\magnification=1200
\input a4new
\hsize=13cm
\auteurcourant={J. ZENG}
\titrecourant={$q$-Stirling numbers}
\def\example{\newparwithcolon
                  \pc Example|
                    \endnewparwithcolon}                          
\def\remark{\newparwithcolon
                        \it Remark
                         \endnewparwithcolon} 
\def\proof{\newparwithpointir
               \pc Proof|
                \endnewparwithpointir}
\def\resume{\bgroup\eightpoint\noindent Abstract\pointir}
\def\endresume{\par\egroup}
\def\cutdisplay{{}\hfill\cr\hfill{}}
\def\preambule[#1]{\tvi\kern#1$##$\kern#1\hfil&&
                              \tvi\kern#1$##$\kern#1\hfil\cr}
\def\S{\cal S}
\def\lrm{\mathop {\rm lrm}}
\def\inv{\mathop {\rm inv}}
\def\Inv{\mathop{\tilde {\rm inv} }}
%\hfill\today
\ctitre
THE $q$-STIRLING NUMBERS, CONTINUED FRACTIONS AND 
THE $q$-CHARLIER AND $q$-LAGUERRE POLYNOMIALS 
 {\eightbf By}
Jiang ZENG
\endctitre
\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. 
\endabstract  
\sectiona 
1. Introduction
\endsection
For $q\in \bb R$ such that $|q|<1$ define 
$$
[x]={q^x-1\over q-1}\cdot
$$
For $n\geq 0$ let $[x]_0=1$, $[x]_n=[x]\,[x-1]\cdots [x-n+1]$
and $[n]!=[n]_n$. Also, for $n\geq k\geq 0$ define 
$$
{n\brack k}={[n]!\over [k]![n-k]!}\cdot
$$ 

Recall that the two related $q$-Stirling
numbers of the second kind [Go, WW] may be defined by
recurrence as :
$$\smalldisplaylines{
S_q(n+1,k+1)=S_q(n,k)+[k+1]S_q(n,k+1),\cr
S_q(0,k)=\delta _{0\,k},\hfil S_q(n,0)=\delta _{n\,0} ;\cr
}\leqno (1.1)
$$
and
$$\smalldisplaylines{
\tilde S_q(n+1,k+1)
=q^k\tilde S_q(n, k)+[k+1]\tilde S_q(n, k+1),\cr
\tilde S_q(0,k)=\delta _{0\,k},\hfil \tilde S_q(n,0)=\delta _{n\,0}.\cr
}\leqno (1.2)
$$  
Note that $\tilde S_q(n, k)=q^{{k\choose 2}}S_q(n, k)$.  

Similarly, the $q$-Stirling numbers of the first kind $c_q(n,k)$
[Go] can be defined by
$$\smalldisplaylines{
c_q(n+1,k+1)
=c_q(n, k)+[n]c_q(n, k+1),\cr
c_q(0,k)=\delta _{0\,k},\hfil c_q(n,0)=\delta _{n\,0}.\cr
}\leqno (1.3)
$$  
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. 
\th Theorem 1
\enonce In the ring of formal power series of $x$, the following
identities hold : 
$$
1+\sum_{n\geq 1}\sum_{k=1}^n S_q[n,k]a^kx^n
={1\over\displaystyle 1-ax- {\strut
aq\,x^2\over {\strut \cdots\over \displaystyle1-b_k\,x-
{\strut \lambda_{k+1}\,x^2\over\cdots
}}}}\raise 2pt\hbox{,} \leqno (1.4)
$$
where $ b_k=aq^k+[k]$,
$\lambda_{k+1}=aq^{k+1}[k+1]$ for $k\geq 0$ ;
$$
1+\sum_{n\geq 1}\sum_{k=1}^n\tilde S_q[n,k]a^kx^n=
{1\over\displaystyle 1-ax-
{\strut a(1+a(q-1))\,x^2\over
{\strut \cdots\over \displaystyle1-b_k\,x-
{\strut \lambda_{k+1}\,x^2\over\cdots
}}}}\virg \leqno (1.5)
$$
where $ b_k=aq^{2k}+[k](1+aq^{k-1}(q-1))$,
$\lambda_{k+1}=aq^{2k}[k+1](1+aq^k(q-1))$ 
for $k\geq 0$ ;
$$
1+\sum_{n\geq 1}\sum_{k=1}^n s_q(n,k)a^k x^n=
{1\over\displaystyle 1-ax-
{\strut aq\,x^2\over
{\strut \cdots\over \displaystyle1-b_k\,x-
{\strut \lambda_{k+1}\,x^2\over\cdots
}}}}\virg\leqno (1.6)
$$
where $b_k=(a+q[k])q^k+q^k[k]$,
$\lambda_{k+1}=(a+q[k])[k+1]q^{2k+1}$ for $k\geq 0$. 
\endth
\medskip
We should point out that all the
above formulas are valid only in the {\sl 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. 
\tha Theorem 2
\enonce 
\noindent a) The polynomials $U_n^{(a)}(x; q)$ orthogonal to the
moments $\mu_n^{(1)}(q)$ :
$$
\mu_n^{(1)}(q)=\sum_{k=1}^nS_q[n,k]a^k, \leqno(1.7)
$$
are defined by $U_0^{(a)}(x; q)=1$, 
$U_1^{(a)}(x; q)=x-a$ and for $n\geq 1$ 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
$\mu_n^{(2)}(q)$ :
$$
\mu_n^{(2)}(q)=\sum_{k=1}^n \tilde S_q[n,k]a^k, \leqno(1.8)
$$
are defined by $V_0^{(a)}(x; q)=1$, $V_1^{(a)}(x;
q)=x-a$ and for $n\geq 1$ by
$$
V_{n+1}^{(a)}(x; q)
=(x-b_n)V_n^{(a)}(x; q)-\lambda_n V_{n-1}^{(a)}(x; q), 
$$
where $b_n=aq^{2n}+[n](1+aq^{n-1}(q-1))$, 
and $\lambda_{n+1}=aq^{2n}[n+1](1+aq^n(q-1))$.

c) The polynomials $W_n^{(a)}(x; q)$ orthogonal to the moments 
$\nu_n(q)$ :
$$
\nu_n(q)=\sum_{k=1}^n s_q(n,k)a^k, \leqno(1.9)
$$
are defined by $ W_0^{(a)}(x; q)=1$, $W_1^{(a)}(x; q)=x-a$ and
for $n\geq 1$ by
$$
W_{n+1}^{(a)}(x; q)
=(x-b_n)W_n^{(a)}(x; q)-\lambda_n W_{n-1}^{(a)}(x; q),
$$
where $b_n=(a+q+\cdots +q^n)q^n+q^n[n]$, 
and $\lambda_n=(a+q+\cdots +q^{n-1})[n]q^{2n-1}$.
\endth

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, \pc Ismail| and \pc 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  S\'eminaire 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, \pc De M\'edicis| and \pc 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 \pc 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, \pc 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 \pc 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 \pc Touchard| [To] and \pc
Milne| [Mi1]. These polynomials turn out to be a rescaled version
of the classical $q$-Charlier polynomials [GR, p. 187].


\sectiona
2. Continued fractions expansions
\endsection
We first expand the ordinary generating functions of the
$q$-Stirling numbers as Stieltjes continued fractions. 
\th Lemma 3 
\enonce The following identities hold :
$$
1+\sum_{n,k\geq 1} S_q(n,k)a^kx^n=
{1\over\displaystyle 1- 
{\strut a\cdot x\over\displaystyle 1-
{\strut 1\cdot x\over\displaystyle 
{\strut \cdots\over\displaystyle 1-
{\strut \lambda _n\cdot x\over\cdots}}}}}\virg\leqno(2.1) 
$$
where $\lambda _{2n-1}=aq^{n-1}$, $\lambda _{2n}=[n]$ for $n\geq 1$ ; 
$$
1+\sum_{n,k\geq 1}\tilde S_q(n,k)a^kx^n
={1\over\displaystyle 1- 
{\strut a\cdot x\over\displaystyle 1-
{\strut (1+a(q-1))\cdot x\over\displaystyle
{\strut \cdots\over\displaystyle 1-
{\strut \lambda _n\cdot x\over\cdots}}}}}\virg\leqno(2.2)
$$
where $\lambda _{2n-1}=aq^{2n}$, 
$\lambda _{2n}=[n](1+aq^{n-1}(q-1))$ for $n\geq 1$ ;
 $$
1+\sum_{n,k\geq 1}s_q(n,k)a^kx^n
={1\over\displaystyle 1- 
{\strut a\cdot x\over\displaystyle 1-
{\strut q\cdot x\over\displaystyle 
{\strut \cdots\over\displaystyle 1-
{\strut \lambda _n\cdot x\over\cdots}}}}}\virg\leqno(2.3)
$$
where $\lambda _{2n-1}=(a+q+\cdots +q^{n-1})q^{n-1}$,
$\lambda _{2n}=[n]\, q^n$ for $n\geq 1$.
\endth
\proof
Let $f(a, x)$ be the left-hand side of (2.1). It then follows
from (1.1) that
$$\leqalignno{
f(a, x)&=\sum_{k\geq 0}
\frac{a^kx^k}{(1-[1]x)(1-[2]x)\cdots(1-[k]x)}&(2.4)\cr
&=1+\frac{ax}{1-x}f\left({a\over q}, {qx\over
1-x}\right)\cdot\cr  
}$$
Assume that
$$ 
f(a, x)={1\over\displaystyle 1-
{\strut c_1(a)\,x\over\displaystyle 1-
{\strut c_2(a)\,x\over\displaystyle 1-
{\strut c_3(a)\,x\over\displaystyle \cdots
}}}}\cdot\leqno(2.5)
$$

Contracting the continued fraction (2.5) starting from
the first row and the second row yields respectively
$$ 
f(a, x)=1+
{c_1(a)x\over\displaystyle 1-(c_1(a)+c_2(a))x-
{\strut c_2(a)c_3(a)\,x^2\over\displaystyle 1-(c_3(a)+c_4(a))x-
{\strut c_4(a)c_5(a)\,x^2\over\displaystyle\cdots
}}}\virg\leqno(2.6)
$$
and
$$
f(a, x)={1\over\displaystyle 1-c_1(a)x-
{\strut c_1(a)c_2(a)\,x^2\over\displaystyle 1-(c_2(a)+c_3(a))x-
{\strut c_3(a)c_4(a)\,x^2\over\displaystyle  \cdots
}}}\cdot\leqno(2.7)
$$
Therefore
$$\displaylines{
(2.8)\qquad\frac{ax}{1-x}f\left(a, {qx\over 1-x}\right)
=\hfill\cr
\hfill{ax\over\displaystyle 1-(1+qc_1(a))x-
{\strut c_1(a)c_2(a)q^2\,x^2\over\displaystyle 1
-(1+qc_2(a)+qc_3(a))x-
{\strut c_3(a)c_4(a)q^2\,x^2\over\displaystyle \cdots
}}}\cdot\cr
}$$

Now substitute (2.6) and (2.8) in the functional equation (2.4)
and identify the corresponding terms. We successively obtain
 that 
$$\eqalign{
&c_1(a)=a\cr
&c_1(a)+c_2(a)=1+qc_1(a/q)\hfill &\Rightarrow c_2(a)&=1\cr
&c_2(a)c_3(a)=c_1(a/q)c_2(a/q)q^2\hfill 
&\Rightarrow c_3(a)&=aq\cr
&c_3(a)+c_4(a)=1+qc_2(a/q)+qc_3(a/q)
\hfill &\Rightarrow c_4(a)&=1+q\cr
&\ldots\ldots&\ldots\ldots\cr
}\leqno (\hbox{I})
$$
One can show by induction
$$
c_{2n-1}(a)=aq^n,\quad
c_{2n}(a)=[n],\quad n\geq 1. \leqno(2.9)
$$
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
$$\leqalignno{
g(a, x)&=\sum_{k\geq 0}\frac{a^kq^{k\choose 2}x^k}
{(1-[1]x)(1-[2]x)\cdots (1-[k]x)} &(2.10)\cr
&=1+\frac{ax}{1-x}g\left(a, {qx\over 1-x}\right)\cdot\cr
}$$
If we replace $f(a, x)$ by $g(a, x)$ in (2.5)-(2.8), we get
from (2.10) that
$$\eqalign{
&c_1(a)=a\cr
&c_1(a)+c_2(a)=1+qc_1(a)\hfill &\Rightarrow
c_2(a)&=1+a(q-1)\cr 
&c_2(a)c_3(a)=c_1(a)c_2(a/q)q^2\hfill 
&\Rightarrow c_3(a)&=aq^2\cr
&c_3(a)+c_4(a)=1+qc_2(a)+qc_3(a)
\hfill &\Rightarrow c_4(a)&=(1+q)(1+aq(q-1))\cr
&\ldots\ldots&\ldots\ldots\cr
}\leqno (\hbox{II})
$$
and more generally
$$
c_{2n-1}(a)=aq^{2(n-1)},\quad
c_{2n}(a)=[n](1+aq^{n-1}(q-1)),\quad n\geq 1. \leqno(2.11)
$$
Finally, it follows from (1.3) that
$$
\sum_{k=0}^ns_q(n,k)a^k=a(a+q)\cdots (a+q+\cdots
+q^{n-1}).
$$
Therefore, if we let $h(a, x)$ be the left-hand
side of (2.3), we have
$$\leqalignno{
h(a, x)&=1+ax+a(a+q)x^2+
a(a+q)(a+q+q^2)x^3+\cdots&(2.11)\cr
&=1+ax\left(1+(1+a/q)(qx)+(1+a/q)
(1+a/q+q)(qx)^2\right)\cr
&=1+axh(1+a/q, qx)\cr 
}$$
Similarly, if we replace $f(a, x)$ by $h(a, x)$ in (2.5)-(2.7), we
get from (2.11) that
$$\eqalign{
&c_1(a)=a\cr
&c_1(a)+c_2(a)=c_1(1+a/q)q\hfill &\Rightarrow
c_2(a)&=q\cr 
&c_2(a)c_3(a)=c_1(1+a/q)c_2(1+a/q)q^2\hfill 
&\Rightarrow c_3(a)&=(a+q)q\cr
&c_3(a)+c_4(a)=qc_2(1+a/q)+qc_3(1+a/q)
\hfill &\Rightarrow c_4(a)&=(1+q)q^2\cr
&\ldots\ldots&\ldots\ldots\cr
}\leqno (\hbox{III})
$$
and more generally
$$
c_{2n-1}(a)=(a+q+\cdots +q^{n-1})q^{n-1},\quad
c_{2n}(a)=[n]q^n,\quad n\geq 1. \leqno(2.12)
$$ 
Thus we have completed the proof of the theorem. 
\qed 
\medskip
\remark
\pc Rogers| [Ro] seems to be the first to have used the
``contracting"  and ``functional equation" techniques to
derive continued fraction expansions of power series. 
\pc 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.


\sectiona
3. Remarks on the combinatorial interpretations of the moments
\endsection
Let $\S_n$ be the set of permutations of $\{1, 2, \ldots, n\}$.
For any permutation $\sigma =\sigma (1)\sigma (2)\cdots \sigma (n)$, a 
{\sl left-right maximum} is a $\sigma (i)$ such that $\sigma (i)>\sigma (j)$
for all $j<i$, and an {\sl inversion} is a pair
$(\sigma (i), \sigma (j))$ where $\sigma (i)>\sigma (j)$ for all pairs
$(i, j)$ such that $1\leq i<j\leq n$.
Denote by $\lrm \sigma $ and $\inv \sigma $ the numbers of the left-right
maxima and inversions of $\sigma $. From the 
{\sl inversion table} we immediately obtain
$$
\sum_{\sigma \in \S_n}a^{lrm \sigma }q^{\inv \sigma }=a(a+q)\cdots
(a+q+\cdots +q^{n-1}).\leqno (3.1)
$$
Thus from (1.10) 
$$
s_q(n, k)=\sum_{\sigma \in \S_n(k)}q^{\inv \sigma },\leqno (3.2)
$$
where ${\goth S}_n(k)$ denotes the set of permutations of 
$\{1, 2, \ldots, n\}$ 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 $\{1, 2, \ldots, n\}$, \ie., 
the blocks of each partition are arranged
in increasing order of their minima. Let 
$\pi =(B_1, B_2, \ldots, B_k)$ be a such partition. An 
{\sl inversion} of $\pi $ is a pair $(b_i, B_j)$ such that 
$b_i\in B_i$ where $i<j$ and $b_i>min\, B_j$. A 
{\sl dual inversion} of $\pi $ is a pair $(B_i, b_j)$ such that
$b_j\in B_j$ where $i<j$ and $min\, B_i<b_j$. Let $\inv \sigma $
and $\Inv \sigma $ be the number of inversions and dual inversions
of $\sigma $. It is easy to see, by verifying the
recursions (1.7) and (1.8), that 
$$\leqalignno{
S_q(n, k)&=\sum_{\pi \in �_n(k)}q^{\inv \pi },&(3.3)\cr
\tilde S_q(n, k)&=\sum_{\pi \in �_n(k)}q^{\Inv \pi }.&(3.4)\cr
}$$

These {\sl inv} interpretations are due to \pd Milne [Mi2]. 
As pointed out by \pc Wachs| and \pc White| [WW],
but calling $\inv (\pi )=lb(\pi )$ and $\Inv \pi =ls(\pi )$, 
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. \pc Stanton| raised the question how this ``hard''
statistics could be the same as the easy ones, and 
\pc Wachs| and \pc 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 \pd Flajolet
[Fl] we can rewrite the left-hand side of (1.4)
as the generating functions of certain {\sl Motzkin paths}
 with respect to some weights. This leads to the {\sl
rs} statistics of partitions of [WW] via a classical bijection 
due to \pd Flajolet [Fl], \pc Fran\c con| 
and \pc 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.


\sectiona
4. The classical $q$-analog of Charlier polynomials
\endsection
The classical $q$-Charlier polynomials 
[GR, p.187] are defined by
$$
C_n(x;a,q)={}_2\varphi_1
\left({q^{-n}, x\atop 0};q, -{q^{n+1}\over a} 
\right),\leqno (4.1)
$$
and satisfy the orthogonality 
$$\ldisplaylinesno{
\sum_{x=0}^\infty C_m(q^{-x};a,q)C_n(q^{-x};a,q){a^x\over 
(q;q)_x}q^{{x\choose 2}} &(4.2)\cr
=(-a;q)_\infty(-qa^{-1};q)_n(q;q)_nq^{-n}\delta_{mn}.\cr
}$$

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 
\pd Touchard [To] and \pd Milne [Mi] did in some special cases.

We define the linear functional $\varphi$ on 
the vector space $\bb C[q^x]$ by
$$
\varphi ([x]^n)=\sum_{k=0}^n\tilde S_q(n,k)\,a^k.\leqno (4.3)
$$

It is easy to see that
the $q$-Stirling numbers $\tilde S_q(n,k)$ satisfy
$$
[x]^n=\sum_{k=0}^n\tilde S_q(n,k)\,[x]_k,
$$
and $\tilde S_q(n,n)=1$. 
Since $\{[x]_n\}_{n\geq 0}$ and $\{[x]^n\}_{n\geq 0}$
are two bases of the vector space $\bb C[q^x]$,
if we define
$$
[x]_n=\sum_{k=0}^ns^*_q(n,k)\,[x]^k,
$$
we should have 
$$
\sum_{k=0}^n\tilde S_q(n,k)s^*_q(k, m)=\delta_{m\,n}\qquad 
\hbox{for}\quad m,n\in \bb N.
$$
It follows that
$$
\varphi([x]_n)=\sum_{k=0}^ns^*_q(n,k)\,\varphi([x]^k)
=\sum_{k=0}^ns^*_q(n,k)\sum_{l=0}^kS_q(k,l)a^l=a^n. 
\leqno (4.4) 
$$

Recall that the two classical $q$-analogs of
the exponential $e^x$ [An, GR] are defined by  
$$
e_q(x)=\sum_{k=0}^\infty{x^k\over [k]!}\qquad
{\textstyle and}\qquad
E_q(x)=\sum_{k=0}^\infty{q^{k\choose 2}x^k\over [k]!}\cdot
$$
Note that $(e_q(x))^{-1}=E_q(-x)$. Hence
$$
\varphi([x]_n)=a^n={1\over e_q(a)}\sum_{k=0}^\infty {a^{n+k}\over [k]!}
     ={1\over e_q(a)}
\sum_{k=0}^\infty {[k]_n\over [k]!}a^k\cdot\leqno (4.5)
$$
Since $\{[x]_n\}$ is a basis of $\bb C [q^x]$ and
$\varphi$ linear, we obtain the following result.
\th Proposition 6
\enonce For any polynomial $P(x)$ of $q^x$, we have
$$
\varphi(P(x))={1\over e_q(a)}\sum_{k=0}^\infty
{P(k)\over [k]!}\,a^k\cdot
$$
\endth
\medskip
 
Setting $P(x)=[x]^n$ in the above
proposition we get then a $q$-analog of 
Touchard's formula [To]. 
\th Corollary 7
\enonce We have
$$
B_{q, n}(a)=\sum_{k=0}^n\tilde S_q(n, k)a^k={1\over e_q(a)}
\sum_{k=0}^\infty {[k]^n\over [k]!}a^k\cdot
$$
\endth

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

\th Lemma 8
\enonce Let $P(x)$ be a polynomial of $q^x$ and 
 $k\geq 0$, then
$$
\varphi([x]_k\,P(x))=a^k\varphi(P(x+k)).
$$
\endth
\proof
We first remark that
$\varphi([x][x-1]_n)=a^{n+1}=a\varphi([x]_n)$. So
$\varphi([x]P(x-1))=a\varphi(P(x))$ for any polynomial
$P(x)$ of $q^x$. Therefore
$\varphi([x]_kP(x))=\varphi([x][x-1]_{k-1}P(x))
=a\varphi([x]_{k-1}P(x))$, and the proof is complete by
induction. \qed 
\medskip
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).
$$
\prod_{i=0}^{n-1}(X-Zq^i)=\sum_{k=0}^n{n\brack k}
\prod_{j=0}^{k-1}(Y-Zq^i)\prod_{l=0}^{n-k-1}(X-Yq^l).
\leqno (4.6)
$$

Putting $X=q^x$, $Y=1$ and $Z=0$ in (4.6) yields
$$
(q^x)^n=\sum_{k=0}^n{n\brack k}
(q-1)^kq^{k\choose 2}[x]_k.\leqno (4.7)
$$
Applying $\varphi$ to (4.7) and then applying (4.6)
with $X=1$, $Y=0$ and $Z=(q-1)a$ leads to
$$
\varphi((q^x)^n)=\sum_{k=0}^n{n\brack k}
(q-1)^kq^{k\choose 2}a^k=(a(q-1); q)_n.\leqno (4.8)
$$

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
$$
\Delta_q^0\,f(x)=f(x), \qquad
\Delta_q^{n+1}\,f(x)=(E-q^n\,I)\Delta_q^nf(x)=
\Delta^n\,f(x+1)-q^n\Delta_q^nf(x).
$$
Note that 
$$
\Delta _q^nf(x)=(E-q^{n-1}I)(E-q^{n-2}I)\cdots
(E-I)f(x).\leqno (4.9)
$$
It follows from (4.6) that 
$$
\Delta_q^n\,f(x)=\sum_{k=0}^n
(-1)^k{n\brack k}q^{k\choose 2}f(x+n-k).\leqno(4.10)
$$

We require the following easily verified formula :
$$
\Delta_q^n[x]_m=\cases{
[m]_n\,[x]_{m-n}\,q^{n(x+n-m)} & if $n\leq m$, \cr
0 & otherwise.\cr}\leqno (4.11)
$$

\th Theorem 9
\enonce Let 
$$
H_n(q^x; a, q)=\sum_{k=0}^n(-a)^k{n\brack k}
q^{k(k-1)/2}[x]_{n-k}.\leqno (4.12) 
$$
We have the orthogonality
$$
\varphi(H_m(q^x; a, q)H_n(q^x; a, q))=
a^n[n]!(a(1-q);q)_n\delta_{m\,n}. \leqno (4.13)
$$
\endth
\proof
Assume that $m\leq n$, then (4.11)
reduces to $\Delta_q^n\,[x]_m=[n]!\,q^{nx}\delta_{m\,n}$. 
By lemma 8 and (4.10) we have
$$\eqalign{
\varphi([x]_m\,H_n(q^x;a,q))
&=\sum_{k=0}^n(-a)^k{n\brack k}
q^{k(k-1)/2}a^{n-k}\varphi([x+n-k]_m)\cr
&=a^n\varphi(\Delta_q^n\,[x]_m)\cr
&=a^n[n]!\varphi(q^{nx})\delta_{m\,n}.\cr
}$$
Therefore
$$
\varphi(H_m(q^x;a, q)H_n(q^x; a,q))
=\varphi([x]_mH_n(q^x;a, q))
=a^n[n]!\varphi(q^{nx})\delta_{m\,n}.
$$ 
The proof is complete in virtue of (4.8). \qed
\medskip
\remark
The special cases of Theorem 9 have been proved respectively
by \pc Touchard| [Tou] for $q=1$ and
\pd Milne [Mi1] for $a=1$ by similar methods. 
Note that our right-hand side of (4.13) is simpler than that of
\pd Milne [Mi1] even in the case $a=1$. 

In (4.12) if we set $z=[x]$ then we get the explicit expression
for $V_n^{(a)}(x;q)$ :
$$
V_n^{(a)}(z;q)=q^{n\choose 2}H_n((q-1)z+1; a, q)
=\sum_{k=0}^n(-aq^{n-1})^{n-k}
{n\brack k}\prod_{i=0}^{k-1}(z-[i]). \leqno (4.14)
$$

The polynomials $H_n(q^x; a, q)$
may also be written in terms of hypergeometric functions as
$$
H_n(q^x;a,q)={(q^{-x};q)_n\over (q-1)^n}q^{nx-{n\choose 2}}
{}_1\varphi_1\left({q^{-n}\atop q^{1-n+x}};q, aq^n(1-q)
\right). \leqno (4.15)
$$ 

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
$$
H_n\left(x;{a\over 1-q}, {1\over q}\right)=\left({-a\over
1-q}\right)^nq^{-{n\choose 2}}{}_2\varphi_1
\left({q^{-n}, x\atop 0};q, -{q^{n+1}\over a}
\right).\leqno (4.16)
$$ 

\section
Acknowledgement
\endsection
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.

\eightpoint
\ctitre
REFERENCES
\endctitre
%_________________
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\endref
%_________________________________________________
\vskip 40pt
\vbox{\sl
\leftline{D\'epartement de Math\'ematiques}
\leftline{Universit\'e Louis-Pasteur}
\leftline{7, rue Ren\'e Descartes}
\leftline{67084 Strasbourg Cedex, France}
\leftline{Email : jzeng@math.u-strasbg.fr}}
\vfill
\bye

