THE q-STIRLING NUMBERS, CONTINUED FRACTIONS AND

THE q-CHARLIER AND q-LAGUERRE POLYNOMIALS

By

Jiang ZENG

Abstract. — We give a simple proof of the continued fraction expansions of the ordinary generating functions of the q-Stirling numbers of both kinds. By generalizing the method of Touchard [To] and Milne [Mi1] we obtain the explicit formulas and measure of one set of the polynomials whose moments are the q-Stirling numbers.

1. Introduction

For q ∈ such that |q| < 1 define

[x] = \frac{q^{x} − 1}{q − 1}·

For n ≥ 0 let [x]_{0} = 1, [x]_{n} = [x] [x − 1] ⋯ [x − n + 1] and [n]! = [n]_{n}. Also, for n ≥ k ≥ 0 define

\left[{n \atop k}\right] = \frac{[n]!}{[k]![n − k]!}·

Recall that the two related q-Stirling numbers of the second kind [Go, WW] may be defined by recurrence as :

(1.1)  \begin{matrix}S_{q}(n + 1, k + 1) = S_{q}(n, k) + [k + 1]S_{q}(n, k + 1), \\ S_{q}(0, k) = δ_{0 k},  S_{q}(n, 0) = δ_{n 0}; \end{matrix}

and

(1.2)  \begin{matrix}S̃_{q}(n + 1, k + 1) = q^{k}S̃_{q}(n, k) + [k + 1]S̃_{q}(n, k + 1), \\ S̃_{q}(0, k) = δ_{0 k},  S̃_{q}(n, 0) = δ_{n 0}. \end{matrix}

Note that S̃_{q}(n, k) = q^{\binom{k}{2}}S_{q}(n, k).

Similarly, the q-Stirling numbers of the first kind c_{q}(n, k) [Go] can be defined by

(1.3)  \begin{matrix}c_{q}(n + 1, k + 1) = c_{q}(n, k) + [n]c_{q}(n, k + 1), \\ c_{q}(0, k) = δ_{0 k},  c_{q}(n, 0) = δ_{n 0}. \end{matrix}

For convenience, we shall take s_{q}(n, k) = q^{n−k}c_{q}(n, k) as our q-Stirling numbers of the first kind.

Our first purpose is to give a short proof of the continued fraction expansions of the ordinary generating functions of the forementioned three q-Stirling numbers.

THEOREM 1. — In the ring of formal power series of x, the following identities hold :

(1.4)  1 + ∑_{n≥1} ∑_{k=1}^{n} S_{q}[n, k]a^{k}x^{n} = \frac{1}{1 − ax − \frac{aq x^{2}}{\frac{···}{1 − b_{k} x − \frac{λ_{k+1} x^{2}}{⋯}}}} ,

where b_{k} = aq^{k} + [k], λ_{k+1} = aq^{k+1}[k + 1] for k ≥ 0 ;

(1.5)  1 + ∑_{n≥1} ∑_{k=1}^{n} S̃_{q}[n, k]a^{k}x^{n} = \frac{1}{1 − ax − \frac{a(1 + a(q − 1)) x^{2}}{\frac{···}{1 − b_{k} x − \frac{λ_{k+1} x^{2}}{⋯}}}} ,

where b_{k} = aq^{2k} + [k](1 + aq^{k−1}(q − 1)), λ_{k+1} = aq^{2k}[k + 1](1 + aq^{k}(q − 1)) for k ≥ 0 ;

(1.6)  1 + ∑_{n≥1} ∑_{k=1}^{n} s_{q}(n, k)a^{k}x^{n} = \frac{1}{1 − ax − \frac{aq x^{2}}{\frac{···}{1 − b_{k} x − \frac{λ_{k+1} x^{2}}{⋯}}}} ,

where b_{k} = (a + q[k])q^{k} + q^{k}[k], λ_{k+1} = (a + q[k])[k + 1]q^{2k+1} for k ≥ 0.

We should point out that all the above formulas are valid only in the formal sense (cf. [Fl]). Actually the Hamburger moment problems assocated with the orthogonal polynomials corresponding to the above continued fractions are not always determinated, see for exemple [Ch, p. 197-198] or [Is]. We refer the reader to [Ch, p. 6-10] for the formal defintion of orthogonality.

It is well-known [Ch, p. 85] that Theorem 1 can be restated in terms of orthogonal polynomials as follows.

THEOREM 2

a) The polynomials U_{n}^{(a)}(x; q) orthogonal to the moments μ_{n}^{(1)}(q) :

(1.7)  μ_{n}^{(1)}(q) = ∑_{k=1}^{n} S_{q}[n, k]a^{k},

q-Stirling numbers 3

are defined by U_{0}^{(a)}(x; q) = 1, U_{1}^{(a)}(x; q) = x − a and for n ≥ 1 by

U_{n+1}^{(a)}(x; q) = (x − aq^{n} − [n])U_{n}^{(a)}(x; q) − aq^{n−1}[n]U_{n−1}^{(a)}(x; q).

b) The polynomials V_{n}^{(a)}(x; q) orthogonal to the moments μ_{n}^{(2)}(q) :

(1.8)  μ_{n}^{(2)}(q) = ∑_{k=1}^{n} S̃_{q}[n, k]a^{k},

are defined by V_{0}^{(a)}(x; q) = 1, V_{1}^{(a)}(x; q) = x − a and for n ≥ 1 by

V_{n+1}^{(a)}(x; q) = (x − b_{n})V_{n}^{(a)}(x; q) − λ_{n}V_{n−1}^{(a)}(x; q),

where b_{n} = aq^{2n} + [n](1 + aq^{n−1}(q − 1)), and λ_{n+1} = aq^{2n}[n + 1](1 + aq^{n}(q − 1)). c) The polynomials W_{n}^{(a)}(x; q) orthogonal to the moments ν_{n}(q) :

(1.9)  ν_{n}(q) = ∑_{k=1}^{n} s_{q}(n, k)a^{k},

are defined by W_{0}^{(a)}(x; q) = 1, W_{1}^{(a)}(x; q) = x − a and for n ≥ 1 by

W_{n+1}^{(a)}(x; q) = (x − b_{n})W_{n}^{(a)}(x; q) − λ_{n}W_{n−1}^{(a)}(x; q),

where b_{n} = (a + q + ⋯ + q^{n})q^{n} + q^{n}[n], and λ_{n} = (a + q + ⋯ + q^{n−1})[n]q^{2n−1}.

We note that if q = 1, the polynomials in a) and b) reduce to the Charlier polynomials, and the polynomials in c) reduce to the Laguerre polynomials [Ch]. The polynomials in a) and b) can then be regarded as two q-analogs of the Charlier polynomials, while the polynomials in c) as q-analogs of the Laguerre polynomials.

Conversely, if we first establish Theorem 2, we automatically get Theorem 1. Actually, ISMAIL and STANTON [St] have earlier noticed that the polynomials in part a) are a rescaled version of the Al-Salam-Carlitz polynomials [Ch, p. 197-198]. So we can also prove part a) by using the moments of the latter polynomials.

Parts b) and c) have been presented by the author at the 27th session of the Séminaire Lotharingien in 1991. Part b) was proved by using the methods developed in this paper, while part c) was derived from a more general result in [Ze].

Recently, DE M^{}ÉDICIS and VIENNOT [DV] have given a bijective proof of part c) and noticed that the polynomials in part c) are a special case of the “little” q-Jacobi polynomials introduced by HAHN (see [GR, p. 166]). Hence part c) can also be derived from the known measure of the “little q-Jacobi” polynomials as the q-Charlier polynomials. Finally, after writing an earlier version of this paper, STANTON [St] informed us that the polynomials in b) are a rescaled version of the classical q-Charlier polynomials (see [GR, p. 187]). So part b) can also be derived from the explicit measure of the classical q-Charlier polynomials.

In contrast with the other proofs, the three expansions of Theorem 1 are proved from scrath by means of the same method, inspired by the work of ROGERS [Ro]. The continued fraction method used to proving Theorem 2 has the merit to be short and elementary.

In section 2 we shall first prove Theorem 1 and Theorem 2 from scratch. In section 3 we comments on the combinatorial interpretations of the q-Stirling numbers. In section 4 we give an explicit formula and measure of the polynomials V_{n}^{(a)}(x; q) by generalizing the method of TOUCHARD [To] and MILNE [Mi1]. These polynomials turn out to be a rescaled version of the classical q-Charlier polynomials [GR, p. 187].

2. Continued fractions expansions

We first expand the ordinary generating functions of the q-Stirling numbers as Stieltjes continued fractions.

LEMMA 3. — The following identities hold :

(2.1)  1 + ∑_{n,k≥1} S_{q}(n, k)a^{k}x^{n} = \frac{1}{1 − \frac{a · x}{1 − \frac{1 · x}{\frac{⋯}{1 − \frac{λ_{n} · x}{⋯}}}}} ,

where λ_{2n−1} = aq^{n−1}, λ_{2n} = [n] for n ≥ 1 ;

(2.2)  1 + ∑_{n,k≥1} S̃_{q}(n, k)a^{k}x^{n} = \frac{1}{1 − \frac{a · x}{1 − \frac{(1 + a(q − 1)) · x}{\frac{⋯}{1 − \frac{λ_{n} · x}{⋯}}}}} ,

where λ_{2n−1} = aq^{2n}, λ_{2n} = [n](1 + aq^{n−1}(q − 1)) for n ≥ 1 ;

(2.3)  1 + ∑_{n,k≥1} s_{q}(n, k)a^{k}x^{n} = \frac{1}{1 − \frac{a · x}{1 − \frac{q · x}{\frac{⋯}{1 − \frac{λ_{n} · x}{⋯}}}}} ,

q-Stirling numbers 5

where λ_{2n−1} = (a + q + ⋯ + q^{n−1})q^{n−1}, λ_{2n} = [n] q^{n} for n ≥ 1.

PROOF. — Let f(a, x) be the left-hand side of (2.1). It then follows from (1.1) that

(2.4)  f(a, x) = ∑_{k≥0} \frac{a^{k}x^{k}}{(1 − [1]x)(1 − [2]x) ⋯ (1 − [k]x)}

= 1 + \frac{ax}{1 − x} f (\frac{a}{q}, \frac{qx}{1 − x}) ·

Assume that

(2.5)  f(a, x) = \frac{1}{1 − \frac{c_{1}(a) x}{1 − \frac{c_{2}(a) x}{1 − \frac{c_{3}(a) x}{⋯}}}} ·

Contracting the continued fraction (2.5) starting from the first row and the second row yields respectively

(2.6) f(a, x) = 1 + \frac{c_{1}(a)x}{1 − (c_{1}(a) + c_{2}(a))x − \frac{c_{2}(a)c_{3}(a) x^{2}}{1 − (c_{3}(a) + c_{4}(a))x − \frac{c_{4}(a)c_{5}(a) x^{2}}{⋯}}} ,

and

(2.7)  f(a, x) = \frac{1}{1 − c_{1}(a)x − \frac{c_{1}(a)c_{2}(a) x^{2}}{1 − (c_{2}(a) + c_{3}(a))x − \frac{c_{3}(a)c_{4}(a) x^{2}}{⋯}}} ·

Therefore

(2.8) \frac{ax}{1 − x} f (a, \frac{qx}{1 − x}) =

\frac{ax}{1 − (1 + qc_{1}(a))x − \frac{c_{1}(a)c_{2}(a)q^{2} x^{2}}{1 − (1 + qc_{2}(a) + qc_{3}(a))x − \frac{c_{3}(a)c_{4}(a)q^{2} x^{2}}{⋯}}} ·

Now substitute (2.6) and (2.8) in the functional equation (2.4) and identify the corresponding terms. We successively obtain that

(I)  \begin{aligned}c_{1}(a) = a \\ c_{1}(a) + c_{2}(a) = 1 + qc_{1}(a/q) & ⇒ c_{2}(a) & = 1 \\ c_{2}(a)c_{3}(a) = c_{1}(a/q)c_{2}(a/q)q^{2} & ⇒ c_{3}(a) & = aq \\ c_{3}(a) + c_{4}(a) = 1 + qc_{2}(a/q) + qc_{3}(a/q) & ⇒ c_{4}(a) & = 1 + q \\ . . . . . . & . . . . . . \end{aligned}

One can show by induction

(2.9)  c_{2n−1}(a) = aq^{n}, c_{2n}(a) = [n], n ≥ 1.

Putting the above values in (2.3) yields formula (2.1). Now let g(a, x) be the left-hand side of (2.2). It then follows from (1.2) that

(2.10)  g(a, x) = ∑_{k≥0} \frac{a^{k}q^{\binom{k}{2}}x^{k}}{(1 − [1]x)(1 − [2]x) ⋯ (1 − [k]x)}

= 1 + \frac{ax}{1 − x} g (a, \frac{qx}{1 − x}) ·

If we replace f(a, x) by g(a, x) in (2.5)-(2.8), we get from (2.10) that

(II) \begin{aligned}c_{1}(a) = a \\ c_{1}(a) + c_{2}(a) = 1 + qc_{1}(a) & ⇒ c_{2}(a) & = 1 + a(q − 1) \\ c_{2}(a)c_{3}(a) = c_{1}(a)c_{2}(a/q)q^{2} & ⇒ c_{3}(a) & = aq^{2} \\ c_{3}(a) + c_{4}(a) = 1 + qc_{2}(a) + qc_{3}(a) & ⇒ c_{4}(a) & = (1 + q)(1 + aq(q − 1)) \\ . . . . . . & . . . . . . \end{aligned}

and more generally

(2.11)  c_{2n−1}(a) = aq^{2(n−1)}, c_{2n}(a) = [n](1 + aq^{n−1}(q − 1)), n ≥ 1.

Finally, it follows from (1.3) that

∑_{k=0}^{n} s_{q}(n, k)a^{k} = a(a + q) ⋯ (a + q + ⋯ + q^{n−1}).

Therefore, if we let h(a, x) be the left-hand side of (2.3), we have

(2.11) h(a, x) = 1 + ax + a(a + q)x^{2} + a(a + q)(a + q + q^{2})x^{3} + ⋯
= 1 + ax (1 + (1 + a/q)(qx) + (1 + a/q)(1 + a/q + q)(qx)^{2})
= 1 + axh(1 + a/q, qx)

Similarly, if we replace f(a, x) by h(a, x) in (2.5)-(2.7), we get from (2.11) that

(III)  \begin{aligned}c_{1}(a) = a \\ c_{1}(a) + c_{2}(a) = c_{1}(1 + a/q)q & ⇒ c_{2}(a) & = q \\ c_{2}(a)c_{3}(a) = c_{1}(1 + a/q)c_{2}(1 + a/q)q^{2} & ⇒ c_{3}(a) & = (a + q)q \\ c_{3}(a) + c_{4}(a) = qc_{2}(1 + a/q) + qc_{3}(1 + a/q) & ⇒ c_{4}(a) & = (1 + q)q^{2} \\ . . . . . . & . . . . . . \end{aligned}

and more generally

(2.12)  c_{2n−1}(a) = (a + q + ⋯ + q^{n−1})q^{n−1}, c_{2n}(a) = [n]q^{n}, n ≥ 1.

Thus we have completed the proof of the theorem.

Remark : ROGERS [Ro] seems to be the first to have used the “contracting” and “functional equation” techniques to derive continued fraction expansions of power series. DUMONT [Du] has proved Lemma 3 in the case q = 1.

By contraction of the continued fractions in lemma 3 (cf. (2.7)) we get immediately Theorem 1.

3. Remarks on the combinatorial interpretations of the moments Let S_{n} be the set of permutations of {1, 2, …, n}. For any permutation

σ = σ(1)σ(2) ⋯ σ(n), a left-right maximum is a σ(i) such that σ(i) > σ(j) for all j < i, and an inversion is a pair (σ(i), σ(j)) where σ(i) > σ(j) for all pairs (i, j) such that 1 ≤ i < j ≤ n. Denote by lrm σ and inv σ the numbers of the left-right maxima and inversions of σ. From the inversion table we immediately obtain

(3.1)  ∑_{σ∈S_{n}} a^{lrmσ}q^{inv σ} = a(a + q) ⋯ (a + q + ⋯ + q^{n−1}).

Thus from (1.10)

(3.2)  s_{q}(n, k) = ∑_{σ∈S_{n}(k)} q^{inv σ},

where 𝔖_{n}(k) denotes the set of permutations of {1, 2, …, n} with k left-right maxima. The interpretation (3.1) seems to appear first in [Ge].

Let _{n}(k) be the set of ordered partitions into k blocks of {1, 2, …, n}, i.e., the blocks of each partition are arranged in increasing order of their minima. Let π = (B_{1}, B_{2}, …, B_{k}) be a such partition. An inversion of π is a pair (b_{i}, B_{j}) such that b_{i} ∈ B_{i} where i < j and b_{i} > min B_{j}. A dual inversion of π is a pair (B_{i}, b_{j}) such that b_{j} ∈ B_{j} where i < j and min B_{i} < b_{j}. Let inv σ and ĩnv σ be the number of inversions and dual inversions of σ. It is easy to see, by verifying the recursions (1.7) and (1.8), that

(3.3)  S_{q}(n, k) = ∑_{π∈_{n}(k)} q^{inv π},
(3.4)  S̃_{q}(n, k) = ∑_{π∈_{n}(k)} q^{_{inv} π}.

These inv interpretations are due to MILNE [Mi2]. As pointed out by WACHS and WHITE [WW], but calling inv(π) = lb(π) and ĩnv π = ls(π), the above combinatorial interpretations of the q-Stirling numbers of the second kind are “easy” and there are also some “hard” statistics on the set of partitions, which also have the q-Stirling numbers of the second kind as their generating functions. However it is not easy to verify this fact. STANTON raised the question how this “hard” statistics could be the same as the easy ones, and WACHS and WHITE [WW] proved it by constructing an explicit bijection between these “hard” statistics and the easy ones. Once established Theorem 1, we can also derive this result as follows. According to a theorem due to FLAJOLET [Fl] we can rewrite the left-hand side of (1.4) as the generating functions of certain Motzkin paths with respect to some weights. This leads to the rs statistics of partitions of [WW] via a classical bijection due to FLAJOLET [Fl], FRANÇON and VIENNOT (see [Vi, II-14] or [Fl]). Similarly, a combinatorial interpretation of (1.6) in terms of Motzkin paths also leads to hard interpretations of the q-Stirling numbers of the first kind on permutations, see [DV]. Note that one of the motivations of this paper is due to these “hard” statistics.

4. The classical q-analog of Charlier polynomials

The classical q-Charlier polynomials [GR, p.187] are defined by

(4.1)  C_{n}(x; a, q) = _{2}φ_{1} ({q^{−n}, x \atop 0}; q, −\frac{q^{n+1}}{a} ) ,

and satisfy the orthogonality

∑_{x=0}^{∞} C_{m}(q^{−x}; a, q)C_{n}(q^{−x}; a, q)\frac{a^{x}}{(q; q)_{x}} q^{\binom{x}{2}} (4.2)

= (−a; q)_{∞}(−qa^{−1}; q)_{n}(q; q)_{n}q^{−n}δ_{mn}.

Although it is possible to verify directly that the polynomials V_{n}^{(a)}(x; q) are actually a rescaled version of C_{n}(x; a, q) by checking the three terms recurrence satisfied by these polynomials, we prefer to give an alternative argument to derive naturally the explicit expression and measure from the q-Stirling numbers as TOUCHARD [To] and MILNE [Mi] did in some special cases.

We define the linear functional φ on the vector space [q^{x}] by

(4.3)  φ([x]^{n}) = ∑_{k=0}^{n} S̃_{q}(n, k) a^{k}.

It is easy to see that the q-Stirling numbers S̃_{q}(n, k) satisfy

[x]^{n} = ∑_{k=0}^{n} S̃_{q}(n, k) [x]_{k},

and S̃_{q}(n, n) = 1. Since {[x]_{n}}_{n≥0} and {[x]^{n}}_{n≥0} are two bases of the vector space [q^{x}], if we define

[x]_{n} = ∑_{k=0}^{n} s_{q}^{∗}(n, k) [x]^{k},

we should have

∑_{k=0}^{n} S̃_{q}(n, k)s_{q}^{∗}(k, m) = δ_{m n} for m, n ∈ .

It follows that

(4.4)  φ([x]_{n}) = ∑_{k=0}^{n} s_{q}^{∗}(n, k) φ([x]^{k}) = ∑_{k=0}^{n} s_{q}^{∗}(n, k) ∑_{l=0}^{k} S_{q}(k, l)a^{l} = a^{n}.

Recall that the two classical q-analogs of the exponential e^{x} [An, GR] are defined by

e_{q}(x) = ∑_{k=0}^{∞} \frac{x^{k}}{[k]!}  and  E_{q}(x) = ∑_{k=0}^{∞} \frac{q^{\binom{k}{2}}x^{k}}{[k]!} ·

Note that (e_{q}(x))^{−1} = E_{q}(−x). Hence

(4.5)  φ([x]_{n}) = a^{n} = \frac{1}{e_{q}(a)} ∑_{k=0}^{∞} \frac{a^{n+k}}{[k]!} = \frac{1}{e_{q}(a)} ∑_{k=0}^{∞} \frac{[k]_{n}}{[k]!} a^{k}·

Since {[x]_{n}} is a basis of [q^{x}] and φ linear, we obtain the following result.

PROPOSITION 6. — For any polynomial P (x) of q^{x}, we have

φ(P (x)) = \frac{1}{e_{q}(a)} ∑_{k=0}^{∞} \frac{P (k)}{[k]!} a^{k}·

Setting P (x) = [x]^{n} in the above proposition we get then a q-analog of Touchard’s formula [To].

COROLLARY 7. — We have

B_{q,n}(a) = ∑_{k=0}^{n} S̃_{q}(n, k)a^{k} = \frac{1}{e_{q}(a)} ∑_{k=0}^{∞} \frac{[k]^{n}}{[k]!} a^{k}·

Note that the above formula generalizes Milne’s q-analog of Dobinski’s identity [Mi1], which corresponds to the a = 1 case.

LEMMA 8. — Let P (x) be a polynomial of q^{x} and k ≥ 0, then

φ([x]_{k} P (x)) = a^{k}φ(P (x + k)).

PROOF. — We first remark that φ([x][x − 1]_{n}) = a^{n+1} = aφ([x]_{n}). So φ([x]P (x − 1)) = aφ(P (x)) for any polynomial P (x) of q^{x}. Therefore φ([x]_{k}P (x)) = φ([x][x − 1]_{k−1}P (x)) = aφ([x]_{k−1}P (x)), and the proof is complete by induction.

We need the following version of the q-binomial theorem (see [An, p. 225] for a combinatorial proof using vector spaces over a finite field).

(4.6)  ∏_{i=0}^{n−1}(X − Zq^{i}) = ∑_{k=0}^{n} \left[{n \atop k}\right] ∏_{j=0}^{k−1}(Y − Zq^{i}) ∏_{l=0}^{n−k−1}(X − Y q^{l}).

Putting X = q^{x}, Y = 1 and Z = 0 in (4.6) yields

(4.7)  (q^{x})^{n} = ∑_{k=0}^{n} \left[{n \atop k}\right](q − 1)^{k}q^{\binom{k}{2}}[x]_{k}.

Applying φ to (4.7) and then applying (4.6) with X = 1, Y = 0 and Z = (q − 1)a leads to

(4.8)  φ((q^{x})^{n}) = ∑_{k=0}^{n} \left[{n \atop k}\right](q − 1)^{k}q^{\binom{k}{2}}a^{k} = (a(q − 1); q)_{n}.

As usual we define, for any function f(x) of x, the shift operator E : E f(x) = f(x + 1), the identity operator I : I f(x) = f(x) and the q-difference operator by means of

Δ_{q}^{0} f(x) = f(x), Δ_{q}^{n+1} f(x) = (E − q^{n} I)Δ_{q}^{n}f(x) = Δ^{n} f(x + 1) − q^{n}Δ_{q}^{n}f(x).

Note that

(4.9)  Δ_{q}^{n}f(x) = (E − q^{n−1}I)(E − q^{n−2}I) ⋯ (E − I)f(x).

It follows from (4.6) that

(4.10)  Δ_{q}^{n} f(x) = ∑_{k=0}^{n}(−1)^{k}\left[{n \atop k}\right]q^{\binom{k}{2}}f(x + n − k).

We require the following easily verified formula :

(4.11)  Δ_{q}^{n}[x]_{m} = \{ \begin{matrix}[m]_{n} [x]_{m−n} q^{n(x+n−m)} & if n ≤ m, \\ 0 & otherwise. \end{matrix}

THEOREM 9. — Let

(4.12)  H_{n}(q^{x}; a, q) = ∑_{k=0}^{n}(−a)^{k}\left[{n \atop k}\right]q^{k(k−1)/2}[x]_{n−k}.

We have the orthogonality

(4.13)  φ(H_{m}(q^{x}; a, q)H_{n}(q^{x}; a, q)) = a^{n}[n]!(a(1 − q); q)_{n}δ_{m n}.

PROOF. — Assume that m ≤ n, then (4.11) reduces to Δ_{q}^{n} [x]_{m} = [n]! q^{nx}δ_{m n}. By lemma 8 and (4.10) we have

φ([x]_{m} H_{n}(q^{x}; a, q)) = ∑_{k=0}^{n}(−a)^{k}\left[{n \atop k}\right]q^{k(k−1)/2}a^{n−k}φ([x + n − k]_{m})
= a^{n}φ(Δ_{q}^{n} [x]_{m})
= a^{n}[n]!φ(q^{nx})δ_{m n}.

Therefore

φ(H_{m}(q^{x}; a, q)H_{n}(q^{x}; a, q)) = φ([x]_{m}H_{n}(q^{x}; a, q)) = a^{n}[n]!φ(q^{nx})δ_{m n}.

The proof is complete in virtue of (4.8).

Remark : The special cases of Theorem 9 have been proved respectively by TOUCHARD [Tou] for q = 1 and MILNE [Mi1] for a = 1 by similar methods. Note that our right-hand side of (4.13) is simpler than that of MILNE [Mi1] even in the case a = 1.

12 J. ZENG

In (4.12) if we set z = [x] then we get the explicit expression for V_{n}^{(a)}(x; q) :

(4.14) V_{n}^{(a)}(z; q) = q^{\binom{n}{2}}H_{n}((q − 1)z + 1; a, q) = ∑_{k=0}^{n}(−aq^{n−1})^{n−k}\left[{n \atop k}\right] ∏_{i=0}^{k−1}(z − [i]).

The polynomials H_{n}(q^{x}; a, q) may also be written in terms of hypergeometric functions as

(4.15)  H_{n}(q^{x}; a, q) = \frac{(q^{−x}; q)_{n}}{(q − 1)^{n}} q^{nx−\binom{n}{2}}_{1}φ_{1} ({q^{−n} \atop q^{1−n+x}}; q, aq^{n}(1 − q)) .

If we replace q by 1/q and then q^{−x} by x in the above formula, we obtain the classical q-Charlier polynomials [GR, p.187], that is

(4.16)  H_{n} (x; \frac{a}{1 − q} , \frac{1}{q}) = (\frac{−a}{1 − q} )^{n} q^{−\binom{n}{2}}_{2}φ_{1} ({q^{−n}, x \atop 0}; q, −\frac{q^{n+1}}{a} ) .

Acknowledgement. — We would like to thank Professor D. Stanton for his careful and critical reading of the manuscript and the two anonymous referees for their helpeful comments on the first version of this paper.

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Département de Mathématiques Université Louis-Pasteur 7, rue René Descartes 67084 Strasbourg Cedex, France Email : jzeng@math.u-strasbg.fr