Barnes-type Narumi polynomials

Kim, Dae San; Kim, Taekyun

doi:10.1186/1687-1847-2014-182

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Published: 22 July 2014

Barnes-type Narumi polynomials

Dae San Kim¹ &
Taekyun Kim²

Advances in Difference Equations volume 2014, Article number: 182 (2014) Cite this article

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Abstract

In this paper, we study the Barnes-type Narumi polynomials with umbral calculus viewpoint. From our study, we derive various identities of the Barnes-type Narumi polynomials.

MSC:05A19, 05A40, 11B68.

1 Introduction

As is well known, the Narumi polynomials of order α are defined by the generating function to be

{(\frac{t}{log (1 + t)})}^{α} {(1 + t)}^{x} = \sum_{n = 0}^{\infty} N_{n}^{(α)} (x) \frac{t^{n}}{n!} (see [1]) .

(1)

Let $r \in Z_{> 0}$ . We consider the polynomials $N_{n} (x | a_{1}, \dots, a_{r})$ and ${\hat{N}}_{n} (x | a_{1}, \dots, a_{r})$ , respectively, called the Barnes-type Narumi polynomials of the first kind and those of the second kind and respectively given by

\prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(1 + t)}^{x} = \sum_{n = 0}^{\infty} N_{n} (x | a_{1}, \dots, a_{r}) \frac{t^{n}}{n!}

(2)

and

\prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t) {(1 + t)}^{a_{j}}}) {(1 + t)}^{x} = \sum_{n = 0}^{\infty} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) \frac{t^{n}}{n!},

(3)

where $a_{1}, a_{2}, \dots, a_{r} \neq 0$ .

When $x = 0$ ,

N_{n} (a_{1}, \dots, a_{r}) = N_{n} (0 | a_{1}, \dots, a_{r})

and

{\hat{N}}_{n} (a_{1}, \dots, a_{r}) = {\hat{N}}_{n} (0 | a_{1}, \dots, a_{r})

are respectively called the Barnes-type Narumi numbers of the first kind and those of the second kind.

Note that

\begin{matrix} N_{n} (x | \underset{r}{\underset{⏟}{1, \dots, 1}}) = N_{n}^{(r)} (x), \\ {\hat{N}}_{n} (x | \underset{r}{\underset{⏟}{1, \dots, 1}}) = {\hat{N}}_{n}^{(r)} (x) \end{matrix}

and

{\hat{N}}_{n} (x | \underset{r}{\underset{⏟}{1, \dots, 1}}) = N_{n}^{(r)} (x - r) .

In the previous paper [2], $N_{n}^{(α)} (x)$ was denoted by $N_{n}^{(- α)}$ and called the Narumi polynomial of order α.

The Bernoulli polynomials are defined by the generating function to be

\frac{t}{e^{t} - 1} e^{x t} = \sum_{n = 0}^{\infty} B_{n} (x) \frac{t^{n}}{n!} (see [3–6]) .

(4)

When $x = 0$ , $B_{n} = B_{n} (0)$ are called the Bernoulli numbers. In [7], it is known that the Cauchy numbers are given by

\frac{t}{log (1 + t)} = \sum_{n = 0}^{\infty} C_{n} \frac{t^{n}}{n!} .

(5)

It is well known that the Stirling number of the first kind is given by

{(x)}_{n} = x (x - 1) \dots (x - n + 1) = \sum_{l = 0}^{\infty} S_{1} (n, l) x^{l} (n \geq 0) (see [1, 2, 7–11]) .

(6)

From (6), we have

{(log (1 + t))}^{n} = n! \sum_{l = n}^{\infty} S_{1} (l, n) \frac{t^{l}}{l!} (n \geq 0) .

(7)

Let ℂ be the complex number field and let ℱ be the set of all formal power series in the variable t:

F = {f (t) = \sum_{k = 0}^{\infty} a_{k} \frac{t^{k}}{k!} | a_{k} \in C} .

Let $P = C [x]$ and let $P^{*}$ be the vector space of all linear functionals on ℙ. $〈 L | p (x) 〉$ denotes the action of the linear functional L on $p (x)$ which satisfies $〈 L + M | p (x) 〉 = 〈 L | p (x) 〉 + 〈 M | p (x) 〉$ , and $〈 c L | p (x) 〉 = c 〈 L | p (x) 〉$ , where c is a complex constant. The linear functional $〈 f (t) | \cdot 〉$ on ℙ is defined by $〈 f (t) | x^{n} 〉 = a_{n}$ ( $n \geq 0$ ), where $f (t) \in F$ . Thus, we note that

〈 t^{k} | x^{n} 〉 = n! δ_{n, k} (n, k \geq 0),

(8)

where $δ_{n, k}$ is the Kronecker symbol (see [12–18]).

For $f_{L} (t) = \sum_{k = 0}^{\infty} \frac{〈 L | x^{k} 〉}{k!} t^{k}$ , we have $〈 f_{L} (t) | x^{n} 〉 = 〈 L | x^{n} 〉$ . So, the map $L \mapsto f_{L} (t)$ is a vector space isomorphism from $P^{*}$ onto ℱ. Henceforth, 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. We call ℱ the umbral algebra. The order $o (f (t))$ of a power series $f (t) \neq 0$ is the smallest integer for which the coefficient of $t^{k}$ does not vanish. If $o (f (t)) = 1$ , then $f (t)$ is called a delta series; if $o (f (t)) = 0$ , then $g (t)$ is called an invertible series. Let $f (t), g (t) \in F$ with $o (f (t)) = 1$ and $o (g (t)) = 0$ . Then there exists a unique sequence $s_{n} (x)$ ( $deg s_{n} (x) = n$ ) such that $〈 g (t) f {(t)}^{k} | s_{n} (x) 〉 = n! δ_{n, k}$ for $n, k \geq 0$ . The sequence $s_{n} (x)$ is called the Sheffer sequence for $(g (t), f (t))$ which is denoted by $s_{n} (x) \sim (g (t), f (t))$ .

For $f (t), g (t) \in F$ and $p (x) \in P$ , we have

〈 f (t) g (t) | p (x) 〉 = 〈 f (t) | g (t) p (x) 〉 = 〈 g (t) | f (t) p (x) 〉

(9)

and

\begin{aligned} f (t) = \sum_{k = 0}^{\infty} 〈 f (t) | x^{k} 〉 \frac{t^{k}}{k!}, \\ p (x) = \sum_{k = 0}^{\infty} 〈 t^{k} | p (x) 〉 \frac{x^{k}}{k!} . \end{aligned}

(10)

From (10), we can derive the following equation (11):

t^{k} p (x) = p^{(k)} (x) = \frac{d^{k} p (x)}{d x^{k}}, e^{y t} p (x) = p (x + y) (see [1]) .

(11)

Let $s_{n} (x) \sim (g (t), f (t))$ . Then the following will be used:

\frac{d s_{n} (x)}{d x} = \sum_{l = 0}^{n - 1} (\binom{n}{l}) 〈 \bar{f} (t) | x^{n - l} 〉 s_{l} (x),

(12)

where $\bar{f} (t)$ is the compositional inverse of $f (t)$ with $\bar{f} (f (t)) = f (\bar{f} (t)) = t$ ,

\frac{1}{g (\bar{f} (t))} e^{x \bar{f} (t)} = \sum_{n = 0}^{\infty} s_{n} (x) \frac{t^{n}}{n!} for all x \in C,

(13)

f (t) s_{n} (x) = n s_{n - 1} (x) (n \geq 1), s_{n} (x) = \sum_{j = 0}^{n} \frac{〈 g {(\bar{f} (t))}^{- 1} \bar{f} {(t)}^{j} | x^{n} 〉}{j!} x^{j},

(14)

s_{n} (x + y) = \sum_{j = 0}^{n} (\binom{n}{j}) s_{j} (x) p_{n - j} (y), where p_{n} (x) = g (t) s_{n} (x),

(15)

〈 f (t) | x p (x) 〉 = 〈 \partial_{t} f (t) | p (x) 〉, where \partial_{t} f (t) = \frac{d f (t)}{d t}

(16)

and

s_{n + 1} (x) = (x - \frac{g^{'} (t)}{g (t)}) \frac{1}{f^{'} (t)} s_{n} (x) (n \geq 0) (see [1, 19]) .

(17)

Let us assume that $s_{n} (x) \sim (g (t), f (t))$ and $r_{n} (x) \sim (h (t), l (t))$ . Then we have

s_{n} (x) = \sum_{m = 0}^{n} c_{n, m} r_{m} (x) (n \geq 0),

(18)

where

c_{n, m} = \frac{1}{m!} 〈 \frac{h (\bar{f} (t))}{g (\bar{f} (t))} l {(\bar{f} (t))}^{m} | x^{n} 〉 (see [1, 5]) .

(19)

From (2), (3) and (13), we note that

N_{n} (x | a_{1}, \dots, a_{r}) \sim (\prod_{j = 1}^{r} (\frac{t}{e^{a_{j} t} - 1}), e^{t} - 1)

(20)

and

{\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) \sim (\prod_{j = 1}^{r} (\frac{t e^{a_{j} t}}{e^{a_{j} t} - 1}), e^{t} - 1) .

(21)

In this paper, we study the Barnes-type Narumi polynomials with umbral calculus viewpoint. From our study, we derive various identities of the Barnes-type Narumi polynomials.

2 Barnes-type Narumi polynomials

From (21), we note that

\prod_{j = 1}^{r} (\frac{t}{e^{a_{j} t} - 1}) N_{n} (x | a_{1}, \dots, a_{r}) \sim (1, e^{t} - 1)

(22)

and

{(x)}_{n} \sim (1, e^{t} - 1) .

(23)

Thus, by (22) and (23), we get

\begin{aligned} N_{n} (x | a_{1}, \dots, a_{r}) & = \prod_{j = 1}^{r} (\frac{e^{a_{j} t} - 1}{t}) {(x)}_{n} \\ = \sum_{m = 0}^{n} S_{1} (n, m) \prod_{j = 1}^{r} (\frac{e^{a_{j} t} - 1}{t}) x^{m} . \end{aligned}

(24)

Note that

\begin{aligned} \prod_{j = 1}^{r} (\frac{e^{a_{j} t} - 1}{t}) & = (\sum_{l_{1} = 0}^{\infty} \frac{a_{1}^{l_{1} + 1}}{(l_{1} + 1)!} t^{l_{1}}) \times \dots \times (\sum_{l_{r} = 0}^{\infty} \frac{a_{r}^{l_{r} + 1}}{(l_{r} + 1)!} t^{l_{r}}) \\ = \sum_{l_{1}, \dots, l_{r} = 0}^{\infty} \sum_{l_{1} + \dots + l_{r} = i} \frac{a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}}{(l_{1} + 1)! \dots (l_{r} + 1)!} t^{i} . \end{aligned}

(25)

From (24) and (25), we have

\begin{array}{rcl} N_{n} (x | a_{1}, \dots, a_{r}) & = & \sum_{m = 0}^{\infty} S_{1} (n, m) \sum_{i = 0}^{m} \sum_{l_{1} + \dots + l_{r} = i} \frac{a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}}{(l_{1} + 1)! \dots (l_{r} + 1)!} t^{i} x^{m} \\ = & \sum_{m = 0}^{n} S_{1} (n, m) \sum_{i = 0}^{m} \frac{i!}{(i + r)!} \sum_{l_{1} + \dots + l_{r} = i} (\binom{i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) \\ \times a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1} x^{m - i} \\ = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} S_{1} (n, m) \frac{(m - i)!}{(m - i + r)!} \\ \times (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} x^{i} . \end{array}

(26)

By (21), we see that

\prod_{j = 1}^{r} (\frac{t e^{a_{j} t}}{e^{a_{j} t} - 1}) {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) \sim (1, e^{t} - 1),

(27)

and we recall (23).

Thus, we have

\begin{array}{rcl} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) & = & \prod_{j = 1}^{r} (\frac{e^{a_{j} t} - 1}{t e^{a_{j} t}}) {(x)}_{n} = e^{- \sum_{j = 1}^{r} a_{j} t} \prod_{j = 1}^{r} (\frac{e^{a_{j} t} - 1}{t}) {(x)}_{n} \\ = & e^{- \sum_{j = 1}^{r} a_{j} t} N_{n} (x | a_{1}, \dots, a_{r}) \\ = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} S_{1} (n, m) \frac{(m - i)!}{(m - i + r)!} \\ \times (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} e^{- \sum_{j = 1}^{r} a_{j} t} x^{i} \\ = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} S_{1} (n, m) \frac{(m - i)!}{(m - i + r)!} \\ \times (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} {(x - \sum_{j = 1}^{r} a_{j})}^{i} \end{array}

(28)

or

\begin{array}{rcl} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) & = & \prod_{j = 1}^{r} (\frac{e^{a_{j} t} - 1}{t e^{a_{j} t}}) {(x)}_{n} = \prod_{j = 1}^{r} (\frac{e^{- a_{j} t} - 1}{- t}) {(x)}_{n} \\ = & \sum_{m = 0}^{n} S_{1} (n, m) \sum_{i = 0}^{m} {(- 1)}^{i} \frac{i!}{(i + r)!} \\ \times \sum_{l_{1} + \dots + l_{r} = i} (\binom{i + r}{i_{1} + 1, \dots, i_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1} x^{m - i} \\ = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} {(- 1)}^{m - i} S_{1} (n, m) \frac{(m - i)!}{(m - i + r)!} \\ \times (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} x^{i} . \end{array}

(29)

Therefore, by (26), (28) and (29), we obtain the following theorem.

Theorem 1 For $n \geq 0$ , we have

\begin{array}{rcl} N_{n} (x | a_{1}, \dots, a_{r}) & = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} S_{1} (n, m) \frac{(m - i)!}{(m - i + r)!} \\ \times (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} x^{i} \end{array}

and

\begin{array}{rcl} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) & = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} S_{1} (n, m) \frac{(m - i)!}{(m - i + r)!} \\ \times (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{i}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} {(x - \sum_{j = 1}^{r} a_{j})}^{i} \\ = & \sum_{i = 0}^{n} {\sum_{m = i}^{n} \sum_{l_{1} + \dots + l_{r} = m - i} {(- 1)}^{m - i} S_{1} (n, m) \\ \times \frac{(m - i)!}{(m - i + r)!} (\binom{m - i + r}{l_{1} + 1, \dots, l_{r} + 1}) (\binom{m}{j}) a_{1}^{l_{1} + 1} \dots a_{r}^{l_{r} + 1}} x^{i} . \end{array}

From (14), we can derive the following equation (30):

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{j = 0}^{n} \frac{1}{j!} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{j} | x^{n} 〉 x^{j},

(30)

where

\begin{matrix} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{j} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | {(log (1 + t))}^{j} x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | j! \sum_{l = j}^{\infty} S_{1} (l, j) \frac{t^{l}}{l!} x^{n} 〉 \\ = j! \sum_{l = j}^{n} (\binom{n}{l}) S_{1} (l, j) N_{n - l} (a_{1}, \dots, a_{r}) . \end{matrix}

(31)

Thus, by (30) and (31), we obtain the following theorem.

Theorem 2 For $n \geq 0$ , we have

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{j = 0}^{n} {\sum_{l = j}^{n} (\binom{n}{l}) S_{1} (l, j) N_{n - l} (a_{1}, \dots, a_{r})} x^{j} .

By the same methods as in (28), (29) and (30), we get

\begin{matrix} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) \\ = \sum_{j = 0}^{n} {\sum_{l = j}^{n} (\binom{n}{l}) S_{1} (l, j) N_{n - l} (a_{1}, \dots, a_{r})} {(x - \sum_{i = 1}^{r} a_{i})}^{j} \\ = \sum_{j = 0}^{n} {\sum_{l = j}^{n} (\binom{n}{l}) S_{1} (l, j) {\hat{N}}_{n - l} (a_{1}, \dots, a_{r})} x^{j} . \end{matrix}

(32)

By (8), we get

\begin{aligned} N_{n} (y | a_{1}, \dots, a_{r}) & = 〈 \sum_{i = 0}^{\infty} N_{i} (y | a_{1}, \dots, a_{r}) \frac{t^{i}}{i!} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(1 + t)}^{y} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | \sum_{m = 0}^{\infty} {(y)}_{m} \frac{t^{m}}{m!} x^{m} 〉 \\ = \sum_{m = 0}^{n} {(y)}_{m} (\binom{n}{m}) 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | x^{n - m} 〉 \\ = \sum_{m = 0}^{n} {(y)}_{m} (\binom{n}{m}) N_{n - m} (a_{1}, \dots, a_{r}) \end{aligned}

(33)

and

\begin{aligned} {\hat{N}}_{n} (y | a_{1}, \dots, a_{r}) & = 〈 \sum_{i = 0}^{\infty} {\hat{N}}_{i} (y | a_{1}, \dots, a_{r}) \frac{t^{i}}{i!} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t) {(1 + t)}^{a_{j}}}) {(1 + t)}^{y} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t) {(1 + t)}^{a_{j}}}) | \sum_{m = 0}^{\infty} {(y)}_{m} \frac{t^{m}}{m!} x^{n} 〉 \\ = \sum_{m = 0}^{n} {(y)}_{m} (\binom{n}{m}) 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t) {(1 + t)}^{a_{j}}}) | x^{n - m} 〉 \\ = \sum_{m = 0}^{n} (\binom{n}{m}) {(y)}_{m} {\hat{N}}_{n - m} (a_{1}, \dots, a_{r}) . \end{aligned}

(34)

Therefore, by (33) and (34), we obtain the following theorem.

Theorem 3 For $n \geq 0$ , we have

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} (\binom{n}{m}) N_{n - m} (a_{1}, \dots, a_{r}) {(x)}_{m}

and

{\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} (\binom{n}{m}) {\hat{N}}_{n - m} (a_{1}, \dots, a_{r}) {(x)}_{m} .

From (15), we note that

N_{n} (x + y | a_{1}, \dots, a_{r}) = \sum_{j = 0}^{n} (\binom{n}{j}) N_{j} (x | a_{1}, \dots, a_{r}) {(y)}_{n - j}

(35)

and

{\hat{N}}_{n} (x + y | a_{1}, \dots, a_{r}) = \sum_{j = 0}^{n} (\binom{n}{j}) {\hat{N}}_{j} (x | a_{1}, \dots, a_{r}) {(y)}_{n - j} .

(36)

By (14), we get

(e^{t} - 1) N_{n} (x | a_{1}, \dots, a_{r}) = n N_{n - 1} (x | a_{1}, \dots, a_{r})

(37)

and

\begin{aligned} (e^{t} - 1) N_{n} (x | a_{1}, \dots, a_{r}) & = e^{t} N_{n} (x | a_{1}, \dots, a_{r}) - N_{n} (x | a_{1}, \dots, a_{r}) \\ = N_{n} (x + 1 | a_{1}, \dots, a_{r}) - N (x | a_{1}, \dots, a_{r}) . \end{aligned}

(38)

From (37) and (38), we have

N_{n} (x + 1 | a_{1}, \dots, a_{r}) - N_{n} (x | a_{1}, \dots, a_{r}) = n N_{n - 1} (x | a_{1}, \dots, a_{r}) .

(39)

By the same method as (39), we get

{\hat{N}}_{n} (x + 1 | a_{1}, \dots, a_{r}) - {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) = n {\hat{N}}_{n - 1} (x | a_{1}, \dots, a_{r}) .

(40)

Recall that $N_{n} (x | a_{1}, \dots, a_{r}) \sim (\prod_{j = 1}^{r} (\frac{t}{e^{a_{j} t} - 1}), e^{t} - 1)$ .

From (17), we can derive the following equation (41):

N_{n + 1} (x | a_{1}, \dots, a_{r}) = x N_{n} (x - 1 | a_{1}, \dots, a_{r}) - e^{- t} \frac{g^{'} (t)}{g (t)} N_{n} (x | a_{1}, \dots, a_{r}) .

(41)

Now, we observe that

\begin{aligned} \frac{g^{'} (t)}{g (t)} & = {(log g (t))}^{'} = {(r log t - \sum_{j = 1}^{r} log (e^{a_{j} t} - 1))}^{'} = \frac{r}{t} - \sum_{j = 1}^{r} \frac{a_{j} e^{a_{j} t}}{e^{a_{j} t} - 1} \\ = \frac{\sum_{j = 1}^{r} \prod_{i \neq j}^{r} (e^{a_{i} t} - 1) {e^{a_{j} t} - 1 - t a_{j} e^{a_{j} t}}}{t \prod_{j = 1}^{r} (e^{a_{j} t} - 1)}, \end{aligned}

(42)

where

\begin{aligned} r - \sum_{j = 1}^{r} \frac{a_{j} t e^{a_{j} t}}{e^{a_{j} t} - 1} & = \frac{\sum_{j = 1}^{r} \prod_{i \neq j} (e^{a_{i} t} - 1) {e^{a_{j} t} - 1 - a_{j} t e^{a_{j} t}}}{\prod_{j = 1}^{r} (e^{a_{j} t} - 1)} \\ = \frac{- \frac{1}{2} (\sum_{j = 1}^{r} a_{1} a_{2} \dots a_{j - 1} a_{j}^{2} a_{j + 1} \dots a_{r}) t^{r + 1} + \dots}{(a_{1} a_{2} \dots a_{r}) t^{r}} \\ = - \frac{1}{2} (\sum_{j = 1}^{r} a_{j}) t + \dots \end{aligned}

(43)

has at least the order 1.

By (42) and (43), we get

\begin{matrix} \frac{g^{'} (t)}{g (t)} N_{n} (x | a_{1}, \dots, a_{r}) \\ = \frac{r - \sum_{j = 1}^{r} \frac{a_{j} t e^{a_{j} t}}{e^{a_{j} t} - 1}}{t} (\sum_{i = 0}^{n} {\sum_{l = i}^{n} (\binom{n}{l}) S_{1} (l, i) N_{n - l} (a_{1}, \dots, a_{r})} x^{i}) \\ = \sum_{i = 0}^{n} {\sum_{l = i}^{n} (\binom{n}{l}) S_{1} (l, i) N_{n - l} (a_{1}, \dots, a_{r})} \frac{r - \sum_{j = 1}^{r} \frac{a_{j} t e^{a_{j} t}}{e^{a_{j} t} - 1}}{t} x^{i} \\ = \sum_{i = 0}^{n} {\sum_{l = i}^{n} (\binom{n}{l}) S_{1} (l, i) N_{n - l} (a_{1}, \dots, a_{r})} (r - \sum_{j = 1}^{r} \frac{a_{j} t e^{a_{j} t}}{e^{a_{j} t} - 1}) \frac{x^{i + 1}}{i + 1} \\ = \sum_{i = 0}^{n} \frac{1}{i + 1} {\sum_{l = i}^{n} (\binom{n}{l}) S_{1} (l, i) N_{n - l} (a_{1}, \dots, a_{r})} (r x^{i + 1} - \sum_{j = 1}^{r} \sum_{m = 0}^{\infty} B_{m} \frac{{(- a_{j} t)}^{m}}{m!} x^{i + 1}) \\ = - \sum_{i = 0}^{n} \frac{1}{i + 1} {\sum_{l = i}^{n} (\binom{n}{l}) S_{1} (l, i) N_{n - l} (a_{1}, \dots, a_{r})} \\ \times \sum_{j = 1}^{r} \sum_{m = 1}^{i + 1} {(- 1)}^{m} (\binom{i + 1}{m}) B_{m} a_{j}^{m} x^{i + 1 - m} \\ = - \sum_{i = 0}^{n} \frac{1}{i + 1} {\sum_{l = i}^{n} (\binom{n}{l}) S_{1} (l, i) N_{n - l} (a_{1}, \dots, a_{r})} \\ \times \sum_{j = 1}^{r} \sum_{m = 0}^{i} {(- 1)}^{i + 1 - m} (\binom{i + 1}{m}) a_{j}^{i + 1 - m} B_{i + 1 - m} x^{m} . \end{matrix}

(44)

Therefore, by (41) and (44), we obtain the following theorem.

Theorem 4 For $n \geq 0$ , we have

\begin{aligned} N_{n + 1} (x | a_{1}, \dots, a_{r}) \\ = x N_{n} (x - 1 | a_{1}, \dots, a_{r}) + \sum_{m = 0}^{n} {\sum_{i = m}^{n} \sum_{l = i}^{n} \sum_{j = 1}^{r} \frac{1}{i + 1} (\binom{n}{l}) (\binom{i + 1}{m}) S_{1} (l, i) \\ \times B_{i + 1 - m} {(- a_{j})}^{i + 1 - m} N_{n - l} (a_{1}, \dots, a_{r})} {(x - 1)}^{m} . \end{aligned}

By the same method as the proof of Theorem 4, we get

\begin{array}{rcl} {\hat{N}}_{n + 1} (x | a_{1}, \dots, a_{r}) & = & (x - \sum_{j = 1}^{r} a_{j}) {\hat{N}}_{n} (x - 1 | a_{1}, \dots, a_{r}) \\ + \sum_{m = 0}^{n} {\sum_{i = m}^{n} \sum_{l = i}^{n} \sum_{j = 1}^{r} \frac{1}{i + 1} (\binom{n}{l}) (\binom{i + 1}{m}) S_{1} (l, i) \\ \times B_{i + 1 - m} {(- a_{j})}^{i + 1 - m} {\hat{N}}_{n - l} (a_{1}, \dots, a_{r})} {(x - 1)}^{m} . \end{array}

(45)

From (12) and (20), we can derive the following equation (46):

〈 \bar{f} (t) | x^{n - l} 〉 = 〈 log (1 + t) | x^{n - l} 〉 = 〈 \sum_{m = 1}^{\infty} \frac{{(- 1)}^{m - 1}}{m} t^{m} | x^{n - l} 〉 = {(- 1)}^{n - l - 1} (n - l - 1)! .

(46)

Thus, by (46), we get

\begin{aligned} \frac{d}{d x} N_{n} (x | a_{1}, \dots, a_{r}) & = \sum_{l = 0}^{n - 1} (\binom{n}{l}) {(- 1)}^{n - l - 1} (n - l - 1)! N_{l} (x | a_{1}, \dots, a_{r}) \\ = n! \sum_{l = 0}^{n - 1} \frac{{(- 1)}^{n - l - 1}}{l! (n - l)} N_{l} (x | a_{1}, \dots, a_{r}) . \end{aligned}

(47)

By the same method as (47), we get

\frac{d}{d x} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) = n! \sum_{l = 0}^{n - 1} \frac{{(- 1)}^{n - l - 1}}{l! (n - l)} {\hat{N}}_{l} (x | a_{1}, \dots, a_{r}) .

(48)

From (8), we note that, for $n \geq 1$ ,

\begin{matrix} N_{n} (y | a_{1}, \dots, a_{r}) \\ = 〈 \sum_{i = 0}^{\infty} N_{i} (y | a_{1}, \dots, a_{r}) \frac{t^{i}}{i!} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(1 + t)}^{y} | x^{n} 〉 \\ = 〈 \partial_{t} (\prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(1 + t)}^{y}) | x^{n - 1} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) (\partial_{t} {(1 + t)}^{y}) | x^{n - 1} 〉 \\ + 〈 (\partial_{t} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)})) {(1 + t)}^{y} | x^{n - 1} 〉 \\ = y N_{n - 1} (y - 1 | a_{1}, \dots, a_{r}) + 〈 (\partial_{t} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)})) {(1 + t)}^{y} | x^{n - 1} 〉 . \end{matrix}

(49)

Now, we observe that

\begin{matrix} \partial_{t} \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) \\ = \sum_{j = 1}^{r} \prod_{i \neq j} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) \frac{a_{j} {(1 + t)}^{a_{j} - 1} log (1 + t) - ({(1 + t)}^{a_{j}} - 1) \frac{1}{1 + t}}{{(log (1 + t))}^{2}} \\ = \frac{1}{1 + t} \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) \sum_{j = 1}^{r} {\frac{a_{j} {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{1}{log (1 + t)}} \\ = \frac{1}{1 + t} \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) \frac{\sum_{j = 1}^{r} {\frac{a_{j} t {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{t}{log (1 + t)}}}{t}, \end{matrix}

(50)

where

\sum_{j = 1}^{r} {\frac{a_{j} t {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{t}{log (1 + t)}} = \frac{1}{2} (\sum_{j = 1}^{r} a_{j}) t + \dots

(51)

is a series with order greater than or equal to 1.

By (50) and (51), we get

\begin{matrix} 〈 (\partial_{t} \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)})) {(1 + t)}^{y} | x^{n - 1} 〉 \\ = 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) \frac{{(1 + t)}^{y}}{1 + t} | \frac{\sum_{j = 1}^{r} {\frac{a_{j} t {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{t}{log (1 + t)}}}{t} x^{n - 1} 〉 \\ = \frac{1}{n} 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{y - 1} | \sum_{j = 1}^{r} {\frac{a_{j} t {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{t}{log (1 + t)}} x^{n} 〉 \\ = \frac{1}{n} {\sum_{j = 1}^{r} a_{j} 〈 \frac{log (1 + t)}{{(1 + t)}^{a_{j}} - 1} \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{y + a_{j} - 1} | \frac{t}{log (1 + t)} x^{n} 〉 \\ - r 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{y - 1} | \frac{t}{log (1 + t)} x^{n} 〉} \\ = \frac{1}{n} \sum_{j = 1}^{r} a_{j} \sum_{l = 0}^{n} (\binom{n}{l}) C_{l} 〈 \frac{log (1 + t)}{{(1 + t)}^{a_{j}} - 1} \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{y + a_{j} - 1} | x^{n - l} 〉 \\ - \frac{r}{n} \sum_{l = 0}^{n} (\binom{n}{l}) C_{l} 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{y - 1} | x^{n - l} 〉 \\ = \frac{1}{n} \sum_{j = 1}^{r} \sum_{l = 0}^{n} (\binom{n}{l}) a_{j} C_{l} N_{n - l} (y + a_{j} - 1 | a_{1}, \dots, {\hat{a}}_{j}, \dots, a_{r}) \\ - \frac{r}{n} \sum_{l = 0}^{n} (\binom{n}{l}) C_{l} N_{n - l} (y - 1 | a_{1}, \dots, a_{r}), \end{matrix}

(52)

where ${\hat{a}}_{j}$ means that $a_{j}$ is omitted.

Therefore, by (49) and (52), we obtain the following theorem.

Theorem 5 For $n \geq 1$ , we have

\begin{array}{rcl} N_{n} (x | a_{1}, \dots, a_{r}) & = & x N_{n - 1} (x - 1 | a_{1}, \dots, a_{r}) \\ + \frac{1}{n} \sum_{j = 1}^{r} \sum_{l = 0}^{n} (\binom{n}{l}) a_{j} C_{l} N_{n - l} (x + a_{j} - 1 | a_{1}, \dots, {\hat{a}}_{j}, \dots, a_{r}) \\ - \frac{r}{n} \sum_{l = 0}^{r} (\binom{n}{l}) C_{l} N_{n - l} (x - 1 | a_{1}, \dots, a_{r}), \end{array}

where $C_{n}$ are the Cauchy numbers with the generating function given by

\frac{t}{log (1 + t)} = \sum_{n = 0}^{\infty} C_{n} \frac{t^{n}}{n!} .

By the same method as the proof of Theorem 5, we get

\begin{array}{rcl} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) & = & (x - \sum_{j = 1}^{r} a_{j}) {\hat{N}}_{n - 1} (x - 1 | a_{1}, \dots, a_{r}) \\ - \frac{r}{n} \sum_{l = 0}^{n} (\binom{n}{l}) C_{l} {\hat{N}}_{n - l} (x - 1 | a_{1}, \dots, a_{r}) \\ - \frac{1}{n} \sum_{j = 1}^{r} \sum_{l = 0}^{n} (\binom{n}{l}) a_{j} C_{l} {\hat{N}}_{n - l} (x - 1 | a_{1}, \dots, {\hat{a}}_{j}, \dots, a_{r}) . \end{array}

(53)

Now we compute the following formula (54) in two different ways:

〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | x^{n} 〉 .

(54)

On the one hand,

\begin{matrix} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | x^{n} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | m! \sum_{l = m}^{\infty} S_{1} (l, m) \frac{t^{l}}{l!} x^{n} 〉 \\ = m! \sum_{l = m}^{n} (\binom{n}{l}) S_{1} (l, m) 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | x^{n - l} 〉 \\ = m! \sum_{l = m}^{n} (\binom{n}{l}) S_{1} (l, m) N_{n - l} (a_{1}, \dots, a_{r}) \\ = m! \sum_{l = 0}^{n - m} (\binom{n}{l}) S_{1} (n - l, m) N_{l} (a_{1}, \dots, a_{r}) . \end{matrix}

(55)

On the other hand,

\begin{matrix} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | x^{n} 〉 \\ = 〈 \partial_{t} (\prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m}) | x^{n - 1} 〉 \\ = 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) (\partial_{t} ({(log (1 + t))}^{m})) | x^{n - 1} 〉 \\ + 〈 (\partial_{t} \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)})) {(log (1 + t))}^{m} | x^{n - 1} 〉 . \end{matrix}

(56)

Note that

\begin{matrix} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) (\partial_{t} ({(log (1 + t))}^{m})) | x^{n - 1} 〉 \\ = m 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) \frac{1}{1 + t} | log {(1 + t)}^{m - 1} x^{n - 1} 〉 \\ = m 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(1 + t)}^{- 1} | (m - 1)! \sum_{l = m - 1}^{\infty} S_{1} (l, m - 1) \frac{t^{l}}{l!} x^{n - 1} 〉 \\ = m! \sum_{l = m - 1}^{n - 1} (\binom{n - 1}{l}) S_{1} (l, m - 1) 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(1 + t)}^{- 1} | x^{n - 1 - l} 〉 \\ = m! \sum_{l = m - 1}^{n - 1} (\binom{n - 1}{l}) S_{1} (l, m - 1) N_{n - 1 - l} (- 1 | a_{1}, \dots, a_{r}) \\ = m! \sum_{l = 0}^{n - m} (\binom{n - 1}{l}) S_{1} (n - 1 - l, m - 1) N_{l} (- 1 | a_{1}, \dots, a_{r}) \end{matrix}

(57)

and

\begin{matrix} 〈 (\partial_{t} \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)})) {(log (1 + t))}^{m} | x^{n - 1} 〉 \\ = 〈 (\partial_{t} \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)})) | m! \sum_{l = m}^{\infty} S_{1} (l, m) \frac{t^{l}}{l!} x^{n - 1} 〉 \\ = m! \sum_{l = m}^{n - 1} (\binom{n - 1}{l}) S_{1} (l, m) \\ \times 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{- 1} | \frac{\sum_{j = 1}^{r} {\frac{a_{j} t {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{t}{log (1 + t)}}}{t} x^{n - 1 - l} 〉 \\ = m! \sum_{l = m}^{n - 1} (\binom{n - 1}{l}) \frac{S_{1} (l, m)}{n - l} \\ \times 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{- 1} | \sum_{j = 1}^{r} {\frac{a_{j} t {(1 + t)}^{a_{j}}}{{(1 + t)}^{a_{j}} - 1} - \frac{t}{log (1 + t)}} x^{n - l} 〉 \\ = \frac{m!}{n} \sum_{l = m}^{n - 1} (\binom{n}{l}) S_{1} (l, m) \\ \times {\sum_{j = 1}^{r} a_{j} 〈 \frac{log (1 + t)}{{(1 + t)}^{a_{j}} - 1} \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{a_{j} - 1} | \frac{t}{log (1 + t)} x^{n - l} 〉 \\ - r 〈 \prod_{i = 1}^{r} (\frac{{(1 + t)}^{a_{i}} - 1}{log (1 + t)}) {(1 + t)}^{- 1} | \frac{t}{log (1 + t)} x^{n - l} 〉} \\ = \frac{m!}{n} \sum_{l = m}^{n - 1} (\binom{n}{l}) S_{1} (l, m) {\sum_{j = 1}^{r} a_{j} \sum_{k = 0}^{n - l} C_{k} (\binom{n - l}{k}) N_{n - l - k} (a_{j} - 1 | a_{1}, \dots, {\hat{a}}_{j}, \dots, a_{r}) \\ - r \sum_{k = 0}^{n - l} C_{k} (\binom{n - l}{k}) N_{n - l - k} (- 1 | a_{1}, \dots, a_{r})} . \end{matrix}

(58)

Therefore, by (55), (56), (57) and (58), we obtain the following theorem.

Theorem 6 For $n - 1 \geq m \geq 1$ , we have

\begin{matrix} \sum_{l = 0}^{n - m} (\binom{n}{l}) S_{1} (n - l, m) N_{l} (a_{1}, \dots, a_{r}) \\ = \sum_{l = 0}^{n - m} (\binom{n - 1}{l}) S_{1} (n - l - 1, m - 1) N_{l} (- 1 | a_{1}, \dots, a_{r}) \\ + \frac{1}{n} \sum_{l = m}^{n - 1} \sum_{k = 0}^{n - l} \sum_{j = 1}^{r} (\binom{n}{l}) (\binom{n - l}{k}) a_{j} C_{n - l - k} S_{1} (l, m) N_{k} (a_{j} - 1 | a_{1}, \dots, {\hat{a}}_{j}, \dots, a_{r}) \\ - \frac{r}{n} \sum_{l = m}^{n - 1} \sum_{k = 0}^{n - l} (\binom{n}{l}) (\binom{n - l}{k}) C_{n - l - k} S_{1} (l, m) N_{k} (- 1 | a_{1}, \dots, a_{r}) . \end{matrix}

By the same method as the proof of Theorem 6, we get

\begin{matrix} \sum_{l = 0}^{n - m} (\binom{n}{l}) S_{1} (n - l, m) {\hat{N}}_{l} (a_{1}, \dots, a_{r}) \\ = \sum_{l = 0}^{n - m} (\binom{n - 1}{l}) S_{1} (n - l - 1, m - 1) {\hat{N}}_{l} (- 1 | a_{1}, \dots, a_{r}) \\ + \frac{1}{n} \sum_{j = 1}^{r} \sum_{l = m}^{n - 1} \sum_{k = 0}^{n - l} (\binom{n}{l}) (\binom{n - l}{k}) a_{j} C_{n - l - k} S_{1} (l, m) {\hat{N}}_{k} (- 1 | a_{1}, \dots, {\hat{a}}_{j}, \dots, a_{r}) \\ - \frac{r}{n} \sum_{l = m}^{n - 1} \sum_{k = 0}^{n - l} (\binom{n}{l}) (\binom{n - l}{k}) C_{n - l - k} S_{1} (l, m) {\hat{N}}_{k} (- 1 | a_{1}, \dots, a_{r}) \\ - \sum_{j = 1}^{r} a_{j} \sum_{l = 0}^{n - m - 1} (\binom{n - 1}{l}) S_{1} (n - l - 1, m) {\hat{N}}_{k} (- 1 | a_{1}, \dots, a_{r}), \end{matrix}

(59)

where $n - 1 \geq m \geq 1$ .

Let us consider the following two Sheffer sequences:

N_{n} (x | a_{1}, \dots, a_{r}) \sim (\prod_{j = 1}^{r} (\frac{t}{e^{a_{j} t} - 1}), e^{t} - 1)

(60)

and (23).

We let

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} C_{n, m} {(x)}_{m} .

(61)

From (18) and (19), we note that

\begin{aligned} C_{n, m} & = \frac{1}{m!} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) t^{m} | x^{n} 〉 \\ = (\binom{n}{m}) 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) | x^{n - m} 〉 \\ = (\binom{n}{m}) N_{n - m} (a_{1}, \dots, a_{r}) . \end{aligned}

(62)

Therefore, by (61) and (62), we obtain the following theorem.

Theorem 7 For $n \geq 0$ , we have

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} (\binom{n}{m}) N_{n - m} (a_{1}, \dots, a_{r}) {(x)}_{m} .

By the same method as the proof of Theorem 7, we get

{\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} (\binom{n}{m}) {\hat{N}}_{n - m} (a_{1}, \dots, a_{r}) {(x)}_{m} .

(63)

For

N_{n} (x | a_{1}, \dots, a_{r}) \sim (\prod_{j = 1}^{r} (\frac{t}{e^{a_{j} t} - 1}), e^{t} - 1)

and

H_{n}^{(s)} (x | λ) \sim ({(\frac{e^{t} - λ}{1 - λ})}^{s}, t), λ \in C with λ \neq 1,

let us assume that

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} C_{n, m} H_{m}^{(s)} (x | λ),

(64)

where $H_{m}^{(s)} (x | λ)$ are the Frobenius-Euler polynomials of order s defined by the generating function as

{(\frac{1 - λ}{e^{t} - λ})}^{s} e^{x t} = \sum_{n = 0}^{\infty} H_{n}^{(s)} (x | λ) \frac{t^{n}}{n!} .

From (18) and (19), we note that

\begin{array}{rcl} C_{n, m} & = & \frac{1}{m! {(1 - λ)}^{s}} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} {(1 - λ + t)}^{s} | x^{n} 〉 \\ = & \frac{1}{m! {(1 - λ)}^{s}} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | \sum_{j = 0}^{min {s, n}} (\binom{s}{j}) {(1 - λ)}^{s - j} t^{j} x^{n} 〉 \\ = & \frac{1}{m! {(1 - λ)}^{s}} \sum_{j = 0}^{n - m} (\binom{s}{j}) {(1 - λ)}^{s - j} {(n)}_{j} \\ \times 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1_{t})}) {(log (1 + t))}^{m} | x^{n - j} 〉 \\ = & \frac{1}{m! {(1 - λ)}^{s}} \sum_{j = 0}^{n - m} (\binom{s}{j}) {(1 - λ)}^{s - j} {(n)}_{j} m! \\ \times \sum_{l = 0}^{n - j - m} (\binom{n - j}{l}) S_{1} (n - j - l, m) N_{l} (a_{1}, \dots, a_{r}) \\ = & \sum_{j = 0}^{n - m} \sum_{l = 0}^{n - m - j} (\binom{s}{j}) (\binom{n - j}{l}) \\ \times {(n)}_{j} {(1 - λ)}^{- j} S_{1} (n - j - l, m) N_{l} (a_{1}, \dots, a_{r}) . \end{array}

(65)

Therefore, by (64) and (65), we obtain the following theorem.

Theorem 8 For $n \geq 0$ , we have

\begin{matrix} N_{n} (x | a_{1}, \dots, a_{r}) \\ = \sum_{m = 0}^{n} {\sum_{j = 0}^{n - m} \sum_{l = 0}^{n - m - j} (\binom{s}{j}) (\binom{n - j}{l}) {(n)}_{j} {(1 - λ)}^{- j} \\ \times S_{1} (n - j - l, m) N_{l} (a_{1}, \dots, a_{r})} H_{m}^{(s)} (x | λ) . \end{matrix}

By the same method as the proof of Theorem 8, we get

\begin{aligned} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) \\ = \sum_{m = 0}^{n} {\sum_{j = 0}^{n - m} \sum_{l = 0}^{n - m - j} (\binom{s}{j}) (\binom{n - j}{l}) {(n)}_{j} \\ \times {(1 - λ)}^{- j} S_{1} (n - j - l, m) {\hat{N}}_{l} (a_{1}, \dots, a_{r})} H_{m}^{(s)} (x | λ) . \end{aligned}

(66)

Now, we consider the following two Sheffer sequences:

N_{n} (x | a_{1}, \dots, a_{r}) \sim (\prod_{j = 1}^{r} (\frac{t}{e^{a_{j} t} - 1}), e^{t} - 1)

(67)

and

B_{n}^{(s)} (x) \sim ({(\frac{e^{t} - 1}{t})}^{s}, e^{t} - 1),

where $B_{n}^{(s)} (x)$ are the Bernoulli polynomials of order s given by the generating function as

{(\frac{t}{e^{t} - 1})}^{s} e^{x t} = \sum_{n = 0}^{\infty} B_{n}^{(s)} (x) \frac{t^{n}}{n!} .

Let us assume that

N_{n} (x | a_{1}, \dots, a_{r}) = \sum_{m = 0}^{n} C_{n, m} B_{m}^{(s)} (x) .

(68)

By (18) and (19), we get

\begin{aligned} C_{n, m} & = \frac{1}{m!} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} {(\frac{t}{log (1 + t)})}^{s} | x^{n} 〉 \\ = \frac{1}{m!} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | {(\frac{t}{log (1 + t)})}^{s} x^{n} 〉 \\ = \frac{1}{m!} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | \sum_{k = 0}^{\infty} C_{k}^{(s)} \frac{t^{k}}{k!} x^{n} 〉 \\ = \frac{1}{m!} \sum_{k = 0}^{n - m} (\binom{n}{k}) C_{k}^{(s)} 〈 \prod_{j = 1}^{r} (\frac{{(1 + t)}^{a_{j}} - 1}{log (1 + t)}) {(log (1 + t))}^{m} | x^{n - k} 〉 \\ = \frac{1}{m!} \sum_{k = 0}^{n - m} (\binom{n}{k}) C_{k}^{(s)} m! \sum_{l = 0}^{n - m - k} (\binom{n - k}{l}) S_{1} (n - l - k, m) N_{l} (a_{1}, \dots, a_{r}) \\ = \sum_{k = 0}^{n - m} \sum_{l = 0}^{n - m - k} (\binom{n}{k}) (\binom{n - k}{l}) C_{k}^{(s)} S_{1} (n - k - l, m) N_{l} (a_{1}, \dots, a_{r}), \end{aligned}

(69)

where $C_{k}^{(s)}$ are the Cauchy numbers of the first kind of order s defined by the generating function as

{(\frac{t}{log (1 + t)})}^{s} = \sum_{n = 0}^{\infty} C_{n}^{(s)} \frac{t^{n}}{n!} .

Therefore, by (68) and (69), we obtain the following theorem.

Theorem 9 For $n \geq 0$ , we have

\begin{array}{rcl} N_{n} (x | a_{1}, \dots, a_{r}) & = & \sum_{m = 0}^{n} {\sum_{k = 0}^{n - m} \sum_{l = 0}^{n - m - k} (\binom{n}{k}) (\binom{n - k}{l}) C_{k}^{(s)} \\ \times S_{1} (n - k - l, m) N_{l} (a_{1}, \dots, a_{r})} B_{m}^{(s)} (x) . \end{array}

By the same method as the proof of Theorem 9, we get

\begin{array}{rcl} {\hat{N}}_{n} (x | a_{1}, \dots, a_{r}) & = & \sum_{m = 0}^{n} {\sum_{k = 0}^{n - m} \sum_{l = 0}^{n - m - k} (\binom{n}{k}) (\binom{n - k}{l}) C_{k}^{(s)} \\ \times S_{1} (n - k - l, m) {\hat{N}}_{l} (a_{1}, \dots, a_{r})} B_{m}^{(s)} (x) . \end{array}

Recently, several authors have studied umbral calculus (see [1–5, 7–18, 20]).

References

Roman S: The Umbral Calculus. Dover, New York; 2005.
Google Scholar
Kim T, Kim DS: Some identities involving associated sequences of special polynomials. J. Comput. Anal. Appl. 2014, 16: 626-642.
MathSciNet Google Scholar
Kim DS, Kim T, Lee S-H, Rim S-H: Umbral calculus and Euler polynomials. Ars Comb. 2013, 112: 293-306.
MathSciNet Google Scholar
Kim DS, Kim T: Some identities of Bernoulli and Euler polynomials arising from umbral calculus. Adv. Stud. Contemp. Math. (Kyungshang) 2013, 23(1):159-171.
MathSciNet Google Scholar
Kim DS, Kim T, Dolgy DV, Rim S-H: Some new identities of Bernoulli, Euler and Hermite polynomials arising from umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 73
Google Scholar
Kim T, Mansour T, Rim S-H, Lee S-H: Apostol-Euler polynomials arising from umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 301
Google Scholar
Kim DS, Kim T: Higher-order Cauchy of the first kind and poly-Cauchy of the first kind mixed type polynomials. Adv. Stud. Contemp. Math. (Kyungshang) 2013, 23(21):621-636.
MathSciNet Google Scholar
Kim T, Kim DS, Mansour T, Rim S-H, Schork M: Umbral calculus and Sheffer sequences of polynomials. J. Math. Phys. 2013., 54(8): Article ID 083504
Google Scholar
Kim T: Identities involving Laguerre polynomials derived from umbral calculus. Russ. J. Math. Phys. 2014, 21(1):36-45. 10.1134/S1061920814010038
Article MathSciNet Google Scholar
Lu D-Q, Xiang Q-M, Luo C-H: Some results for Apostol-type polynomials associated with umbral algebra. Adv. Differ. Equ. 2013., 2013: Article ID 201
Google Scholar
Maldonado M, Prada J, Senosiain MJ: Appell bases on sequence spaces. J. Nonlinear Math. Phys. 2011, 18: 189-194. suppl. 1 10.1142/S1402925111001362
Article MathSciNet Google Scholar
Araci S, Kong X, Acikgoz M, Şen E: A new approach to multivariate q -Euler polynomials using the umbral calculus. J. Integer Seq. 2014., 17(1): Article ID 14.1.2
Google Scholar
Dere R, Simsek Y: Applications of umbral algebra to some special polynomials. Adv. Stud. Contemp. Math. (Kyungshang) 2012, 22(3):433-438.
MathSciNet Google Scholar
Ernst T: Examples of a q -umbral calculus. Adv. Stud. Contemp. Math. (Kyungshang) 2008, 16(1):1-22.
MathSciNet Google Scholar
Fang Q, Wang T: Umbral calculus and invariant sequences. Ars Comb. 2011, 101: 257-264.
MathSciNet Google Scholar
Gessel IM: Applications of the classical umbral calculus. Algebra Univers. 2003, 49(4):397-434. Dedicated to the memory of Gian-Carlo Rota 10.1007/s00012-003-1813-5
Article MathSciNet Google Scholar
Kim DS, Kim T: Some identities of Frobenius-Euler polynomials arising from umbral calculus. Adv. Differ. Equ. 2012., 2012: Article ID 196
Google Scholar
Kim DS, Kim T, Ryoo CS: Sheffer sequences for the powers of Sheffer pairs under umbral composition. Adv. Stud. Contemp. Math. (Kyungshang) 2013, 23(2):275-285.
MathSciNet Google Scholar
Ryoo CS, Song H, Agarwal RP: On the roots of the q -analogue of Euler-Barnes’ polynomials. Adv. Stud. Contemp. Math. 2004, 9(2):153-163.
MathSciNet Google Scholar
Taylor BD: Umbral presentations for polynomial sequences. Comput. Math. Appl. 2001, 41(9):1085-1098. Umbral calculus and its applications (Cortona, 1998) 10.1016/S0898-1221(01)00083-9
Article MathSciNet Google Scholar

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Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MOE) (No. 2012R1A1A2003786). The authors would like to thank the referees for their valuable comments.

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Department of Mathematics, Sogang University, Seoul, 121-742, Republic of Korea
Dae San Kim
Department of Mathematics, Kwangwoon University, Seoul, 139-701, Republic of Korea
Taekyun Kim

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Kim, D.S., Kim, T. Barnes-type Narumi polynomials. Adv Differ Equ 2014, 182 (2014). https://doi.org/10.1186/1687-1847-2014-182

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Received: 30 April 2014
Accepted: 09 June 2014
Published: 22 July 2014
DOI: https://doi.org/10.1186/1687-1847-2014-182

Barnes-type Narumi polynomials

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1 Introduction

2 Barnes-type Narumi polynomials

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