On a Class of Polynomials Generated by F (xt -- R(t))

TL;DR

This paper studies polynomial sets generated by specific functions, revealing conditions under which the generating series are polynomials of bounded degree and connecting them to hypergeometric series.

Contribution

It characterizes the form of the generating series R(t) and the structure of the polynomials for certain recursive and symmetry conditions, linking to hypergeometric functions.

Findings

R(t) is a polynomial of degree at most d+1 under certain conditions

In the d-symmetric case, R(t) is a monomial of degree d+1

F(t) can be expressed as a hypergeometric series

Abstract

We investigate polynomial sets {P n } n$\ge$0 with generating power series of the form F (xt -- R(t)) and satisfying, for n $\ge$ 0, the (d + 1)-order recursion xP\_ n (x) = P\_{ n+1 }(x) +\sum\_{ l=0}^{d} \gamma^{l}\_{n} P\_{ n--l} (x), where \ {\gamma ^{l}\_{ n}\ } is a complex sequence for 0 $\le$ l $\le$ d, P \_0 (x) = 1 and P \_n (x) = 0 for all negative integer n. We show that the formal power series R(t) is a polynomial of degree at most d + 1 if certain coefficients of R(t) are null or if F (t) is a generalized hypergeometric series. Moreover, for the d-symmetric case we demonstrate that R(t) is the monomial of degree d + 1 and F (t) is expressed by hypergeometric series.

Equations

$$F(xt-R(t))=\sum _{n\ge 0}{\alpha }_n{P}_n(x)t^n.$$

$$\displaystyle\left\{\begin{array}[]{l}xP_{n}(x)=P_{n+1}(x)+\beta_{n}P_{n}(x)+\omega_{n}P_{n-1}(x),\quad n\geq 0,\\ P_{-1}(x)=0,\;\;P_{0}(x)=1\end{array}\right.$$

$$\displaystyle\left\{\begin{array}[]{l}xP_{n}(x)=P_{n+1}(x)+\sum_{l=0}^{d}\gamma_{n}^{l}P_{n-l}(x),\quad n\geq 0,\\ P_{0}(x)=1,\;\;P_{-l}(x)=0,\;\;1\leq l\leq d\end{array}\right.$$

$$\{{\gamma }_n^l{\}}_{n\ge 0},0\le l\le d,\text{are complex sequences}$$

$$\{{\gamma }_n^d{\}}_{n\ge d}\text{is not the null sequence, for}d\ge 1.$$

$$P_n(\omega x)={\omega }^n{P}_n(x),n\ge 0.$$

$$\exp \left({\int }_0^t\frac{{\rho }_d(s)}{{\sigma }_{d+1}(s)}ds\right)\exp \left(x{\int }_0^t\frac{1}{{\sigma }_{d+1}(s)}ds\right)=\sum _{k=0}^{\infty }{P}_n(x)\frac{t^n}{n!}$$

$${\tilde{\sigma }}_0(n{\tilde{\sigma }}_0-{\tilde{\rho }}_d)\ne 0,n\ge 1.$$

$${\alpha }_{n}x{P}_n^{\prime }(x)-\sum _{k=1}^{n-1}{R}_{k+1}{\alpha }_{n-k}{P}_{n-k}^{\prime }(x)=n{\alpha }_n{P}_n(x),n\ge 1.$$

$$a_n=\frac{{\alpha }_n}{{\alpha }_{n+1}},(n\ge 0)\text{ and }{c}_n^l=\frac{{\alpha }_n}{{\alpha }_{n-l}}{\gamma }_n^l,(1\le l\le d,n\ge l),$$

$${\gamma }_n^0=0,\text{ for }n\ge 0.$$

$${\gamma }_n^1=\frac{R_2}{2}(n{a}_n-(n-1)a_{n-1}),\text{ for }n\ge 1.$$

$$c_n^1=\frac{R_2}{2}\left(n\frac{a_n}{a_{n-1}}-(n-1)\right),\text{ for }n\ge 1.$$

$$c_n^2=\frac{R_3}{3}\left((n-1)\frac{a_n}{a_{n-2}}-(n-2)\right),\text{ for }n\ge 2.$$

$$\frac{k+1}{n-k+1}{a}_{n-k}{c}_n^k$$

$$R_{k+1}\left(a_n-\frac{n-k}{n-k+1}{a}_{n-k}\right)+\sum _{l=1}^d{R}_{k-l}\left(\frac{n+2}{n-l+1}{c}_n^l-\frac{n-k+l+1}{n-k+l+2}{c}_{n-k+l+1}^l\right)$$

$$-\sum _{l=k-1-d}^d\frac{l+1}{n-l+1}{c}_n^l{c}_{n-l}^{k-l-1}=\sum _{l=1}^{k-2}\frac{R_{l+1}{R}_{k-l}}{n-l+1},d+1\le k\le 2d+1,n\ge k.$$

$$R_{k+1}\left(a_n-\frac{n-k}{n-k+1}{a}_{n-k}\right)+\sum _{l=1}^d{R}_{k-l}\left(\frac{n+2}{n-l+1}{c}_n^l-\frac{n-k+l+1}{n-k+l+2}{c}_{n-k+l+1}^l\right)$$

$$=\sum _{l=1}^{k-2}\frac{R_{l+1}{R}_{k-l}}{n-l+1},k\ge 2d+2,n\ge k.$$

$$x{P}_n^{\prime }(x)+P_n(x)=P_{n+1}^{\prime }(x)+\sum _{l=0}^d{\gamma }_n^l{P}_{n-l}^{\prime }(x).$$

$$(n+1){\alpha }_{n}x{P}_n^{\prime }(x)=n{\alpha }_n\left(P_{n+1}^{\prime }(x)+\sum _{l=0}^d{\gamma }_n^l{P}_{n-l}^{\prime }(x)\right)+\sum _{k=1}^{n-1}{R}_{k+1}{\alpha }_{n-k}{P}_{n-k}^{\prime }(x)$$

$$(n+1){\alpha }_n{P}_n(x)={\alpha }_n\left(P_{n+1}^{\prime }(x)+\sum _{l=0}^d{\gamma }_n^l{P}_{n-l}^{\prime }(x)\right)-\sum _{k=1}^{n-1}{R}_{k+1}{\alpha }_{n-k}{P}_{n-k}^{\prime }(x).$$

$$(n+1){\alpha }_n\left(P_{n+1}+\sum _{l=0}^d{\gamma }_n^l{P}_{n-l}(x)\right)={\alpha }_n\left(x{P}_{n+1}^{\prime }(x)+\sum _{l=0}^d{\gamma }_n^{l}x{P}_{n-l}^{\prime }(x)\right)-\sum _{k=1}^{n-1}{R}_{k+1}{\alpha }_{n-k}x{P}_{n-k}^{\prime }(x).$$

$$\frac{n+1}{n+2}{\alpha }_n\left(P_{n+2}^{\prime }(x)+\sum _{l=0}^d{\gamma }_{n+1}^l{P}_{n+1-l}^{\prime }(x)\right)-\frac{n+1}{n+2}\frac{{\alpha }_n}{{\alpha }_{n+1}}\sum _{k=1}^n{R}_{k+1}{\alpha }_{n+1-k}{P}_{n+1-k}^{\prime }(x)$$

$$+(n+1){\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{1}{n+1-l}{P}_{n+1-l}^{\prime }(x)+(n+1){\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{1}{n+1-l}\sum _{l^{\prime }=0}^d{\gamma }_{n-l}^{l^{\prime }}{P}_{n-l-l^{\prime }}^{\prime }(x)$$

$$-(n+1){\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{1}{n+1-l}\frac{1}{{\alpha }_{n-l}}\sum _{k=1}^{n-l-1}{R}_{k+1}{\alpha }_{n-l-k}{P}_{n-k-l}^{\prime }(x)=\frac{n+1}{n+2}{\alpha }_n\left(P_{n+2}^{\prime }(x)+\sum _{l=0}^d{\gamma }_{n+1}^l{P}_{n+1-l}^{\prime }(x)\right)$$

$$+\frac{1}{n+2}\frac{{\alpha }_n}{{\alpha }_{n+1}}\sum _{k=1}^n{R}_{k+1}{\alpha }_{n+1-k}{P}_{n+1-k}^{\prime }(x)+{\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{n-l}{n+1-l}{P}_{n+1-l}^{\prime }(x)$$

$$+{\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{n-l}{n+1-l}\sum _{l^{\prime }=0}^d{\gamma }_{n-l}^{l^{\prime }}{P}_{n-l-l^{\prime }}^{\prime }(x)+{\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{1}{n+1-l}\frac{1}{{\alpha }_{n-l}}\sum _{k=1}^{n-l-1}{R}_{k+1}{\alpha }_{n-l-k}{P}_{n-l-k}^{\prime }(x)$$

$$-\sum _{k=1}^{n-1}{R}_{k+1}\frac{n-k}{n+1-k}{\alpha }_{n-k}{P}_{n+1-k}^{\prime }(x)-\sum _{k=1}^{n-1}{R}_{k+1}\frac{n-k}{n+1-k}{\alpha }_{n-k}\sum _{l=0}^d{\gamma }_{n-k}^l{P}_{n-k-l}^{\prime }(x)$$

$$-\sum _{k=1}^{n-1}{R}_{k+1}\frac{1}{n+1-k}\sum _{k^{\prime }=1}^{n-k-1}{R}_{k^{\prime }+1}{\alpha }_{n-k-k^{\prime }}{P}_{n-k-k^{\prime }}^{\prime }(x).$$

$$-\frac{{\alpha }_n}{{\alpha }_{n+1}}\sum _{k=1}^n{R}_{k+1}{\alpha }_{n+1-k}{P}_{n+1-k}^{\prime }(x)+{\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{l+1}{n+1-l}{P}_{n+1-l}^{\prime }(x)+{\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{l+1}{n+1-l}\sum _{l^{\prime }=0}^d{\gamma }_{n-l}^{l^{\prime }}{P}_{n-l-l^{\prime }}^{\prime }(x)$$

$$-(n+2){\alpha }_n\sum _{l=0}^d{\gamma }_n^l\frac{1}{n+1-l}\frac{1}{{\alpha }_{n-l}}\sum _{k=1}^{n-l-1}{R}_{k+1}{\alpha }_{n-l-k}{P}_{n-k-l}^{\prime }(x)=-\sum _{k=1}^{n-1}{R}_{k+1}\frac{n-k}{n+1-k}{\alpha }_{n-k}{P}_{n+1-k}^{\prime }(x)$$

$$-\sum _{k=1}^{n-1}{R}_{k+1}\frac{n-k}{n+1-k}{\alpha }_{n-k}\sum _{l=0}^d{\gamma }_{n-k}^l{P}_{n-k-l}^{\prime }(x)-\sum _{k=1}^{n-1}{R}_{k+1}\frac{1}{n+1-k}\sum _{k^{\prime }=1}^{n-k-1}{R}_{k^{\prime }+1}{\alpha }_{n-k-k^{\prime }}{P}_{n-k-k^{\prime }}^{\prime }(x)$$

$$\frac{1}{n+1}{\alpha }_n{\gamma }_n^0=0,\text{ for }n\ge 0,$$

$${\gamma }_n^0=0,\text{ for }n\ge 0.$$

Full text

**Keywords: ** Generating functions, $d$-orthogonal polynomials; recurrence relations; generalized hypergeometric series.

**AMS Subject Classification: ** 12E10, 33C47; 33C20

1 Introduction

In [1, 3, 4] the authors used different methods to show that the orthogonal polynomials defined by a generating function of the form $F(xt-\alpha t^{2})$ are the ultraspherical and Hermite polynomials. On the other hand, the author in [2] found (even if $F$ is a formal power series) that the orthogonal polynomials are the ultraspherical, Hermite and Chebychev polynomials of the first kind. Motivated by the problem, posed in [2], of describing (all or just orthogonal) polynomials with generating functions $F(xU(t)-R(t))$ we have generalized in [14] the above results by proving the following:

Theorem 1

[14]** Let $F(t)=\sum_{n\geq 0}\alpha_{n}t^{n}$ and $R(t)=\sum_{n\geq 1}R_{n}t^{n}/n$ be formal power series where $\{\alpha_{n}\}$ and $\{R_{n}\}$ are complex sequences with $\alpha_{0}=1$ and $R_{1}=0$. Define the polynomial set $\{P_{n}\}_{n\geq 0}$ by

[TABLE]

If this polynomial set (which is automatically monic) satisfies the three-term recursion relation

[TABLE]

*where $\{\beta_{n}\}$ and $\{\omega_{n}\}$ are complex sequences, then we have:

a) If $R_{2}=0$ and $\alpha_{n}\neq 0$ for $n\geq 1$, then $R(t)=0$, $F(t)$ is arbitrary and $F(xt)=\sum_{n\geq 0}\alpha_{n}x^{n}t^{n}$ generates the monomials $\{x^{n}\}_{n\geq 0}$.

b) If $\alpha_{1}R_{2}\neq 0,$ then $R(t)=R_{2}t^{2}/2$ and the polynomial sets $\{P_{n}\}_{n\geq 0}$ are the rescaled ultraspherical, Hermite and Chebychev polynomials of the first kind.*

Note that, the polynomials in Theorem 1 which satisfy a three term recursion with complex coefficients are not necessary orthogonal with respect to a moment functional $\mathcal{L}$, i.e. for all non negative integers $m,n$; $\mathcal{L}(P_{m}(x)P_{n}(x))=0$ if $m\neq n$ and $\mathcal{L}(P^{2}_{n}(x))\neq 0$, see Definition 2.2 in [7].

Remark 1

**

*a) A polynomial set PS, $\{P_{n}\}_{n\geq 0}$, is such that $degree(P_{n})=n$,  $n\geq 0$.

b) A PS is called a monic PS if $P_{n}(x)=x^{n}+\cdots$, for $\;n\geq 0$.

c) The choice $\alpha_{0}=1$ and $R_{1}=0$ comes from the fact that the generating function $\gamma_{1}+\gamma_{2}F((x+R_{1})t-R(t))=\gamma_{1}+\gamma_{2}\sum_{n\geq 0}\alpha_{n}P_{n}(x+R_{1})t^{n}$, with $\gamma_{1}$ and $\gamma_{2}$ constants, is also of type (1).*

In the present paper, we are interested in monic PSs generated by (1) (with $F(t)$ and $R(t)$ as in Theorem 1) and satisfying higher order recurrence relations (7). For this purpose, we adopt the following definitions:

Definition 1

Let $d\in\mathbb{N}$. A PS $\{Q_{n}\}_{n\geq 0}$ is called a $d$-polynomial set $d$-PS if its corresponding monic PS $\{P_{n}\}_{n\geq 0}$, defined by $P_{n}(x)=(\lim_{x\rightarrow+\infty}x^{-n}Q_{n}(x))^{-1}Q_{n}(x)$, $n\geq 0$, satisfies the $(d+1)$-order recurrence relation:

[TABLE]

where

[TABLE]

and

[TABLE]

Definition 2

Let $\{P_{n}\}_{n\geq 0}$ be a $d$-PS. If the PS of the derivatives $\{(n+1)^{-1}P_{n+1}^{{}^{\prime}}\}_{n\geq 0}$ is also a $d$-PS, then $\{P_{n}\}_{n\geq 0}$ is called a classical $d$-PS.

Definition 3

[9]** Let $\omega=\exp(2i\pi/(d+1))$, where $i^{2}=-1$. The PS $\{P_{n}\}_{n\geq 0}$ is called $d$-symmetric if it fulfils:

[TABLE]

Remark 2

In Definition 1:

*a) For $d\geq 1$, the first terms $\{\gamma_{n}^{l}\}_{0\leq n<l\leq d}$ of the sequences $\{\gamma_{n}^{l}\}_{n\geq 0}$, $1\leq l\leq d$, are arbitrary.

b) For $d=0$, (7) becomes $xP_{n}(x)=P_{n+1}(x)+\gamma_{n}^{0}P_{n}(x),n\geq 0,$ with $P_{0}(x)=1$. Here $\{\gamma_{n}^{0}\}_{n\geq 0}$ can be the null sequence, so the set of monomials is a [math]-PS.*

An interesting class of $d$-PSs characterized by (7), with the additional condition $\gamma_{n}^{d}\neq 0$ for $n\geq d$, are the $d$-orthogonal polynomial sets $d$-OPSs [12, 15]. In this context, the authors in [4] generalized the result stated in [1, 3] by showing the following:

Theorem 2

[4]** The only $d$-OPSs generated by $G((d+1)xt-t^{d+1})$ are the classical d-symmetric polynomials.

Another contribution concerns $d$-OPSs with generating functions of Sheffer type, i.e. of the form $A(t)\exp(xH(t))$. We have

Theorem 3

[16]** Let $\rho_{d}(t)=\sum_{k=0}^{d}\tilde{\rho}_{k}t^{k}$ be a polynomial of degree d $(\tilde{\rho}_{d}\neq 0)$ and $\sigma_{d+1}(t)=\sum_{k=0}^{d+1}\tilde{\sigma}_{k}t^{k}$ be a polynomial of degree less than or equal to $d+1$. The only PSs, which are $d$-orthogonal and also Sheffer PS, are generated by

[TABLE]

with the conditions

[TABLE]

Note that Theorem 3 characterizes also the $d$-OPSs with generating functions of the form $F(xH(t)-R(t))$ with $F(t)=\exp(t)$, since $A(t)\exp(xH(t))=F(xH(t)-R(t))$ where $R(t)=-\ln(A(t))$. For $H(t)=t$ we meet the Appell case with the Hermite $d$-OPSs [8] generated by $\exp(xt-\bar{\rho}_{d+1}(t))$ where $\bar{\rho}_{d+1}(t)$ is a polynomial of degree $d+1$.

As a consequence of the results obtained in this paper, we give some generalizations of Theorems 1, 2 and the Appell case in Theorem 3 (see also [8]) to $d$-PS generated by (1).

After this short introduction, we give in section 2 some results for $d$-PSs generated by (1). Then in section 3 we show that the only $d$-symmetric $d$-PSs generated by (1) are the classical d-symmetric polynomials. For this later case, we give in section 4 the $(d+1)$-order recurrence relation (7) and the expression of $F(t)$ by means of hypergeometric functions.

2 Some general results

The results in this section concern all $d$-PSs generated by (1). The central result is Proposition 2 below from which the other results arise. First we have

Proposition 1

[14]**

Let $\{P_{n}\}_{n\geq 0}$ be a PS generated by (1). Then we have

[TABLE]

Secondly

Proposition 2

Let $\{P_{n}\}$ be a $d$-PS generated by (1) and satisfying (7), (8) and (9), with $\alpha_{n}\neq 0$ for $n\geq 1$. Putting

[TABLE]

then we have:

a)

[TABLE]

b)

[TABLE]

or equivalently

[TABLE]

c)

[TABLE]

d.i)

[TABLE]

d.ii)

[TABLE]

d.iii)

[TABLE]

Proof:  By differentiating (7) we get

[TABLE]

Then by making the operations $n\alpha_{n}Eq\eqref{gf2}+Eq\eqref{gf1}$ and $Eq\eqref{gf1}-\alpha_{n}Eq\eqref{gf2}$ we obtain, respectively,

[TABLE]

and

[TABLE]

Inserting (13) in the left-hand side of the equation $Eq\eqref{gf4}$ multiplied by $x$ we obtain

[TABLE]

Using (23) and (22) respectively in the left hand side and right hand side of (24) we get

[TABLE]

It follows that

[TABLE]

a) By comparing the coefficients of $P_{n+1}^{\prime}(x)$ in the both sides of (2) we obtain

[TABLE]

and then

[TABLE]

b) Equating the coefficients of $P^{\prime}_{n}(x)$ in the both sides of the equation (2) gives

[TABLE]

which can be written as

[TABLE]

Now by equating the coefficients of $P^{\prime}_{n+1-k}(x)$ for $k\geq 2$ in the both sides of the equation (2) we obtain

[TABLE]

then by taking $k=2$ in (2) we retrieve c) and by considering $3\leq k\leq d$, $d+1\leq k\leq 2d+1$ and $k\geq 2d+2$ we obtain d.i.), d.ii.) and d.iii.) respectively.

■

In the following corollaries we adopt the same conditions and notations as in Proposition 2.

Corollary 1

If $R_{2}=R_{3}=\cdots=R_{d+1}=0$ and $\alpha_{n}\neq 0$ for $n\geq 1$, then $R(t)=0$, $F(t)$ is arbitrary and $F(xt)=\sum_{n\geq 0}\alpha_{n}x^{n}t^{n}$ generates the monomials $\{x^{n}\}_{n\geq 0}$.

Proof:  As $R_{1}=R_{2}=\cdots=R_{d+1}=0$, it is enough to show by induction that $R_{n}=0$ for $n\geq d+2$. For $n=1,2,...,d+2$, the equation (13) gives $P_{1}(x)=x$, $P_{2}(x)=x^{2},...,P_{d+1}(x)=x^{d+1}$ and $P_{d+2}(0)=\frac{-R_{d+2}\alpha_{1}}{(d+2)\alpha_{d+2}}$. But according to equation (7), for $n=d+1$, $P_{d+2}(0)=0$ and then $R_{d+2}=0$.

Now assume that $R_{k}=0$ for $d+2\leq k\leq n-1$. According to (13) we have, for $d+2\leq k\leq n-1$, $P_{k}(0)=0$ and $P_{n}(0)=-\frac{R_{n}\alpha_{1}}{n\alpha_{n}}$. On other hand, by the shift $n\rightarrow n-1$ in (7) we have $P_{n}(0)=0$ and thus $R_{n}=0$. As $R(t)=0$, the generating function (1) reduces to $F(xt)=\sum_{n\geq 0}\alpha_{n}x^{n}t^{n}$ which generates the monomials with $F(t)$ arbitrary. ■

Corollary 2

If $R_{d+2}=R_{d+3}=\cdots=R_{2d+2}=0$ then $R(t)=R_{2}t^{2}/2+R_{3}t^{3}/3+\cdots+R_{d+1}t^{d+1}/(d+1)$.

Proof:  We will use (2) and proceed by induction on $k$ to show that $R_{k}=0$ for $k\geq 2d+3$. Indeed $k=2d+2$ and $n=2d+2$ in (2) leads to $a_{2d+2}R_{2d+3}=0$ and since $a_{n}\neq 0$ we get $R_{2d+3}=0$. Suppose that $R_{2d+3}=R_{2d+4}=\cdots=R_{k}=0$, then for $n=k$ the equation (2) gives $a_{k}R_{k+1}=0$ and finally $R_{k+1}=0$. ■

Corollary 3

If $R_{\kappa+d+1}=\cdots=R_{\kappa+1}=R_{\kappa}=R_{\kappa-1}=\cdots=R_{\kappa-d}=0$ for some $\kappa\geq 3d+3$, then $R_{d+2}=R_{d+3}=\cdots=R_{2d+2}=0$.

Proof:

$\bullet$ Let $k=\kappa$ in (2), then for $n\geq\kappa$ the fraction $\sum_{l=1}^{\kappa-2}\frac{R_{l+1}R_{\kappa-l}}{n-l+1}$, as function of integer $n$, is null even for real $n$. So,

[TABLE]

which is $R_{d+2}R_{\kappa-d-1}=0$ when $l=d+1$. Supposing $R_{d+2}\neq 0$ leads to $R_{\kappa-d-1}=0$. So $R_{\kappa+d}=R_{\kappa+d-1}=\cdots=R_{\kappa-d}=R_{\kappa-d-1}=0$ and with the same procedure we find $R_{\kappa-d-2}=0$. Going so on till we arrive at $R_{d+2}=0$ which contradicts $R_{d+2}\neq 0$.

$\bullet$ By taking successively $k=\kappa+r,\kappa+r-1,...,\kappa$ in (2), for $1\leq r\leq d$, we find

[TABLE]

If $R_{d+2+r}\neq 0$ then by taking $l=d+1+r$ we get $R_{\kappa-d-1}=R_{\kappa-d-2}=\cdots=R_{\kappa-d-r-1}=0$. So $R_{\kappa+d-r}=R_{\kappa+d-r-1}=\cdots=R_{\kappa-d-r-1}=0$ and with the same procedure we find $R_{\kappa-d-r-2}=R_{\kappa-d-r-3}=\cdots=R_{\kappa-d-2r-2}=0$. Going so on till we arrive at $R_{d+2+r}=0$ which contradicts $R_{d+2+r}\neq 0$. ■

Corollary 4

If $a_{n}$ is a rational function of $n$ then $R_{d+2}=R_{d+3}=\cdots=R_{2d+2}=0$.

Proof:  From (16), (17) and (18) observe that $c_{n}^{l}$ will also be a rational function of $n$. Then it follows that, in (2), two fractions are equal for natural numbers $n\geq k$, $k\geq 2d+2$, and consequently will be for real numbers $n$. If we denote by $N_{s}(G(x))$ the number of singularities of a rational function $G(x)$ then we can easily verify, for all rational functions $G$ and $\tilde{G}$ of $x$ and a constant $a\neq 0$, that:

a) $N_{s}(G(x+a))=N_{s}(G(x))$,

b) $N_{s}(aG(x))=N_{s}(G(x))$,

c) $N_{s}(G(x)+\tilde{G}(x))\leq N_{s}(G(x))+N_{s}(\tilde{G}(x))$.

Using property a) of $N_{s}$ we have

[TABLE]

According to properties b) and c) of $N_{s}$, the $N_{s}$ of the left-hand side of (2) is finite and independent of $k$. Thus, the right-hand side of (2) has a finite number of singularities which is independent of $k$. As consequence there exists a $k_{1}\geq 3d+3$ for which $R_{l+1}R_{k-l}=0$ for all $k\geq k_{1}-d-1$ and $k_{1}-d-1\leq l\leq k$. According to Corollary 1, there exists a $k_{0}$ such that $2\leq k_{0}\leq d+1$ and $R_{k_{0}}\neq 0$. So, taking successively $k=k_{0}+l$ with $l=k_{1}+d,k_{1}+d-1,...,k_{1}-d-1$ we get $R_{k_{1}+d+1}=R_{k_{1}+d}=\cdots=R_{k_{1}-d}=0$. Then, by Corollary 3 we have $R_{d+2}=R_{d+3}=\cdots=R_{2d+2}=0$.

■

The fact that $a_{n}$ is a rational function of $n$ means that $F(\epsilon z)=\sum_{n\geq 0}\alpha_{n}(\epsilon z)^{n}$ (where $\epsilon$ is the quotient of the leading coefficients of the numerator and the denominator of $a_{n}$) is a generalized hypergeometric series, i.e. of the form:

[TABLE]

where $\left(\mu_{l}\right)_{l=k}^{p}$ denotes the array of complex parameters $\mu_{k},\mu_{k+1},...,\mu_{p}$, and if $k>p$ we take the convention that $\left(\mu_{l}\right)_{l=k}^{p}$ is the empty array. The symbol $(\mu)_{n}$ stands for the shifted factorials, i.e.

[TABLE]

As an interesting consequence, from Corollary 4 and Corollary 2 we state the following result, which can be interpreted as a generalization of the Appell case in the above Theorem 3 (see also [8]):

Theorem 4

Let $\{P_{n}\}$ be a $d$-PS generated by (1) with $F(z)$ a generalized hypergeometric series. Then $R(t)=R_{2}t^{2}/2+R_{3}t^{3}/3+\cdots+R_{d+1}t^{d+1}/(d+1)$.

Proof:  $F(z)=\sum_{n\geq 0}\alpha_{n}z^{n}$ has the form (33). Then $a_{n}=\alpha_{n}/\alpha_{n+1}$ is a rational function of $n$, since $(\mu)_{n+1}/(\mu)_{n}=n+\mu$. The use of Corollary 4 and Corollary 2 completes the proof. ■

Corollary 5

*Let $R(t)=R_{2}t^{2}/2+R_{3}t^{3}/3+\cdots+R_{d+1}t^{d+1}/(d+1)$. Then,

i) If $R_{d+1}=0$ we have $c_{n+d}^{d}c_{n}^{d}=0$, for $n\geq d+1$.

ii) If $R_{d+1}\neq 0$ then*

[TABLE]

*where $b_{md+r}=(b_{d+r}-b_{r})m+b_{r},\;for\;m\geq 0,\;1\leq r\leq d$.

iii) The $\{c_{n}^{m}\}_{1\leq m\leq d-1}$ can be calculated recursively by solving the following $d$-order linear difference equations:*

[TABLE]

Proof:  

The proof of i)

Put $k=2d+1$ in (2) to get the following Riccati equation for $\{c_{n}^{d}\}$:

[TABLE]

By taking $R_{d+1}=0$ in (37), i) follows immediately.

The proof of ii)

Substituting (35) in (37) we find the $2d$-linear homogeneous equation

[TABLE]

By writing $n=md+r$, where $m,r$ are natural numbers with $1\leq r\leq d$, the equation (38) can be solved by summing twice to find that

[TABLE]

The proof of iii)

Since $d+1\leq k\leq 2d+1$ we have $R_{k+1}=0$ and $R_{k-l}=0$ for $l\leq k-2-d$. So, we can write (2) as

[TABLE]

Putting $m=k-d-1\neq 0$ in (2) and rearranging we obtain (5). ■

Remark 3

In the case of $d$-OPSs, in Corollary 5 the polynomial $R(t)$ is of degree $d+1$. Otherwise (i.e. $R_{d+1}=0$), we have a contradiction with the regularity conditions $\gamma_{n}^{d}\neq 0$, for $n\geq d$.

Corollary 6

The $d$-PS is classical if and only if $R(t)=R_{2}t^{2}/2+R_{3}t^{3}/3+\cdots+R_{d+1}t^{d+1}/(d+1)$ with $R_{d+1}\neq 0$.

Proof:  

1. Assume that the $d$-PS is classical. From (22) and Definition 2 we have $R_{k+1}=0$ for $d+1\leq k\leq n-1$. We get $R(t)$ by taking $n\geq 2d+2$ and using Corollary 2. Now we show that $R_{d+1}\neq 0$. Equation (22) becomes

[TABLE]

where $Q_{n}(x)=(n+1)^{-1}P^{\prime}_{n+1}(x)$ and

[TABLE]

From (35), if $R_{d+1}=0$ then $\gamma_{n+1}^{d}=0$, and (41) gives $\tilde{\gamma}_{n}^{d}=0$, for $n\geq d$. So, $\{Q_{n}\}$ is not a $d$-PS which contradicts the fact that $\{P_{n}\}$ is classical (see Definition 2).

2. Assume that $R(t)=R_{2}t^{2}/2+R_{3}t^{3}/3+\cdots+R_{d+1}t^{d+1}/(d+1)$ with $R_{d+1}\neq 0$ , then the PS of the derivatives $\{Q_{n}\}$ satisfy (40) and are generated by $F^{\prime}(xt-R(t))=\sum_{n\geq 0}(n+1)\alpha_{n+1}Q_{n}(x)t^{n}$. Using Corollary 5 we find that $\tilde{c}_{n}^{d}:=(n+1)\alpha_{n+1}\tilde{\gamma}_{n}^{d}/((n-d+1)\alpha_{n-d+1})$ satisfies (37). And according to the same expression (37), we should have, if $c_{n}^{d}=0$ or $\gamma_{n}^{d}=0$ (for $n\geq d+1$), $R_{d+1}=0$. Therefore, there exists for $\tilde{c}_{n}^{d}$, since $R_{d+1}\neq 0$, a $n_{0}\geq d+1$ such that $\tilde{c}_{n_{0}}^{d}\neq 0$ or $\tilde{\gamma}_{n_{0}}^{d}\neq 0$. This means that $\{P_{n}\}$ is classical.

■

3 The $d$-symmetric case

The main result of this section is the following:

Theorem 5

If $\{P_{n}\}$ is a $d$-symmetric $d$-PS generated by (1) then $R(t)=R_{d+1}t^{d+1}/(d+1)$.

Theorem 5 generalizes Theorem 1 and Theorem 2 mentioned above. Its proof is quite similar to that of Theorem 1 in [14] and it requires the following Lemmas.

Lemma 1

If $\{P_{n}\}$ is a $d$-symmetric $d$-PS generated by (1) then

[TABLE]

Proof:  Let $\{P_{n}\}$ be a $d$-symmetric $d$-PS satisfying (7) and generated by (1). Then it has, according to Definition 3, the property

[TABLE]

where $\omega=\exp(2\pi i/(d+1))$. It follows that (7) becomes [9]

[TABLE]

Let us show that $R_{k}=0$ when $k$ is not a multiple of $d+1$. First we replace $x$ by $\omega x$ in (13) and use (43) with $P^{\prime}_{n}(\omega x)=\omega^{n-1}P^{\prime}_{n}(x)$ to get

[TABLE]

Subtracting (47) from (13) gives

[TABLE]

which leads to

[TABLE]

Since $\omega^{k}\neq 1$, provided $k$ is not a multiple of $d+1$, gives the result. ■

By Lemma 1 and putting $T_{k}=R_{k(d+1)}$ for $k\geq 0$, the equations in Proposition 2 simplify to particular forms. Indeed, from (16), (17) and (18) we get

[TABLE]

The equation (2), with $k=2d+1$, becomes

[TABLE]

which by (49) takes the form

[TABLE]

Finally, the equation (2) simplifies to

[TABLE]

This equation will be denoted by $E_{k}(n)$ in below.

Lemma 2

If $T_{2}=0$ then $R(t)=T_{1}t^{d+1}/(d+1)$.

Proof:  According to Corollary 2, if $T_{2}=R_{2(d+1)}=0$ then $R(t)=T_{1}t^{d+1}/(d+1)$, since in this case we have $R_{d+2}=R_{d+3}=\cdots=R_{2d+1}=0$. ■

Lemma 3

If $T_{m}=T_{m+1}=0$ for some $m\geq 3$, then $T_{2}=0$.

Proof:  $T_{m}=T_{m+1}=0$ means that $R_{(d+1)m}=R_{(d+1)(m+1)}=0$. Also by Lemma 1, we have $R_{(d+1)m+d}=R_{(d+1)m+d-1}=\cdots=R_{(d+1)m+1}=0$ and $R_{(d+1)m-1}=R_{(d+1)m-2}=\cdots=R_{(d+1)m-d}=0$ which represents the condition of corollary 3 with $\kappa=(d+1)m\geq 3(d+1)$ and therefore gives $R_{2d+2}=T_{2}=0$. ■

Lemma 4

If $T_{\kappa}=T_{m}=0$ for some $\kappa\neq m\geq 3$, then $T_{2}=0$.

Proof:  The proof is similar to that of Corollary 7 in [14]. Let assume that $T_{\kappa+1}\neq 0$ and $T_{m+1}\neq 0$, since if not, we apply Corollary 3. When $m>\kappa$ and by using (3), the following operations

[TABLE]

give

[TABLE]

where $Q(n)$ is a rational function of $n$. Consequently, $a_{n}$ is a rational function of $n$ and by Corollary 4 we have $T_{2}=0$. ■

Lemma 5

The following equality is true for $k\geq 3\hbox{ and }n\geq k(d+1)+2d+1.$

[TABLE]

where

• •

$D_{k,l}=T_{k}T_{k-l+1}-T_{k+1}T_{k-l}$.

• •

$D_{k}=D_{k,1}=T_{k}^{2}-T_{k+1}T_{k-1}$.

• •

$V_{k,l}=\frac{T_{1}}{2}\left(T_{l}T_{k+1}D_{k-1,l-1}-T_{l+1}T_{k-1}D_{k,l}\right)$.

• •

$\tilde{a}_{n}=\frac{n}{n+1}a_{n}$.

Proof:  Just by making the following combinations it is easy to get (55):

[TABLE]

■

To prove Theorem 5 it is sufficient, according to Lemma 2, to show that $T_{2}=0$. To this end, we will consider three cases:

Case 1: There exists $k_{0}\geq 3$ such that $D_{k}\neq 0$ for $k\geq k_{0}$.

Considering Lemma 4, we can choose $\tilde{k}\geq k_{0}$ such that $T_{k}\neq 0$ for $k\geq\tilde{k}-1$. Let define, for $k\geq\tilde{k}$, $\bar{D}_{k}=\frac{D_{k}}{T_{k-1}T_{k}}$ and $\bar{E}_{k}(n)$ be the equation (55) divided by $T_{k-1}T_{k}T_{k+1}$. By making the operations

[TABLE]

we get, for $k\geq\tilde{k}+1$, the equation

[TABLE]

where $W_{k,l}$ is independent of $n$ and

[TABLE]

Similarly, by the operations

[TABLE]

and the shift $n\rightarrow n+(d+1)k-1$ in (57) we obtain

[TABLE]

where $\widetilde{W}_{k,l}$ is independent of $n$. Now, for $k\neq\kappa\geq\tilde{k}+1$, the equations (56) and (58) give, respectively,

[TABLE]

and

[TABLE]

If $\tilde{D}_{k}\neq\tilde{D}_{\kappa}$ for some $k\neq\kappa\geq\tilde{k}+1$, then by (59) and (60) we can eliminate $a_{n-d-1}$ to get that $a_{n}$ is a rational function of $n$. So, by Corollary 4, we have $T_{2}=0$.

If $\tilde{D}_{k}=D$ for $k\geq\tilde{k}+1$, then (56) and (58) become, respectively,

[TABLE]

and

[TABLE]

The combinations $\left((n+d+2)Eq\eqref{Dk66}-(n+d+1)Eq\eqref{Dk6}\right)/(d+1)$ and $((n-2d-1)Eq\eqref{Dk66}-(n-2d-2)Eq\eqref{Dk6})/(d+1)$ give, respectively,

[TABLE]

and

[TABLE]

By shifting $n\rightarrow n+d+1$ in (63) we obtain

[TABLE]

The coefficients $a_{n+d+1}$ and $a_{n-d-1}$ can be eliminated by the operations $D\times Eq\eqref{Dk9}-3\,Eq\eqref{Dk10}$ and $3\,Eq\eqref{Dk9}-D\times Eq\eqref{Dk10}$ leaving us with

[TABLE]

and

[TABLE]

Finally, the shifting $n\rightarrow n-d-1$ in (67) leads to

[TABLE]

and the operation $(6D-2D^{2})Eq\eqref{Dk11}+(D^{2}-9)Eq\eqref{Dk13}$ gives

[TABLE]

According to manipulations made above, $Q_{k}^{(7)}(n)$ is a rational function of $n$. As consequence, if $D\neq 3$, $a_{n}$ is a rational function of $n$ and then $T_{2}=0$.

Now, we explore the case $D=3$. According to the left-hand sides of (64) and (65), we have

[TABLE]

which can be written as

[TABLE]

By using, from (56) and (58), the expressions of $Q_{k}^{(1)}(n)$ and $\widetilde{Q}_{k}^{(1)}(n)$ with $W_{k,k+1}=\widetilde{W}_{k,k+1}=W_{k,0}=\widetilde{W}_{k,0}=0$ we obtain

[TABLE]

Observe that in (71) the singularities of the left hand side are different from those of the right hand side. So,

[TABLE]

and by induction on $l$, all the $W_{k,l}$ and $\widetilde{W}_{k,l}$ are null. Thus, (56) reads

[TABLE]

For $n=(d+1)m+r$, $m\geq 0$ and $0\leq r\leq d$, the solutions of (73) have the form

[TABLE]

where $C_{0,r},C_{1,r}$ and $C_{2,r}$ are constants. So, by Corollary 4 we get $T_{2}=0$.

Case 2: There exists $k_{0}\geq 3$ such that $D_{k}=0$ for $k\geq k_{0}$.

Suppose that $D_{k}=T_{k}^{2}-T_{k-1}T_{k+1}=0$ for all $k\geq k_{0}$. First, notice that if there exists a $k_{1}\geq k_{0}$ such that $T_{k_{1}}=0$, then $T_{k_{1}-1}T_{k_{1}+1}=0$. Then, $T_{k_{1}-1}=0$ or $T_{k_{1}+1}=0$ and by Corollary 3, $T_{2}=0$. We have also $T_{k_{0}-1}\neq 0$, otherwise $T_{k_{0}}=0$ and by Corollary 3, $T_{2}=0$.

Now, for $T_{k}\neq 0$ $(k\geq k_{0}-1)$, we have

[TABLE]

This means that

[TABLE]

where $a=T_{k_{0}-1}^{k_{0}}/T_{k_{0}}^{k_{0}-1}\neq 0$ and $b=T_{k_{0}}/T_{k_{0}-1}\neq 0$.

The substitution $T_{k}=ab^{k}$ in (3) for $k\geq k_{0}$ leads to the equation

[TABLE]

Let denote (77) by $\widetilde{E}\left(k,n\right)$ and make the subtraction $\widetilde{E}\left(k+1,n+d+1\right)-\widetilde{E}\left(k,n\right)$ to get

[TABLE]

On the right hand side of (78) we have, for $k\geq k_{0}$, the expression

[TABLE]

from which we deduce

[TABLE]

Now since the left hand side of equation (78) is independent of $k$, it follows

[TABLE]

As a result, for $2\leq l\leq k$ and $k\geq k_{0}$, we have

[TABLE]

Let take $k=2\left(k_{0}-2\right)-1$ and $l=k_{0}-2$ to get $\left(T_{k_{0}-1}-bT_{k_{0}-2}\right)^{2}=0$ and then $T_{k_{0}-1}=bT_{k_{0}-2},$ (or equivalently $D_{k_{0}-1}=0$). Thus, the equations (75) and (76) are valid for $k=k_{0}-1$ and by induction we arrive at $T_{4}=bT_{3},$ (or equivalently $D_{4}=0$). For $k=4$, the right-hand side of (55) is null. Consequently, $V_{4,2}=0$ and using $T_{5}=T_{4}^{2}/T_{3}$ (from $D_{4}=0$) we get $D_{3}=0$. On the other side (when $T_{2}\neq 0$) we can write

[TABLE]

where $b=T_{3}/T_{2}\neq 0$ and $a=T_{2}^{3}/T_{3}^{2}\neq 0$. Therefore, the equation (77) reads

[TABLE]

When $n=(d+1)k+d$ and $n=(d+1)(k+1)$, the equation (83) gives

[TABLE]

and

[TABLE]

respectively. Let take $n=(d+1)N+d$ in (83) and use (84) to obtain the expression

[TABLE]

In this last equality let put $N-k$ instead of $k$ to get

[TABLE]

After defining $A_{1}=\frac{a_{1}}{d+1}+\frac{d+2}{(d+3)b}c_{d+2}^{d}+\frac{T_{1}}{(d+3)b}$ , $A_{2}=-\frac{d+1}{d+2}\frac{c_{d+1}^{d}}{b}-\frac{T_{1}}{(d+2)b}$ and $A_{3}=\frac{T_{1}}{b}$, the operation

[TABLE]

leads to

[TABLE]

where the Digamma function $\Psi(x)$ as well as the short notations $B_{1}=\frac{2{A_{1}}-{A_{3}}}{d+1}-\frac{2a}{(d+1)^{2}}\Psi\left(2+\frac{2}{d+1}\right)$, $B_{2}=\frac{-dA_{1}+(d-1)A_{3}}{d(d+1)}+\frac{a}{(d+1)^{2}}\Psi\left(2+\frac{2}{d+1}\right)$ and $B_{3}=\frac{dA_{2}+A_{3}}{d(d+1)}+\frac{a}{(d+1)^{2}}\Psi\left(2+\frac{1}{d+1}\right)$ are introduced. Taking

[TABLE]

and

[TABLE]

then (87) can be written in compact form as

[TABLE]

The later recurrence is easily solved to give

[TABLE]

By using the formula $\Psi(j+1)=\Psi(j)+1/j$ and the relations [13, Theorems 3.1 and 3.2]

[TABLE]

[TABLE]

we obtain

[TABLE]

where

[TABLE]

From (91) we deduce the asymptotic behaviour of $a_{(d+1)k}$ as $k\to\infty$:

[TABLE]

where coefficients $\delta_{i}$ are defined by (higher terms are omitted)

[TABLE]

At this level we should remark that $\lim_{k\to\infty}a_{2k}=\infty$ for all $\delta_{i}$, $i=1,2,3,...,$ since $a\neq 0$. Recall that $c_{n}^{d}=\frac{T_{1}}{d+1}\left((n-d+1)\frac{a_{n}}{a_{n-d}}-(n-d)\right),\;\;\text{for }n\geq d$, then the equation (84) can be written as

[TABLE]

where

[TABLE]

and the equation (85) can be written

[TABLE]

where

[TABLE]

From (93) and (94) we have

[TABLE]

and

[TABLE]

which give an explicit formula for $a_{(d+1)(k+1)+d}$ and $a_{(d+1)k+1}$.

$\bullet$ If we suppose $\lim_{k\to\infty}\frac{a_{(d+1)k}}{(d+1)k}=\infty,$ then from (95) and (96) we deduce on one side

[TABLE]

and

[TABLE]

On the other side, for $n=(d+1)k+2d+1$, (51) reads

[TABLE]

Under the assumption $T_{2}\neq 0$, (99) admits the limit $\infty=0$, as $k\to\infty$, which exhibit a contradiction.

$\bullet$ Now if $\lim_{k\to\infty}\frac{a_{(d+1)k}}{(d+1)k}=\eta_{1}\neq 0,$ then from (95) and (96) we have

[TABLE]

and

[TABLE]

By taking the limit in (99) we obtain,

[TABLE]

If $\eta_{2}=0,$ we have the contradiction $T_{2}=0$. But if $\eta_{2}\neq 0,$ then $\eta_{2}=1/\eta_{1}$. From (100) we get $(d+1)b/T_{1}+1/\eta_{1}=\eta_{2}$, which gives the contradiction $b=0$.

$\bullet$ Finally if $\lim_{k\to\infty}\frac{a_{(d+1)k}}{(d+1)k}=0,$ then according to (92), we have $\delta_{1}=\delta_{2}=0$ and

[TABLE]

Let write the equation (99) as

[TABLE]

After multiplying the both sides of the equation (103) by $\phi(k+1)a_{(d+1)(k+1)}$ and using (102), (95) and (96), we get, as $k\to\infty$,

[TABLE]

where

[TABLE]

and

[TABLE]

So, we must have $\lambda_{i}=0$, $i=1,2,3,...$. As $a\neq 0$ and $b\neq 0$, from $\lambda_{1}=0$ we get

[TABLE]

and by replacing $a=T_{2}/b^{2}$ and (105) in the equation $\lambda_{2}=0$ we obtain

[TABLE]

If $d\neq 1$, (106) gives $T_{1}=0$ which is a contradiction. The case $d=1$ is already treated, [14, Theorem 1].

Case 3: For every $k_{0}\geq 3$, there exists infinitely many $k,\kappa\geq k_{0}$ such that: $D_{k}=0$ and $D_{\kappa}\neq 0$.

We take $k_{1}$ and $k_{2}$, $k_{1}\neq k_{2}$, with $D_{k_{1}}=0,D_{k_{1}+1}\neq 0$, $D_{k_{2}}=0$ and $D_{k_{2}+1}\neq 0$ to get from (55) that

[TABLE]

are two rational functions of $n$. Consequently, an analogous reasoning to that of Lemma 4 completes the proof.

In the next subsection we give some expressions concerning the sequence $\{\gamma_{n}^{d}\}_{n\geq d}$ and the power series $F(t)$. The later is expressed by hypergeometric series (33).

3.1 Expressions for $\gamma_{n}^{d}$ and $F(t)$

Proposition 3

The $d$-symmetric $d$-PS, $\{P_{n}\}$, generated by (1) satisfies

[TABLE]

with

[TABLE]

and for $n=dm+r$, $m\geq 1$, $1\leq r\leq d$ we have

[TABLE]

with $\gamma_{d}^{d}=T_{1}d!/\prod_{l=1}^{d}b_{l}$, $b_{l}=(l+1)\alpha_{l+1}/\alpha_{l}$ and $\beta_{l}=b_{d+l}-b_{l}$, $1\leq l\leq d$.

Proof:  The equation (49) is

[TABLE]

As $a_{n}=\alpha_{n}/\alpha_{n+1}$ we get (111).

According to Theorem 5 we have $R(t)=T_{1}t^{d+1}/(d+1)$. So, by Corollary 5 we obtain

[TABLE]

with $b_{n}=(n+1)/a_{n}=(n+1)\alpha_{n+1}/\alpha_{n}$ and

[TABLE]

where $\beta_{r}=b_{d+r}-b_{r}$, for $1\leq r\leq d$.

The equation (114) gives

[TABLE]

We calculate $\alpha_{n}b_{n}/\alpha_{n-d}$ by using the relation $\alpha_{n}=b_{n-1}\alpha_{n-1}/n=\prod_{l=0}^{n-1}b_{l}/n!$ to find

[TABLE]

Now for $n=md+r$, we can write

[TABLE]

and

[TABLE]

Finally, (111) gives

[TABLE]

and (112) follows by combining (117), (118) and (119). ■

Proposition 4

If $\prod_{l=1}^{d}\beta_{l}\neq 0$ then $F(t)=1+F_{1}(t)+F_{2}(t)$, where

[TABLE]

and

[TABLE]

Furthermore, if $\tilde{\beta}_{d}:=b_{d}-\beta_{d}=2b_{d}-b_{2d}\neq 0$, then $F_{1}(t)$ in (122) can be written as

[TABLE]

Proof:

Recall that $\alpha_{n}=\prod_{l=0}^{n-1}b_{l}/n!$. If $\prod_{l=1}^{d}\beta_{l}\neq 0$, then for $n=md+r$ and using the expression of $b_{md+r}$, we obtain

[TABLE]

We have also, for $m\geq 0$ and $0\leq r\leq d-1$, the expressions [6, Lemma 3.3]

[TABLE]

and

[TABLE]

So,

[TABLE]

[TABLE]

and, if $\tilde{\beta}_{d}\neq 0$,

[TABLE]

Now, expanding $F(t)$ as

[TABLE]

the expressions (122), (125) and (126) follow from (128), (129) and (130), respectively. ■

Remark 4

In proposition 4 two expressions of $F(t)$ are given. The first, when $\prod_{l=1}^{d}\beta_{l}\neq 0$, Equations (122) and (125), from which we can deduce the other limiting cases by tending to zero at least a constant $\beta_{r}$, $1\leq r\leq d$. So, we can enumerate $2^{d}$ expressions of $F(t)$ similar to that given in [6, Theorem 3.1]. The second, when $\prod_{l=1}^{d}\beta_{l}\neq 0$ and $\tilde{\beta}_{d}=b_{d}-\beta_{d}\neq 0$, Equations (126) and (125), seems to be a new representation of $F(t)$. Similarly, the other limiting cases can be obtained by tending to zero at least a constant $\beta_{r}$, $1\leq r\leq d$, and $\tilde{\beta}_{d}$. So, in this second representation, we can enumerate $3\cdot 2^{d-1}(=2^{d}+2^{d-1})$ expressions of $F(t)$. We note that, the resulting $2^{d-1}$ expressions when $\tilde{\beta}_{d}\to 0$, i.e. $\beta_{d}\to b_{d}$, are special cases of the first representation when $\beta_{d}=b_{d}$. See the illustrative examples given below.

Example 1

*If $d=1$ then $R(t)=T_{1}t^{2}/2$, $b_{n}=(b_{2}-b_{1})n+2b_{1}-b_{2}=\beta_{1}n+\tilde{\beta}_{1}$, for $n\geq 1$, and $F(t)$, $\gamma_{n}^{d}$ have the expressions:

1. If $\beta_{1}\neq 0$ we have*

[TABLE]

with $\gamma_{1}^{1}=T_{1}/b_{1}$ and for $n\geq 2$,

[TABLE]

The limiting case is

[TABLE]

*with $\gamma_{n}^{1}=T_{1}n/b_{1}$ for $n\geq 1$.

2. If $\tilde{\beta}_{1}\beta_{1}\neq 0$, [2, 14],*

[TABLE]

with

[TABLE]

The limiting cases are

[TABLE]

and

[TABLE]

Remark that (133) and (135) are special cases of (131) for $\tilde{\beta}_{1}\beta_{1}\neq 0$ and $\tilde{\beta}_{1}=0$,i.e. $\beta_{1}=b_{1}$, respectively. Also, (132) is exactly (134), since in this case $\beta_{1}\to b_{1}$.

Example 2

For $d\geq 1$ we take $b_{n}=\alpha n+\beta$ for all $n\geq 1$. So, from (115) we have $\beta_{r}=d\alpha$ and, of course, $b_{r}=\alpha r+\beta$, for $1\leq r\leq d$. Thus, for $\alpha\beta\neq 0$, $\gamma_{n}^{d}$ becomes

[TABLE]

and for $F(t)$, since $\alpha_{n}=\prod_{l=0}^{n-1}b_{l}/n!$, we obtain (see [2] for calculations)

[TABLE]

*Let $\lambda=\beta/\alpha$. Then for $T_{1}=\alpha^{d+1}(d+1)^{-d}$ and with the change of variable $t\to(d+1)t/\alpha$ in the generating function $\left(1-\alpha(xt-T_{1}t^{d+1}/(d+1))\right)^{-\lambda}$, we meet the Humbert polynomials [11] generated by $\left(1-(d+1)xt+t^{d+1}\right)^{-\lambda}$. For $d=1$ we have the ultraspherical polynomials.

*The limiting cases are

1. $\alpha\to 0$ and $\beta\neq 0$:

[TABLE]

*with $\gamma_{n}^{d}=T_{1}\beta^{-d}n!/(n-d)!$, for $n\geq d$. In the generating function $\exp\left(\beta\left(xt-T_{1}t^{d+1}/(d+1)\right)\right)$, with $T_{1}=\beta^{d}\left((d+1)d!\right)^{-1}$ and the change of variable $t\to t/\beta$, we find the generating function $\exp\left(xt-(d+1)^{-2}t^{d+1}/d!\right)$ of the Gould-Hopper polynomials [10] .

2. $\beta\to 0$ and $\alpha\neq 0$

[TABLE]

*with $\gamma_{d}^{d}=T_{1}\alpha^{-d}$ and $\gamma_{n}^{d}=T_{1}\alpha^{-d}/(d+1)$ for $n\geq d+1$.

Let $b=T_{1}\alpha^{-d}/(d+1)$. Then by the shift $n\to n+d$ in (110), these polynomials satisfy*

[TABLE]

where we recognise the monic Chebyshev $d$-OPS of the first kind generated by (see [5, Theorem 5.1])

[TABLE]

Remark that

[TABLE]

Then, by changing the variable $t\to\alpha t$, multiplying by $\alpha_{1}/\alpha$ and adding 1 in (144), we get the generating function (with $F(t)$ as in (139)),

[TABLE]

Example 3

For $d=2$ we have $R(t)=T_{1}t^{3}/3$ and from (112) we get, for $m\geq 1$, the two expressions

[TABLE]

and

[TABLE]

*with $\gamma_{2}^{2}=2T_{1}/(b_{1}b_{2})$, $\beta_{1}=b_{3}-b_{1}$, $\beta_{2}=b_{4}-b_{2}$ and $\tilde{\beta}_{2}=2b_{2}-b_{4}$.

We enumerate the following forms of $F(t)$:*

**A. The first representation by (122) and (125).

**

If $\beta_{1}\beta_{2}\neq 0$, then

[TABLE]

The limiting cases are obtained when: $\beta_{1}\to 0$, $\beta_{2}\to 0$ or $\left(\beta_{1},\beta_{2}\right)\to(0,0)$.

**B. The second representation, by (125) and (126), with its limiting cases.

**

1. If $\tilde{\beta}_{2}\beta_{1}\beta_{2}\neq 0$ we have

[TABLE]

2. If $\beta_{1}\to 0$:

[TABLE]

with

[TABLE]

3. If $\beta_{2}\to 0$:

[TABLE]

with

[TABLE]

4. If $\tilde{\beta}_{2}\to 0$:

[TABLE]

with $\gamma_{2}^{2}=2T_{1}/(b_{1}b_{2})$,

[TABLE]

*Clearly (176) is (152) with $\beta_{2}=b_{2}$.

5. If $\beta_{1}\to 0$ and $\beta_{2}\to 0$:

[TABLE]

with

[TABLE]

6. If $\beta_{1}\to 0$ and $\tilde{\beta}_{2}\to 0$:

[TABLE]

with $\gamma_{2}^{2}=2T_{1}/(b_{1}b_{2})$,

[TABLE]

Acknowledgements: We would like to thank Dr. Yanallah Abdelkader for precious help and useful discussions.