Some new identities of Frobenius-Euler numbers and polynomials

Abstract

In this paper, we give some new and interesting identities which are derived from the basis of Frobenius-Euler. Recently, several authors have studied some identities of Frobenius-Euler polynomials. From the methods of our paper, we can also derive many interesting identities of Frobenius-Euler numbers and polynomials.

1 Introduction

Let $\lambda (\ne 1)\in \mathbf{C}$. As is well known, the Frobienius-Euler polynomials are defined by the generating function to be

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

(1)

with the usual convention about replacing $H^n(x|\lambda )$ by $H_n(x|\lambda )$ (see [1–6]).

In the special case, $x=0$, $H_n(0|\lambda )=H_n(\lambda )$ are called the n th Frobenius-Euler numbers.

Thus, by (1), we get

$${(H(\lambda )+1)}^n-\lambda H_n(\lambda )=H_n(1|\lambda )-\lambda H_n(\lambda )=(1-\lambda ){\delta }_{0,n},$$

(2)

where ${\delta }_{0,n}$ is the Kronecker symbol.

From (1), we can derive the following equation:

$$H_n(x|\lambda )={(H(\lambda )+x)}^n=\sum _{0\le l\le n}\left(\begin{array}{c}n\\ l\end{array}\right)H_{n-l}(\lambda )x^l(\text{see [6–16]}).$$

(3)

Thus, by (3), we easily see that the leading coefficient of $H_n(x|\lambda )$ is $H_0(\lambda )=1$. So, $H_n(x|\lambda )$ are monic polynomials of degree n with coefficients in $\mathbf{Q}(\lambda )$.

From (1), we have

$$\sum _{n=0}^{\mathrm{\infty }}(H_n(x+1|\lambda )-\lambda H_n(x|\lambda ))\frac{t^n}{n!}=\frac{(1-\lambda )e^{(x+1)t}}{e^t-\lambda }-\lambda \frac{1-\lambda }{e^t-\lambda }{e}^{xt}.$$

(4)

Thus, by (4), we get

$$H_n(x+1|\lambda )-\lambda H_n(x|\lambda )=(1-\lambda )x^n,\text{for }n\in {\mathbf{Z}}_{+}.$$

(5)

It is easy to show that

$$\frac{d}{dx}{H}_n(x|\lambda )=\frac{d}{dx}{(H(\lambda )+x)}^n=n{H}_{n-1}(x|\lambda )(n\in \mathbf{N}).$$

(6)

From (6), we have

$${\int }_0^1{H}_n(x|\lambda )dx=\frac{1}{n+1}(H_{n+1}(1|\lambda )-H_{n+1}(\lambda ))=\frac{\lambda -1}{n+1}{H}_{n+1}(\lambda ).$$

(7)

Let ${\mathbb{P}}_n(\lambda )=\{p(x)\in \mathbf{Q}(\lambda )[x]\mid degp(x)\le n\}$ be a vector space over $\mathbf{Q}(\lambda )$. Then we note that $\{H_0(x|\lambda ),H_1(x|\lambda ),\dots ,H_n(x|\lambda )\}$ is a good basis for ${\mathbb{P}}_n(\lambda )$.

In this paper, we develop some new methods to obtain some new identities and properties of Frobenius-Euler polynomials which are derived from the basis of Frobenius-Euler polynomials. Those methods are useful in studying the identities of Frobenius-Euler polynomials.

2 Some identities of Frobenius-Euler polynomials

Let us take $p(x)\in {\mathbb{P}}_n(\lambda )$. Then $p(x)$ can be expressed as a $\mathbf{Q}(\lambda )$-linear combination of $H_0(x|\lambda ),\dots ,H_n(x|\lambda )$ as follows:

$$p(x)=b_0{H}_0(x|\lambda )+b_1{H}_1(x|\lambda )+\cdots +b_n{H}_n(x|\lambda )=\sum _{0\le k\le n}{b}_k{H}_k(x|\lambda ).$$

(8)

Let us define the operator ${\mathrm{△}}_{\lambda }$ by

$$g(x)={\mathrm{△}}_{\lambda }p(x)=p(x+1)-\lambda p(x).$$

(9)

From (9), we can derive the following equation (10):

$$g(x)={\mathrm{△}}_{\lambda }p(x)=\sum _{0\le k\le n}{b}_k(H_k(x+1|\lambda )-\lambda H_k(x|\lambda ))=(1-\lambda )\sum _{0\le k\le n}{b}_k{x}^k.$$

(10)

For $r\in {\mathbf{Z}}_{+}$, let us take the r th derivative of $g(x)$ in (10) as follows:

$$g^{(r)}(x)=(1-\lambda )\sum _{r\le k\le n}k(k-1)\cdots (k-r+1)b_k{x}^{k-r},\text{where }{g}^{(r)}(x)=\frac{d^{r}g(x)}{d{x}^r}.$$

(11)

Thus, by (11), we get

$$g^r(0)=\frac{d^{r}g(x)}{d{x}^r}{|}_{x=0}=(1-\lambda )r!b_r.$$

(12)

From (12), we have

$$b_r=\frac{g^{(r)}(0)}{(1-\lambda )r!}=\frac{1}{(1-\lambda )r!}(p^{(r)}(1)-\lambda p^{(r)}(0)),$$

(13)

where $r\in {\mathbf{Z}}_{+}$ and $p^{(r)}(0)=\frac{d^{r}p(x)}{d{x}^r}{|}_{x=0}$. Therefore, by (13), we obtain the following theorem.

Theorem 1 For $\lambda (\ne 1)\in \mathbf{C}$, $n\in {\mathbf{Z}}_{+}$, let $p(x)\in {\mathbb{P}}_n(\lambda )$ with $p(x)={\sum }_{0\le k\le n}{b}_k{H}_k(x|\lambda )$. Then we have

$$b_k=\frac{1}{(1-\lambda )k!}{g}^{(k)}(0)=\frac{1}{(1-\lambda )k!}(p^{(k)}(1)-\lambda p^{(k)}(0)).$$

Let us take $p(x)=H_n(x|{\lambda }^{-1})$. Then, by Theorem 1, we get

$$H_n(x|{\lambda }^{-1})=\sum _{0\le k\le n}{b}_k{H}_k(x|\lambda ),$$

(14)

where

$$\begin{array}{rl}{b}_k& =\frac{1}{(1-\lambda )k!}\frac{n!}{(n-k)!}\{H_{n-k}(1|{\lambda }^{-1})-\lambda H_{n-k}\left({\lambda }^{-1}\right)\}\\ =\frac{1}{1-\lambda }\left(\begin{array}{c}n\\ k\end{array}\right)\{H_{n-k}(1|{\lambda }^{-1})-\lambda H_{n-k}\left({\lambda }^{-1}\right)\}\\ =\frac{1}{1-\lambda }\left(\begin{array}{c}n\\ k\end{array}\right)\{(1-{\lambda }^{-1})0^{n-k}+\frac{1}{\lambda }{H}_{n-k}\left({\lambda }^{-1}\right)-\lambda H_{n-k}\left({\lambda }^{-1}\right)\}.\end{array}$$

(15)

By (14) and (15), we get

(16)

Therefore, by (16), we obtain the following theorem.

Theorem 2 For $n\in {\mathbf{Z}}_{+}$, we have

$$\lambda H_n(x|{\lambda }^{-1})+H_n(x|\lambda )=(1+\lambda )\sum _{0\le k\le n}\left(\begin{array}{c}n\\ k\end{array}\right)H_{n-k}\left({\lambda }^{-1}\right)H_k(x|\lambda ).$$

Let

$$p(x)=\sum _{0\le k\le n}{H}_k(x|\lambda )H_{n-k}(x|\lambda )\in {\mathbb{P}}_n(\lambda ).$$

(17)

From Theorem 2, we note that $p(x)$ can be generated by $\{H_0(x|\lambda ),H_1(x|\lambda ),\dots ,H_n(x|\lambda )\}$ as follows:

$$p(x)=\sum _{0\le k\le n}{H}_k(x|\lambda )H_{n-k}(x|\lambda )=\sum _{0\le k\le n}{b}_k{H}_k(x|\lambda ).$$

(18)

By (17), we get

$$p^{(k)}(x)=\frac{(n+1)!}{(n-k+1)!}\sum _{k\le l\le n}{H}_{l-k}(x|\lambda )H_{n-k}(x|\lambda ),$$

(19)

and

(20)

From (18) and (20), we have

$$\begin{array}{rcl}\sum _{0\le k\le n}{H}_k(x|\lambda )H_{n-k}(x|\lambda )& =& (n+1)\sum _{0\le k\le n-1}\frac{\left(\begin{array}{c}n\\ k\end{array}\right)}{n-k+1}\sum _{k\le l\le n}\{(-\lambda )H_{l-k}(\lambda )H_{n-l}(\lambda )\\ +2\lambda H_{n-k}(\lambda )\}{H}_k(x|\lambda )+(n+1)H_n(x|\lambda ).\end{array}$$

(21)

Therefore, by (21), we obtain the following theorem.

Theorem 3 For $n\in {\mathbf{Z}}_{+}$, we have

Let us consider

$$p(x)=\sum _{k=0}^n\frac{1}{k!(n-k)!}{H}_k(x|\lambda )H_{n-k}(x|\lambda )\in {\mathbb{P}}_n(\lambda ).$$

(22)

By Theorem 1, $p(x)$ can be expressed by

$$p(x)=\sum _{k=0}^n{b}_k{H}_k(x|\lambda ).$$

(23)

From (22), we have

$$p^{(r)}(x)=2^r\sum _{k=r}^n\frac{H_{k-r}(x|\lambda )H_{n-k}(x|\lambda )}{(k-r)!(n-k)!}(r\in {\mathbf{Z}}_{+}).$$

(24)

By Theorem 1, we get

$$\begin{array}{rcl}{b}_k& =& \frac{1}{2k!}\{p^{(k)}(1)-p^{(k)}(0)\}\\ =& \frac{2^{k-1}}{k!}\sum _{l=k}^n\frac{1}{(l-k)!(n-l)!}\{H_{l-k}(1|\lambda )H_{n-l}(1|\lambda )-\lambda H_{l-k}(\lambda )H_{n-l}(\lambda )\}\\ =& \frac{2^{k-1}}{k!}\sum _{l=k}^n\frac{1}{(l-k)!(n-l)!}\{(\lambda H_{l-k}(\lambda )+(1-\lambda ){\delta }_{0,l-k})(\lambda H_{n-l}(\lambda )+(1-\lambda ){\delta }_{0,n-l})\\ -\lambda H_{l-k}(\lambda )H_{n-l}(\lambda )\}\\ =& \frac{2^{k-1}}{k!}\{\sum _{l=k}^n\frac{\lambda (\lambda -1)H_{l-k}(\lambda )H_{n-l}(\lambda )}{(l-k)!(n-l)!}+\frac{2\lambda (1-\lambda )H_{n-k}(\lambda )}{(n-k)!}+{(1-\lambda )}^2{\delta }_{n,k}\}\\ =& \{\begin{array}{cc}\frac{2^{k-1}}{k!}{\sum }_{l=k}^n\{\frac{\lambda (\lambda -1)H_{l-k}(\lambda )H_{n-l}(\lambda )}{(l-k)!(n-l)!}+\frac{2\lambda (1-\lambda )H_{n-k}(\lambda )}{(n-k)!}\},\hfill & \text{if }k\ne n,\hfill \\ \frac{2^{n-1}(1-\lambda )}{n!},\hfill & \text{if }k=n.\hfill \end{array}\end{array}$$

(25)

Therefore, by (25), we obtain the following theorem.

Theorem 4 For $n\in {\mathbf{Z}}_{+}$, we have

3 Higher-order Frobenius-Euler polynomials

For $n\in {\mathbf{Z}}_{+}$, the Frobenius-Euler polynomials of order r are defined by the generating function to be

$$\begin{array}{rl}{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r{e}^{xt}& =e^{H^{(r)}(x|\lambda )t}\\ =\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!},\end{array}$$

(26)

with the usual convention about replacing ${(H^{(r)}(x|\lambda ))}^n$ by $H_n^{(r)}(x|\lambda )$ (see [1–10]). In the special case, $x=0$, $H_n^{(r)}(0|\lambda )=H_n^{(r)}(\lambda )$ are called the n th Frobenius-Euler numbers of order r (see [8, 9]).

From (26), we have

$$H_n^{(r)}(x|\lambda )={(H^{(r)}(\lambda )+x)}^n=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)H_{n-l}^{(r)}(\lambda )x^l,$$

(27)

with the usual convention about replacing ${(H^{(r)}(\lambda ))}^n$ by $H_n^{(r)}(\lambda )$.

By (26), we get

$$H_n^{(r)}(\lambda )=\sum _{n_1+\cdots +n_r=n}\left(\begin{array}{c}n\\ n_1,n_2,\dots ,n_r\end{array}\right)H_{n_1}(\lambda )\cdots H_{n_r}(\lambda ),$$

(28)

where $\left(\begin{array}{c}n\\ n_1,n_2,\dots ,n_r\end{array}\right)=\frac{n!}{n_1!n_2!\cdots n_r!}$. From (27) and (28), we note that the leading coefficient of $H_n^{(r)}(x|\lambda )$ is given by

$$\begin{array}{rl}{H}_0^{(r)}(\lambda )& =\sum _{n_1+\cdots +n_r=0}\left(\begin{array}{c}n\\ n_1,n_2,\dots ,n_r\end{array}\right)H_{n_1}(\lambda )\cdots H_{n_r}(\lambda )\\ =H_0(\lambda )\cdots H_0(\lambda )=1.\end{array}$$

(29)

Thus, by (29), we see that $H_n^{(r)}$ is a monic polynomial of degree n with coefficients in $\mathbf{Q}(\lambda )$. From (26), we have

$$H_n^{(0)}(x|\lambda )=x^n,\text{for }n\in {\mathbf{Z}}_{+},$$

(30)

and

$$\frac{\partial }{\partial x}{H}_n^{(r)}(x|\lambda )=\frac{\partial }{\partial x}{(H^{(r)}(\lambda )+x)}^n=n{H}_{n-1}^{(r)}(x|\lambda )(r\ge 0).$$

(31)

It is not difficult to show that

$$H_n^{(r)}(x+1|\lambda )-\lambda H_n^{(r)}(x|\lambda )=(1-\lambda )H_n^{(r-1)}(x|\lambda ).$$

(32)

Now, we note that $\{H_0^{(r)}(x|\lambda ),H_1^{(r)}(x|\lambda ),\dots ,H_n^{(r)}(x|\lambda )\}$ is also a good basis for ${\mathbb{P}}_n(\lambda )$.

Let us define the operator D as $Df(x)=\frac{df(x)}{dx}$ and let $p(x)\in {\mathbb{P}}_n(\lambda )$. Then $p(x)$ can be written as

$$p(x)=\sum _{k=0}^n{C}_k{H}_k^{(r)}(x|\lambda ).$$

(33)

From (9) and (32), we have

$${\mathrm{△}}_{\lambda }{H}_n^{(r)}(x|\lambda )=H_n^{(r)}(x+1|\lambda )-\lambda H_n^{(r)}(x|\lambda )=(1-\lambda )H_n^{(r-1)}(x|\lambda ).$$

(34)

Thus, by (33) and (34), we get

$${\mathrm{△}}_{\lambda }^{r}p(x)={(1-\lambda )}^r\sum _{k=0}^n{C}_k{H}_k^{(0)}(x|\lambda )={(1-\lambda )}^r\sum _{k=0}^n{C}_k{x}^k.$$

(35)

Let us take the k th derivative of ${\mathrm{△}}_{\lambda }^{r}p(x)$ in (35).

Then we have

$$D^k({\mathrm{△}}_{\lambda }^{r}p(x))={(1-\lambda )}^r\sum _{l=k}^n\frac{l!}{(l-k)!}{C}_l{x}^{l-k}.$$

(36)

Thus, from (36), we have

$$D^k({\mathrm{△}}_{\lambda }^{r}p(0))={(1-\lambda )}^r\sum _{l=k}^n\frac{l!C_l}{(l-k)!}{0}^{l-k}={(1-\lambda )}^{r}k!C_k.$$

(37)

Thus, by (37), we get

$$\begin{array}{rl}{C}_k& =\frac{D^k({\mathrm{△}}_{\lambda }^{r}p(0))}{{(1-\lambda )}^{r}k!}\\ =\frac{{\mathrm{△}}_{\lambda }^r(D^{k}p(0))}{{(1-\lambda )}^{r}k!}=\frac{1}{{(1-\lambda )}^{r}k!}\sum _{j=0}^r\left(\begin{array}{c}r\\ j\end{array}\right){(-\lambda )}^{(r-j)}{D}^{k}p(j).\end{array}$$

(38)

Therefore, by (33) and (38), we obtain the following theorem.

Theorem 5 For $r\in {\mathbf{Z}}_{+}$, let $p(x)\in {\mathbb{P}}_n(\lambda )$ with

$$p(x)=\frac{1}{{(1-\lambda )}^r}\sum _{0\le k\le n}{C}_k{H}_k^{(r)}(x|\lambda )(C_k\in \mathbf{Q}(\lambda )).$$

Then we have

$$C_k=\frac{1}{{(1-\lambda )}^{r}k!}\sum _{0\le j\le r}\left(\begin{array}{c}r\\ j\end{array}\right){(-\lambda )}^{r-j}{D}^{k}p(j).$$

That is,

$$p(x)=\frac{1}{{(1-\lambda )}^r}\sum _{0\le k\le n}(\sum _{0\le j\le r}\frac{1}{k!}\left(\begin{array}{c}r\\ j\end{array}\right){(-\lambda )}^{r-j}{D}^{k}p(j))H_k^{(r)}(x|\lambda ).$$

Let us take $p(x)=H_n(x|\lambda )\in {\mathbf{P}}_n(\lambda )$. Then, by Theorem 5, $p(x)=H_n(x|\lambda )$ can be generated by $\{H_0^{(r)}(x|\lambda ),H_1^{(r)}(\lambda ),\dots ,H_n^{(r)}(x|\lambda )\}$ as follows:

$$H_n(x|\lambda )=\sum _{0\le k\le n}{C}_k{H}_k^{(r)}(x|\lambda ),$$

(39)

where

$$C_k=\frac{1}{{(1-\lambda )}^r}\frac{1}{k!}\sum _{0\le j\le r}\left(\begin{array}{c}r\\ j\end{array}\right){(-\lambda )}^{r-j}{D}^{k}p(j),$$

(40)

and

$$p^{(k)}(x)=D^{k}p(x)=n(n-1)\cdots (n-k+1)H_{n-k}(x|\lambda )=\frac{n!}{(n-k)!}{H}_{n-k}(x|\lambda ).$$

(41)

By (40) and (41), we get

$$C_k=\frac{1}{{(1-\lambda )}^r}\left(\begin{array}{c}n\\ k\end{array}\right)\sum _{0\le j\le r}\left(\begin{array}{c}r\\ j\end{array}\right){(-\lambda )}^{r-j}{H}_{n-k}(j|\lambda ).$$

(42)

Therefore, by (39) and (42), we obtain the following theorem.

Theorem 6 For $n\in {\mathbf{Z}}_{+}$, we have

$$H_n(x|\lambda )=\frac{1}{{(1-\lambda )}^r}\sum _{0\le k\le n}\left(\begin{array}{c}n\\ k\end{array}\right)(\sum _{0\le j\le r}\left(\begin{array}{c}r\\ j\end{array}\right){(-\lambda )}^{r-j}{H}_{n-k}(j|\lambda ))H_k^{(r)}(x|\lambda ).$$

Let us assume that $p(x)=H_n^{(r)}(x|\lambda )$.

Then we have

$$\begin{array}{rl}{p}^k(x)& =n(n-1)\cdots (n-k+1)H_{n-k}^{(r)}(x|\lambda )\\ =\frac{n!}{(n-k)!}{H}_{n-k}^{(r)}(x|\lambda ).\end{array}$$

(43)

From Theorem 1, we note that $p(x)=H_n^{(r)}(x|\lambda )$ can be expressed as a linear combination of $H_0(x|\lambda ),H_1(x|\lambda ),\dots ,H_n(x|\lambda )$

$$H_n^{(r)}(x|\lambda )=\sum _{0\le k\le n}{b}_k{H}_k(x|\lambda ),$$

(44)

where

$$\begin{array}{rl}{b}_k& =\frac{1}{(1-\lambda )k!}\{p^k(1)-\lambda p^{(k)}(0)\}\\ =\frac{n!}{(1-\lambda )k!(n-k)!}\{H_{n-k}^{(r)}(1|\lambda )-\lambda H_{n-k}^{(r)}(\lambda )\}.\end{array}$$

(45)

By (34) and (45), we get

$$b_k=\left(\begin{array}{c}n\\ k\end{array}\right)H_{n-k}^{(r-1)}(\lambda ).$$

(46)

Therefore, by (44) and (46), we obtain the following theorem.

Theorem 7 For $n\in {\mathbf{Z}}_{+}$, we have

$$H_n^{(r)}(x|\lambda )=\sum _{0\le k\le n}\left(\begin{array}{c}n\\ k\end{array}\right)H_{n-k}^{(r-1)}(\lambda )H_k(x|\lambda ).$$

Remark From (2) and (37), we note that

$$\begin{array}{rl}\frac{d}{d\lambda }\left(\frac{1-\lambda }{e^t-\lambda }\right)& =\frac{1-e^t}{{(e^t-\lambda )}^2}=\frac{1}{{(1-\lambda )}^2}(\frac{{(1-\lambda )}^2}{{(e^t-\lambda )}^2}-\frac{{(1-\lambda )}^2}{{(e^t-\lambda )}^2}{e}^t)\\ =\frac{1}{{(1-\lambda )}^2}(\frac{{(1-\lambda )}^2}{{(e^t-\lambda )}^2}-\frac{{(1-\lambda )}^2}{{(e^t-\lambda )}^2}(e^t-\lambda +\lambda ))\\ =\frac{1}{1-\lambda }(\frac{{(1-\lambda )}^2}{{(e^t-\lambda )}^2}-\frac{1-\lambda }{e^t-\lambda })\\ =\frac{1}{1-\lambda }\sum _{n=0}^{\mathrm{\infty }}(H_n^{(2)}(\lambda )-H_n(\lambda ))\frac{t^n}{n!},\end{array}$$

(47)

and

$$\begin{array}{rl}\frac{d^2}{d{\lambda }^2}\left(\frac{1-\lambda }{e^t-\lambda }\right)& =2!\frac{1-e^t}{{(e^t-\lambda )}^3}=\frac{2!}{{(1-\lambda )}^3}(\frac{{(1-\lambda )}^3}{{(e^t-\lambda )}^3}-\frac{{(1-\lambda )}^3}{{(e^t-\lambda )}^3}{e}^t)\\ =\frac{2!}{{(1-\lambda )}^3}(\frac{{(1-\lambda )}^3}{{(e^t-\lambda )}^3}-\frac{{(1-\lambda )}^3}{{(e^t-\lambda )}^3}(e^t-\lambda +\lambda ))\\ =\frac{2!}{{(1-\lambda )}^2}(\frac{{(1-\lambda )}^3}{{(e^t-\lambda )}^3}-\frac{{(1-\lambda )}^2}{{(e^t-\lambda )}^2})\\ =\frac{2!}{{(1-\lambda )}^2}\sum _{n=0}^{\mathrm{\infty }}(H_n^{(3)}(\lambda )-H_n^{(2)}(\lambda ))\frac{t^n}{n!}.\end{array}$$

(48)

Continuing this process, we obtain the following equation:

$$\begin{array}{rl}\frac{d^k}{d{\lambda }^k}\left(\frac{1-\lambda }{e^t-\lambda }\right)& =\frac{k!}{{(1-\lambda )}^k}(\frac{{(1-\lambda )}^{k+1}}{{(e^t-\lambda )}^{k+1}}-\frac{{(1-\lambda )}^k}{{(e^t-\lambda )}^k})\\ =\frac{k!}{{(1-\lambda )}^k}\sum _{n=0}^{\mathrm{\infty }}(H_n^{(k+1)}(\lambda )-H_n^{(k)}(\lambda ))\frac{t^n}{n!}\text{(see [8])}.\end{array}$$

(49)

By (1), (2) and (49), we get

$$\frac{d^k}{d{\lambda }^k}{H}_n(\lambda )=\frac{k!}{{(1-\lambda )}^k}(H_n^{(k+1)}(\lambda )-H_n^{(k)}(\lambda )),$$

where k is a positive integer (see [7, 8]).