a convolution formula for the numbers by the following corollary.

Corollary 1.9

$$P_{n,N(M)}^{(k+l)}=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)P_{j,N(M)}^{(k)}{P}_{n-j,N(M)}^{(l)}.$$

By differentiating both sides of equation (2) with respect to the variable x, we obtain the following higher-order differential equation:

$$\frac{{\partial }^j}{\partial x^j}{\mathcal{F}}_{N(M)}(t,x)=t^j{\mathcal{F}}_{N(M)}(t,x).$$

(9)

Remark 1.10 Setting $N(M)=(N_1,1,1,\dots ,1)$, $P_{n,(N_1,1,1,\dots ,1)}$ is reduced $P_{n,N_1}(x)$ (cf. [1]). Therefore $P_{n,N}(x)$ was defined by generalized Bernoulli-Euler polynomials in [1] as follows:

$$\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N}(x)\frac{t^n}{n!}=\{\begin{array}{cc}\frac{t{e}^{tx}}{e^t-1},\hfill & N=1,\hfill \\ \frac{({\xi }_N-1)e^{tx}}{{\xi }_N{e}^t-1},\hfill & N\ge 2,\hfill \end{array}$$

so that, obviously,

$$P_{n,1}(x)=B_n(x)$$

and

$$P_{n,2}(x)=E_n(x).$$

Here $B_n(x)$ and $E_n(x)$ are Bernoulli polynomials and Euler polynomials, respectively (cf. [1–19]).

The Frobenius-Euler polynomial is defined as follows:

Let u be an algebraic number such that $1\ne u\in \mathbb{C}$. Then the Frobenius-Euler polynomial $H_n(x,u)$ is defined by

$$\frac{1-u}{e^t-u}{e}^{tx}=\sum _{n=0}^{\mathrm{\infty }}{H}_n(x,u)\frac{t^n}{n!},$$

where

$$|t+ln\frac{1}{u}|<2\pi$$

(cf. [1–19]).

Remark 1.11 Frobenius-Euler number is denoted by $H_n(u)$ such that $H_n(0,u)=H_n(u)$. Also, $H_n(x,-1)=E_n(x)$ (cf. [1–19]).

By using Frobenius-Euler numbers, one can obtain the Frobenius-Euler polynomials as follows:

$$H_n(x,u)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)x^{n-j}{H}_j(u)$$

(cf. [1–19]).

The Apostol-Bernoulli polynomial is defined as follows.

Definition 1.12 [3, 16]

The Apostol-Bernoulli polynomial ${\mathcal{B}}_n(x,\lambda )$ is defined by

$$\frac{t}{\lambda e^t-1}{e}^{tx}=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{B}}_n(x,\lambda )\frac{t^n}{n!},$$

where λ is the arbitrary real or complex parameter and

$$|t|<|ln\lambda |.$$

Remark 1.13 For $\lambda =1$, we obtain that ${\mathcal{B}}_n(x,1)=B_n(x)$ (cf. [1–19]).

2 A functional equation of generalized Euler-type polynomials

Bayad, Aygunes and Simsek showed that for $a\equiv 1mod(N)$, there exists a unique sequence of monic polynomials ${(P_{n,N})}_{n\in {\mathbb{N}}_0}$ in $\mathbb{Q}({\xi }_N)[x]$ with $deg{P}_{n,N}=n$ such that

$$T_{{\chi }_{a,N}}(P_{n,N}(x))=a^{-n}{P}_{n,N}(x),$$

where $a,N\in \mathbb{N}$ (cf. [1]).

In this section, we give the following theorem.

Theorem 2.1 Let $a,N_1,N_2,\dots ,N_M\in \mathbb{N}$ and $a\equiv 1(mod{N}_1{N}_2\cdots N_M)$. Then there exists a sequence ${(P_{n,N(M)})}_{n\in {\mathbb{N}}_0}$ in

$$\mathbb{Q}(\xi (N_1)\xi (N_2)\cdots \xi (N_M))[x]$$

with

$$deg{P}_{n,N(M)}=n$$

such that

$$T_{{\chi }_{a,N(M)}}(P_{n,N(M)}(x))=a^{-n}{P}_{n,N(M)}(x).$$

(10)

Proof Since $P_{n,N(M)}\in \mathbb{C}[x]$ and $T_{{\chi }_{a,N(M)}}:\mathbb{C}[x]\to \mathbb{C}[x]$, we get

$$T_{{\chi }_{a,N(M)}}(P_{n,N(M)}(x))=\sum _{k=0}^{a-1}{\chi }_{a,N(M)}(k)P_{n,N(M)}\left(\frac{x+k}{a}\right).$$

From the definition of ${\chi }_{a,N(M)}(k)$, we have

$$T_{{\chi }_{a,N(M)}}(P_{n,N(M)}(x))=\sum _{k=0}^{a-1}\left(\prod _{j=1}^M{e}^{\frac{2\pi ik}{N_j}}\right)P_{n,N(M)}\left(\frac{x+k}{a}\right).$$

By using the generating function of $P_{n,N(M)}(x)$, we get

$$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{a-1}\left(\prod _{j=1}^M{e}^{\frac{2\pi ik}{N_j}}\right)P_{n,N(M)}\left(\frac{x+k}{a}\right)\frac{t^n}{n!}\hfill \\ =\sum _{k=0}^{a-1}\left(\prod _{j=1}^M{e}^{\frac{2\pi ik}{N_j}}\right)\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}\left(\frac{x+k}{a}\right)\frac{t^n}{n!}\hfill \\ =\sum _{k=0}^{a-1}\left(\prod _{j=1}^M{e}^{\frac{2\pi ik}{N_j}}\right)\frac{(({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})-1)e^{t(\frac{x+k}{a})}}{({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})e^t-1}\hfill \\ =\frac{(({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})-1)e^{\frac{tx}{a}}}{({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})e^t-1}\sum _{k=0}^{a-1}(exp\left(\sum _{j=1}^M{e}^{\frac{2\pi ik}{N_j}}\right))exp\left(\frac{tk}{a}\right)\hfill \\ =\frac{(({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})-1)e^{\frac{tx}{a}}}{({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})e^t-1}\sum _{k=0}^{a-1}{(exp(\frac{t}{a}+\sum _{j=1}^M\frac{2\pi i}{N_j}))}^k\hfill \\ =\frac{(({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})-1)e^{\frac{tx}{a}}}{({\prod }_{j=1}^M{e}^{\frac{2\pi i}{N_j}})e^t-1}\frac{e^t{(exp({\sum }_{j=1}^M\frac{2\pi i}{N_j}))}^a-1}{e^{\frac{t}{a}}(exp({\sum }_{j=1}^M\frac{2\pi i}{N_j}))-1}.\hfill \end{array}$$

Since $a\equiv 1(mod{N}_1{N}_2\cdots N_M)$, the following relation holds:

$${(exp\left(\sum _{j=1}^M\frac{2\pi i}{N_j}\right))}^a=exp\left(\sum _{j=1}^M\frac{2\pi i}{N_j}\right)=\prod _{j=1}^M{e}^{\frac{2\pi i}{N_j}}.$$

Therefore, we have

$$\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^{a-1}\left(\prod _{j=1}^M{e}^{\frac{2\pi ik}{N_j}}\right)P_{n,N(M)}\left(\frac{x+k}{a}\right)\right)\frac{t^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}{a}^{-n}{P}_{n,N(M)}(x)\frac{t^n}{n!}.$$

By comparing the coefficients of $\frac{t^n}{n!}$ on both sides of the above equation, we get the desired result. □

Remark 2.2 A different proof of (10) is given in [2]. If we take $N_1\ge 2$ and $N_2=N_3=\cdots =N_M=1$, we have the following functional equation:

$$T_{{\chi }_{a,N_1}}(P_{n,N_1}(x))=a^{-n}{P}_{n,N_1}(x)$$

which is satisfied for generalized Bernoulli-Euler polynomials in [1].

3 Some properties of generalized Euler-type polynomials

In this section, we obtain some relations between generalized Euler-type polynomials, Apostol-Bernoulli polynomials and Frobenius-Euler polynomials. Also, we give a formula to obtain the generalized Euler-type polynomials.

Theorem 3.1 Let $n\in \mathbb{N}$. Then we have

$$P_{n+1,N(M)}(x)=P_{n,N(M)}(x)+\frac{{\prod }_{j=1}^M\xi (N_j)}{1-{\prod }_{j=1}^M\xi (N_j)}\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)P_{k,N(M)}^{(2)}(x).$$

Proof By differentiating both sides of equation (2) with respect to the variable t, we have

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{P}_{n+1,N(M)}(x)\frac{t^n}{n!}& =& \frac{\partial }{\partial t}{\mathcal{F}}_{N(M)}(t,x)\\ =& {\mathcal{F}}_{N(M)}(t,x)+\left(\frac{{\prod }_{j=1}^M\xi (N_j)}{1-{\prod }_{j=1}^M\xi (N_j)}\right)e^t{e}^{tx}{({\mathcal{F}}_{N(M)}(t))}^2\\ =& \sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}(x)\frac{t^n}{n!}+\left(\frac{{\prod }_{j=1}^M\xi (N_j)}{1-{\prod }_{j=1}^M\xi (N_j)}\right)e^t(\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}^{(2)}(x)\frac{t^n}{n!}).\end{array}$$

Therefore, we obtain

$$\sum _{n=0}^{\mathrm{\infty }}{P}_{n+1,N(M)}(x)\frac{t^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}(P_{n,N(M)}(x)+\frac{{\prod }_{j=1}^M\xi (N_j)}{1-{\prod }_{j=1}^M\xi (N_j)}\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)P_{k,N(M)}^{(2)}(x))\frac{t^n}{n!}.$$

By comparing the coefficients of $\frac{t^n}{n!}$, we obtain the desired result. □

In the following theorem, we give a relation between the polynomials $P_{n,N(M)}(x)$ and Frobenius-Euler polynomials.

Theorem 3.2 [2]

Let $n\in {\mathbb{N}}_0$. Then we have

$$P_{n,N(M)}(x)=H_n(x,\prod _{j=1}^M\frac{1}{\xi (N_j)}).$$

Proof By using the generating function of $P_{n,N(M)}(x)$, we have

$$\sum _{n=0}^{\mathrm{\infty }}{P}_{n,N(M)}(x)\frac{t^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}{H}_n(x,\prod _{j=1}^M\frac{1}{\xi (N_j)})\frac{t^n}{n!}.$$

In the above equation, if we compare the coefficients of $\frac{t^n}{n!}$, we get the desired result. □

In the following theorem, we give a relation between $P_{n,N(M)}(x)$ and Apostol-Bernoulli polynomials.

Theorem 3.3 [2]

Let $n\in \mathbb{N}$. Then we have

$$P_{n-1,N(M)}(x)=(\prod _{j=1}^M\xi (N_j)-1)\frac{1}{n}{\mathcal{B}}_n(x,\prod _{j=1}^M\xi (N_j)).$$

Proof

We arrange the generating function of generalized Euler-type polynomials as follows:

$$\sum _{n=1}^{\mathrm{\infty }}{P}_{n-1,N(M)}\frac{t^{n-1}}{(n-1)!}=\frac{{\prod }_{j=1}^M\xi (N_j)-1}{e^t{\prod }_{j=1}^M\xi (N_j)-1}{e}^{xt}.$$

Therefore, we have

$$\sum _{n=1}^{\mathrm{\infty }}{P}_{n-1,N(M)}\frac{t^{n-1}}{(n-1)!}=\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{n}(\prod _{j=1}^M\xi (N_j)-1){\mathcal{B}}_n(x,\prod _{j=1}^M\xi (N_j))\right)\frac{t^{n-1}}{(n-1)!}.$$

In the above equation, if we compare the coefficients of $\frac{t^{n-1}}{(n-1)!}$, we get the desired result. □

In the following theorem, it is possible to find the generalized Euler-type polynomials.

Theorem 3.4 Let $n\in {\mathbb{N}}_0$. Then we have

$$P_{n,N(M)}(x)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)x^{n-j}{P}_{j,N(M)}.$$

(11)

Proof of (11) is the same as that of (5), so we omit it [2].

$$\begin{array}{c}{P}_{1,N(M)}=\frac{1}{{\chi }_{a,N(M)}^{-1}-1},\hfill \\ P_{2,N(M)}=\frac{2}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{1}{{\chi }_{a,N(M)}^{-1}-1},\hfill \\ P_{3,N(M)}=\frac{6}{{({\chi }_{a,N(M)}^{-1}-1)}^3}+\frac{6}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{1}{{\chi }_{a,N(M)}^{-1}-1}\hfill \end{array}$$

and

$$P_{4,N(M)}=\frac{24}{{({\chi }_{a,N(M)}^{-1}-1)}^4}+\frac{36}{{({\chi }_{a,N(M)}^{-1}-1)}^3}+\frac{14}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{1}{{\chi }_{a,N(M)}^{-1}-1}.$$

By using (11), we have the following list for the generalized Euler-type polynomials:

$$\begin{array}{c}{P}_{0,N(M)}(x)=1,\hfill \\ P_{1,N(M)}(x)=x+\frac{1}{{\chi }_{a,N(M)}^{-1}-1},\hfill \\ P_{2,N(M)}(x)=x^2+x\left(\frac{2}{{\chi }_{a,N(M)}^{-1}-1}\right)+(\frac{2}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{1}{{\chi }_{a,N(M)}^{-1}-1}),\hfill \\ \begin{array}{rl}{P}_{3,N(M)}(x)=& x^3+x^2\left(\frac{3}{{\chi }_{a,N(M)}^{-1}-1}\right)+x(\frac{6}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{3}{{\chi }_{a,N(M)}^{-1}-1})\\ +(\frac{6}{{({\chi }_{a,N(M)}^{-1}-1)}^3}+\frac{6}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{1}{{\chi }_{a,N(M)}^{-1}-1})\end{array}\hfill \end{array}$$

and

$$\begin{array}{rcl}{P}_{4,N(M)}(x)& =& x^4+x^3\left(\frac{4}{{\chi }_{a,N(M)}^{-1}-1}\right)+x^2(\frac{12}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{6}{{\chi }_{a,N(M)}^{-1}-1})\\ +x(\frac{24}{{({\chi }_{a,N(M)}^{-1}-1)}^3}+\frac{24}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{4}{{\chi }_{a,N(M)}^{-1}-1})\\ +(\frac{24}{{({\chi }_{a,N(M)}^{-1}-1)}^4}+\frac{36}{{({\chi }_{a,N(M)}^{-1}-1)}^3}+\frac{14}{{({\chi }_{a,N(M)}^{-1}-1)}^2}+\frac{1}{{\chi }_{a,N(M)}^{-1}-1}).\end{array}$$

Author’s contributions

The author completed the paper himself. The author read and approved the final manuscript.