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

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

with the usual convention about replacing ${H}^{n}\left(x|\lambda \right)$ by ${H}_{n}\left(x|\lambda \right)$ (see ).

In the special case, $x=0$, ${H}_{n}\left(0|\lambda \right)={H}_{n}\left(\lambda \right)$ are called the n th Frobenius-Euler numbers.

Thus, by (1), we get

${\left(H\left(\lambda \right)+1\right)}^{n}-\lambda {H}_{n}\left(\lambda \right)={H}_{n}\left(1|\lambda \right)-\lambda {H}_{n}\left(\lambda \right)=\left(1-\lambda \right){\delta }_{0,n},$
(2)

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

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

${H}_{n}\left(x|\lambda \right)={\left(H\left(\lambda \right)+x\right)}^{n}=\sum _{0\le l\le n}\left(\begin{array}{c}n\\ l\end{array}\right){H}_{n-l}\left(\lambda \right){x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [6–16]}\right).$
(3)

Thus, by (3), we easily see that the leading coefficient of ${H}_{n}\left(x|\lambda \right)$ is ${H}_{0}\left(\lambda \right)=1$. So, ${H}_{n}\left(x|\lambda \right)$ are monic polynomials of degree n with coefficients in $\mathbf{Q}\left(\lambda \right)$.

From (1), we have

$\sum _{n=0}^{\mathrm{\infty }}\left({H}_{n}\left(x+1|\lambda \right)-\lambda {H}_{n}\left(x|\lambda \right)\right)\frac{{t}^{n}}{n!}=\frac{\left(1-\lambda \right){e}^{\left(x+1\right)t}}{{e}^{t}-\lambda }-\lambda \frac{1-\lambda }{{e}^{t}-\lambda }{e}^{xt}.$
(4)

Thus, by (4), we get

(5)

It is easy to show that

$\frac{d}{dx}{H}_{n}\left(x|\lambda \right)=\frac{d}{dx}{\left(H\left(\lambda \right)+x\right)}^{n}=n{H}_{n-1}\left(x|\lambda \right)\phantom{\rule{1em}{0ex}}\left(n\in \mathbf{N}\right).$
(6)

From (6), we have

${\int }_{0}^{1}{H}_{n}\left(x|\lambda \right)\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{n+1}\left({H}_{n+1}\left(1|\lambda \right)-{H}_{n+1}\left(\lambda \right)\right)=\frac{\lambda -1}{n+1}{H}_{n+1}\left(\lambda \right).$
(7)

Let ${\mathbb{P}}_{n}\left(\lambda \right)=\left\{p\left(x\right)\in \mathbf{Q}\left(\lambda \right)\left[x\right]\mid degp\left(x\right)\le n\right\}$ be a vector space over $\mathbf{Q}\left(\lambda \right)$. Then we note that $\left\{{H}_{0}\left(x|\lambda \right),{H}_{1}\left(x|\lambda \right),\dots ,{H}_{n}\left(x|\lambda \right)\right\}$ is a good basis for ${\mathbb{P}}_{n}\left(\lambda \right)$.

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 $p\left(x\right)\in {\mathbb{P}}_{n}\left(\lambda \right)$. Then $p\left(x\right)$ can be expressed as a $\mathbf{Q}\left(\lambda \right)$-linear combination of ${H}_{0}\left(x|\lambda \right),\dots ,{H}_{n}\left(x|\lambda \right)$ as follows:

$p\left(x\right)={b}_{0}{H}_{0}\left(x|\lambda \right)+{b}_{1}{H}_{1}\left(x|\lambda \right)+\cdots +{b}_{n}{H}_{n}\left(x|\lambda \right)=\sum _{0\le k\le n}{b}_{k}{H}_{k}\left(x|\lambda \right).$
(8)

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

$g\left(x\right)={\mathrm{△}}_{\lambda }p\left(x\right)=p\left(x+1\right)-\lambda p\left(x\right).$
(9)

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

$g\left(x\right)={\mathrm{△}}_{\lambda }p\left(x\right)=\sum _{0\le k\le n}{b}_{k}\left({H}_{k}\left(x+1|\lambda \right)-\lambda {H}_{k}\left(x|\lambda \right)\right)=\left(1-\lambda \right)\sum _{0\le k\le n}{b}_{k}{x}^{k}.$
(10)

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

(11)

Thus, by (11), we get

${g}^{r}\left(0\right)=\frac{{d}^{r}g\left(x\right)}{d{x}^{r}}{|}_{x=0}=\left(1-\lambda \right)r!{b}_{r}.$
(12)

From (12), we have

${b}_{r}=\frac{{g}^{\left(r\right)}\left(0\right)}{\left(1-\lambda \right)r!}=\frac{1}{\left(1-\lambda \right)r!}\left({p}^{\left(r\right)}\left(1\right)-\lambda {p}^{\left(r\right)}\left(0\right)\right),$
(13)

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

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

${b}_{k}=\frac{1}{\left(1-\lambda \right)k!}{g}^{\left(k\right)}\left(0\right)=\frac{1}{\left(1-\lambda \right)k!}\left({p}^{\left(k\right)}\left(1\right)-\lambda {p}^{\left(k\right)}\left(0\right)\right).$

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

${H}_{n}\left(x|{\lambda }^{-1}\right)=\sum _{0\le k\le n}{b}_{k}{H}_{k}\left(x|\lambda \right),$
(14)

where

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

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

Let

$p\left(x\right)=\sum _{0\le k\le n}{H}_{k}\left(x|\lambda \right){H}_{n-k}\left(x|\lambda \right)\in {\mathbb{P}}_{n}\left(\lambda \right).$
(17)

From Theorem 2, we note that $p\left(x\right)$ can be generated by $\left\{{H}_{0}\left(x|\lambda \right),{H}_{1}\left(x|\lambda \right),\dots ,{H}_{n}\left(x|\lambda \right)\right\}$ as follows:

$p\left(x\right)=\sum _{0\le k\le n}{H}_{k}\left(x|\lambda \right){H}_{n-k}\left(x|\lambda \right)=\sum _{0\le k\le n}{b}_{k}{H}_{k}\left(x|\lambda \right).$
(18)

By (17), we get

${p}^{\left(k\right)}\left(x\right)=\frac{\left(n+1\right)!}{\left(n-k+1\right)!}\sum _{k\le l\le n}{H}_{l-k}\left(x|\lambda \right){H}_{n-k}\left(x|\lambda \right),$
(19)

and (20)

From (18) and (20), we have

$\begin{array}{rcl}\sum _{0\le k\le n}{H}_{k}\left(x|\lambda \right){H}_{n-k}\left(x|\lambda \right)& =& \left(n+1\right)\sum _{0\le k\le n-1}\frac{\left(\begin{array}{c}n\\ k\end{array}\right)}{n-k+1}\sum _{k\le l\le n}\left\{\left(-\lambda \right){H}_{l-k}\left(\lambda \right){H}_{n-l}\left(\lambda \right)\\ +2\lambda {H}_{n-k}\left(\lambda \right)\right\}{H}_{k}\left(x|\lambda \right)+\left(n+1\right){H}_{n}\left(x|\lambda \right).\end{array}$
(21)

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

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

$p\left(x\right)=\sum _{k=0}^{n}\frac{1}{k!\left(n-k\right)!}{H}_{k}\left(x|\lambda \right){H}_{n-k}\left(x|\lambda \right)\in {\mathbb{P}}_{n}\left(\lambda \right).$
(22)

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

$p\left(x\right)=\sum _{k=0}^{n}{b}_{k}{H}_{k}\left(x|\lambda \right).$
(23)

From (22), we have

${p}^{\left(r\right)}\left(x\right)={2}^{r}\sum _{k=r}^{n}\frac{{H}_{k-r}\left(x|\lambda \right){H}_{n-k}\left(x|\lambda \right)}{\left(k-r\right)!\left(n-k\right)!}\phantom{\rule{1em}{0ex}}\left(r\in {\mathbf{Z}}_{+}\right).$
(24)

By Theorem 1, we get

(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

$\begin{array}{rl}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}{e}^{xt}& ={e}^{{H}^{\left(r\right)}\left(x|\lambda \right)t}\\ =\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(r\right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!},\end{array}$
(26)

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

From (26), we have

${H}_{n}^{\left(r\right)}\left(x|\lambda \right)={\left({H}^{\left(r\right)}\left(\lambda \right)+x\right)}^{n}=\sum _{l=0}^{n}\left(\begin{array}{c}n\\ l\end{array}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right){x}^{l},$
(27)

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

By (26), we get

${H}_{n}^{\left(r\right)}\left(\lambda \right)=\sum _{{n}_{1}+\cdots +{n}_{r}=n}\left(\begin{array}{c}n\\ {n}_{1},{n}_{2},\dots ,{n}_{r}\end{array}\right){H}_{{n}_{1}}\left(\lambda \right)\cdots {H}_{{n}_{r}}\left(\lambda \right),$
(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 ${H}_{n}^{\left(r\right)}\left(x|\lambda \right)$ is given by

$\begin{array}{rl}{H}_{0}^{\left(r\right)}\left(\lambda \right)& =\sum _{{n}_{1}+\cdots +{n}_{r}=0}\left(\begin{array}{c}n\\ {n}_{1},{n}_{2},\dots ,{n}_{r}\end{array}\right){H}_{{n}_{1}}\left(\lambda \right)\cdots {H}_{{n}_{r}}\left(\lambda \right)\\ ={H}_{0}\left(\lambda \right)\cdots {H}_{0}\left(\lambda \right)=1.\end{array}$
(29)

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

(30)

and

$\frac{\partial }{\partial x}{H}_{n}^{\left(r\right)}\left(x|\lambda \right)=\frac{\partial }{\partial x}{\left({H}^{\left(r\right)}\left(\lambda \right)+x\right)}^{n}=n{H}_{n-1}^{\left(r\right)}\left(x|\lambda \right)\phantom{\rule{1em}{0ex}}\left(r\ge 0\right).$
(31)

It is not difficult to show that

${H}_{n}^{\left(r\right)}\left(x+1|\lambda \right)-\lambda {H}_{n}^{\left(r\right)}\left(x|\lambda \right)=\left(1-\lambda \right){H}_{n}^{\left(r-1\right)}\left(x|\lambda \right).$
(32)

Now, we note that $\left\{{H}_{0}^{\left(r\right)}\left(x|\lambda \right),{H}_{1}^{\left(r\right)}\left(x|\lambda \right),\dots ,{H}_{n}^{\left(r\right)}\left(x|\lambda \right)\right\}$ is also a good basis for ${\mathbb{P}}_{n}\left(\lambda \right)$.

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

$p\left(x\right)=\sum _{k=0}^{n}{C}_{k}{H}_{k}^{\left(r\right)}\left(x|\lambda \right).$
(33)

From (9) and (32), we have

${\mathrm{△}}_{\lambda }{H}_{n}^{\left(r\right)}\left(x|\lambda \right)={H}_{n}^{\left(r\right)}\left(x+1|\lambda \right)-\lambda {H}_{n}^{\left(r\right)}\left(x|\lambda \right)=\left(1-\lambda \right){H}_{n}^{\left(r-1\right)}\left(x|\lambda \right).$
(34)

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

${\mathrm{△}}_{\lambda }^{r}p\left(x\right)={\left(1-\lambda \right)}^{r}\sum _{k=0}^{n}{C}_{k}{H}_{k}^{\left(0\right)}\left(x|\lambda \right)={\left(1-\lambda \right)}^{r}\sum _{k=0}^{n}{C}_{k}{x}^{k}.$
(35)

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

Then we have

${D}^{k}\left({\mathrm{△}}_{\lambda }^{r}p\left(x\right)\right)={\left(1-\lambda \right)}^{r}\sum _{l=k}^{n}\frac{l!}{\left(l-k\right)!}{C}_{l}{x}^{l-k}.$
(36)

Thus, from (36), we have

${D}^{k}\left({\mathrm{△}}_{\lambda }^{r}p\left(0\right)\right)={\left(1-\lambda \right)}^{r}\sum _{l=k}^{n}\frac{l!{C}_{l}}{\left(l-k\right)!}{0}^{l-k}={\left(1-\lambda \right)}^{r}k!{C}_{k}.$
(37)

Thus, by (37), we get

$\begin{array}{rl}{C}_{k}& =\frac{{D}^{k}\left({\mathrm{△}}_{\lambda }^{r}p\left(0\right)\right)}{{\left(1-\lambda \right)}^{r}k!}\\ =\frac{{\mathrm{△}}_{\lambda }^{r}\left({D}^{k}p\left(0\right)\right)}{{\left(1-\lambda \right)}^{r}k!}=\frac{1}{{\left(1-\lambda \right)}^{r}k!}\sum _{j=0}^{r}\left(\begin{array}{c}r\\ j\end{array}\right){\left(-\lambda \right)}^{\left(r-j\right)}{D}^{k}p\left(j\right).\end{array}$
(38)

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

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

$p\left(x\right)=\frac{1}{{\left(1-\lambda \right)}^{r}}\sum _{0\le k\le n}{C}_{k}{H}_{k}^{\left(r\right)}\left(x|\lambda \right)\phantom{\rule{1em}{0ex}}\left({C}_{k}\in \mathbf{Q}\left(\lambda \right)\right).$

Then we have

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

That is,

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

Let us take $p\left(x\right)={H}_{n}\left(x|\lambda \right)\in {\mathbf{P}}_{n}\left(\lambda \right)$. Then, by Theorem 5, $p\left(x\right)={H}_{n}\left(x|\lambda \right)$ can be generated by $\left\{{H}_{0}^{\left(r\right)}\left(x|\lambda \right),{H}_{1}^{\left(r\right)}\left(\lambda \right),\dots ,{H}_{n}^{\left(r\right)}\left(x|\lambda \right)\right\}$ as follows:

${H}_{n}\left(x|\lambda \right)=\sum _{0\le k\le n}{C}_{k}{H}_{k}^{\left(r\right)}\left(x|\lambda \right),$
(39)

where

${C}_{k}=\frac{1}{{\left(1-\lambda \right)}^{r}}\frac{1}{k!}\sum _{0\le j\le r}\left(\begin{array}{c}r\\ j\end{array}\right){\left(-\lambda \right)}^{r-j}{D}^{k}p\left(j\right),$
(40)

and

${p}^{\left(k\right)}\left(x\right)={D}^{k}p\left(x\right)=n\left(n-1\right)\cdots \left(n-k+1\right){H}_{n-k}\left(x|\lambda \right)=\frac{n!}{\left(n-k\right)!}{H}_{n-k}\left(x|\lambda \right).$
(41)

By (40) and (41), we get

${C}_{k}=\frac{1}{{\left(1-\lambda \right)}^{r}}\left(\begin{array}{c}n\\ k\end{array}\right)\sum _{0\le j\le r}\left(\begin{array}{c}r\\ j\end{array}\right){\left(-\lambda \right)}^{r-j}{H}_{n-k}\left(j|\lambda \right).$
(42)

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

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

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

Let us assume that $p\left(x\right)={H}_{n}^{\left(r\right)}\left(x|\lambda \right)$.

Then we have

$\begin{array}{rl}{p}^{k}\left(x\right)& =n\left(n-1\right)\cdots \left(n-k+1\right){H}_{n-k}^{\left(r\right)}\left(x|\lambda \right)\\ =\frac{n!}{\left(n-k\right)!}{H}_{n-k}^{\left(r\right)}\left(x|\lambda \right).\end{array}$
(43)

From Theorem 1, we note that $p\left(x\right)={H}_{n}^{\left(r\right)}\left(x|\lambda \right)$ can be expressed as a linear combination of ${H}_{0}\left(x|\lambda \right),{H}_{1}\left(x|\lambda \right),\dots ,{H}_{n}\left(x|\lambda \right)$

${H}_{n}^{\left(r\right)}\left(x|\lambda \right)=\sum _{0\le k\le n}{b}_{k}{H}_{k}\left(x|\lambda \right),$
(44)

where

$\begin{array}{rl}{b}_{k}& =\frac{1}{\left(1-\lambda \right)k!}\left\{{p}^{k}\left(1\right)-\lambda {p}^{\left(k\right)}\left(0\right)\right\}\\ =\frac{n!}{\left(1-\lambda \right)k!\left(n-k\right)!}\left\{{H}_{n-k}^{\left(r\right)}\left(1|\lambda \right)-\lambda {H}_{n-k}^{\left(r\right)}\left(\lambda \right)\right\}.\end{array}$
(45)

By (34) and (45), we get

${b}_{k}=\left(\begin{array}{c}n\\ k\end{array}\right){H}_{n-k}^{\left(r-1\right)}\left(\lambda \right).$
(46)

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

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

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

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

$\begin{array}{rl}\frac{d}{d\lambda }\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)& =\frac{1-{e}^{t}}{{\left({e}^{t}-\lambda \right)}^{2}}=\frac{1}{{\left(1-\lambda \right)}^{2}}\left(\frac{{\left(1-\lambda \right)}^{2}}{{\left({e}^{t}-\lambda \right)}^{2}}-\frac{{\left(1-\lambda \right)}^{2}}{{\left({e}^{t}-\lambda \right)}^{2}}{e}^{t}\right)\\ =\frac{1}{{\left(1-\lambda \right)}^{2}}\left(\frac{{\left(1-\lambda \right)}^{2}}{{\left({e}^{t}-\lambda \right)}^{2}}-\frac{{\left(1-\lambda \right)}^{2}}{{\left({e}^{t}-\lambda \right)}^{2}}\left({e}^{t}-\lambda +\lambda \right)\right)\\ =\frac{1}{1-\lambda }\left(\frac{{\left(1-\lambda \right)}^{2}}{{\left({e}^{t}-\lambda \right)}^{2}}-\frac{1-\lambda }{{e}^{t}-\lambda }\right)\\ =\frac{1}{1-\lambda }\sum _{n=0}^{\mathrm{\infty }}\left({H}_{n}^{\left(2\right)}\left(\lambda \right)-{H}_{n}\left(\lambda \right)\right)\frac{{t}^{n}}{n!},\end{array}$
(47)

and

$\begin{array}{rl}\frac{{d}^{2}}{d{\lambda }^{2}}\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)& =2!\frac{1-{e}^{t}}{{\left({e}^{t}-\lambda \right)}^{3}}=\frac{2!}{{\left(1-\lambda \right)}^{3}}\left(\frac{{\left(1-\lambda \right)}^{3}}{{\left({e}^{t}-\lambda \right)}^{3}}-\frac{{\left(1-\lambda \right)}^{3}}{{\left({e}^{t}-\lambda \right)}^{3}}{e}^{t}\right)\\ =\frac{2!}{{\left(1-\lambda \right)}^{3}}\left(\frac{{\left(1-\lambda \right)}^{3}}{{\left({e}^{t}-\lambda \right)}^{3}}-\frac{{\left(1-\lambda \right)}^{3}}{{\left({e}^{t}-\lambda \right)}^{3}}\left({e}^{t}-\lambda +\lambda \right)\right)\\ =\frac{2!}{{\left(1-\lambda \right)}^{2}}\left(\frac{{\left(1-\lambda \right)}^{3}}{{\left({e}^{t}-\lambda \right)}^{3}}-\frac{{\left(1-\lambda \right)}^{2}}{{\left({e}^{t}-\lambda \right)}^{2}}\right)\\ =\frac{2!}{{\left(1-\lambda \right)}^{2}}\sum _{n=0}^{\mathrm{\infty }}\left({H}_{n}^{\left(3\right)}\left(\lambda \right)-{H}_{n}^{\left(2\right)}\left(\lambda \right)\right)\frac{{t}^{n}}{n!}.\end{array}$
(48)

Continuing this process, we obtain the following equation:

$\begin{array}{rl}\frac{{d}^{k}}{d{\lambda }^{k}}\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)& =\frac{k!}{{\left(1-\lambda \right)}^{k}}\left(\frac{{\left(1-\lambda \right)}^{k+1}}{{\left({e}^{t}-\lambda \right)}^{k+1}}-\frac{{\left(1-\lambda \right)}^{k}}{{\left({e}^{t}-\lambda \right)}^{k}}\right)\\ =\frac{k!}{{\left(1-\lambda \right)}^{k}}\sum _{n=0}^{\mathrm{\infty }}\left({H}_{n}^{\left(k+1\right)}\left(\lambda \right)-{H}_{n}^{\left(k\right)}\left(\lambda \right)\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see )}.\end{array}$
(49)

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

$\frac{{d}^{k}}{d{\lambda }^{k}}{H}_{n}\left(\lambda \right)=\frac{k!}{{\left(1-\lambda \right)}^{k}}\left({H}_{n}^{\left(k+1\right)}\left(\lambda \right)-{H}_{n}^{\left(k\right)}\left(\lambda \right)\right),$

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

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## Acknowledgements

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology 2012R1A1A2003786.

## Author information

Authors

### Corresponding author

Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, D.S., Kim, T. Some new identities of Frobenius-Euler numbers and polynomials. J Inequal Appl 2012, 307 (2012). https://doi.org/10.1186/1029-242X-2012-307

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• DOI: https://doi.org/10.1186/1029-242X-2012-307

### Keywords

• Positive Integer
• Linear Combination
• Vector Space
• Good Basis
• Monic Polynomial 