Symbolic Methods Applied to a Class of Identities Involving Appell Polynomials and Stirling Numbers

Abstract

In this paper, we present two symbolic methods, in particular, the method starting from the source identity, umbra identity, for constructing identities of s-Appell polynomials related to Stirling numbers and binomial coefficients. We discuss some properties of s-Appell polynomial sequences related to Riordan arrays, Sheffer matrices, and their q analogs.

1. Introduction

In the early 17th century, Johann Faulhaber discovered some remarkable representations of sums of powers $1^m$, $2^m,\dots$, and $n^m$ as polynomials in $N=(n^2+n)/2$, and more generally, r-fold summations of powers as polynomials in $n(n+r)$. Rigorous proofs of these formulas were provided by Jacobi in 1834. In Knuth [1], the notions of r-reflective and anti-r-reflective functions are introduced and used to explain Faulhaber’s results. As an amusement, a 360-year-old riddle of Faulhaber is solved. Edwards [2] provides an analogue for matrix forms of the algorithm of Johann Faulhaber from 1631.

Faulhaber’s formula, named after Johann Faulhaber, expresses the sum of the k-th powers of the first n positive integers ${\sum }_{\mathsf{\ell }=1}^n{\mathsf{\ell }}^k$ as a $(k+1)$-th degree polynomial function of n, the coefficients involving Bernoulli numbers $B_j$, in the following form:

$$\sum _{\mathsf{\ell }=1}^n{\mathsf{\ell }}^k={\displaystyle \frac{1}{k+1}}\sum _{\mathsf{\ell }=0}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{\mathsf{\ell }}}\right)B_{\mathsf{\ell }}{n}^{k+1-\mathsf{\ell }},$$

(1)

where we use the Bernoulli number of the second kind $B_1=1/2$. If we use the Bernoulli number of the first kind, $B_1=-1/2$, noting all Bernoulli numbers with an odd index $k\ge 3$ are zero, then Formula (1) deduces the following identity:

$$\begin{array}{cc}\hfill \sum _{\mathsf{\ell }=1}^n{\mathsf{\ell }}^k=& {\displaystyle \frac{1}{k+1}}\sum _{\mathsf{\ell }=0}^k{(-1)}^{\mathsf{\ell }}{B}_{\mathsf{\ell }}\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{\mathsf{\ell }}}\right)n^{k+1-\mathsf{\ell }}\hfill \\ \hfill =& {\displaystyle \frac{1}{k+1}}\left(B_{k+1}(n+1)-B_{k+1}\left(1\right)\right),\hfill \end{array}$$

where $B_1=-1/2$. Shiue and one of the authors [3] applied Faulhaber’s formula for the study of the divisibility of sums of powers of integers.

We can recall that determining combinatorial identities is a very famous branch of combinatorics. In this paper, we present two symbolic methods, in particular, the method for constructing identities starting from source identities (umbra identities), some of which involve Bernoulli numbers and other famous numbers.

It is known that symbolic operations $\mathrm{\Delta }$ (difference), E (displacement), and D (derivative) play an important role in the calculus of finite differences, as well as in certain topics of computational methods. For various classical results, see Jordan [4], Milne-Thomson [5], etc. Certainly, the theoretical basis of symbolic methods could be found within the theory of formal power series, inasmuch as all the symbolic expressions treated are expressible as power series in $\mathrm{\Delta }$, E, or D, and all the operations employed are just the same as those applied to formal power series. For some easily accessible references on formal series, we recommend Bourbaki [6], Comtet [7], Gould [8], Graham, Knuth, and Partashnik [9], and Wilf [10].

Throughout this paper, the theory of formal power series and differential operators will be utilized. $\mathbb{R}$, $\mathbb{C}$, $\mathbb{N}$, ${\mathbb{N}}_0$, and $\mathbb{Z}$ denote, respectively, the sets of real numbers, complex numbers, natural numbers, natural numbers including 0, and integers. Generally, we will use $A\left(t\right), g\left(t\right), f\left(t\right), \phi \left(t\right),$ etc., to denote either the formal power series (fps) in $\mathbb{K}\left[\right[t\left]\right]$, the ring of formal power series in the real field when $\mathbb{K}=\mathbb{R}$, the complex field when $\mathbb{K}=\mathbb{C}$, or the infinitely differentiable functions (members of $C^{\infty }$) defined in $\mathbb{R}$ or $\mathbb{C}.$

We will make use of the ordinary operators $\mathrm{\Delta }$ (difference), D (differentiation), and E (shift operator), which are defined by the following relations:

$$Ef\left(t\right)=f(t+1),\mathrm{\Delta }f\left(t\right)=f(t+1)-f\left(t\right),Df\left(t\right)={\displaystyle \frac{d}{dt}}f\left(t\right).$$

Powers of these operators are defined in the usual way when they are available. In particular, for any real number x, one may define $E^{x}f\left(t\right)=f(t+x)$. Also, the number 1 is defined as an identity operator, i.e., $1f\left(t\right)\equiv f\left(t\right)$. Let $f\in \mathbb{C}\left[t\right]$. Using Taylor’s theorem,

$$f(n+x)=\sum _{k\ge 0}{D}^{k}f\left(n\right){\displaystyle \frac{x^k}{k!}}.$$

Setting $x=1$ yields the following:

$$f(n+1)=\left(\sum _{k\ge 0}{\displaystyle \frac{D^k}{k!}}\right)f\left(n\right)={\mathrm{e}}^{D}f\left(n\right),$$

which implies $\mathrm{\Delta }f\left(n\right)=({\mathrm{e}}^D-1)f\left(n\right)$. Thus, we symbolically obtain the following:

$$E=1+\mathrm{\Delta }={\mathrm{e}}^D,\mathrm{\Delta }=E-1={\mathrm{e}}^D-1,\mathrm{and}D=\log (1+\mathrm{\Delta }).$$

Note that $E^{k}f\left(0\right)={\left[E^{k}f\left(t\right)\right]}_{t=0}=f\left(k\right)$ so that ${\left(xE\right)}^{k}f\left(0\right)=f\left(k\right)x^k$. This means that ${\left(xE\right)}^k$ with x as a parameter may be used to generate a general term of the series ${\sum }_{k=0}^{\infty }f\left(k\right)x^k$. In this paper, we focus on summations and identities arising from the interrelations of a number of operators in common use in combinatorics, number theory, and discrete mathematics.

• Example.

Let $B_n^{\left(a\right)}\left(t\right)$ be the generalized Bernoulli polynomial defined by the following (see [11] (p. 93)):

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

where we enjoy the property

$${\displaystyle \frac{d}{dx}}{B}_n^{\left(a\right)}\left(x\right)=n{B}_{n-1}^{\left(a\right)}\left(x\right)$$

for $a\in \mathbb{Z}$, and let $B_n^{\left(a\right)}=B_n^{\left(a\right)}\left(0\right)$. It is clear that $B_n^{\left(1\right)}\left(x\right)=B_n\left(x\right)$ and $B_n^{\left(1\right)}\left(0\right)=B_n$, which are the classical Bernoulli polynomial and classical Bernoulli number, respectively. Substitute $g\left(x\right)=B_n^{\left(a\right)}\left(x\right)$ into (36), as shown in Section 4, and note $g^{\left(k\right)}\left(0\right)=D_x^k{B}_n^{\left(a\right)}\left(0\right)={\left(n\right)}_k{B}_{n-k}^{\left(a\right)}$ and $g^{\left(k\right)}\left(x\right)={\left(n\right)}_k{B}_{n-k}^{\left(a\right)}\left(x\right)$, where ${\left(n\right)}_k=n(n-1)\cdots (n-k+1)$. Then, we have the following identity, combining generalized Bernoulli polynomials and numbers, the Stirling number of the second kind $\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{r}{k}}\right\}$, and binomial coefficients together as follows:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)B_{n-k}^{\left(a\right)}{k}^r{t}^k=\sum _{k=0}^r\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)B_{n-k}^{\left(a\right)}\left(t\right)k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{r}{k}}\right\}{t}^k.$$

(2)

The particular case $r=0$ provides the following well-known expression:

$$B_n^{\left(a\right)}\left(t\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)B_{n-k}^{\left(a\right)}{t}^k.$$

(3)

The generalized Bernoulli polynomial sequence is an Appell polynomial sequence. In mathematics, an Appell polynomial sequence, named after Paul Émile Appell, is any polynomial sequence ${\left\{p_n\left(x\right)\right\}}_{n\ge 0}$ satisfying the following identity:

$${\displaystyle \frac{d}{dx}}{p}_n\left(x\right)=n{p}_{n-1}\left(x\right),$$

in which $p_0\left(x\right)$ is a non-zero constant.

Among the most notable Appell sequences besides the trivial example $\left\{x^n\right\}$ are Hermite polynomials, Bernoulli polynomials, and Euler polynomials. Every Appell sequence is a Sheffer sequence, but most Sheffer sequences are not Appell sequences. Appell sequences have a probabilistic interpretation as systems of moments.

In this paper, we discuss a generalization of Appell sequences, called an s-Appell sequence ($s\ne 0$) and denoted by ${\left\{p_n(s;x)\right\}}_{n\ge 0}$, which is defined by the following:

$$p(x;t)=\sum _{n\ge 0}{p}_n(s;x){\displaystyle \frac{t^n}{n!}}=g\left(t\right){\mathrm{e}}^{sxt},$$

(4)

where $g\left(t\right)={\sum }_{n\ge 0}{g}_n{\displaystyle \frac{t^n}{n!}}$ is a given exponential series (cf. Munarini [12]). If $g_0\ne 0$, we have a proper sequence. So, $p_n(s;x)$ is a polynomial of degree n and

$$p_n(s;x)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)g_{n-k}{s}^k{x}^k.$$

(5)

When $s=1$, we obtain ordinary Appell polynomials $p_n\left(x\right)=p_n(1;x)$ (cf. Appell [13], Roman [11] (p. 86), and Roman and Rota [14]).

Recall that a Sheffer sequence is a polynomial sequence ${\left\{p_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ defined by the formal exponential series:

$$\sum _{n\ge 0}{p}_n\left(x\right){\displaystyle \frac{t^n}{n!}}=g\left(t\right){\mathrm{e}}^{xf\left(t\right)},$$

(6)

where $g\left(t\right)={\sum }_{n\ge 0}{g}_n{\displaystyle \frac{t^n}{n!}}$ is an exponential series with $g\left(0\right)=g_0=1$, and $f\left(t\right)={\sum }_{n\ge 0}{f}_n\dots$ is an exponential series with $f\left(0\right)=f_0=0$ and $f^{\prime }\left(0\right)=f_1\ne 0$. In particular, $s_n\left(0\right)=g_n$. Particularly, if $f\left(t\right)=st$ and $s\ne 0$, then $p_n\left(x\right)=p_n(s;x)$. A Sheffer sequence ${\left\{p_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ can be represented by its corresponding Riordan array, i.e., $\left[g\right(t),f(t\left)\right]$. More formally, let us consider the set of formal power series rings $\mathcal{F}=\mathbb{K}[[$t$\left]\right]$, where $\mathbb{K}$ is the field $\mathbb{R}$ or $\mathbb{C}$. The order of $f\left(t\right)\in \mathcal{F}$ and $f\left(t\right)={\sum }_{k=0}^{\infty }{f}_k{t}^k/k!$ ($f_k\in \mathbb{K}$) is the minimum number $r\in \mathbb{N}$, such that $f_r\ne 0$, where $\mathbb{N}=\{0,1,2\dots \}$ is a set of all natural numbers. We denote by ${\mathcal{F}}_r$ the set of formal power series of order r. Let $g\left(t\right)\in {\mathcal{F}}_0$ and $f\left(t\right)\in {\mathcal{F}}_1$; the pair $(g,f)$ defines the (proper) Riordan array$R={\left[d_{n,k}\right]}_{n,k\in \mathbb{N}}=[g,f]$ having

$$d_{n,k}=\left[{\displaystyle \frac{t^n}{n!}}\right]g\left(t\right){\displaystyle \frac{f{\left(t\right)}^k}{k!}}$$

(7)

or, in other words, having $g{f}^k$ as the generating function of the kth column of $[g,f]$. Hence,

$$[g,f]:=(g,gf,g{f}^2,g{f}^3,\dots ),$$

(8)

where g, $gf$, $g{f}^2$, and $g{f}^3\cdots$ are the generating functions of the 0th, 1st, 2nd, 3rd, ⋯ columns of the matrix $[g,f]$, respectively. With the matrix multiplication, the set of all exponential Riordan arrays $\mathcal{R}$ forms a group, called the Riordan group (see Shapiro, Sprugnoli, Barry, Cheon, He, Merlini, and Wang [15] (Ch. 6)).

From (6) and (7),

$$\begin{array}{cc}\hfill & \sum _{n\ge 0}{p}_n\left(x\right){\displaystyle \frac{t^n}{n!}}=g\left(t\right){\mathrm{e}}^{xf\left(t\right)}=\sum _{k\ge 0}g\left(t\right){\displaystyle \frac{f{\left(t\right)}^k}{k!}}{x}^k\hfill \\ \hfill =& \sum _{k\ge 0}\sum _{n\ge 0}{d}_{n,k}{x}^k{\displaystyle \frac{t^n}{n!}}=\sum _{n\ge 0}\left(\sum _{k=0}^n{d}_{n,k}{x}^k\right){\displaystyle \frac{t^n}{n!}},\hfill \end{array}$$

which implies

$$p_n\left(x\right)=\sum _{k=0}^n{d}_{n,k}{x}^k,$$

(9)

or equivalently,

$$p_{n,k}=\left[x^k\right]p_n\left(x\right)=d_{n,k}$$

(10)

for $p_n\left(x\right)={\sum }_{k=0}^n{p}_{n,k}{x}^k$.

Let $[g\left(t\right),f\left(t\right)]={\left[d_{n,k}\right]}_{n,k\in bN}$ be an exponential Riordan array, and let $h\left(t\right)={\sum }_{k\ge 0}{h}_k{t}^k/k!$ be the generating function of the sequence ${\left(h_n\right)}_{n\in \mathbb{N}}$. Then, we have the first fundamental theorem of Riordan arrays:

$$\sum _{k=0}^n{d}_{n,k}{h}_k=\left[{\displaystyle \frac{t^n}{n!}}\right]g\left(t\right)h\left(f\left(t\right)\right),$$

which can be abbreviated as follows:

$$\left[g\right(t),f(t\left)\right]h\left(t\right)=g\left(t\right)(h\circ f)\left(t\right),$$

or simplified to $[g,f]h=gh\left(f\right)$. Thus, we immediately see that the usual row-by-column product of two Riordan arrays is also a Riordan array:

$$[g_1,f_1][g_2,f_2]=[g_1{g}_2\left(f_1\right),f_2\left(f_1\right)].$$

(11)

The Riordan array $I=[1,t]$ is the identity matrix because its entries are $d_{n,k}=[t^n/n!]t^k/k!={\delta }_{n,k}$.

Let $[g\left(t\right),f\left(t\right)]$ be a Riordan array. Then, its inverse is the following:

$${[g\left(t\right),f\left(t\right)]}^{-1}=\left[{\displaystyle \frac{1}{g\left(\overline{f}\left(t\right)\right)}},\overline{f}\left(t\right),\right]$$

(12)

where $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$, i.e., $(f\circ \overline{f})\left(t\right)=(\overline{f}\circ f)\left(t\right)=t$. In this way, the set $\mathcal{R}$ of all proper Riordan arrays forms a group (see [16]) called the Riordan group. Hence, the set $\mathcal{L}$ of all Sheffer sequences forms a group with the product defined below called the Sheffer group. If $p_n\left(x\right)$ and $q_n\left(x\right)$ are the Sheffer sequences with Riordan matrices $[g_1,f_1]={\left[d_{n,k}\right]}_{n,k\in \mathbb{N}}$ and $[g_2,f_2]={\left[c_{n,k}\right]}_{n,k\in \mathbb{N}}$, respectively, then the operation, called the product of ${\left\{p_n\right\}}_{n\in \mathbb{N}}$ and ${\left\{q_n\right\}}_{n\in \mathbb{N}}$ and denoted by $p_n\left(x\right)\odot q_n\left(x\right)$, is defined by the following:

$$\sum _{n\ge 0}{p}_n\left(x\right)\odot q_n\left(x\right){\displaystyle \frac{t^n}{n!}}=g_1\left(t\right)g_2\left(f_1\left(t\right)\right){\mathrm{e}}^{x{f}_2\left(f_1\left(t\right)\right)}.$$

(13)

Denote

$$[g_1\left(t\right)g_2\left(f_1\left(t\right)\right),f_2\left(f_1\left(t\right)\right)]={\left[r_{n,k}\right]}_{n,k\in \mathbb{N}},$$

where $r_{n,k}={\sum }_{\mathsf{\ell }=k}^n{d}_{n,\mathsf{\ell }}{c}_{\mathsf{\ell },k}$ using (11). Thus,

$$g_1\left(t\right)g_2\left(f_1\left(t\right)\right){\mathrm{e}}^{x{f}_2\left(f_1\left(t\right)\right)}=\sum _{n\ge 0}{r}_n\left(x\right){\displaystyle \frac{t^n}{n!}},$$

where $r_n\left(x\right)={\sum }_{k=0}^n{r}_{n,k}{x}^k$. Compared to (13), we obtain the following:

$$r_n\left(x\right)=p_n\left(x\right)\odot q_n\left(x\right)$$

(14)

for $p_n\left(x\right)={\sum }_{k=0}^n{p}_{n,k}{x}^k$, $q_n\left(x\right)={\sum }_{k=0}^n{q}_{n,k}{x}^k$, and $r_n\left(x\right)={\sum }_{k=0}^n{r}_{n,k}{x}^k$. In addition, (14) holds if and only if $r_{n,k}={\sum }_{\mathsf{\ell }=k}^n{d}_{n,\mathsf{\ell }}{c}_{\mathsf{\ell },k}$, i.e., ${\left[r_{n,k}\right]}_{n,k\in \mathbb{N}}={\left[d_{n,k}\right]}_{n,k\in \mathbb{N}}{\left[c_{n,k}\right]}_{n,k\in \mathbb{N}}$, which establishes an isomorphism between the Riordan group and the Sheffer group (cf. Hsu, Shiue, and one of the authors [17]). Hence, we use ${\left\{p_n\left(x\right)\right\}}_{n\in \mathbb{N}}\cong [g\left(t\right),f\left(t\right)]$ to represent the isomorphism between ${\left\{p_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ and $\left[g\right(t),f(t\left)\right]$ if ${\sum }_{n\ge 0}{p}_n\left(x\right)t^n/n!=g\left(t\right){\mathrm{e}}^{xf\left(t\right)}$.

An infinite lower triangular matrix ${\left[d_{n,k}\right]}_{n,k\in \mathbb{N}}$ is a Riordan array if and only if a unique sequence $A=(a_0\ne 0,a_1,a_2,\dots )$ exists such that for every $n,k\in \mathbb{N},$

$$\begin{array}{cc}\hfill d_{n+1,k+1}=& {\displaystyle \frac{c_{n+1}{c}_k}{c_n{c}_{k+1}}}{a}_0{d}_{n,k}+{\displaystyle \frac{c_{n+1}{c}_{k+1}}{c_n{c}_{k+1}}}{a}_1{d}_{n,k+1}+\cdots +{\displaystyle \frac{c_{n+1}{c}_n}{c_n{c}_{k+1}}}{a}_n{d}_{n,n}\hfill \end{array}$$

(15)

$$\begin{array}{c}=\sum _{j\ge 0}{\displaystyle \frac{c_{n+1}{c}_{k+j}}{c_n{c}_{k+1}}}{a}_j{d}_{n,k+j}.\hfill \end{array}$$

(16)

This is equivalent to

$$f\left(t\right)=tA\left(f\left(t\right)\right)\mathrm{or}t=\overline{f}\left(t\right)A\left(t\right).$$

(17)

Here, $A\left(t\right)$ is the generating function of the A-sequence. The first formula of (17) is also called the second fundamental theorem of Riordan arrays.

Moreover, there exists a unique sequence $Z=(z_0,z_1,z_2,\dots )$ such that every element in column 0 can be expressed as a linear combination:

$$\begin{array}{cc}\hfill d_{n+1,0}=& {\displaystyle \frac{c_{n+1}{c}_0}{c_n{c}_0}}{z}_0{d}_{n,0}+{\displaystyle \frac{c_{n+1}{c}_1}{c_n{c}_0}}{z}_1{d}_{n,1}+\cdots +{\displaystyle \frac{c_{n+1}{c}_n}{c_n{c}_0}}{z}_n{d}_{n,n},\hfill \end{array}$$

(18)

$$\begin{array}{l}=\sum _{j\ge 0}{\displaystyle \frac{c_{n+1}{c}_j}{c_n{c}_0}}{z}_j{d}_{n,j},\hfill \end{array}$$

(19)

or equivalently,

$$g\left(t\right)={\displaystyle \frac{1}{1-tZ\left(f\right(t\left)\right)}},$$

(20)

in which we always assume $g\left(0\right)=g_0=1$, a usual hypothesis for proper Riordan arrays. From (20), we may obtain the following:

$$Z\left(t\right)={\displaystyle \frac{g\left(\overline{f}\left(t\right)\right)-1}{\overline{f}\left(t\right)g\left(\overline{f}\left(t\right)\right)}}.$$

A- and Z-sequence characterizations of Riordan arrays were introduced, developed, and/or studied in Merlini, Rogers, Sprugnoli, and Verri [18], Roger [19], Sprugnoli and one of the authors [20], Jean-Louis and Nkwanta [21], Luzón, Morón, Prieto-Martinez [22,23], and [24], as well as their references.

In the next section, we provide some properties of s-Appell sequences and construct their identities, including the umbral identity, using the resource identity (36). In Section 3, the umbral identity found in Section 2 is used to obtain a kind of identities related to Riordan and Sheffer-type matrices. The q-analogues of the above identities will also be presented.

s-Appell sequences form Sheffer sequences (cf. Sheffer [25], Roman [11], and Roman and Rota [14]).

Here is a list of six important subgroups of the Riordan group (see [15,16,26,27]). Six subgroups of the Sheffer group can be provided accordingly based on the isomorphism between the Sheffer group and the Riordan group (cf. [17]).

• the s-Appell subgroup $\{[g\left(t\right),st]:g\in {\mathcal{F}}_0\}$ for $s\ne 0$.
• the Lagrange (associated) subgroup $\{[1,f\left(t\right)]:f\in {\mathcal{F}}_1)\}$.
• the k-Bell subgroup $\{[g\left(t\right),t{\left(g\left(t\right)\right)}^k]:g\in {\mathcal{F}}_0\}$, where k is a fixed positive integer.
• the hitting-time subgroup $\{[t{f}^{\prime }\left(t\right)/f\left(t\right),f\left(t\right)]:f\in {\mathcal{F}}_1\}$.
• the derivative subgroup $\{[f^{\prime }\left(t\right),f\left(t\right)]:f\in {\mathcal{F}}_1\}$.
• the checkerboard subgroup $\{[g\left(t\right),f\left(t\right)]:g\in {\mathcal{F}}_0, f\in {\mathcal{F}}_1\},$ where g is an even function, and f is an odd function.

The 1-Appell subgroup is referred to as the Appell subgroup, the 1-Bell subgroup is referred to as the Bell subgroup for short, and the ordinary Appell subgroup can be considered the 0-Bell subgroup if we allow $k=0$ to be included in the definition of the k-Bell subgroup. From the isomorphism between the Sheffer group and the Riordan group, the set ${\mathcal{A}}^{*}$ of all s-Appell polynomial sequences with $s\in \mathbb{R}$ and $s\ne 0$ is a subgroup of the Sheffer group $\mathcal{S}$, where the umbral composition of an $s_1$-Appell polynomial sequence with an $s_2$-Appell polynomial sequence is an $s_1{s}_2$-Appell polynomial sequence.

2. Main Identity

We start by obtaining a simple but fundamental identity, which can be considered the prototype of all other identities considered in this paper since they will be obtained by this “umbral identity” by the suitable application of an “umbral operator”. To obtain such an identity, consider the Euler operator ${\vartheta }_t=t{D}_t$, where $D_t={\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}$ is the usual derivative with respect to t. For this operator, we have the following Grunert formula [28,29] (Formula (4.8)), [9] (p. 310):

$${\vartheta }_t^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{t}^k{D}_t^k.$$

(21)

Notice that by applying this formula, we can find the generating series of the powers $n^m$ [29] (Formula (4.10)):

$$\sum _{k\ge 0}{k}^m{t}^k=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}k!{\displaystyle \frac{t^k}{{(1-t)}^{k+1}}}.$$

(22)

Theorem 1.

For every $m,n\in \mathbb{N}$, we have the following identity:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{x}^{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{(x+1)}^{n-k}$$

(23)

or equivalently,

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{(-1)}^{n-k}{x}^{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{(1-x)}^{n-k},$$

(24)

where $m\wedge n=min\{m,n\}$.

Proof. 

From formula (21), the exponential generating series of the first member of (23) is as follows:

$${\mathrm{e}}^{xt}{\vartheta }_t^m{\mathrm{e}}^t=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{t}^k{\mathrm{e}}^{(x+1)t}=\sum _{n\ge 0}\left(\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{(x+1)}^{n-k}\right){\displaystyle \frac{t^n}{n!}}.$$

This establishes identity (23). Then, replacing x by $-x$ in (23), we obtain (24). □

Remark 1.

Identity (23) can also be obtained as the binomial transform of series (22). Indeed, if we apply the binomial matrix

$$\left({\displaystyle \frac{1}{1-xt}},{\displaystyle \frac{t}{1-xt}}\right)={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)x^{n-k}\right]}_{n,k\ge 0}$$

to both sides of series (22), then we have

$$\sum _{k\ge 0}{k}^m{\displaystyle \frac{t^k}{{(1-xt)}^{k+1}}}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}k!{\displaystyle \frac{t^k}{{(1-(x+1)t)}^{k+1}}}.$$

Finally, taking the coefficients of $t^n$ on both sides, we have our identity.

Identity (23), in addition to the next theorem, allows us to determine numerous new identities, starting with a sequence and its binomial transform.

Theorem 2.

Let ${\left\{A_n\right\}}_{n\ge 0}$ and ${\left\{B_n\right\}}_{n\ge 0}$ be two sequences such that

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)A_k=B_n.$$

Then, for every $m,n\in \mathbb{N}$, we have

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{A}_{n-k}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!B_{n-k}.$$

(25)

Proof. 

Consider the linear isomorphism $\phi :\mathbb{R}\left[x\right]\to \mathbb{R}\left[x\right]$ defined on the canonical basis by $\phi \left(x^n\right)=A_n$. Then, we have

$$\phi \left({(x+1)}^n\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\phi \left(x^k\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)A_k=B_n.$$

Finally, by applying $\phi$ to the umbral identity (23), we have identity (25). □

Remark 2.

Notice that Formula (25) (as well as Formula (23)) can be considered a closed form for the sum appearing on the left-hand side since the sum on the right-hand side has, at most, m terms (for $m>0$), and m is constant. For the first few values of m, we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k{A}_{n-k}=n{B}_{n-1}\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^2{A}_{n-k}=(n-1)n{B}_{n-2}+n{B}_{n-1}\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^3{A}_{n-k}=(n-2)(n-1)n{B}_{n-3}+3(n-1)n{B}_{n-2}+n{B}_{n-1}\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^4{A}_{n-k}=(n-3)(n-2)(n-1)n{B}_{n-4}+6(n-2)(n-1)n{B}_{n-3}\hfill \\ \hfill & +7(n-1)n{B}_{n-2}+n{B}_{n-1}.\hfill \end{array}$$

This remark continues to be true for all other identities of this kind that we will consider in the rest of the paper.

• Examples.

• Consider the generalized Fibonacci numbers $f_n^{\left[s\right]}$, $s\in \mathbb{N}$, $s\ge 1$ [30]. They are defined by the following generating series:

  $$\sum _{n\ge 0}{f}_n^{\left[s\right]}{t}^n={\displaystyle \frac{1}{1-t-t^2-\cdots -t^s}}.$$
  In this case, we have the following binomial identity:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)f_{(s+1)k}^{\left[s\right]}=2^n{f}_{sn}^{\left[s\right]}.$$
  So, by Theorem 2, we have

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{f}_{(s+1)(n-k)}^{\left[s\right]}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!2^{n-k}{f}_{s(n-k)}^{\left[s\right]}.$$
• For the Bell numbers $b_n={\sum }_{k=0}^n\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right\}$ [9], we have the following binomial identities:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)b_k=b_{n+1},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{n-k}{b}_k=b_{n+2}-b_{n+1}.\hfill \end{array}$$
  Hence, by Theorem 2, we have

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{b}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!b_{n-k+1},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{2}^k{b}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!2^k(b_{n-k+2}-b_{n-k+1}).\hfill \end{array}$$
• The generalized derangement numbers $d_n^{\left(\nu \right)}$ with $\nu \in \mathbb{N}$ [31,32] have exponential generating series

  $$\sum _{n\ge 0}{d}_n^{\left(\nu \right)}{\displaystyle \frac{t^n}{n!}}={\displaystyle \frac{\nu !{\mathrm{e}}^{-t}}{{(1-t)}^{\nu +1}}}$$

  and satisfy the binomial identities

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)d_k^{\left(\nu \right)}=(\nu +n)!,\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}(\nu +k)!=d_n^{\left(\nu \right)}.$$
  Therefore, by Theorem 2, we have

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{d}_{n-k}^{\left(s\right)}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!(s+n-k)!\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{(-1)}^k(s+n-k)!=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{(-1)}^k{d}_{n-k}^{\left(s\right)}.\hfill \end{array}$$
• For the harmonic numbers $H_n={\sum }_{k=0}^n{\displaystyle \frac{1}{k}}$ and the harmonic numbers of the second kind $H_n^{\left(2\right)}={\sum }_{k=0}^n{\displaystyle \frac{1}{k^2}},$ we have the following binomial identities [33,34]:

  $$\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{k-1}}{k}}=H_n,\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{k-1}{H}_k={\displaystyle \frac{1}{n}}n\ge 1$$

  and

  $$\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{k-1}}{k}}{H}_k=H_n^{\left(2\right)},\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{k-1}{H}_k^{\left(2\right)}={\displaystyle \frac{H_n}{n}}n\ge 1.$$
  Consequently, by Theorem 2, we have

  $$\begin{array}{cc}\hfill & \sum _{k=0}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\displaystyle \frac{{(-1)}^{n-k-1}}{n-k}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mathcal{H}}_{n-k}\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{(-1)}^{n-k-1}{\mathcal{H}}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{k!}{n-k}}m<n\hfill \end{array}$$

  and

  $$\begin{array}{cc}\hfill & \sum _{k=0}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\displaystyle \frac{{(-1)}^{n-k-1}}{n-k}}{\mathcal{H}}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mathcal{H}}_{n-k}^{\left(2\right)}\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{(-1)}^{n-k-1}{\mathcal{H}}_{n-k}^{\left(2\right)}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{k!}{n-k}}{\mathcal{H}}_{n-k}m<n.\hfill \end{array}$$
  Similarly, for the skew harmonic numbers (or alternating harmonic numbers) $H_n^{-}={\sum }_{k=0}^n{\displaystyle \frac{{(-1)}^{k-1}}{k}}$ we have the following binomial identities:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^k{H}_k^{-}={\displaystyle \frac{1-2^n}{n}}n\ge 1,\sum _{k=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^k}{k}}(1-2^k)=H_n^{-}.$$
  Then,

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{(-1)}^{n-k}{\mathcal{H}}_{n-k}^{-}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\displaystyle \frac{1-2^{n-k}}{n-k}}m<n\hfill \\ \hfill & \sum _{k=0}^{n-1}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\displaystyle \frac{{(-1)}^{n-k}}{n-k}}(1-2^{n-k})=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mathcal{H}}_{n-k}^{-}.\hfill \end{array}$$
• Recall that the beta function is defined by the following:

  $$B(p,q)={\int }_0^1{x}^{p-1}{(1-x)}^{q-1}\mathrm{d}x$$

  for $\Re p,\Re q>0$. In particular, if $p,q\in \mathbb{N}$, then

  $$B(p+1,q+1)={\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{p+q}{p}\right)}}{\displaystyle \frac{1}{p+q+1}}.$$
  Now, if we multiply both sides of (24) by $x^s$ and we integrate from 0 to 1, we have

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{(-1)}^{n-k}{x}^{s+n-k}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!B(s+1,n-k+1).$$
  That is,

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{n-k}{k}^m}{s+n-k+1}}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{k!}{\left(\genfrac{}{}{0pt}{}{s+n-k}{s}\right)}}{\displaystyle \frac{1}{s+n-k+1}}.$$
  In particular, for $s=0$, we have

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{(-1)}^{n-k}{k}^m}{n-k+1}}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{k!}{n-k+1}}.$$

3. Appell Sequences

In this section, we will apply Theorem 2 in order to obtain a general formula involving an arbitrary s-Appell sequence. An s-Appell sequence ($s\ne 0$) is a polynomial sequence ${\left\{a_n\left(x\right)\right\}}_{n\ge 0}$ with exponential generating series:

$$a(x;t)=\sum _{n\ge 0}{a}_n\left(x\right){\displaystyle \frac{t^n}{n!}}=g\left(t\right){\mathrm{e}}^{sxt},$$

(26)

where $g\left(t\right)={\sum }_{n\ge 0}{g}_n{\displaystyle \frac{t^n}{n!}}$ is a given exponential series. If $g_0\ne 0$, we have a proper sequence. In particular, these polynomials can be written as follows:

$$a_n\left(x\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)g_{n-k}{s}^k{x}^k.$$

(27)

All these are equivalent to $a_n^{\prime }\left(x\right)=sn{a}_{n-1}\left(x\right)$ for all $n\in \mathbb{N}$.

In the following examples, we recall some classical s-Appell sequences with their exponential generating series. All sequences are proper except the last one, which is provided by Genocchi polynomials.

• Examples.

• Ordinary powers $x^n$: ${\sum }_{n\ge 0}{x}^n{\displaystyle \frac{t^n}{n!}}={\mathrm{e}}^{xt}$.
• Generalized Hermite polynomials $H_n^{\left(\nu \right)}\left(x\right)$ (cf. [35], (Vol. 2, p. 192)):

  $$\sum _{n\ge 0}{H}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\mathrm{e}}^{2xt-\nu t^2}(s=2).$$
  Generalized Hermite numbers: $H_n^{\left(\nu \right)}\left(x\right)=H_n^{\left(\nu \right)}\left(0\right)$.
• Laguerre polynomials $L_n^{(\nu -n)}\left(x\right)$ (cf. [35] (Vol.2, p.189, Formula(19))):

  $$\sum _{n\ge 0}{L}_n^{(\nu -n)}\left(x\right){\displaystyle \frac{t^n}{n!}}={(1+t)}^{\nu }{\mathrm{e}}^{-xt}(s=-1).$$
• Generalized Rencontre polynomials $D_n^{\left(\nu \right)}\left(x\right)$, $\nu \in \mathbb{N}$ (cf. [31,36]):

  $$\sum _{n\ge 0}{D}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\displaystyle \frac{\nu !{\mathrm{e}}^{(x-1)t}}{{(1-t)}^{\nu +1}}}.$$
  Generalized derangement numbers: $d_n^{\left(\nu \right)}=D_n^{\left(\nu \right)}\left(0\right)$.
• Generalized Bernoulli polynomials $B_n^{\left(\nu \right)}\left(x\right)$ and generalized Euler polynomials $E_n^{\left(\nu \right)}\left(x\right)$ (cf. [11] (p. 93, p. 100) and [35] (Vol. 3, p. 252)):

  $$\begin{array}{cc}\hfill & \sum _{n\ge 0}{B}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\left({\displaystyle \frac{t}{{\mathrm{e}}^t-1}}\right)}^{\nu }{\mathrm{e}}^{xt}\hfill \\ \hfill & \sum _{n\ge 0}{E}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\left({\displaystyle \frac{2}{{\mathrm{e}}^t+1}}\right)}^{\nu }{\mathrm{e}}^{xt}.\hfill \end{array}$$
  Generalized Bernoulli numbers: $B_n^{\left(\nu \right)}=B_n^{\left(\nu \right)}\left(0\right)$ and ${\tilde{E}}_n^{\left(\nu \right)}=E_n^{\left(\nu \right)}\left(0\right)$. Generalized Euler numbers: $E_n^{\left(\nu \right)}=2^n{E}_n^{\left(\nu \right)}(\nu /2)$.
• Genocchi polynomials $G_n\left(x\right)$ (cf. [37,38]):

  $$\sum _{n\ge 0}{G}_n\left(x\right){\displaystyle \frac{t^n}{n!}}={\displaystyle \frac{2t}{{\mathrm{e}}^t+1}}{\mathrm{e}}^{xt}.$$
  Genocchi numbers: $G_n=G_n\left(0\right)$.

Now, we can prove our theorem on Appell polynomials.

Theorem 3.

For any s-Appell sequence ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$, $s\ne 0$, we have the following identity:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{a}_{n-k}\left(x\right)y^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!a_{n-k}(x+y/s)y^k.$$

(28)

In particular, for $y=1$, we have the following identity:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{a}_{n-k}\left(x\right)=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!a_{n-k}(x+1/s).$$

(29)

Proof. 

Just notice that, by the exponential generating series (26), we have

$${\mathrm{e}}^{yt}a(x,t)=g\left(t\right){\mathrm{e}}^{(sx+y)t}=g\left(t\right){\mathrm{e}}^{s(x+y/s)t}=a(x+y/s;t).$$

That is,

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a_k\left(x\right)y^{n-k}=a_n(x+y/s)\mathrm{or}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a_k\left(x\right)y^{-k}=a_n(x+y/s)y^{-n}.$$

So, by applying Theorem 2 and simplifying it, we have identity (28). □

We also have the following result.

Theorem 4.

Let ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ be an Appell sequence with exponential generating series (26). Then, we have

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{g}_{n-k}{\left(sx\right)}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\left(sx\right)}^k{a}_{n-k}\left(x\right).$$

(30)

Proof. 

From formula (27), we can write the following:

$$a_n\left(x\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)g_k{\left(sx\right)}^{n-k}\mathrm{or}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)g_k{\left(sx\right)}^{-k}={\left(sx\right)}^{-n}{a}_n\left(x\right).$$

Therefore, using Theorem 2, we have formula (30). □

• Examples.

• Using (28), we may obtain the following identities for the s-Appell sequences recalled in the initial examples:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{y}^k{x}^{n-k}=\sum _{k=0}^{m\wedge n}k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)y^k{(x+y)}^{n-k},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{H}_{n-k}^{\left(v\right)}\left(x\right)y^k=\sum _{k=0}^{m\wedge n}k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)H_{n-k}^{\left(v\right)}\left(x+{\displaystyle \frac{y}{2}}\right)y^k,\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{L}_{n-k}^{(\alpha -n)}\left(x\right)y^k=\sum _{k=0}^{m\wedge n}k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)L_{n-k}^{(\alpha -n)}(x-y)y^k,\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{B}_{n-k}^{\left(a\right)}\left(x\right)y^k=\sum _{k=0}^{m\wedge n}k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)B_{n-k}^{\left(a\right)}(x+y)y^k,\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{E}_{n-k}^{\left(a\right)}\left(x\right)y^k=\sum _{k=0}^{m\wedge n}k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)E_{n-k}^{\left(a\right)}(x+y)y^k,\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{G}_{n-k}\left(x\right)y^k=\sum _{k=0}^{m\wedge n}k!\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)G_{n-k}(x+y)y^k.\hfill \end{array}$$
  Some of these identities can be rewritten in a different way. Indeed, since $B_n(x+1)=B_n\left(x\right)+n{x}^{n-1}$, from (29) we obtain the following:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{B}_{n-k}\left(x\right)=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!(B_{n-k}\left(x\right)+(n-k)x^{n-k-1}).$$
  In particular, for $x=0$, we have the Bernoulli numbers $B_n=B_n\left(0\right)$, and the last identity becomes

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{B}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!B_{n-k}+\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{n-1}}\right\}n!.$$
  Similarly, since $E_n(x+1)=2x^n-E_n\left(x\right)$, (29) implies

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{E}_{n-k}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!(2x^{n-k}-E_{n-k}\left(x\right)).$$
  In particular, for $x=0$, we have $E_n\left(0\right)=(2-2^{n+2}){\displaystyle \frac{B_{n+1}}{n+1}}$, which changes the last identity to the following:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m(1-2^{n-k+1}){\displaystyle \frac{B_{n-k+1}}{n-k+1}}=\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{n}}\right\}n!-\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!(1-2^{n-k+1}){\displaystyle \frac{B_{n-k+1}}{n-k+1}}.$$
  Moreover, we have the Euler numbers $E_n=2^n{E}_n(1/2)$. So, for $x=1/2$, we have the following identity:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{2}^k{E}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!2^k(2-E_{n-k}).$$
  Finally, since $G_n(x+1)=2n{x}^{n-1}-G_n\left(x\right)$, (29) yields the following identity:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{G}_{n-k}\left(x\right)=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!(2(n-k)x^{n-k-1}-G_{n-k}\left(x\right)).$$
  In particular, for $x=0$, we have $G_n\left(0\right)=(2-2^{n+1})B_n$, and the above identity becomes

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m(1-2^{n-k})B_{n-k}=\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{n-1}}\right\}n!-\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!(1-2^{n-k})B_{n-k}.$$
• Using Formula (30), we have the following:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^k{k}^m{H}_{n-k}^{\left(\nu \right)}{x}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{k}k!x^k{H}_{n-k}^{\left(\nu \right)}\left(x\right),\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^k{k}^m{\nu }^{\underline{n-k}}{x}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{k}k!x^k{L}_{n-k}^{(\nu -n+k)}\left(x\right),\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{d}_{n-k}^{\left(\nu \right)}{x}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!x^k{D}_{n-k}^{\left(\nu \right)}\left(x\right),\hfill \end{array}$$

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{B}_{n-k}^{\left(\nu \right)}{x}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!x^k{B}_{n-k}^{\left(\nu \right)}\left(x\right),\hfill \end{array}$$

  (31)

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\tilde{E}}_{n-k}^{\left(\nu \right)}{x}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!x^k{E}_{n-k}^{\left(\nu \right)}\left(x\right),\hfill \end{array}$$

  (32)

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{G}_{n-k}{x}^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!x^k{G}_{n-k}\left(x\right),\hfill \end{array}$$

  where ${\nu }^{\underline{k}}=\nu (\nu -1)\cdots (\nu -k+1)$ is the falling factorial power.
  From these formulas we can deduce some other interesting identities. For instance, since $B_n\left(1\right)={(-1)}^n{B}_n$, $B_n(1/2)=({\displaystyle \frac{1}{2^{n-1}}}-1)B_n$, and $B_n(1/4)={\displaystyle \frac{1}{2^n}}({\displaystyle \frac{1}{2^{n-1}}}-1)B_n-{\displaystyle \frac{n}{4n}}{E}_{n-1}$. Then, identity (31) for $\nu =1$ becomes the following:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{B}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{(-1)}^{n-k}{B}_{n-k},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{k^m}{2^k}}{B}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!\left({\displaystyle \frac{1}{2^{n-1}}}-{\displaystyle \frac{1}{2^k}}\right)B_{n-k},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{k^m}{4^k}}{B}_{n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!\left({\displaystyle \frac{1}{2^n}}\left({\displaystyle \frac{1}{2^{n-1}}}-{\displaystyle \frac{1}{2^k}}\right)B_{n-k}-{\displaystyle \frac{n-k}{4^n}}{E}_{n-k-1}\right).\hfill \end{array}$$
  We have the Euler number $E_n=2^n{E}_n(1/2)$ and the Springer numbers $S_n={(-1)}^{\lceil n/2\rceil }{4}^n{E}_n(1/4)$ [39]. Since ${\tilde{E}}_n=E_n\left(0\right)=(2-2^{n+2}){\displaystyle \frac{B_{n+1}}{n+1}}$, identity (32) becomes the following:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{2}^{n-k}(2-2^{n-k+2}){\displaystyle \frac{B_{n-k+1}}{n-k+1}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!E_{n-k},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{4}^{n-k}(2-2^{n-k+2}){\displaystyle \frac{B_{n-k+1}}{n-k+1}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{\lceil {\displaystyle \frac{n-k}{2}}k!\rceil }{S}_{n-k}.\hfill \end{array}$$

4. Remarks on the Case of Appell Sequences

Notice that Theorem 1, providing the main umbral identity (together with Theorem 2 for the binomial tranform), implies Theorem 3 in Appell sequences. Vice versa, Theorem 3 implies Theorem 1 because the powers $x^n$ form an Appell sequence. Therefore, Theorems 1 and 3 are equivalent. However, as we already remarked, the umbral identity in Theorem 1 is somewhat more fundamental since it can be considered the root of all other identities of the same nature. Furthermore, as we will see in the last Section 7, this approach can be easily extended in order to obtain a q-analogue of all these results.

Nonetheless, it is interesting to observe that Theorem 3 in Appell sequences can be proved with a different approach based on the following general theorem proven in [40], in order to constructing combinatorial identities in a symbolic way. Recall that the difference operator $\mathrm{\Delta }$ is defined by $\mathrm{\Delta }f\left(x\right)=f(x+1)-f\left(x\right)$.

Theorem 5.

(Ref. [40]) Let ${\left\{f\left(k\right)\right\}}_{n\in \mathbb{N}}$ be a sequence of numbers (real or complex), and let $g\left(t\right)$ and $h\left(t\right)$ be infinitely differentiable functions on $[0,\infty )$. Then, we have the following formal identities:

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }f\left(k\right)g^{\left(k\right)}\left(0\right){\displaystyle \frac{x^k}{k!}}=\sum _{k=0}^{\infty }{\mathrm{\Delta }}^{k}f\left(0\right)g^{\left(k\right)}\left(x\right){\displaystyle \frac{x^k}{k!}},\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }h\left(k\right)g^{\left(k\right)}\left(0\right){\displaystyle \frac{x^k}{k!}}=\sum _{k=0}^{\infty }{\displaystyle \frac{1}{k!}}{h}^{\left(k\right)}\left(0\right)A_k(x,g\left(x\right)),\hfill \end{array}$$

(34)

where $A_m(x,g\left(x\right))$ is an extension of the Euler fraction in terms of $g\left(x\right)$, which is defined as follows:

$$A_m(x,g\left(x\right))=\sum _{j=0}^m\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{j}}\right\}{g}^{\left(j\right)}\left(x\right)x^j,$$

(35)

where $g^{\left(j\right)}\left(x\right)$ is the j-th derivative of $g\left(x\right)$.

At this point, as an immediate application of Theorem 5, we have the following formula.

Theorem 6.

Let $g\left(x\right)$ be an infinitely differentiable function on $[0,\infty )$, and let $m\in \mathbb{N}$. Then, we have the following identity:

$$\sum _{k\ge 0}{k}^m{g}^{\left(k\right)}\left(0\right){\displaystyle \frac{x^k}{k!}}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}{g}^{\left(k\right)}\left(x\right)x^k.$$

(36)

Proof. 

Substituting $f\left(x\right)=x^m$ into (33) and noting $\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{k}{j}}\right\}={\left({\mathrm{\Delta }}^j{x}^k\right)}_{x=0}/j!$, we obtain (36). □

Now, we can obtain our different proof of the formula involving Appell sequences.

Alternative proof of Theorem 3. Given our s-Appell polynomials $a_n\left(x\right)$, consider the function $g\left(x\right)=a_n(y+x/s)$ and place it into (36). Since

$${\displaystyle \frac{{\mathrm{d}}^k}{\mathrm{d}{x}^k}}{a}_n(y+x/s)=n^{\underline{k}}{a}_{n-k}(y+x/s),$$

we have

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{a}_{n-k}\left(y\right)x^k=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0.0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!a_{n-k}(y+x/s)x^k.$$

By exchanging x and y, we obtain the identities of the s-Appell polynomials (28).

5. Other Polynomial Sequences

The results obtained for Appell sequences can be extended to some other polynomial sequences depending on a parameter, as in the following cases.

Theorem 7.

Let ${\left\{s_n^{\left(\nu \right)}\left(x\right)\right\}}_{n\ge 0}$ be a polynomial sequence with exponential generating series:

$$\sum _{n\ge 0}{s}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\mathrm{e}}^{\nu t}F(x;t),$$

(37)

where $F(x;t)={\sum }_{n\ge 0}{F}_n\left(x\right){\displaystyle \frac{t^n}{n!}}$ is a given exponential series. Then, for every $m,n\in \mathbb{N}$, we have the following identities:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\mu }^k{s}_{n-k}^{\left(\nu \right)}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mu }^k{s}_{n-k}^{(\mu +\nu )}\left(x\right)$$

(38)

and

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\nu }^k{F}_{n-k}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\nu }^k{s}_{n-k}^{\left(\nu \right)}\left(x\right).$$

(39)

Proof. 

Notice that polynomials $s_n^{\left(\nu \right)}\left(x\right)$ form an Appell sequence with respect to the parameter $\nu$. Hence, using (34), we obtain identity (38). Furthermore, from series (37), we have

$$s_n^{\left(\nu \right)}\left(x\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\nu }^{n-k}{F}_k\left(x\right)\mathrm{or}{\displaystyle \frac{s_n^{\left(\nu \right)}\left(x\right)}{{\nu }^n}}=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{F_k\left(x\right)}{{\nu }^k}}(\nu \ne 0).$$

So, using Theorem 2, we obtain identity (39). □

Theorem 8.

Let ${\left\{s_n^{\left(\nu \right)}\left(x\right)\right\}}_{n\ge 0}$ be a polynomial sequence with exponential generating series:

$$s^{\left(\nu \right)}(x;t)=\sum _{n\ge 0}{s}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\mathrm{e}}^{\nu t}g\left(t\right){\mathrm{e}}^{(x+\lambda \nu )f\left(t\right)},$$

(40)

where $g\left(t\right)$ and $f\left(t\right)$ are given exponential series. Then, for every $m,n\in \mathbb{N}$, we have the following identities:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\mu }^k{s}_{n-k}^{\left(\nu \right)}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mu }^k{s}_{n-k}^{(\mu +\nu )}(x-\lambda \mu ).$$

(41)

Proof. 

First, from series (40), we have

$$\begin{array}{c}\sum _{n\ge 0}\left(\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\mu }^{n-k}{s}_k^{\left(\nu \right)}\left(x\right)\right){\displaystyle \frac{t^n}{n!}}={\mathrm{e}}^{\mu t}{s}^{\left(\nu \right)}(x;t)=\\ & ={\mathrm{e}}^{(\mu +\nu )t}g\left(t\right){\mathrm{e}}^{(x-\lambda \mu +\lambda (\mu +\nu \left)\right)f\left(t\right)}=s^{(\mu +\nu )}(x-\lambda \mu ;t),\end{array}$$

that is

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\mu }^{-k}{s}_k^{\left(\nu \right)}\left(x\right)={\mu }^{-n}{s}_n^{(\mu +\nu )}(x-\lambda \mu ).$$

Consequently, applying Theorem 2 and simplifying, we obtain identity (41). □

• Examples.

• The generalized exponential polynomials $S_n^{\left(\nu \right)}\left(x\right)$ [41,42] and generalized Fubini polynomials $F_n^{\left(\nu \right)}\left(x\right)$ have exponential generating series:

  $$\sum _{n\ge 0}{S}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\mathrm{e}}^{\nu t}{\mathrm{e}}^{x({\mathrm{e}}^t-1)}\mathrm{and}\sum _{n\ge 0}{F}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\displaystyle \frac{{\mathrm{e}}^{\nu t}}{1-x({\mathrm{e}}^t-1)}}.$$
  So, we can apply Theorem 7, obtaining the following identities:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\mu }^k{S}_{n-k}^{\left(\nu \right)}\left(x\right)=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mu }^k{S}_{n-k}^{(\mu +\nu )}\left(x\right)\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\mu }^k{F}_{n-k}^{\left(\nu \right)}\left(x\right)=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mu }^k{F}_{n-k}^{(\mu +\nu )}\left(x\right),\hfill \end{array}$$

  and

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\nu }^k{S}_{n-k}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\nu }^k{S}_{n-k}^{\left(\nu \right)}\left(x\right)\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\nu }^k{F}_{n-k}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\nu }^k{F}_{n-k}^{\left(\nu \right)}\left(x\right),\hfill \end{array}$$

  where $S_n\left(x\right)=S_n^{\left(0\right)}\left(x\right)$ and $F_n\left(x\right)=F_n^{\left(0\right)}\left(x\right)$ are the ordinary exponential and Fubini polynomials. In particular, for $x=1$, we obtain generalized Bell numbers $b_n^{\left(\nu \right)}$ and generalized Fubini numbers $f_n^{\left(\nu \right)}$. Then, the above two identities become the following:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\nu }^k{b}_{n-k}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\nu }^k{b}_{n-k}^{\left(\nu \right)}\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\nu }^k{f}_{n-k}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\nu }^k{f}_{n-k}^{\left(\nu \right)}.\hfill \end{array}$$
• The Tricomi continuants [43] are defined by the exponential generating series

  $$\sum _{n\ge 0}{T}_n^{\left(\nu \right)}\left(x\right){\displaystyle \frac{t^n}{n!}}={\displaystyle \frac{{\mathrm{e}}^{\nu t}}{{(1-t)}^{x-\nu }}}.$$
  This time, we use the case described in Theorem 8. So,

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{\mu }^k{T}_{n-k}^{\left(\nu \right)}\left(x\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\mu }^k{T}_{n-k}^{(\mu +\nu )}(x+\mu ).$$

6. Riordan and Sheffer-Type Matrices

One of the authors define Riordan-type matrices (or Riordan-type arrays) below, showing that the collocation of all Riordan-type matrices forms a semigroup called the Riordan semigroup, as presented in [44].

Let ${\left(d_{n,k}\right)}_{n,k\in \mathbb{N}}=(g,f)$ be a Riordan-type array, where $g,f\in {\mathcal{F}}_0$ with $g\left(0\right)\ne 0$. Then, ${\left({\tilde{d}}_{n,k}\right)}_{n,k\in \mathbb{N}}=(g,tf)$ is the Riordan array associated with the Riordan-type array, where $g,f\in {\mathcal{F}}_0$ with $g\left(0\right)\ne 0$. If $f_0\ne 0$, $(g,tf)$ is a proper Riordan array. Sheffer-type matrices can be defined similarly. In [44] (Theorem $5.2$), it was shown that different kind of identities can be obtained using Riordan-type matrices based on the umbral identity. Namely, for a given Riordan matrix $R={\left[r_{n,k}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right))$, we consider the associated Riordan-type matrices:

$$R^{♯}={\left[r_{n,k}^{♯}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right)+1)\mathrm{and}{R}^{♮}={\left[r_{n,k}^{♮}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right)-1).$$

Then, for every $m,n,s\in \mathbb{N}$, there exist the following identities:

$$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)r_{s,n-k}{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!r_{s,n-k}^{♯},\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}{r}_{s,n-k}{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}k!r_{s,n-k}^{♮},\hfill \end{array}$$

(43)

where $s\ge n-k$. Formulas (42) and (43) can be proven by replacing x with $f\left(t\right)$ and then multiplying both members by $g\left(t\right)$ in the umbral identities (23) and (24). After equating the coefficients of $t^s$ on both sides of the resulting equations, we obtain the desired solution.

In addition to the example related to Fuss-Catalan numbers shown in [44], we now present more examples of this construction process and will extend the process to Riordan matrices and Sheffer-type matrices in this section.

• Examples.

• Consider the following Riordan matrices:

  $$\begin{array}{cc}\hfill & R=\left({\displaystyle \frac{1}{{(1-t)}^{\alpha +1}}},{\displaystyle \frac{t}{1-t}}\right)={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n+\alpha }{n-k}}\right)\right]}_{n,k\ge 0}\hfill \\ \hfill & R^{♯}=\left({\displaystyle \frac{1}{{(1-t)}^{\alpha +1}}},{\displaystyle \frac{1}{1-t}}\right)={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n+\alpha +k}{n}}\right)\right]}_{n,k\ge 0}.\hfill \end{array}$$
  Then, identity (42) becomes the following:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s+\alpha }{s-n+k}}\right)k^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s+\alpha +n-k}{s}}\right)k!.$$
  In particular, for $\alpha =0$ and $s=n$, we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}^2{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-k}{n-k}}\right)k!.$$
  Similarly, for $\alpha =s=n$, we have

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)k^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3n-k}{2n-k}}\right)k!.$$
  Finally, for $\alpha =n$ and $s=2n$, we have

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3n}{n+k}}\right)k^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{4n-k}{3n-k}}\right)k!.$$
• Let

  $$\begin{array}{cc}\hfill & B\left(t\right)=\sum _{n\ge 0}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)t^n={\displaystyle \frac{1}{\sqrt{1-4t}}},\hfill \\ \hfill & C\left(t\right)=\sum _{n\ge 0}{C}_n{t}^n={\displaystyle \frac{1-\sqrt{1-4t}}{2t}}\hfill \end{array}$$

  be the generating series for the central binomial coefficients and the Catalan numbers, and consider the Riordan matrix and Riordan-type matrix, respectively:

  $$\begin{array}{cc}\hfill & R=(B\left(t\right),C\left(t\right)-1)={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n+k}}\right)\right]}_{n,k\ge 0},\hfill \\ \hfill & R^{♯}=(B\left(t\right),C\left(t\right))={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{2n+k}{n+k}}\right)\right]}_{n,k\ge 0}.\hfill \end{array}$$

Then, identity (42) becomes the following:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2s}{s+n-k}}\right)k^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2s+n-k}{s+n-k}}\right)k!.$$

To avoid using a Riordan-type array in the construction, we establish the following theorem for Riordan matrices.

Theorem 9.

Given a Riordan matrix $R={\left[r_{n,k}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right))$, consider the associated Riordan matrices:

$${\widehat{R}}^{♯}={\left[{\widehat{r}}_{n,k}^{♯}\right]}_{n,k\ge 0}=(g\left(t\right),t(f\left(t\right)+1))\mathrm{and}{\widehat{R}}^{♮}={\left[{\widehat{r}}_{n,k}^{♮}\right]}_{n,k\ge 0}=(g\left(t\right),t(f\left(t\right)-1)),$$

where both ${\widehat{R}}^{♯}$ and ${\widehat{R}}^{♮}$ are Riordan arrays. Then, for every $m,n,s\in \mathbb{N}$, we have the following identities:

$$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)r_{s,n-k}{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\widehat{r}}_{s+n-k,n-k}^{♯},\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}{r}_{s,n-k}{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}k!{\widehat{r}}_{s+n-k,n-k}^{♮},\hfill \end{array}$$

(45)

where $s\ge n-k$.

Proof. 

By replacing x with $f\left(t\right)$ and then multiply both members by $g\left(t\right)$, we may rewrite (23) as follows:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^{m}g\left(t\right)f{\left(t\right)}^{n-k}=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\displaystyle \frac{1}{t^{n-k}}}g\left(t\right){\left(t(f\left(t\right)+1)\right)}^{n-k},$$

and taking the coefficients of $t^s$ on both sides of the above equation, we may have

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)r_{s,n-k}{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!{\widehat{r}}_{s+n-k,n-k}^{♯}$$

for $s+n-k\ge n-k$, i.e., $s\ge 0$, which is (44). In addition, $s\ge 0$ implies that $s\ge n-k$ because $n\ge k$. Similarly, we can prove (45). □

• Example.

Consider the following Riordan matrices:

$$\begin{array}{cc}\hfill & R=\left({\displaystyle \frac{1}{{(1-t)}^{\alpha +1}}},{\displaystyle \frac{t}{1-t}}\right)={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n+\alpha }{n-k}}\right)\right]}_{n,k\ge 0}\hfill \\ \hfill & {\widehat{R}}^{♯}=\left({\displaystyle \frac{1}{{(1-t)}^{\alpha +1}}},{\displaystyle \frac{t}{1-t}}\right)={\left[\left({\displaystyle \genfrac{}{}{0pt}{}{n+\alpha }{n-k}}\right)\right]}_{n,k\ge 0}.\hfill \end{array}$$

Then, identity (44) becomes the following:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s+\alpha }{s-n+k}}\right)k^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{s+\alpha +n-k}{s}}\right)k!.$$

In particular, for $\alpha =0$ and $s=n$, we have

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}^2{k}^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-k}{n-k}}\right)k!.$$

Similarly, for $\alpha =s=n$, we have

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)k^m=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{3n-k}{2n-k}}\right)k!.$$

Theorem 9 can be extended to Sheffer-type matrices. Here, Sheffer-type matrices are defined similar to Riordan-type matrices.

Theorem 10.

Given a Sheffer matrix $S={\left[s_{n,k}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right))$, consider the associated Sheffer-type matrices:

$$\begin{array}{cc}\hfill & S^{♯}={\left[s_{n,k}^{♯}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right)+1),\hfill \\ \hfill & S^{♮}={\left[s_{n,k}^{♮}\right]}_{n,k\ge 0}=(g\left(t\right),f\left(t\right)-1).\hfill \end{array}$$

Then, for every $m,n,r\in \mathbb{N}$, we have the following identities:

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{\displaystyle \frac{k^m}{k!}}{s}_{r,n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{s}_{r,n-k}^{♯},\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^{n-k}{\displaystyle \frac{k^m}{k!}}{s}_{r,n-k}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{(-1)}^{n-k}{s}_{r,n-k}^{♮}.\hfill \end{array}$$

(47)

Proof. 

In umbral identities and (47), we replace x with $f\left(t\right)$ and then multiply both members by $g\left(t\right)$. So, we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{\displaystyle \frac{k^m}{k!}}g\left(t\right){\displaystyle \frac{f{\left(t\right)}^{n-k}}{(n-k)!}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}g\left(t\right){\displaystyle \frac{{(f\left(t\right)+1)}^{n-k}}{(n-k)!}}\hfill \\ \hfill & \sum _{k=0}^n{(-1)}^{n-k}{\displaystyle \frac{k^m}{k!}}g\left(t\right){\displaystyle \frac{f{\left(t\right)}^{n-k}}{(n-k)!}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{(-1)}^{n-k}g\left(t\right){\displaystyle \frac{{(f\left(t\right)-1)}^{n-k}}{(n-k)!}}.\hfill \end{array}$$

Now, by taking the coefficient of ${\displaystyle \frac{t^r}{r!}}$ in both members (and in both cases), we have identities (46) and (24). □

• Examples.

• Consider the following Sheffer matrices:

  $$\begin{array}{cc}\hfill & S=\left({\displaystyle \frac{1}{{(1-t)}^{2\nu }}},{\displaystyle \frac{t}{1-t}}\right)={\left[L_{n,k}^{\left(\nu \right)}\right]}_{n,k\ge 0},\hfill \\ \hfill & S^{♯}=\left({\displaystyle \frac{1}{{(1-t)}^{2\nu }}},{\displaystyle \frac{1}{1-t}}\right)={\left[{\displaystyle \frac{{(\nu +k)}_n}{k!}}\right]}_{n,k\ge 0},\hfill \end{array}$$

  where the coefficient $L_{n,k}^{\left(\nu \right)}$ is the generalized Lah number. Then, the umbral identity becomes the following:

  $$\sum _{k=0}^n{\displaystyle \frac{k^m}{k!}}{L}_{r,n-k}^{\left(\nu \right)}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{\displaystyle \frac{{(\nu +n-k)}_r}{(n-k)!}}.$$
  In particular, for $\nu =0$ and $\nu =1$, we have the following identities:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left|{\displaystyle \genfrac{}{}{0pt}{}{r}{n-k}}\right|{\displaystyle \frac{k^m}{k!}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{\displaystyle \frac{{(n-k)}_r}{(n-k)!}},\hfill \\ \hfill & \sum _{k=0}^n\left|{\displaystyle \genfrac{}{}{0pt}{}{r+1}{n-k+1}}\right|{\displaystyle \frac{k^m}{k!}}=\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{\displaystyle \frac{{(n-k+1)}_r}{(n-k)!}}.\hfill \end{array}$$
• Consider the following Sheffer matrices:

  $$\begin{array}{cc}\hfill & S=({\mathrm{e}}^{\nu t},{\mathrm{e}}^t)={\left[{\displaystyle \frac{{(\nu +k)}^n}{k!}}\right]}_{n,k\ge 0},\hfill \\ \hfill & S^{♮}=({\mathrm{e}}^{\nu t},{\mathrm{e}}^t-1)={\left[S_{n,k}^{\left(\nu \right)}\right]}_{n,k\ge 0},\hfill \end{array}$$

  where the coefficients $S_{n,k}^{\left(\nu \right)}$ are the generalized Stirling numbers of the second kind. Then, identity (24) becomes the following:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}{k}^m{(\nu +n-k)}^r=n!\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}{(-1)}^{n-k}{S}_{r,n-k}^{\left(\nu \right)}.$$
  In particular, for $\nu =0$ and $\nu =1$, we have the following identities:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}{k}^m{(n-k)}^r=n!\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left\{{\displaystyle \genfrac{}{}{0pt}{}{r}{n-k}}\right\}{(-1)}^{n-k},\hfill \\ \hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(-1)}^{n-k}{k}^m{(n-k+1)}^r=n!\sum _{k=0}^{m\wedge n}\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left\{{\displaystyle \genfrac{}{}{0pt}{}{r+1}{n-k+1}}\right\}{(-1)}^{n-k}.\hfill \end{array}$$

7. q-Analogues

In this section, we will show that it is possible to obtain a q-analogue of the umbral identity (23). First, however, we recall some basic definitions of q-calculus. For every $n\in \mathbb{N}$, we have the q-natural number${\left[n\right]}_q=1+q+q^2+\cdots +q^{n-1}$ and the q-factorial number${\left[n\right]}_q!={\left[n\right]}_q{[n-1]}_q\cdots {\left[2\right]}_q{\left[1\right]}_q$. Then, for every $n,k\in \mathbb{N}$, the q-binomial coefficients (or Gaussian coefficients) are defined by

$${\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q={\displaystyle \frac{{\left[n\right]}_q!}{{\left[k\right]}_q!{[n-k]}_q!}}\mathrm{for}k=0,1,\dots ,n$$

and by 0 otherwise. The q-binomial inversion theorem says that for any two sequences ${\left\{a_n\right\}}_{n\in \mathbb{N}}$ and ${\left\{b_n\right\}}_{n\in \mathbb{N}}$, we have

$$a_n=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{b}_k\iff b_n=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{a}_k.$$

A q-exponential generating series is a formal series of the following form:

$$f\left(t\right)=\sum _{n\ge 0}{f}_n{\displaystyle \frac{t^n}{{\left[n\right]}_q!}}.$$

The q-exponential series is the following series:

$$E_q\left(t\right)=\sum _{n\ge 0}{\displaystyle \frac{t^n}{{\left[n\right]}_q!}}.$$

q-Hermite polynomials are defined by the following:

$$H_n(q;x)={\left(x{\oplus }_{q}1\right)}^n=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{x}^k,$$

and their q-exponential generating series is provided as follows:

$$H_q(x;t)=\sum _{n\ge 0}{H}_n(q;x){\displaystyle \frac{t^n}{{\left[n\right]}_q!}}=E_q\left(t\right)E_q(x;t).$$

(48)

The q-derivative (Jackson’s derivative) $D_q$ of a q-exponential generating series $f\left(t\right)={\sum }_{n\ge 0}{f}_n{\displaystyle \frac{t^n}{{\left[n\right]}_q!}}$ is defined [45,46] by the formula

$$D_{q}f\left(t\right)={\displaystyle \frac{f\left(qt\right)-f\left(t\right)}{(q-1)t}}=\sum _{n\ge 0}{f}_{n+1}{\displaystyle \frac{t^n}{{\left[n\right]}_q!}}.$$

In particular, for the exponential series we have $D_q^m{E}_q\left(t\right)=E_q\left(t\right)$.

The operator ${\vartheta }_q$ is defined by ${\vartheta }_q=t{D}_q$. Consequently, we have

$${\vartheta }_{q}f\left(t\right)=\sum _{n\ge 0}{\left[n\right]}_q{f}_n{\displaystyle \frac{t^n}{{\left[n\right]}_q!}},$$

and more generally, for every $m\in \mathbb{N}$,

$${\vartheta }_q^{m}f\left(t\right)=\sum _{n\ge 0}{\left[n\right]}_q^m{f}_n{\displaystyle \frac{t^n}{{\left[n\right]}_q!}}.$$

(49)

For this operator, we have the following q-analogue of the Grünert formula:

$${\vartheta }_q^m=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{t}^k{D}_q^k,$$

(50)

where the coefficients ${\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q$ are the q-Stirling numbers of the second kind [47]. These coefficients are defined as the connection constants [48,49] between the ordinary powers $x^n$ and the q-falling factorial polynomials $x_q^{\underline{n}}=x(x-{\left[1\right]}_q)(x-{\left[2\right]}_q)\cdots (x-{[n-1]}_q)$:

$$x^n=\sum _{k=0}^n{\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right\}}_q{x}_q^{\underline{k}}.$$

Equivalently, they are defined by the following recurrence:

$${\left\{{\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right\}}_q={\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right\}}_q+{[k+1]}_q{\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{k+1}}\right\}}_q,$$

with the initial values ${\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right\}}_q={\delta }_{n,0}$ and ${\left\{{\displaystyle \genfrac{}{}{0pt}{}{0}{k}}\right\}}_q={\delta }_{0,k}$.

They can also be expressed explicitly as follows:

$${\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right\}}_q={\displaystyle \frac{1}{{\left[k\right]}_q!q^{\left(\genfrac{}{}{0pt}{}{k}{2}\right)}}}\sum _{i=0}^k{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)}_q{(-1)}^i{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{i}{2}}\right)}{[k-i]}_q^n.$$

Now, we can prove the q-analogue of the umbral identity (23).

Theorem 11.

For every $m,n\in \mathbb{N}$, we have the following identity:

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{x}^{n-k}=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{\left(x{\oplus }_{q}1\right)}^{n-k},$$

(51)

where the ${\left(x{\oplus }_{q}1\right)}^n=H_n(q;x)$ are the q-Hermite polynomials.

Proof. 

From Formulas (50) and (49) and series (48), the q-exponential generating series of the first member of (51) is as follows:

$$E_q(x;t){\vartheta }_q^m{E}_q\left(t\right)=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{t}^k{E}_q\left(t\right)E_q(x;t),=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{t}^k{H}_q(x;t),$$

from which it is straightforward to obtain identity (51). □

• Examples.

• For $x=0$, identity (51) reduces to the formula

  $${\left[n\right]}_q^m=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}.$$
• For $x=1$, identity (51) becomes the following:

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{G}_{n-k}\left(q\right),$$

  where $G_n\left(q\right)={\sum }_{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q$ are the Galois numbers.

The q-umbral identity (51) implies the following theorem, which is a q-analogous of Theorem 2.

Theorem 12.

Let ${\left\{A_n\left(q\right)\right\}}_{n\ge 0}$ and ${\left\{B_n\left(q\right)\right\}}_{n\ge 0}$ be two sequences such that

$$B_n\left(q\right)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{A}_k\left(q\right).$$

Then, for every $m,n\in \mathbb{N}$, we have

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{A}_{n-k}\left(q\right)=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{B}_{n-k}\left(q\right).$$

(52)

Proof. 

Consider the linear isomorphism $\phi :\mathbb{R}\left[q\right]\left[x\right]\to \mathbb{R}\left[q\right]\left[x\right]$ defined on the canonical basis by $\phi \left(x^n\right)=B_n\left(q\right)$. Then, we have

$$\phi \left({\left(x{\oplus }_{q}1\right)}^n\right)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q\phi \left(x^k\right)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{A}_k\left(q\right)=B_n\left(q\right).$$

Finally, by applying $\phi$ to the umbral identity (51), we have identity (52). □

• Examples.

• The q-Pochhammer symbol ${(x;q)}_n$ and the Gaussian polynomials $g_n(q;x)$ are defined by the following:

  $$\begin{array}{cc}\hfill & {(x;q)}_n=(1-x)(1-qx)(1-q^{2}x)\cdots (1-q^{n-1}x)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^k{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{x}^k,\hfill \\ \hfill & g_n(q;x)=(x-1)(x-q)(x-q^2)\cdots (x-q^{n-1})=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{x}^k.\hfill \end{array}$$
  Hence, using the q-binomial inversion theorem, we have

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^k{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{x}^k={(x;q)}_n,\hfill \\ \hfill & \sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{g}_k(q;x)=x^n,\hfill \end{array}$$

  and, consequently, by Theorem (12), we have

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{x}^{n-k}=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{(x;q)}_{n-k},\hfill \\ \hfill & \sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{g}_{n-k}(q;x)=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{x}^{n-k}.\hfill \end{array}$$
• The q-derangement numbers [50,51,52,53] are defined by the following formula:

  $$d_n\left(q\right)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{[n-k]}_q!{(-1)}^k{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}.$$
  Equivalently, we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{d}_k\left(q\right)={\left[n\right]}_q!,$$

  and so, by Theorem (12), we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{d}_{n-k}\left(q\right)={\left[n\right]}_q!\sum _{k=0}^{m\wedge n}{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}.$$
  In particular, if we consider the q-Bell numbers [52,54] defined by

  $$b_n\left(q\right)=\sum _{k=0}^n{\left\{{\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right\}}_q{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)},$$

  then, from the above identity for $m=n$, we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^n{d}_{n-k}\left(q\right)={\left[n\right]}_q!b_n\left(q\right).$$
  In this way, we have the q-Bell numbers expressed in terms of the q-derangement numbers. Notice that similar results have been obtained in [52], where the formulas involve the q-Stirling numbers of the first and second kind:

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n{\left[{\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right]}_q{(-1)}^k{b}_k\left(q\right)=d_n\left(q\right),\hfill \\ \hfill & \sum _{k=0}^n{\left\{{\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right\}}_q{(-1)}^k{d}_k\left(q\right)=b_n\left(q\right).\hfill \end{array}$$
  Notice also that these q-Bell numbers satisfy [54] the ordinary binomial identity as follows:

  $$b_{n+1}\left(q\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)q^k{b}_k\left(q\right).$$
  This means that in this case, we have to apply Theorem 2:

  $$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k^m{q}^{n-k}{b}_{n-k}\left(q\right)=\sum _{k=0}^m\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)k!b_{n-k+1}\left(q\right).$$
• For q-harmonic numbers ${\mathcal{H}}_n\left(q\right)$ and q-harmonic numbers of the second kind ${\tilde{\mathcal{H}}}_n\left(q\right)$, defined by

  $${\mathcal{H}}_n\left(q\right)=\sum _{k=1}^n{\displaystyle \frac{1}{{\left[k\right]}_q}}\mathrm{and}{\tilde{\mathcal{H}}}_n\left(q\right)=\sum _{k=1}^n{\displaystyle \frac{q^k}{{\left[k\right]}_q}},$$

  we have the identities (see [34] and [55], respectively)

  $$\begin{array}{c}\hfill \sum _{k=1}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^{k-1}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{\mathcal{H}}_k\left(q\right)={\displaystyle \frac{1}{{\left[k\right]}_q}}\\ \hfill \sum _{k=1}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\displaystyle \frac{{(-1)}^{k-1}}{{\left[k\right]}_q}}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{2}}\right)}={\tilde{\mathcal{H}}}_n\left(q\right).\end{array}$$
  Hence, by Theorem (12), we have

  $$\begin{array}{cc}\hfill & \sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{(-1)}^{n-k-1}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{\mathcal{H}}_{n-k}\left(q\right)=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\displaystyle \frac{{\left[k\right]}_q!}{{[n-k]}_q}}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}m<n,\hfill \\ \hfill & \sum _{k=0}^{n-1}{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\displaystyle \frac{{(-1)}^{n-k-1}}{{[n-k]}_q}}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k+1}{2}}\right)}{\left[k\right]}_q^m=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{\tilde{\mathcal{H}}}_{n-k}\left(q\right).\hfill \end{array}$$

We conclude by extending Theorem 3 (for $s=1$). Recall that a q-Appell sequence [56,57] is a polynomial sequence ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$, where $a_n\left(x\right)$ has degree $n$ and $D_q{a}_n\left(x\right)={\left[n\right]}_q{a}_{n-1}\left(x\right)$ for every $n\in \mathbb{N}$. This is equivalent to asking that the sequence ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ has q-exponential generating series:

$$A(x;t)=\sum _{n\ge 0}{a}_n\left(x\right){\displaystyle \frac{t^n}{{\left[n\right]}_q!}}=g\left(t\right)E_q\left(xt\right),$$

(53)

where $g\left(t\right)={\sum }_{n\ge 0}{g}_n{\displaystyle \frac{t^n}{{\left[n\right]}_q!}}$ is a q-exponential series with $g_0\ne 0$.

First of all, we have the following result.

Lemma 1.

For any q-Appell sequence ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$, we have the following binomial identity:

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{a}_k\left(x\right)=a_n\left(x{\oplus }_{q}1\right).$$

(54)

Proof. 

The q-exponential generating series of the left-hand side of (55) is as follows:

$$\begin{array}{c}{E}_q\left(t\right)A(x;t)=g\left(t\right)E_q\left(t\right)E_q\left(xt\right)=g\left(t\right)H_q(x;t)\\ =\sum _{n\ge 0}\left(\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{g}_{n-k}{\left(x{\oplus }_{q}1\right)}^k\right){\displaystyle \frac{t^n}{{\left[n\right]}_q!}}=\sum _{n\ge 0}{a}_n\left(x{\oplus }_{q}1\right){\displaystyle \frac{t^n}{{\left[n\right]}_q!}}.\end{array}$$

Hence, we have identity (55). □

Now, by applying Theorem 12 and Lemma 1, we have the following q-analogue of the theorem.

Theorem 13.

For any q-Appell sequence ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$, we have

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{a}_{n-k}\left(x\right)=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{a}_{n-k}\left(x{\oplus }_{q}1\right).$$

(55)

Finally, we have the following q-analogue of Theorem 4.

Theorem 14.

Let ${\left\{a_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ be a q-Appell sequence with q-exponential generating series (53). Then, we have

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{g}_{n-k}{x}^k=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{x}^k{a}_{n-k}\left(x\right).$$

(56)

Proof. 

From series (53), we know that the polynomials $a_n\left(q\right)$ can be written as follows:

$$a_n\left(x\right)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{g}_{n-k}{x}^k=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{g}_k{x}^{n-k}.$$

Hence, we have

$$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{g}_k{x}^{-k}=x^{-n}{a}_n\left(x\right).$$

Then, applying Theorem 12 and simplifying, we obtain identity (56). □

• Examples.

• The Gaussian polynomials form a q-Appell sequence as follows:

  $$g_n(q;x)=\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{x}^k.$$
  So, by Theorem 14, we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{\left[k\right]}_q^m{x}^k=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{x}^k{g}_{n-k}^{\left(\nu \right)}(q;x).$$
  In particular, for $x=1$ and $m=n$, we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{(-1)}^{n-k}{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{2}}\right)}{\left[k\right]}_q^n={\left[n\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}.$$
• The generalized q-Bernoulli numbers $B_n^{\left(\nu \right)}\left(q\right)$ and generalized q-Bernoulli polynomials $B_n^{\left(\nu \right)}(q;x)$ are defined, respectively, by the q-exponential generating series [57,58]:

  $$\begin{array}{cc}\hfill & B_q^{\left(\nu \right)}\left(t\right)=\sum _{n\ge 0}{B}_n^{\left(\nu \right)}\left(q\right){\displaystyle \frac{t^n}{{\left[n\right]}_q!}}={\left({\displaystyle \frac{t}{E_q\left(t\right)-1}}\right)}^{\nu }\hfill \\ \hfill & B_q^{\left(\nu \right)}(x;t)=\sum _{n\ge 0}{B}_n^{\left(\nu \right)}(q;x){\displaystyle \frac{t^n}{{\left[n\right]}_q!}}={\left({\displaystyle \frac{t}{E_q\left(t\right)-1}}\right)}^{\nu }{E}_q\left(xt\right).\hfill \end{array}$$
  Then, using Theorem 14, we have

  $$\sum _{k=0}^n{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q^m{B}_{n-k}^{\left(\nu \right)}\left(q\right)x^k=\sum _{k=0}^m{\left\{{\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right\}}_q{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}_q{\left[k\right]}_q!q^{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{2}}\right)}{x}^k{B}_{n-k}^{\left(\nu \right)}(q;x).$$