Frobenius-Euler polynomials and umbral calculus in the p-adic case

Abstract

In this paper, we study some p-adic Frobenius-Euler measure related to umbral calculus in the p-adic case. Finally, we derive some identities of Frobenius-Euler polynomials from our study.

MSC:05A10, 05A19.

1 Introduction

Let p be a fixed prime number. Throughout this paper ${\mathbb{Z}}_p$, ${\mathbb{Q}}_p$ and ${\mathbb{C}}_p$ will denote the ring of p-adic integers, the field of p-adic rational numbers and the completion of algebraic closure of ${\mathbb{Q}}_p$, respectively.

For $f\in \mathbb{N}$ with $(f,p)=1$, let

Note that the natural map induces

$$\pi :X\to {\mathbb{Z}}_p.$$

If g is a function on ${\mathbb{Z}}_p$, we denote by the same g the function $g\circ \pi$ on X. Namely, we can consider g as a function on X.

For $k\ge 0$ and $\lambda \in {\mathbb{C}}_p$ with ${|1-\lambda |}_p>1$, the Frobenius-Euler measure on X is defined by

$${\mu }_{\lambda }(x+f{p}^N{\mathbb{Z}}_p)=\frac{{\lambda }^{f{p}^N-x}}{1-{\lambda }^{f{p}^N}}(\text{see [5, 8])},$$

(1.1)

where the p-adic absolute value on ${\mathbb{C}}_p$ is normalized by ${|p|}_p=\frac{1}{p}$.

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

$$\left(\frac{1-\lambda }{e^t-\lambda }\right)e^{xt}=e^{H(x|\lambda )t}=\sum _{n=0}^{\mathrm{\infty }}{H}_n(x|\lambda )\frac{t^n}{n!}(\text{see [5, 7, 9])},$$

(1.2)

with the usual convention about replacing $H^n(x|\lambda )$ by $H_n(x|\lambda )$. In the special case, $x=0$, $H_n(0|\lambda )=H_n(\lambda )$ are called the n th Frobenius-Euler numbers

$$H_n(x|\lambda )={(H(\lambda )+x)}^n=\sum _{l=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l(\lambda )x^{n-l}\text{(see [6, 9, 10])}.$$

(1.3)

Thus, by (1.2) and (1.3), we easily get

$${(H(\lambda )+1)}^n-\lambda H_n(\lambda )=(1-\lambda ){\delta }_{0,n}(\text{see [1–19])},$$

(1.4)

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

For $r\in \mathbb{N}$, the Frobenius-Euler polynomials of order r are defined by the generating function

$$\begin{array}{rl}{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r{e}^{xt}& =\underset{r\text{-times}}{\underbrace{\left(\frac{1-\lambda }{e^t-\lambda }\right)\times \cdots \times \left(\frac{1-\lambda }{e^t-\lambda }\right)}}{e}^{xt}\\ =\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!}(\text{see [5, 9])}.\end{array}$$

(1.5)

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. The n th Frobenius-Euler polynomials can be represented by (1.1) as follows:

$$\begin{array}{rl}\frac{\lambda H_n(x|\lambda )}{1-\lambda }& ={\int }_X{(x+y)}^{n}d{\mu }_{\lambda }(y)={\int }_{{\mathbb{Z}}_p}{(x+y)}^{n}d{\mu }_{\lambda }(y)\\ =\underset{N\to \mathrm{\infty }}{lim}\frac{1}{1-{\lambda }^{p^N}}\sum _{y=0}^{p^N-1}{(x+y)}^n{\lambda }^{p^N-y}(\text{see [6, 7])}.\end{array}$$

(1.6)

Let ℱ be the set of all formal power series in the variable t over ${\mathbb{C}}_p$ with

$$\mathcal{F}=\{f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{k!}{t}^k|a_k\in {\mathbb{C}}_p\}.$$

(1.7)

Let $\mathbb{P}={\mathbb{C}}_p[x]$ and ${\mathbb{P}}^{\ast }$ denote the vector space of all linear functionals on ℙ.

The formal power series

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{k!}{t}^k\in \mathcal{F}(\text{see [11, 15])}$$

(1.8)

defines a linear functional on ℙ by setting

$$〈f(t)|x^n〉=a_n\text{for all }n\ge 0.$$

(1.9)

From (1.8) and (1.9), we have

$$〈t^n|x^n〉=n!{\delta }_{n,k}(n,k\ge 0).$$

(1.10)

Here, ℱ denotes both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and so an element $f(t)$ of ℱ will be thought of as both a formal power series and a linear functional (see [11, 15]). We will call ℱ the umbral algebra. The umbral calculus is the study of umbral algebra (see [11, 15]).

The order $o(f(t))$ of power series $f(t)$ (≠0) is the smallest integer k for which $a_k$ does not vanish (see [11, 15]). A series $f(t)$ for which $o(f(t))=1$ is called a delta series. If a series $f(t)$ has $o(f(t))=0$, then $f(t)$ is called an invertible series (see [11, 15]). Let $f(t),g(t)\in \mathcal{F}$. Then we easily see that $〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉=〈g(t)|f(t)p(x)〉$. From (1.10), we note that

$$〈e^{yt}|x^n〉=y^n,〈e^{yt}|p(x)〉=p(y),$$

(1.11)

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f(t)|x^k〉}{k!}{t}^k,f(t)\in \mathcal{F},$$

(1.12)

and

$$p(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈t^k|p(x)〉}{k!}{x}^k,p(x)\in \mathbb{P}\text{ (see [15])}.$$

(1.13)

For $f_1(t),f_2(t),\dots ,f_m(t)\in \mathcal{F}$, we have

$$〈f_1(t)f_2(t)\cdots f_m(t)|x^n〉=\sum _{i_1+\cdots +i_m=n}\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_m}\right)〈f_1(t)|x^{i_1}〉\cdots 〈f_m(t)|x^{i_m}〉.$$

(1.14)

By (1.13), we get

$$p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}=\sum _{l=k}^n〈t^l|p(x)〉\left(\genfrac{}{}{0ex}{}{l}{k}\right)\frac{k!}{l!}{x}^{l-k}$$

(1.15)

and

$$p^{(k)}(0)=〈t^k|p(x)〉=〈1|p^{(k)}(x)〉.$$

Thus, by (1.15), we get

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}\text{(see [11, 15])}.$$

(1.16)

By (1.16), we easily see that

$$e^{yt}p(x)=p(x+y)\text{(see [15])}.$$

(1.17)

Let $S_n(x)$ denote a polynomial of degree n. Suppose that $f(t),g(t)\in \mathcal{F}$ with $o(f(t))=1$ and $o(g(t))=0$. Then there exists a unique sequence $S_n(x)$ of polynomials satisfying $〈g(t)f{(t)}^k|S_n(x)〉=n!{\delta }_{n,k}$ for all $n,k\ge 0$. The sequence $S_n(x)$ is called the Sheffer sequence for $(g(t),f(t))$, which is denoted by $S_n(x)\sim (g(t),f(t))$. If $S_n(x)\sim (g(t),t)$, then $S_n(x)$ is called the Appell sequence for $g(t)$ (see [15]).

For $p(x)\in \mathbb{P}$, we have

$$〈f(t)|xp(x)〉=〈{\partial }_{t}f(t)|p(x)〉=〈f^{\mathrm{\prime }}(t)|p(x)〉,$$

(1.18)

$$〈e^{yt}-1|p(x)〉=p(y)-p(0).$$

(1.19)

If $S_n(x)\sim (g(t),f(t))$, then we have

$$h(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈h(t)|S_k(x)〉}{k!}g(t)f{(t)}^k,h(t)\in \mathcal{F},$$

(1.20)

$$p(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈g(t)f{(t)}^k|p(x)〉}{k!}{S}_k(x),p(x)\in \mathbb{P},$$

(1.21)

$$f(t)S_n(x)=n{S}_{n-1}(x),$$

(1.22)

and

$$\frac{1}{g(\overline{f}(t))}{e}^{y\overline{f}(t)}=\sum _{k=0}^{\mathrm{\infty }}\frac{S_k(y)}{k!}{t}^k\text{for all }y\in {\mathbb{C}}_p,$$

(1.23)

where $\overline{f}(t)$ is compositional inverse of $f(t)$ (see [11, 15]). In [9], Kim and Kim have studied some identities of Frobenius-Euler polynomials arising from umbral calculus. In this paper, we study some p-adic Frobenius-Euler integral on related to umbral calculus in the p-adic case. Finally, we derive some new and interesting identities of Frobenius-Euler polynomials from our study.

2 Frobenius-Euler polynomials associated with umbral calculus

Let

$$g(t;\lambda )=\frac{e^t-\lambda }{1-\lambda }\in \mathcal{F}.$$

(2.1)

Then we see that $g(t;\lambda )$ is an invertible series. From (1.2), we have

$$\sum _{k=0}^{\mathrm{\infty }}{H}_k(x|\lambda )\frac{t^k}{k!}=\frac{1}{g(t;\lambda )}{e}^{xt}.$$

(2.2)

Hence, by (2.2), we get

$$\left(\frac{1-\lambda }{e^t-\lambda }\right)x^n=\frac{1}{g(t;\lambda )}{x}^n=H_n(x|\lambda ).$$

(2.3)

By (2.2) and (2.3), we get

$$H_n(x|\lambda )\sim (g(t;\lambda ),t).$$

From (1.6), we have

$${\int }_{{\mathbb{Z}}_p}{e}^{(x+y)t}d{\mu }_{\lambda }(y)=\frac{\lambda }{e^t-\lambda }{e}^{xt},$$

(2.4)

and

$${\int }_{{\mathbb{Z}}_p}{e}^{(x+y+1)t}d{\mu }_{\lambda }(y)-\lambda {\int }_{{\mathbb{Z}}_p}{e}^{(x+y)t}d{\mu }_{\lambda }(y)=\lambda e^{xt}.$$

(2.5)

By (2.5), we get

$${\int }_{{\mathbb{Z}}_p}{(x+y+1)}^{n}d{\mu }_{\lambda }(y)-\lambda {\int }_{{\mathbb{Z}}_p}{(x+y)}^{n}d{\mu }_{\lambda }(y)=\lambda x^n.$$

(2.6)

From (1.6) and (2.6), we have

$$\frac{\lambda }{1-\lambda }{H}_n(x+1|\lambda )-\frac{{\lambda }^2}{1-\lambda }{H}_n(x|\lambda )=\lambda x^n.$$

(2.7)

From (2.2), we can easily derive

$$H_{n+1}(x|\lambda )=(x-\frac{g^{\prime }(t;\lambda )}{g(t;\lambda )})H_n(x|\lambda ).$$

(2.8)

By (2.8), we get

$$g(t;\lambda )H_{n+1}(x|\lambda )=g(t;\lambda )x{H}_n(x|\lambda )-g^{\prime }(t;\lambda )H_n(x|\lambda ).$$

(2.9)

Thus, from (2.9), we have

$$(e^t-\lambda )H_{n+1}(x|\lambda )=(e^t-\lambda )x{H}_n(x|\lambda )-e^t{H}_n(x|\lambda ).$$

(2.10)

By (2.10), we get

$$H_{n+1}(x+1|\lambda )-\lambda H_{n+1}(x|\lambda )=x(H_n(x+1|\lambda )-\lambda H_n(x|\lambda )).$$

(2.11)

From (2.11), we note that

$$\begin{array}{rl}{H}_n(x+1|\lambda )-\lambda H_n(x|\lambda )& =x(H_{n-1}(x+1|\lambda )-\lambda H_{n-1}(x|\lambda ))\\ =x^2(H_{n-2}(x+1|\lambda )-\lambda H_{n-2}(x|\lambda ))=\cdots \\ =x^n(H_0(x+1|\lambda )-\lambda H_0(x|\lambda ))=x^n(1-\lambda ).\end{array}$$

(2.12)

Let us consider the linear functional $f(t)$ such that

$$〈f(t)|p(x)〉={\int }_{{\mathbb{Z}}_p}p(u)d{\mu }_{\lambda }(u)$$

(2.13)

for all polynomials $p(x)$ can be determined from (1.12) to be

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f(t)|x^k〉}{k!}{t}^k=\sum _{k=0}^{\mathrm{\infty }}{\int }_{{\mathbb{Z}}_p}{u}^{k}d{\mu }_{\lambda }(u)\frac{t^k}{k!}={\int }_{{\mathbb{Z}}_p}{e}^{ut}d{\mu }_{\lambda }(u).$$

(2.14)

By (2.4) and (2.14), we get

$$f(t)={\int }_{{\mathbb{Z}}_p}{e}^{ut}d{\mu }_{\lambda }(u)=\frac{\lambda }{e^t-\lambda }.$$

(2.15)

Therefore, by (2.15), we obtain the following theorem.

Theorem 2.1 For $p(x)\in \mathbb{P}$, we have

$$〈\frac{\lambda }{e^t-\lambda }|p(x)〉={\int }_{{\mathbb{Z}}_p}p(u)d{\mu }_{\lambda }(u).$$

In particular,

$$\frac{\lambda }{1-\lambda }{H}_n(\lambda )=〈{\int }_{{\mathbb{Z}}_p}{e}^{yt}d{\mu }_{\lambda }(y)|x^n〉.$$

From (1.6), we have

$$\sum _{n=0}^{\mathrm{\infty }}{\int }_{{\mathbb{Z}}_p}{(x+y)}^{n}d{\mu }_{\lambda }(y)\frac{t^n}{n!}={\int }_{{\mathbb{Z}}_p}{e}^{(x+y)t}d{\mu }_{\lambda }(y)=\sum _{n=0}^{\mathrm{\infty }}{\int }_{{\mathbb{Z}}_p}{e}^{yt}d{\mu }_{\lambda }(y)x^n\frac{t^n}{n!}.$$

(2.16)

By (1.6), (2.4) and (2.16), we get

$$\frac{\lambda }{1-\lambda }{H}_n(x|\lambda )={\int }_{{\mathbb{Z}}_p}{e}^{yt}d{\mu }_{\lambda }(y)x^n=\frac{\lambda }{e^t-\lambda }{x}^n,\text{for }n\ge 0.$$

(2.17)

Therefore, by (2.17), we obtain the following theorem.

Theorem 2.2 For $p(x)\in \mathbb{P}$, we have

$${\int }_{{\mathbb{Z}}_p}p(x+y)d{\mu }_{\lambda }(y)={\int }_{{\mathbb{Z}}_p}{e}^{yt}d{\mu }_{\lambda }(y)p(x)=\frac{\lambda }{e^t-\lambda }p(x).$$

In particular,

$$\frac{\lambda }{1-\lambda }{H}_n(x|\lambda )={\int }_{{\mathbb{Z}}_p}{e}^{yt}d{\mu }_{\lambda }(y)x^n=\frac{\lambda }{e^t-\lambda }{x}^n(n\ge 0).$$

By (1.6) and (2.16), we get

$$\frac{\lambda }{1-\lambda }{H}_n(x|\lambda )\sim (\frac{e^t-\lambda }{\lambda },t).$$

(2.18)

From Appell identity and (2.18), we can derive the following identities:

$$H_n(x+y|\lambda )=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)H_k(x|\lambda )y^{n-k}.$$

(2.19)

Let

$$g^r(t;\lambda )={\left(\frac{e^t-\lambda }{\lambda }\right)}^r=\underset{r\text{-times}}{\underbrace{\left(\frac{e^t-\lambda }{\lambda }\right)\times \cdots \times \left(\frac{e^t-\lambda }{\lambda }\right)}}\in \mathcal{F}.$$

(2.20)

Then $g^r(t;\lambda )$ is an invertible functional in ℱ. By (1.5) and (2.20), we get

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

(2.21)

Thus, from (2.21), we have

$$\frac{1}{g^r(t;\lambda )}{x}^n={\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda ),$$

(2.22)

and

$${\left(\frac{\lambda }{1-\lambda }\right)}^{r}t{H}_n^{(r)}(x|\lambda )=\frac{n}{g^r(t;\lambda )}{x}^{n-1}=n{\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_{n-1}^{(r)}(x|\lambda ).$$

(2.23)

By (2.22) and (2.23), we see that

$${\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda )\sim (g^r(t;\lambda ),t).$$

(2.24)

From (2.4), we can derive the following identity:

$$\begin{array}{r}\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r+x)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)\\ ={\left(\frac{\lambda }{e^t-\lambda }\right)}^r{e}^{xt}={\left(\frac{\lambda }{1-\lambda }\right)}^r\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!}.\end{array}$$

(2.25)

By (1.10) and (2.25), we get

$$\begin{array}{r}{\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda )\\ =〈\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r+x)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)|x^n〉.\end{array}$$

(2.26)

From (1.14), we have

$$\begin{array}{r}〈\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)|x^n〉\\ =\sum _{n=i_1+\cdots +i_r}\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)〈{\int }_{{\mathbb{Z}}_p}{e}^{x_{1}t}d{\mu }_{\lambda }(x_1)|x^{i_1}〉\times \cdots \\ \times 〈{\int }_{{\mathbb{Z}}_p}{e}^{x_{r}t}d{\mu }_{\lambda }(x_r)|x^{i_r}〉\\ =\sum _{n=i_1+\cdots +i_r}\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right){\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_{i_1}(x|\lambda )\cdots H_{i_r}(x|\lambda ).\end{array}$$

(2.27)

By (2.26) and (2.27), we get

$$H_n^{(r)}(x|\lambda )=\sum _{n=i_1+\cdots +i_r}\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)H_{i_1}(x|\lambda )\cdots H_{i_r}(x|\lambda ),$$

where $\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)=\frac{n!}{i_1!\cdots i_r!}$. From (2.25), we note that

$$g^r(t;\lambda )=\frac{1}{\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)}={\left(\frac{e^t-\lambda }{\lambda }\right)}^r.$$

(2.28)

Thus, by (2.28), we get

$$\begin{array}{rl}\frac{1}{g^r(t;\lambda )}{e}^{xt}& ={\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)e^{xt}\\ ={\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}{e}^{(x_1+x_2+\cdots +x_r+x)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)\\ ={\left(\frac{\lambda }{1-\lambda }\right)}^r\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!}.\end{array}$$

(2.29)

By (2.29), we see that

$$\begin{array}{r}{\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda )\\ =\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{(x+x_1+\cdots +x_r)}^{n}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)\\ =\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)x^n\\ =\frac{1}{g^r(t;\lambda )}{x}^n.\end{array}$$

(2.30)

Therefore, by (2.30), we obtain the following theorem.

Theorem 2.3 For $p(x)\in \mathbb{P}$ and $r\in \mathbb{N}$, we have

$$\underset{r\mathit{\text{-times}}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}p(x_1+x_2+\cdots +x_r+x)d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)={\left(\frac{\lambda }{e^t-\lambda }\right)}^{r}p(x).$$

In particular,

$${\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda )={\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)x^n.$$

Moreover,

$${\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda )\sim (\frac{1}{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)},t).$$

Let us consider the function $f^{\ast }(t)$ in ℱ such that

$$〈f^{\ast }(t)|p(x)〉={\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}p(x_1+x_2+\cdots +x_r)d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)$$

(2.31)

for all polynomials $p(x)$ can be determined from (1.12) to be

$$\begin{array}{rl}{f}^{\ast }(t)& =\sum _{k=0}^{\mathrm{\infty }}\frac{〈f^{\ast }(t)|x^k〉}{k!}{t}^k\\ =\sum _{k=0}^{\mathrm{\infty }}\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{(x_1+\cdots +x_r)}^{k}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)\frac{t^k}{k!}\\ =\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r).\end{array}$$

(2.32)

Therefore, by (2.31) and (2.32), we obtain the following theorem.

Theorem 2.4 For $p(x)\in \mathbb{P}$, we have

$$\begin{array}{r}〈\underset{r\mathit{\text{-times}}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)|p(x)〉\\ =\underset{r\mathit{\text{-times}}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}p(x_1+x_2+\cdots +x_r)d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r).\end{array}$$

In particular,

$$〈{\left(\frac{\lambda }{e^t-\lambda }\right)}^r|p(x)〉=\underset{r\mathit{\text{-times}}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}p(x_1+x_2+\cdots +x_r)d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r).$$

Indeed, the n th Frobenius-Euler number of order r is given by

$${\left(\frac{\lambda }{1-\lambda }\right)}^r{H}_n^{(r)}(x|\lambda )=〈\underset{r\text{-times}}{\underbrace{{\int }_{{\mathbb{Z}}_p}\cdots {\int }_{{\mathbb{Z}}_p}}}{e}^{(x_1+x_2+\cdots +x_r)t}d{\mu }_{\lambda }(x_1)\cdots d{\mu }_{\lambda }(x_r)|x^n〉,$$

where $n\ge 0$.

Remark From (1.2) and (1.5), we note that

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

(2.33)

and

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

(2.34)

Continuing this process, we obtain the following equation:

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

(2.35)

By (1.2), (1.5) and (2.35), 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.