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# 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 . As is well known, the Frobienius-Euler polynomials are defined by the generating function to be

$\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}{e}^{xt}={e}^{H\left(x|\mathrm{Î»}\right)t}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}\left(x|\mathrm{Î»}\right)\frac{{t}^{n}}{n!},$
(1)

with the usual convention about replacing ${H}^{n}\left(x|\mathrm{Î»}\right)$ by ${H}_{n}\left(x|\mathrm{Î»}\right)$ (see [1â€“6]).

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

Thus, by (1), we get

${\left(H\left(\mathrm{Î»}\right)+1\right)}^{n}âˆ’\mathrm{Î»}{H}_{n}\left(\mathrm{Î»}\right)={H}_{n}\left(1|\mathrm{Î»}\right)âˆ’\mathrm{Î»}{H}_{n}\left(\mathrm{Î»}\right)=\left(1âˆ’\mathrm{Î»}\right){\mathrm{Î´}}_{0,n},$
(2)

where ${\mathrm{Î´}}_{0,n}$ is the Kronecker symbol.

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

${H}_{n}\left(x|\mathrm{Î»}\right)={\left(H\left(\mathrm{Î»}\right)+x\right)}^{n}=\underset{0â‰¤lâ‰¤n}{âˆ‘}\left(\begin{array}{c}n\\ l\end{array}\right){H}_{nâˆ’l}\left(\mathrm{Î»}\right){x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [6â€“16]}\right).$
(3)

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

From (1), we have

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({H}_{n}\left(x+1|\mathrm{Î»}\right)âˆ’\mathrm{Î»}{H}_{n}\left(x|\mathrm{Î»}\right)\right)\frac{{t}^{n}}{n!}=\frac{\left(1âˆ’\mathrm{Î»}\right){e}^{\left(x+1\right)t}}{{e}^{t}âˆ’\mathrm{Î»}}âˆ’\mathrm{Î»}\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}{e}^{xt}.$
(4)

Thus, by (4), we get

(5)

It is easy to show that

$\frac{d}{dx}{H}_{n}\left(x|\mathrm{Î»}\right)=\frac{d}{dx}{\left(H\left(\mathrm{Î»}\right)+x\right)}^{n}=n{H}_{nâˆ’1}\left(x|\mathrm{Î»}\right)\phantom{\rule{1em}{0ex}}\left(nâˆˆ\mathbf{N}\right).$
(6)

From (6), we have

${âˆ«}_{0}^{1}{H}_{n}\left(x|\mathrm{Î»}\right)\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{n+1}\left({H}_{n+1}\left(1|\mathrm{Î»}\right)âˆ’{H}_{n+1}\left(\mathrm{Î»}\right)\right)=\frac{\mathrm{Î»}âˆ’1}{n+1}{H}_{n+1}\left(\mathrm{Î»}\right).$
(7)

Let ${\mathbb{P}}_{n}\left(\mathrm{Î»}\right)=\left\{p\left(x\right)âˆˆ\mathbf{Q}\left(\mathrm{Î»}\right)\left[x\right]âˆ£degp\left(x\right)â‰¤n\right\}$ be a vector space over $\mathbf{Q}\left(\mathrm{Î»}\right)$. Then we note that $\left\{{H}_{0}\left(x|\mathrm{Î»}\right),{H}_{1}\left(x|\mathrm{Î»}\right),â€¦,{H}_{n}\left(x|\mathrm{Î»}\right)\right\}$ is a good basis for ${\mathbb{P}}_{n}\left(\mathrm{Î»}\right)$.

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 $p\left(x\right)âˆˆ{\mathbb{P}}_{n}\left(\mathrm{Î»}\right)$. Then $p\left(x\right)$ can be expressed as a $\mathbf{Q}\left(\mathrm{Î»}\right)$-linear combination of ${H}_{0}\left(x|\mathrm{Î»}\right),â€¦,{H}_{n}\left(x|\mathrm{Î»}\right)$ as follows:

$p\left(x\right)={b}_{0}{H}_{0}\left(x|\mathrm{Î»}\right)+{b}_{1}{H}_{1}\left(x|\mathrm{Î»}\right)+â‹¯+{b}_{n}{H}_{n}\left(x|\mathrm{Î»}\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{b}_{k}{H}_{k}\left(x|\mathrm{Î»}\right).$
(8)

Let us define the operator ${\mathrm{â–³}}_{\mathrm{Î»}}$ by

$g\left(x\right)={\mathrm{â–³}}_{\mathrm{Î»}}p\left(x\right)=p\left(x+1\right)âˆ’\mathrm{Î»}p\left(x\right).$
(9)

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

$g\left(x\right)={\mathrm{â–³}}_{\mathrm{Î»}}p\left(x\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{b}_{k}\left({H}_{k}\left(x+1|\mathrm{Î»}\right)âˆ’\mathrm{Î»}{H}_{k}\left(x|\mathrm{Î»}\right)\right)=\left(1âˆ’\mathrm{Î»}\right)\underset{0â‰¤kâ‰¤n}{âˆ‘}{b}_{k}{x}^{k}.$
(10)

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

(11)

Thus, by (11), we get

${g}^{r}\left(0\right)=\frac{{d}^{r}g\left(x\right)}{d{x}^{r}}{|}_{x=0}=\left(1âˆ’\mathrm{Î»}\right)r!{b}_{r}.$
(12)

From (12), we have

${b}_{r}=\frac{{g}^{\left(r\right)}\left(0\right)}{\left(1âˆ’\mathrm{Î»}\right)r!}=\frac{1}{\left(1âˆ’\mathrm{Î»}\right)r!}\left({p}^{\left(r\right)}\left(1\right)âˆ’\mathrm{Î»}{p}^{\left(r\right)}\left(0\right)\right),$
(13)

where $râˆˆ{\mathbf{Z}}_{+}$ 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 , $nâˆˆ{\mathbf{Z}}_{+}$, let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}\left(\mathrm{Î»}\right)$ with $p\left(x\right)={âˆ‘}_{0â‰¤kâ‰¤n}{b}_{k}{H}_{k}\left(x|\mathrm{Î»}\right)$. Then we have

${b}_{k}=\frac{1}{\left(1âˆ’\mathrm{Î»}\right)k!}{g}^{\left(k\right)}\left(0\right)=\frac{1}{\left(1âˆ’\mathrm{Î»}\right)k!}\left({p}^{\left(k\right)}\left(1\right)âˆ’\mathrm{Î»}{p}^{\left(k\right)}\left(0\right)\right).$

Let us take $p\left(x\right)={H}_{n}\left(x|{\mathrm{Î»}}^{âˆ’1}\right)$. Then, by Theorem 1, we get

${H}_{n}\left(x|{\mathrm{Î»}}^{âˆ’1}\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{b}_{k}{H}_{k}\left(x|\mathrm{Î»}\right),$
(14)

where

$\begin{array}{rl}{b}_{k}& =\frac{1}{\left(1âˆ’\mathrm{Î»}\right)k!}\frac{n!}{\left(nâˆ’k\right)!}\left\{{H}_{nâˆ’k}\left(1|{\mathrm{Î»}}^{âˆ’1}\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}\left({\mathrm{Î»}}^{âˆ’1}\right)\right\}\\ =\frac{1}{1âˆ’\mathrm{Î»}}\left(\begin{array}{c}n\\ k\end{array}\right)\left\{{H}_{nâˆ’k}\left(1|{\mathrm{Î»}}^{âˆ’1}\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}\left({\mathrm{Î»}}^{âˆ’1}\right)\right\}\\ =\frac{1}{1âˆ’\mathrm{Î»}}\left(\begin{array}{c}n\\ k\end{array}\right)\left\{\left(1âˆ’{\mathrm{Î»}}^{âˆ’1}\right){0}^{nâˆ’k}+\frac{1}{\mathrm{Î»}}{H}_{nâˆ’k}\left({\mathrm{Î»}}^{âˆ’1}\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}\left({\mathrm{Î»}}^{âˆ’1}\right)\right\}.\end{array}$
(15)

By (14) and (15), we get

(16)

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

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

$\mathrm{Î»}{H}_{n}\left(x|{\mathrm{Î»}}^{âˆ’1}\right)+{H}_{n}\left(x|\mathrm{Î»}\right)=\left(1+\mathrm{Î»}\right)\underset{0â‰¤kâ‰¤n}{âˆ‘}\left(\begin{array}{c}n\\ k\end{array}\right){H}_{nâˆ’k}\left({\mathrm{Î»}}^{âˆ’1}\right){H}_{k}\left(x|\mathrm{Î»}\right).$

Let

$p\left(x\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{H}_{k}\left(x|\mathrm{Î»}\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right)âˆˆ{\mathbb{P}}_{n}\left(\mathrm{Î»}\right).$
(17)

From Theorem 2, we note that $p\left(x\right)$ can be generated by $\left\{{H}_{0}\left(x|\mathrm{Î»}\right),{H}_{1}\left(x|\mathrm{Î»}\right),â€¦,{H}_{n}\left(x|\mathrm{Î»}\right)\right\}$ as follows:

$p\left(x\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{H}_{k}\left(x|\mathrm{Î»}\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{b}_{k}{H}_{k}\left(x|\mathrm{Î»}\right).$
(18)

By (17), we get

${p}^{\left(k\right)}\left(x\right)=\frac{\left(n+1\right)!}{\left(nâˆ’k+1\right)!}\underset{kâ‰¤lâ‰¤n}{âˆ‘}{H}_{lâˆ’k}\left(x|\mathrm{Î»}\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right),$
(19)

and

(20)

From (18) and (20), we have

$\begin{array}{rcl}\underset{0â‰¤kâ‰¤n}{âˆ‘}{H}_{k}\left(x|\mathrm{Î»}\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right)& =& \left(n+1\right)\underset{0â‰¤kâ‰¤nâˆ’1}{âˆ‘}\frac{\left(\begin{array}{c}n\\ k\end{array}\right)}{nâˆ’k+1}\underset{kâ‰¤lâ‰¤n}{âˆ‘}\left\{\left(âˆ’\mathrm{Î»}\right){H}_{lâˆ’k}\left(\mathrm{Î»}\right){H}_{nâˆ’l}\left(\mathrm{Î»}\right)\\ +2\mathrm{Î»}{H}_{nâˆ’k}\left(\mathrm{Î»}\right)\right\}{H}_{k}\left(x|\mathrm{Î»}\right)+\left(n+1\right){H}_{n}\left(x|\mathrm{Î»}\right).\end{array}$
(21)

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

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

Let us consider

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\frac{1}{k!\left(nâˆ’k\right)!}{H}_{k}\left(x|\mathrm{Î»}\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right)âˆˆ{\mathbb{P}}_{n}\left(\mathrm{Î»}\right).$
(22)

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

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{b}_{k}{H}_{k}\left(x|\mathrm{Î»}\right).$
(23)

From (22), we have

${p}^{\left(r\right)}\left(x\right)={2}^{r}\underset{k=r}{\overset{n}{âˆ‘}}\frac{{H}_{kâˆ’r}\left(x|\mathrm{Î»}\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right)}{\left(kâˆ’r\right)!\left(nâˆ’k\right)!}\phantom{\rule{1em}{0ex}}\left(râˆˆ{\mathbf{Z}}_{+}\right).$
(24)

By Theorem 1, we get

(25)

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

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

## 3 Higher-order Frobenius-Euler polynomials

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

$\begin{array}{rl}{\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)}^{r}{e}^{xt}& ={e}^{{H}^{\left(r\right)}\left(x|\mathrm{Î»}\right)t}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\frac{{t}^{n}}{n!},\end{array}$
(26)

with the usual convention about replacing ${\left({H}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\right)}^{n}$ by ${H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)$ (see [1â€“10]). In the special case, $x=0$, ${H}_{n}^{\left(r\right)}\left(0|\mathrm{Î»}\right)={H}_{n}^{\left(r\right)}\left(\mathrm{Î»}\right)$ are called the n th Frobenius-Euler numbers of order r (see [8, 9]).

From (26), we have

${H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)={\left({H}^{\left(r\right)}\left(\mathrm{Î»}\right)+x\right)}^{n}=\underset{l=0}{\overset{n}{âˆ‘}}\left(\begin{array}{c}n\\ l\end{array}\right){H}_{nâˆ’l}^{\left(r\right)}\left(\mathrm{Î»}\right){x}^{l},$
(27)

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

By (26), we get

${H}_{n}^{\left(r\right)}\left(\mathrm{Î»}\right)=\underset{{n}_{1}+â‹¯+{n}_{r}=n}{âˆ‘}\left(\begin{array}{c}n\\ {n}_{1},{n}_{2},â€¦,{n}_{r}\end{array}\right){H}_{{n}_{1}}\left(\mathrm{Î»}\right)â‹¯{H}_{{n}_{r}}\left(\mathrm{Î»}\right),$
(28)

where $\left(\begin{array}{c}n\\ {n}_{1},{n}_{2},â€¦,{n}_{r}\end{array}\right)=\frac{n!}{{n}_{1}!{n}_{2}!â‹¯{n}_{r}!}$. From (27) and (28), we note that the leading coefficient of ${H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)$ is given by

$\begin{array}{rl}{H}_{0}^{\left(r\right)}\left(\mathrm{Î»}\right)& =\underset{{n}_{1}+â‹¯+{n}_{r}=0}{âˆ‘}\left(\begin{array}{c}n\\ {n}_{1},{n}_{2},â€¦,{n}_{r}\end{array}\right){H}_{{n}_{1}}\left(\mathrm{Î»}\right)â‹¯{H}_{{n}_{r}}\left(\mathrm{Î»}\right)\\ ={H}_{0}\left(\mathrm{Î»}\right)â‹¯{H}_{0}\left(\mathrm{Î»}\right)=1.\end{array}$
(29)

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

(30)

and

$\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}x}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)=\frac{\mathrm{âˆ‚}}{\mathrm{âˆ‚}x}{\left({H}^{\left(r\right)}\left(\mathrm{Î»}\right)+x\right)}^{n}=n{H}_{nâˆ’1}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\phantom{\rule{1em}{0ex}}\left(râ‰¥0\right).$
(31)

It is not difficult to show that

${H}_{n}^{\left(r\right)}\left(x+1|\mathrm{Î»}\right)âˆ’\mathrm{Î»}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)=\left(1âˆ’\mathrm{Î»}\right){H}_{n}^{\left(râˆ’1\right)}\left(x|\mathrm{Î»}\right).$
(32)

Now, we note that $\left\{{H}_{0}^{\left(r\right)}\left(x|\mathrm{Î»}\right),{H}_{1}^{\left(r\right)}\left(x|\mathrm{Î»}\right),â€¦,{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\right\}$ is also a good basis for ${\mathbb{P}}_{n}\left(\mathrm{Î»}\right)$.

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

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{C}_{k}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$
(33)

From (9) and (32), we have

${\mathrm{â–³}}_{\mathrm{Î»}}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)={H}_{n}^{\left(r\right)}\left(x+1|\mathrm{Î»}\right)âˆ’\mathrm{Î»}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)=\left(1âˆ’\mathrm{Î»}\right){H}_{n}^{\left(râˆ’1\right)}\left(x|\mathrm{Î»}\right).$
(34)

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

${\mathrm{â–³}}_{\mathrm{Î»}}^{r}p\left(x\right)={\left(1âˆ’\mathrm{Î»}\right)}^{r}\underset{k=0}{\overset{n}{âˆ‘}}{C}_{k}{H}_{k}^{\left(0\right)}\left(x|\mathrm{Î»}\right)={\left(1âˆ’\mathrm{Î»}\right)}^{r}\underset{k=0}{\overset{n}{âˆ‘}}{C}_{k}{x}^{k}.$
(35)

Let us take the k th derivative of ${\mathrm{â–³}}_{\mathrm{Î»}}^{r}p\left(x\right)$ in (35).

Then we have

${D}^{k}\left({\mathrm{â–³}}_{\mathrm{Î»}}^{r}p\left(x\right)\right)={\left(1âˆ’\mathrm{Î»}\right)}^{r}\underset{l=k}{\overset{n}{âˆ‘}}\frac{l!}{\left(lâˆ’k\right)!}{C}_{l}{x}^{lâˆ’k}.$
(36)

Thus, from (36), we have

${D}^{k}\left({\mathrm{â–³}}_{\mathrm{Î»}}^{r}p\left(0\right)\right)={\left(1âˆ’\mathrm{Î»}\right)}^{r}\underset{l=k}{\overset{n}{âˆ‘}}\frac{l!{C}_{l}}{\left(lâˆ’k\right)!}{0}^{lâˆ’k}={\left(1âˆ’\mathrm{Î»}\right)}^{r}k!{C}_{k}.$
(37)

Thus, by (37), we get

$\begin{array}{rl}{C}_{k}& =\frac{{D}^{k}\left({\mathrm{â–³}}_{\mathrm{Î»}}^{r}p\left(0\right)\right)}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}k!}\\ =\frac{{\mathrm{â–³}}_{\mathrm{Î»}}^{r}\left({D}^{k}p\left(0\right)\right)}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}k!}=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}k!}\underset{j=0}{\overset{r}{âˆ‘}}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’\mathrm{Î»}\right)}^{\left(râˆ’j\right)}{D}^{k}p\left(j\right).\end{array}$
(38)

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

Theorem 5 For $râˆˆ{\mathbf{Z}}_{+}$, let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}\left(\mathrm{Î»}\right)$ with

$p\left(x\right)=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{0â‰¤kâ‰¤n}{âˆ‘}{C}_{k}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\phantom{\rule{1em}{0ex}}\left({C}_{k}âˆˆ\mathbf{Q}\left(\mathrm{Î»}\right)\right).$

Then we have

${C}_{k}=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}k!}\underset{0â‰¤jâ‰¤r}{âˆ‘}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{D}^{k}p\left(j\right).$

That is,

$p\left(x\right)=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{0â‰¤kâ‰¤n}{âˆ‘}\left(\underset{0â‰¤jâ‰¤r}{âˆ‘}\frac{1}{k!}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{D}^{k}p\left(j\right)\right){H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$

Let us take $p\left(x\right)={H}_{n}\left(x|\mathrm{Î»}\right)âˆˆ{\mathbf{P}}_{n}\left(\mathrm{Î»}\right)$. Then, by Theorem 5, $p\left(x\right)={H}_{n}\left(x|\mathrm{Î»}\right)$ can be generated by $\left\{{H}_{0}^{\left(r\right)}\left(x|\mathrm{Î»}\right),{H}_{1}^{\left(r\right)}\left(\mathrm{Î»}\right),â€¦,{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\right\}$ as follows:

${H}_{n}\left(x|\mathrm{Î»}\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{C}_{k}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right),$
(39)

where

${C}_{k}=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\frac{1}{k!}\underset{0â‰¤jâ‰¤r}{âˆ‘}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{D}^{k}p\left(j\right),$
(40)

and

${p}^{\left(k\right)}\left(x\right)={D}^{k}p\left(x\right)=n\left(nâˆ’1\right)â‹¯\left(nâˆ’k+1\right){H}_{nâˆ’k}\left(x|\mathrm{Î»}\right)=\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}\left(x|\mathrm{Î»}\right).$
(41)

By (40) and (41), we get

${C}_{k}=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\left(\begin{array}{c}n\\ k\end{array}\right)\underset{0â‰¤jâ‰¤r}{âˆ‘}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{H}_{nâˆ’k}\left(j|\mathrm{Î»}\right).$
(42)

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

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

${H}_{n}\left(x|\mathrm{Î»}\right)=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{0â‰¤kâ‰¤n}{âˆ‘}\left(\begin{array}{c}n\\ k\end{array}\right)\left(\underset{0â‰¤jâ‰¤r}{âˆ‘}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{H}_{nâˆ’k}\left(j|\mathrm{Î»}\right)\right){H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$

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

Then we have

$\begin{array}{rl}{p}^{k}\left(x\right)& =n\left(nâˆ’1\right)â‹¯\left(nâˆ’k+1\right){H}_{nâˆ’k}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\\ =\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).\end{array}$
(43)

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

${H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}{b}_{k}{H}_{k}\left(x|\mathrm{Î»}\right),$
(44)

where

$\begin{array}{rl}{b}_{k}& =\frac{1}{\left(1âˆ’\mathrm{Î»}\right)k!}\left\{{p}^{k}\left(1\right)âˆ’\mathrm{Î»}{p}^{\left(k\right)}\left(0\right)\right\}\\ =\frac{n!}{\left(1âˆ’\mathrm{Î»}\right)k!\left(nâˆ’k\right)!}\left\{{H}_{nâˆ’k}^{\left(r\right)}\left(1|\mathrm{Î»}\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}^{\left(r\right)}\left(\mathrm{Î»}\right)\right\}.\end{array}$
(45)

By (34) and (45), we get

${b}_{k}=\left(\begin{array}{c}n\\ k\end{array}\right){H}_{nâˆ’k}^{\left(râˆ’1\right)}\left(\mathrm{Î»}\right).$
(46)

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

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

${H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)=\underset{0â‰¤kâ‰¤n}{âˆ‘}\left(\begin{array}{c}n\\ k\end{array}\right){H}_{nâˆ’k}^{\left(râˆ’1\right)}\left(\mathrm{Î»}\right){H}_{k}\left(x|\mathrm{Î»}\right).$

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

$\begin{array}{rl}\frac{d}{d\mathrm{Î»}}\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)& =\frac{1âˆ’{e}^{t}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}âˆ’\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}{e}^{t}\right)\\ =\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}âˆ’\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}\left({e}^{t}âˆ’\mathrm{Î»}+\mathrm{Î»}\right)\right)\\ =\frac{1}{1âˆ’\mathrm{Î»}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}âˆ’\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)\\ =\frac{1}{1âˆ’\mathrm{Î»}}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({H}_{n}^{\left(2\right)}\left(\mathrm{Î»}\right)âˆ’{H}_{n}\left(\mathrm{Î»}\right)\right)\frac{{t}^{n}}{n!},\end{array}$
(47)

and

$\begin{array}{rl}\frac{{d}^{2}}{d{\mathrm{Î»}}^{2}}\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)& =2!\frac{1âˆ’{e}^{t}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{3}}=\frac{2!}{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{3}}âˆ’\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{3}}{e}^{t}\right)\\ =\frac{2!}{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{3}}âˆ’\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{3}}\left({e}^{t}âˆ’\mathrm{Î»}+\mathrm{Î»}\right)\right)\\ =\frac{2!}{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{3}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{3}}âˆ’\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{2}}\right)\\ =\frac{2!}{{\left(1âˆ’\mathrm{Î»}\right)}^{2}}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({H}_{n}^{\left(3\right)}\left(\mathrm{Î»}\right)âˆ’{H}_{n}^{\left(2\right)}\left(\mathrm{Î»}\right)\right)\frac{{t}^{n}}{n!}.\end{array}$
(48)

Continuing this process, we obtain the following equation:

$\begin{array}{rl}\frac{{d}^{k}}{d{\mathrm{Î»}}^{k}}\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)& =\frac{k!}{{\left(1âˆ’\mathrm{Î»}\right)}^{k}}\left(\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{k+1}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{k+1}}âˆ’\frac{{\left(1âˆ’\mathrm{Î»}\right)}^{k}}{{\left({e}^{t}âˆ’\mathrm{Î»}\right)}^{k}}\right)\\ =\frac{k!}{{\left(1âˆ’\mathrm{Î»}\right)}^{k}}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({H}_{n}^{\left(k+1\right)}\left(\mathrm{Î»}\right)âˆ’{H}_{n}^{\left(k\right)}\left(\mathrm{Î»}\right)\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [8])}.\end{array}$
(49)

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

$\frac{{d}^{k}}{d{\mathrm{Î»}}^{k}}{H}_{n}\left(\mathrm{Î»}\right)=\frac{k!}{{\left(1âˆ’\mathrm{Î»}\right)}^{k}}\left({H}_{n}^{\left(k+1\right)}\left(\mathrm{Î»}\right)âˆ’{H}_{n}^{\left(k\right)}\left(\mathrm{Î»}\right)\right),$

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.

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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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