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\auteurcourant={DOMINIQUE FOATA AND GUO-NIU HAN}
\titrecourant={JACOBI AND WATSON IDENTITIES REVISITED}

\vglue2cm
\centerline{JACOBI AND WATSON IDENTITIES}
\smallskip
\centerline{COMBINATORIALLY REVISITED}
\bigskip
\centerline{\bf Dominique Foata and Guo-Niu Han}
\bigskip
\abstract{This paper takes up again the study of the Jacobi and
Watson identities that have been derived combinatorially in
several manners in the classical literature.}
\bigskip

\section 1. Introduction|In the classical literature the Jacobi triple
product appears in one of the following two forms
$$\displaylines{(1.1)\quad
\prod_{n=1}^\infty (1-x^{-1}q^{n-1})(1-xq^n)
=\prod_{i=1}^\infty {1\over (1-q^i)}
\sum_{k=-\infty}^{+\infty} (-1)^k\,x^k\,q^{k(k+1)/2},\hfill\cr
(1.2)\quad
\prod_{n=1}^\infty
(1-x^{-1}\,q^{2n-1})(1-x\,q^{2n-1})
=\prod_{i=1}^\infty {1\over (1-q^{2i})}
\sum_{k=-\infty}^{+\infty} (-1)^k\,x^{k}
q^{k^2},\hfill\cr
\noalign{\hbox{while the Watson quintuple product reads}}
(1.3)\quad
\prod_{n=1}^\infty 
(1-x^{-1}q^{n-1})(1-xq^n)
(1-x^{-2}\,q^{2n-1})(1-x^2\,q^{2n-1})
\hfill\cr
\hfill{}
=\prod_{i=1}^\infty {1\over (1-q^i)}
\sum_{k=-\infty}^{+\infty} q^{(3k^2+k)/2}
(x^{3k}-x^{-3k-1}).\quad\cr
}
$$
The letters $x$ and $q$ may be regarded as complex variables
with $|q|<1$ and $x\not=0$ or as simple indeterminates. In the
latter case consider the ring $\Omega[x,x^{-1}]$ of the
polynomials in the variables~$x$ and~$x^{-1}$ such that
$xx^{-1}=1$ with coefficients in a ring~$\Omega$.
Then the identities hold in the algebra of formal power series in
the variable~$q$ with coefficients in $\Omega[x,x^{-1}]$. 

As usual, let $(a;q)_n$ denote the $q$-ascending factorial 
$$\leqalignno{
(a;q)_n&=\cases{1,&if $n=0$;\cr
                 (1-a)(1-aq)\ldots (1-aq^{n-1}),&if $n\ge 1$;\cr}\cr
(a;q)_\infty&=\prod_{n\ge 0}(1-aq^n);\cr
\noalign{\hbox{and adopt the following classical notation for
the $q$-binomial coefficient:}}
{n\brack k}_q&={(q;q)_n\over  (q;q)_{n-k}(q;q)_k}\quad (0\le k\le
n).\cr}$$
Then identities (1.1) and (1.2) have two finite
versions given by
$$
\leqalignno{\noalign{\vskip-5pt}
(x^{-1};q)_n\,(xq;q)_m&=\sum_{j=-n}^m{n+m\brack j+n}_q(-x)^j\,
q^{j(j+1)/2};&(1.4)\cr
(x^{-2};q^2)_n\,(x^2q;q^2)_m&=\sum_{j=-n}^m{n+m\brack
j+n}_{q^2}(-x^2)^j\, q^{j^2}.&(1.5)\cr
\noalign{\vskip-5pt}}
$$
Those two versions with $n$ and $m$ not necessarily equal are
apparently due to MacMahon ([Ma15], vol.~2, \S\kern2pt 323). He proved
(1.5) by using Sylvester's [Sy82] ``quasi-geometrical method of
demonstration" and notes that to obtain (1.4) the variable $x$ is to be
replaced by $xq$ and then $q^2$ by~$q$. With similar substitutions
(1.5) can be derived from (1.4). As those substitutions are made within
finite expressions the derivations are straightforward.

On the other hand, as shown to us by Andrews [An98], and as it is
well-known in the case $m=n$, identity (1.4) can be proved by 
means of the $q$-binomial identity in its finite form. Proceed 
as follows:
$$
\leqalignno{
(x^{-1};q)_n\,(xq;q)_m\!
&=\!(-1)^nx^{-n}q^{n(n-1)/2}(xq^{1-n};q)_n\,(xq;q)_m\cr
&=\!(-1)^nx^{-n}q^{n(n-1)/2}(xq^{1-n};q)_{n+m}\cr
&=\!(-1)^nx^{-n}q^{n(n-1)/2}\sum_{j=0}^{n+m}{n+m\brack j}_q\!\!
(-xq^{1-n})^jq^{j(j-1)/2}\cr
\noalign{\vskip-10pt}
&=\!\sum_{j=0}^{n+m}{n+m\brack j}_q
(-x)^{j-n} q^{(j-n)(j-n+1)/2}\cr
&=\!\sum_{j=-n}^{m}{n+m\brack j+n}_q
(-x)^{j} q^{j(j+1)/2}.\cr
}
$$
Now to deduce the ``infinite" versions (1.1), (1.2) from the finite ones we
only have to let $n$ and $m$ tend to infinity. 
Using (1.4) for $n=m$ the product
$(x^{-1};q)_m\,(xq;q)_m\,(q;q)_\infty$
can be expressed as
$$
\sum_{j=-m}^m
(q^{m-j+1};q)_{m+j}\,(q^{m+j+1};q)_{\infty}\,(-x)^j\,
q^{j(j+1)/2}.
$$
In that sum the running term is equal to
$(-x)^j\,
q^{j(j+1)/2}(1-q^{m-|j|+1}a_j)$, with $a_j$ a series in~$q$, so that
$(x^{-1};q)_m\,(xq;q)_m\,(q;q)_\infty
=b_m+q^{m}c$, where $b_m$ is the series
$b_m=\sum\limits_{j=-m}^m (-x)^j\,q^{j(j+1)/2}$
and $c$ is a non-null series. Hence
$(x^{-1};q)_\infty\,(xq;q)_\infty\,(q;q)_\infty=\lim_mb_m
=\smash{\sum\limits_{j=-\infty}^{\infty} (-x)^j\,q^{j(j+1)/2}}$, which
is simply~(1.1).

\goodbreak
The MacMahon finite versions (1.4) and (1.5) could be
regarded as the ``fundamental" triple product identities and still, they
are derived by means of the $q$-binomial identity in its finite form. Here we face
one of the mysteries of mathematical tradition: explain why 
so many proofs of those identities can be found in the literature, as will
be recalled below.

On the other hand, a glance at the left-hand sides of  identities
(1.1), (1.2), (1.3) shows that (1.3) must be a consequence of
(1.1) and (1.2) once we know how to handle the product of the
right-hand sides of the first two identities, when $x$ is replaced
by $x^2$ in~(1.2). It is true, as was first derived by Carlitz and
Subbarao [Ca72]. In section~5 we shall reprove (1.3) using an
argument very close to theirs. 

In fact, our first aim was to give a {\it combinatorial proof} of
(1.3), that could be deduced from the combinatorial study of (1.1)
and (1.2). This program was only partially fulfilled, because
(1.3) is an easy consequence of both triple product identities and
changing the ``manipulatorics" needed into some combinatorial
construction would have been a useless task. As will be seen in
section~5, besides the two triple product identities, we only need
the Euler pentagonal number formula, another special case of those
two identities, and a simple summation manipulation.

Before imagining our own combinatorial proofs of (1.1) and (1.2) we
tried to go through the classical literature. The first
combinatorial proofs go back to Sylvester~[Sy82] and have been the
sources of inspiration of several subsequent ones, by
Wright~[Wr65], Sudler~[Su66], Ewell~[Ew81], Lewis~[Le84]. Joichi
and Stanton~[Jo89] discuss the various merits of those proofs.
They are mostly interested in building natural involutions
for proving partition identities; they also compare the approaches
due to  Zolnowsky~[79] and Cheema~[Ch64].

The other proofs are of formal nature, as in
MacMahon~([Ma15], vol.~2, \S\kern2pt 327,
see our discussion of that proof in section~2), 
Bressoud~[Br97] or of
analytical nature, as in Andrews [An65], [An74],  [An84], or in the
classical treatises by Hardy and Wright~[Ha38], Andrews~[An76],
Gupta~[Gu87]. A fairly complete bibliography can be found in Gasper and
Rahman~[Ga90].

The quintuple product identity is originally due to 
Watson~[Wa29]. Other proofs were given by Gordon~[Go61],
Carlitz and Subbarao~[Ca72], Subbarao and Vidyasagar~[Su70].
Hirschhorn~[Hi88] proposes a generalisation of that identity
and claims that there are ``no fewer than twelve proofs of the
quintuple product identity," in particular by Bailey~[Ba51],
Sears~[Se52], Atkin and Swinnerton-Dyer~[At54],
Andrews~[An74].

Finally, those identities are found in classical topics in Number
Theory or Lie Algebra, as in Adiga, Berndt, Bhargava and
Watson~[Ad85], Gustafson~[Gu87], Kac~[Ka78], [Ka85], Lepowsky
and Milne~[Le78], Macdonald~[Ma82], Menon~[Me65], Milne~[Mi85].

\vfill\eject
The combinatorial construction we are using can be seen on Fig.~2. Let
us call it the {\it moving rectangle construction}: rectangle~$B$ after
being transposed is placed under rectangle~$A$, while the
staircase~$E$ is deleted. This combinatorial pattern is used to prove
{\it both} identities (1.1) and (1.2) (see Propositions~1 and~2 below).
The only difference between the proofs of the two propositions is the
fillings of the diagrammes with pluses and crosses in the first case
(see Fig.~1) and crosses and bullets in the second case (see Fig.~3).

As pointed out by an anonymous referee, a similar construction had
already been made, though  only for proving (1.1), by Garvan [Ga86] in his unpublished Ph.D. thesis. We then
called our construction the ``modified Garvan method." Finally, our
friends Zeng [Ze98] and Krattenthaler [Kr99] mentioned to us that our
diagramme-filling method could also be used in the very first model
imagined by Sylvester himself.

The organization of the paper is the following. In the next section we
go back to the classical proof of (1.1) originated by MacMahon and
show how it can be derived using only the topology of formal power
series. Then in sections~3 and~4 we describe our rectangle moving method. We
also show that the finite version identities (1.4) and (1.5) can also be
proved by means of those combinatorial methods. We end the paper with
the derivation of the quintuple product identity. In the sequel the set of
all partitions of the nonnegative integer~$n$ will be denoted by~${\cal
P}(n)$ and $p(n)$ will denote the number of partitions in~${\cal P}(n)$. 


\section 2. The classical proof|
Let $S$ be the set of all the pairs $(i,j)$ with
$i\in\{-1,+1\}$, $j\in{\bboard N}$ and $(i,j)\not=(1,0)$.
Associate with each $(i,j)\in S$ the polynomial
$1-x^iq^j$ in the variable~$q$ with coefficients in
$\Omega[x,x^{-1}]$. The family $(1-x^iq^j)$ $((i,j)\in S)$ is
multipliable. Denote by~$a$ the product of this family:
$$\eqalignno{
a=\prod_{(i,j)\in S} (1-x^iq^j)
&=(1-x^{-1})(1-x^{-1}q)\ldots (1-x^{-1}q^n)\ldots\cr
\noalign{\vskip-8pt}
&\qquad\qquad{}\times(1-xq)(1-xq^2)\ldots (1-xq^n)\ldots\cr}
$$
We can always write the series $a$ as the sum of the summable family
$$
\displaylines{
a=\sum_{i\in {\bboard Z}} x^i\, c(i,\star),\cr
\noalign{\hbox{where}}
c(i,\star):=\cases{\displaystyle\sum_{j\ge i} q^j\,c(i,j),&if $i\ge
0$;\cr
\noalign{\smallskip}
\displaystyle\sum_{j\ge |i|-1} q^j\,c(i,j),&if $i\le -1$.\cr}\cr
}
$$
As derived by MacMahon ([Ma15], vol.~2, \S\kern2pt 327) the (classical)
proof of (1.1) consists of proving first the identity
$$\leqalignno{
a&=c(0,\star)\sum_{i\in {\bboard Z}} x^i\,q^{i(i+1)/2}&(2.1)\cr
\noalign{\vskip-9pt}
\noalign{\hbox{and then}}
c(0,\star)&=1/(q;q)_\infty.&(2.2)\cr}
$$
For proving (2.1) MacMahon ({\it op. cit.}) did not bother using any kind of analytical
argument, but still got the formula. His purpose was ``to give some of the
methods and processes of Gauss in the study of partition series." He believed
that ``they are highly ingenious and no student would find his equipment
complete without them." As for (2.2) instead of using 
what we now call the Frobenius notation for partitions, he
invented a clever but lengthy formal argument.

In modern times, see for instance [An65], [An74], [An84], the
missing argument in the proof of (2.1) is easily provided and of
course the Frobenius notation is a classic.

The purpose of this section is to derive (2.1) using only the topology of
formal power series, although our Krattenthaler [Kr99] finds the
present derivation somehow pedantic! We proceed as follows.
First,
$1-x^{-1}$ is the constant coefficient of~$a$ and for
$n\ge 1$ its coefficient of $q^n$ is equal to the coefficient of~$q^n$ in
the {\it finite} product $b_n$ defined by
$$\displaylines{\qquad
b_n:=\prod_{(i,j)\in S,\,j\le n} a_{i,j}
=(1-x^{-1})(1-x^{-1}q)\ldots (1-x^{-1}q^n)\hfill\cr
\noalign{\vskip-8pt}
\hfill{}\times(1-xq)(1-xq^2)\ldots (1-xq^n).\quad\cr
\noalign{\vskip-3pt}
\noalign{\hbox{Rewrite $b_n$ in the form}}
b_n=\sum_{-(n+1)\le i\le n} x^i\, b_n(i,\star),\cr
\noalign{\hbox{where}}
\noalign{\vskip-9pt}
b_n(i,\star):=\cases{\displaystyle\sum_{i\le j\le 2n-i}
q^j\,b_n(i,j),&if $i\ge 0$ ;\cr
\noalign{\smallskip}
\displaystyle\sum_{|i|-1\le j\le 2n+1-|i|} 
q^j\,b_n(i,j),&if $i\le -1$.\cr}\cr}
$$
The coefficients $b_n(i,j)$ are zero outside the hexagon 
$H_n$ based on the vertices $(0,0)$, $(n,n)$, $(0,2n)$, $(-1,2n)$,
$(-n-1,n)$, $(-1,0)$, while the coefficients $c(i,j)$ are zero for all
points $(i,j)$ $(j\ge 0)$ not contained in the angle~$A$ delimited by the
lines $i+j=-1$, $i=j$. In particular, for all $n\ge 0$
$$\leqalignno{\noalign{\vskip-8pt}
\sum_{-(n+1)\le i\le n} x^i c(i,n)
&=\sum_{-(n+1)\le i\le n} x^i \,b_n(i,n).&(2.3)\cr}
$$
On the other hand,  $b_{n+1}=b_n\times (1+x^{-1}q^{n+1})
(1+xq^{n+1})$ implies that for $0\le j\le n$ and all $i\in {\bboard Z}$
we have 
$$
b_{n+1}(i,j)=b_n(i,j).\leqno(2.4)
$$
From (2.3) et (2.4) it follows that for $-(n+1)\le i\le n$ we have
$b_n(i,n)=c(i,n)$; therefore, for $-n\le i\le (n-1)$ we have
$b_n(i,n-1)=b_{n-1}(i,n-1)=c(i,n-1)$, so that for
$0\le j\le n$ and $-(j+1)\le i\le j$, that is, for
$0\le j\le n$ and $(i,j)\in A\cap H_n$ the following identity holds:
$$
b_n(i,j)=c(i,j).\leqno(2.5)
$$
Relation (2.5) implies that for $n\ge i\ge 0$ and $i\le -1$, $n\ge |i|-1$
the series $c(i,\star)-b_n(i,\star)$ has an order in~$q$ at least equal
to $(n+1)$, a result that we express as
$$
b_n(i,\star)\equiv c(i,\star)\pmod{q^{n+1}}.\leqno(2.6)
$$
In $b_n=b_n(x^{-1},x;q)=
(1+x^{-1})\cdots(1+x^{-1}q^n)
(1+xq)\cdots(1+xq^n)$,
make the substitution $x\leftarrow xq^{-1}$,
$x^{-1}\leftarrow x^{-1}q$, so that
$b_n(x^{-1}q,x\,q^{-1};q)
=\break (1+x^{-1}q)\cdots(1+x^{-1}q^{n+1})
(1+x)\cdots(1+xq^{n-1})$.
Then
$$(1+xq^n)\,b_n(x^{-1}q,x\,q^{-1};q)=x\,b_n(x^{-1},x;q),$$
since $x(1+x^{-1})=(1+x)$. We then get
$$
(1+xq^n)\sum_{-(n+1)\le i\le n} x^i\, q^{-i}\,b_n(i,\star)=
\sum_{-(n+1)\le i\le n} x^{i+1}b_n(i,\star).
$$
Comparing the coefficients of~$x^i$ yields the formula
$$q^{-i}\,b_n(i,\star)+q^{n-i+1}b_n(i-1,\star)=b_n(i-1,\star),$$
for $1\le i\le n$, i.e.,
$$
b_n(i,\star)-q^i\,b_n(i-1,\star) \equiv 0\pmod{q^{n+1}}.
$$
By induction on~$i$ we get
$$
b_n(i,\star)\equiv q^{i(i+1)/2}\,b_n(0,\star)
\pmod{q^{n+1}}.
\leqno(2.7)
$$
for $1\le i\le n$.
In the same manner, from
$b_n(0,\star)+q^{n+1}b_n(-1,\star)=b_n(-1,\star)$
we deduce that for all $i\le -1$, $n\ge |i|-1$ the relation
$$
b(-i,\star)\equiv q^{-i(-i+1)/2}\, b(0,\star)\pmod{q^{n+1}}.
\leqno(2.8)
$$
It follows from (2.6), (2.7), (2.8) that for $0\le i\le n$
and $i\le -1$, $|i|\le n+1$ we have
$c(i,\star)\equiv q^{i(i+1)/2}c(0,\star)\pmod{q^{n+1}}$
and then for all $i\in {\bboard Z}$ the relation
$c(i,\star)=q^{i(i+1)/2}c(0,\star)$ and therefore identity~(2.1).

Now evaluating
$c(0,\star)=\sum\limits_{n\ge 0} q^n\,c(0,n)$ is a classic argument. We
reproduce it for the sake of completeness.
The coefficient $c(0,n)$ is the coefficient of~$x^0\,q^n$
in the finite product
$$
(1+x^{-1})(1+x^{-1}q)\ldots
(1+x^{-1}q^n)\times(1+xq)(1+xq^2)\ldots (1+xq^n).
$$
Consequently, for $n\ge 1$ the coefficient $c(0,n)$ is equal to the
number of sequences
$(i_1,i_2,\ldots,i_k,j_1,j_2,\ldots,j_k)$ such that

(i) $1\le k\le n$ ;

(ii) $0\le i_1<i_2<\cdots<i_k\le n$,
$1\le j_1<j_2<\cdots <j_k\le n$ ;

(iii) $i_1+i_2+\cdots+i_k+j_1+j_2+\cdots+j_k=n$.

In such a sequence we can never have $i_k=n$, for
condition~(iii) would be violated. We can then replace conditions (ii) and
(iii) by

${\rm (ii)}'$ $0\le i_1<i_2<\cdots <i_k\le n-1$,
$0\le j_1<j_2<\cdots <j_k\le n-1$ ;

${\rm (iii)}'$ 
$i_1+i_2+\cdots+i_k+j_1+j_2+\cdots+j_k=n-k$.

But a two-row matrix
$\displaystyle {i_k\, \ldots\, i_2\,i_1\choose j_k\,\ldots\,
j_2\,j_1}$ satisfying conditions (i), ${\rm (ii)}'$, ${\rm
(iii)}'$ is nothing but the Frobenius notation for a partition of~$n$, of
rank~$k$. Therefore,
$c(0,n)=p(n)$, the number of partitions of~$n$ and
$c(0,\star)=1/(q;q)_\infty$.

\goodbreak

\section 3. The rectangle moving method|Consider the infinite series
$$
a=\prod_{(i,j)\in S}(1-x^iq^j)=\sum_{n\ge 0} a(n)q^n
$$
and for each $k\in {\bboard Z}$ and $n\in {\bboard N}$ let ${\cal
B}(k,n)$ be the set of ordered pairs of sequences of integers 
$(i_1,\ldots,i_l)$, $(j_1,\ldots, j_m)$ such that

(i) $0\le i_1<\cdots <i_l$;\quad
$1\le j_1<\cdots <j_m$;

(ii) $i_1+\cdots +i_l+j_1+\cdots +j_m=n$;

(iii) $m-l=k$.

\noindent
Let $b(k,n)$ be the cardinality of ${\cal B}(k,n)$. Clearly, for
${\cal B}(n,k)$ to be non-empty, the inequality
$n\ge k(k+1)/2$ must hold. Therefore, for each $k\in {\bboard Z}$
$$
a(n)=\sum_{\scriptstyle n\ge k(k+1)/2
\atop \scriptstyle k\in {\bboard Z}}
(-1)^k\,x^k\, b(k,n).
$$

\proclaim Proposition 1. For each pair
$(k,n)\in {\bboard Z}\times {\bboard N}$ there is a bijection
$\phi:{\cal B}(k,n)\rightarrow {\cal P}(n-k(k+1)/2)$. In
particular, $b(k,n)=p(n-k(k+1)/2)$.

The construction of such a bijection~$\phi$ is made as follows.
First let $ k\le -1$ and consider the pair
$\pi=(i_1,\ldots,i_l),(j_1,\ldots, j_m)$ belonging to
${\cal B}(k,n)$, so that $l=m-k\ge 1$ and
$i_1+\cdots+i_{m-k}+j_1+\cdots+j_m=n$. Define
$\pi'=(j_1-1,\ldots,j_m-1),(i_1+1,\ldots,i_{m-k}+1)$. The
sequence~$\pi'$ has the following properties:
$$\displaylines{0\le (j_1-1)<\cdots<(j_m-1),\cr
1\le (i_1+1)<\cdots <(i_{m-k}+1),\cr
(j_1-1)+\cdots+(j_m-1)+(i_1+1)+\cdots
+(i_{m-k}+1)=n-k.\cr}
$$
Thus $\pi'$ belongs to
${\cal B}(-k,n-k)$. The mapping from $\pi'$ to~$\pi$ is defined
in an analogous manner. Consequently,
$\pi\mapsto \pi'$ is a bijection of 
${\cal B}(k,n)$ onto ${\cal B}(-k,n-k)$ and then
$b(k,n)=b(-k,n-k)$ for $k\le -1$.
On the other hand, 
${\cal P}(n-k(k+1)/2)={\cal P}((n-k)-(-k)(-k+1)/2)$.

\medskip
It then suffices to construct the bijection~$\phi$ for $k\ge 0$.
Each sequence $\pi=(i_1,\ldots,i_l),(j_1,\ldots, j_m)$ in the set
${\cal B}(k,n)$ can be represented by a top-rectified
diagramme containing $n$ {\it crosses} and {\it plus signs}
displayed on the lattice ${\bboard Z}^2$ in the following manner.
On the horizontal axis starting at and to the right of
$(1,-1)$ place $j_m$ crosses, on the axis just under it starting
at $(2,-2)$ place $j_{m-1}$ crosses, \dots~, on the axis of
ordinate~$-m$ place $j_1$ crosses starting at $(m,-m)$. 

Next on the vertical axis starting from the point
$(1,-2)$ and going down place $i_l$ plus signs, on the axis of
abscissa~2, starting from $(2,-3)$, place
$i_{l-1}$ plus signs, \dots~, on the vertical axis of abscissa~$l$,
starting from $(l,-(l+1))$, place $i_1$ plus signs.
$$\hskip2.35cm
\raise 2.75cm\hbox{\vbox{${\scriptstyle l}\left\{\vrule height
.94cm depth 0pt width 0pt\right.$}}\hss
\vbox{\offinterlineskip\halign{\strut\vrule\ $#$\ 
&\ $#$\ &\ $#$\ &\ $#$\ \vrule&\ $#$\ &\ $#$\ &\ $#$\ \vrule&\
$#$\ &\
$#$\  &\ $#$\ \vrule\cr
\multispan{4}$\overbrace{\kern2cm}^{l}$&\multispan3\kern-.5pt
$\overbrace{\kern1.6cm}^{k}$\cr
\noalign{\hrule}
\times&\times&\times&\times&\times&\times&\times&\times&\times
&\times\cr
+&\times&\times&\times&\times&\times&\times&\times&\times
&\times\cr
\multispan8&\omit\hrulefill&\omit\hrulefill\cr
+&+&\times&\times&\times&\times&\times&\omit
\ $\times$\ \vrule\cr
+&+&+&\times&\times&\times&\times&\omit
\ $\times$\ \vrule\cr
\multispan7\hrulefill\cr
+&+&\omit\ $+$\ \vrule&&\omit\ 
$\times$\ &\times&\times&\omit\  $\times$\ \vrule\cr
\multispan2&\omit\hrulefill&&\omit\hrulefill&\cr
+&\omit\ $+$\ \vrule&&\omit&&\omit\vrule\ 
$\times$\ &\times&\omit\  $\times$\ \vrule\cr
\multispan5&\omit\hrulefill&&\omit\hrulefill\cr
+&\omit\ $+$\ \vrule&\multispan4&\omit\vrule\ 
$\times$\ \vrule\cr
\multispan2\hrulefill&\multispan4&\omit\hrulefill\cr
\noalign{\vskip 12pt}
\multispan{10}\hfil Fig. 1\hfil\cr
}}\hskip 2.8cm
$$
For example, for $n=51$, $k=3$, $l=4$, $m=l+k=7$, the sequence
$\pi=(i_1=0,i_2=2,i_3=5,i_4=6)$,
$(j_1=1,j_2=3,j_3=4,j_4=5,j_5=6,  j_6=9,j_7=10)$
has the representation shown in Fig.~1. The diagramme consists
of five parts: (1) a {\it square}~$A$ of side~$l=4$ having
crosses on and above its diagonal, and plus signs below its
diagonal; (2) a {\it rectangle}~$B$ of width~$k=3$ and of
height~$l=4$; (3) a diagramme~$C$ containing crosses which is
necessary a {\it Ferrers diagramme} because the sequence 
$(i_m,\ldots,i_1)$ is strictly decreasing; (4) under $A$ another
{\it Ferrers diagramme}~$D$ containing plus signs; (5) under~$B$ a
{\it staircase}~$E$ containing $k(k+1)/2$ crosses.

$$\quad
\vtop{\offinterlineskip\halign{\strut\vrule\ #\ 
&\ #\ &\ #\ &\ #\ \vrule&\ #\ &\ #\ &\ #\ \vrule&\
#\ &\ #\ &\ #\ \vrule\cr
\multispan{4}$\overbrace{\kern2cm}^{l}$&\multispan3\kern-.5pt
$\overbrace{\kern1.6cm}^{k}$\cr
\noalign{\hrule}
\hphantom{$\times$}&\hphantom{$\times$}&\hphantom{$\times$}%
&\hphantom{$\times$}&\hphantom{$\times$}&\hphantom{$\times$}%
&\hphantom{$\times$}&\hphantom{$\times$}&\hphantom{$\times$}%
&\hphantom{$\times$}\cr
&\multispan2\hfil$A$\hfil&&&\omit\hfil$B$\hfil&&&\omit
\hfil$C$\hfil&\cr
\multispan8&\omit\hrulefill&\omit\hrulefill\cr
&&&&&&&\omit
\ \ \vrule\cr
&&&&&&&\omit
\ \ \vrule&\cr
\multispan7\hrulefill\cr
&&\omit\ \ \vrule&&\omit\ 
\ &&&\omit\  \ \vrule\cr
\multispan2&\omit\hrulefill&&\omit\hrulefill&\cr
&\omit$D$ \vrule&&\omit&&\omit\vrule\ 
\ $E$&&\omit\  \ \vrule\cr
\multispan5&\omit\hrulefill&&\omit\hrulefill\cr
&\omit\ \ \vrule&\multispan4&\omit\vrule\ 
\ \vrule\cr
\multispan2\hrulefill&\multispan4&\omit\hrulefill\cr
}}
\kern.5cm
\vtop{\kern2cm
\hbox{$\mapsto$}}
%\kern.5cm
%\hskip2.35cm
\lower 2.5cm
\hbox{\vbox{${\scriptstyle k}\left\{\vrule height
.72cm depth 0pt width 0pt\right.$}}\hss
\vtop{\offinterlineskip\halign{\strut\vrule\ #\ 
&\ #\ &\ #\ &\ #\ \vrule&\ #\ &\ #\ &\ #\ \vrule\cr
\multispan{4}$\overbrace{\kern2cm}^{l}$\cr
\noalign{\hrule}
\hphantom{$\times$}&\hphantom{$\times$}&\hphantom{$\times$}%
&\hphantom{$\times$}&\hphantom{$\times$}&\hphantom{$\times$}%
&\hphantom{$\times$}\cr
&\multispan2\hfil$A$\hfil&&&\omit
\hfil$C$\hfil&\cr
\multispan5&\omit\hrulefill&\omit\hrulefill\cr
&&&&\omit
\ \ \vrule\cr
&&&&\omit
\ \ \vrule\cr
\multispan4\hrulefill\cr
&&&&\omit\  \ \vrule\cr
&\multispan2\hfil$B^T$\hfil&&\omit\  \ \vrule\cr
\multispan4&\omit\hrulefill\cr
&&&$Q$&\cr
\multispan4\hrulefill\cr
&&\omit\ \vrule\cr
\multispan2&\omit\hrulefill\cr
&\omit$D$ \vrule&\cr
&\omit\ \ \vrule\cr
\multispan2\hrulefill\cr
}
}\hskip.3cm
$$

\centerline{Fig. 2}

\medskip

The bijection $\phi$ is completely described in Fig.~2:
remove the staircase~$E$; then move out the rectangle~$B$ and
insert it between the square~$A$ and the Ferrers diagramme~$D$
in such a way that its side of length~$l$ coincide with the side
of length~$l$ of the square~$A$; finally, push the diagramme~$E$
to the left so that its left rim coincide with the figure
constructed with $A$, the transpose~$B^T$ of the
rectangle~$B$, and~$D$. The conditions $k\ge 1$, $m=l+k$ imply
that the configuration thereby derived---call it~$F$---is a
{\it Ferrers diagramme} containing $n-k(k+1)/2$ crosses or
plus signs.

Note that when $l=0$, the square~$A$, the rectangle~$B$ and the
diagramme~$D$ are empty. The initial configuration
is reduced to the staircase~$E$ and the Ferrers diagramme~$C$,
itself reduced to its part contiguous to~$E$. Then the
configuration~$F$ is simply the Ferrers diagramme~$C$. In
particular, $F$ has most~$k$ rows.

When $k=0$, the rectangle~$B$ and the staircase~$E$ are empty
and $E$ is reduced to its part contiguous to~$A$. The
configuration reduced to~$A$, to~$D$ and to the part of~$E$
contiguous to~$A$ is a Ferrers diagramme which remains alike
under the transform~$\phi$.

When $k\ge 1$ and $l\ge 1$, the Ferrers diagramme $F$ has at
least $(k+1)$ rows, since $k$ rows have been added to the
square~$A$ which itself has~$l\ge 1$ rows.

Conversely, how can we get back the original representation
of~$\pi$ from the diagramme~$F$, i.e., from any Ferrers
diagramme with $n-k(k+1)/2$ points?

First, if $k=0$, we know that $\pi\mapsto F$ is the identity
mapping. If $k\ge 1$ and if the diagramme~$F$ has {\it at
most}~$k$ rows, the representation of~$\pi$ is obtained by
putting the diagramme~$F$ to the right of a staircase~$E$ with
$k(k+1)/2$ crosses.

It remains to study the case where $k\ge 1$ and $F$ has at least
$(k+1)$ rows. When building the diagramme~$F$ (see~Fig.~2) the
rightmost bottom point of the rectangle~$B^T$, in position
$(l,-l-k)$, always belongs to the rim of the diagramme~$F$. Its
right neighbor belongs to~$F$ if $C$ has $(l+k)$ rows; in the
same manner, the point just above it belongs to~$F$ if the
greatest part of~$D$ is equal to~$l$. But, in every case, the
diagramme~$F$ has {\it no} point in position
$(l+1,-l-k-1)$. The point $(l,-l-k)$ can then be characterized
as being the unique point~$Q$ in~$F$ located on the line of
equation $y=-x-k$ {\it and} on the rim of~$F$. 

To reconstruct $\pi$ from~$F$ we just have to find the location
of the point~$Q$. Its position $(l,-l-k)$ uniquely
determines~$l$. We can then redraw the rectangle~$B^T$,
the square~$A$ and reverse the construction.\qed


From Proposition~1 it follows that
$$\eqalignno{a&=\sum_{n\ge 0}a(n)\,q^n\cr
&=\sum_{\scriptstyle k\in {\bboard Z}\atop
\scriptstyle n\ge k(k+1)/2}(-1)^k\,x^k\,
p(n-k(k+1)/2)\,q^n\cr
&=\sum_{n\ge 0,\,k\in {\bboard Z}}(-1)^k\,x^k\,
p(n)\, q^{n+k(k+1)/2}\cr
&=\sum_{n\ge 0}p(n)\, q^n \sum_{k\in {\bboard Z}}
(-1)^k x^k\, q^{k(k+1)/2},\cr
}
$$
which is precisely the Jacobi triple product identity (1.1).

\section 4. Crosses and bullets| 
To prove identity (1.2) we consider the
formal power series:
$$
d=\prod_{n=1}^\infty(1-x^{-1}q^{2n-1})(1-xq^{2n-1})
=\sum_{n\ge 0} q^{n}\,d(n)
$$
and try to express it as the product of the two series forming
the right-hand side of~(1.2). For each $k\in {\bboard Z}$ and 
$n\ge 0$ let ${\cal B}'(k,n)$ denote the set of all ordered pairs
of {\it strictly increasing} sequences of {\it odd} integers
$(i_1,\ldots,i_l),(j_1,\ldots, j_m)$ such that
$l-m=k$ and
$i_1+\cdots+i_l+j_1+\cdots +j_m=n$. Let $b'(k,n)$ be the
cardinality of
${\cal B}'(k,n)$.  Then
$$
d(n)=\sum_k (-1)^{k}\,x^{2k}\,b'(k,n).
$$
The combinatorial result to be proved is the following
proposition which is parallel to Proposition~1. It was already
stated by Lewis [Le84], but proved differently, as that author
makes use of a different bijection.

\proclaim Proposition 2. For each pair
$(k,n)\in {\bboard Z}\times {\bboard N}$ there is a bijection
$\phi':{\cal B}'(k,n)\rightarrow {\cal P}((n-k^2)/2)$. In
particular, $b'(k,n)=p((n-k^2)/2)$.

As we shall see, the bijection $\phi'$ will make use of the same
geometric moves that entered the definition of~$\phi$. Only
the displaying of {\it crosses} and now {\it bullets} will be
different. In the following construction whenever a cross is
deposited onto a point of the lattice~${\bboard Z}^2$ on which a
previous cross has been deposited, the superimposition of the
two crosses will be indicated by a bullet. For convenience let us
write: $\times\;\&\; \times=\bullet$~.

Let $\pi=(i_1,\ldots,i_l), (j_1,\ldots, j_m)$ be an element of
${\cal B}'(k,n)$. It can be represented by a top-rectified
diagramme in ${\bboard Z}^2$ in the following manner. On
$(1,-1)$ place a single cross and to its right place $(i_l-1)/2$
bullets, then on 
$(2,-2)$ place a single cross and to its right $(i_{l-1}-1)/2$
bullets,  \dots~, on the row of ordinate~$-m$ place a single
cross on $(m,-m)$ and $(i_1-1)/2$ bullets to its right.

In the same way, place crosses and bullets on the vertical axis
starting with $(-1,1)$. If a cross is deposited on a point
already occupied by another cross, change the superimposition
of those two crosses by a bullet. Place a cross on $(-1,1)$ and
$(j_m-1)/2$ bullets under that point, then a cross on $(-2,2)$ 
and $(j_{m-1}-1)/2$ bullets under it, \dots~, a cross on
$(-m,m,)$ and $(j_1-1)/2$ bullets under it.

$$
\vbox{\offinterlineskip\halign{\strut\vrule\ $#$\ 
&\ $#$\ &\ $#$\ &\ $#$\ \vrule&\ $#$\ &\ $#$\ &\ $#$\ \vrule&\
$#$\ &\
$#$\  &\ $#$\ \vrule\cr
\noalign{\hrule}
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet
&\bullet\cr
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet
&\bullet\cr
\multispan8&\omit\hrulefill&\omit\hrulefill\cr
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\omit
\ $\bullet$\ \vrule\cr
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\omit
\ $\bullet$\ \vrule\cr
\multispan7\hrulefill\cr
\bullet&\bullet&\omit\ $\bullet$\ \vrule&&\omit\ 
$\times$\ &\bullet&\bullet&\omit\  $\bullet$\ \vrule\cr
\multispan2&\omit\hrulefill&&\omit\hrulefill&\cr
\bullet&\omit\ $\bullet$\ \vrule&&\omit&&\omit\vrule\ 
$\times$\ &\bullet&\omit\  $\bullet$\ \vrule\cr
\multispan5&\omit\hrulefill&&\omit\hrulefill\cr
\bullet&\omit\ $\bullet$\ \vrule&\multispan4&\omit\vrule\ 
$\times$\ \vrule\cr
\multispan2\hrulefill&\multispan4&\omit\hrulefill\cr
\noalign{\vskip 12pt}
\multispan{10}\hfil Fig. 3\hfil\cr
}}
$$
With the above convention the configuration will exactly
have\hfil\break
$(i_l-1)/2+\cdots+(i_1-1)/2+(j_m-1)/2+\cdots
+(j_1-1)/2+\min(l,m)$
${}=n/2-(l+m)/2+\min(l,m)=n/2-k/2\ {\rm bullets}$
and $|k|=|l-m|$ crosses. 

\goodbreak
For example, for $l=7$, $m=4$, $k=l-m=3$, the sequence
$\pi=(i_1=1,i_2=5,i_3=7,i_4=9,
i_5=11, i_6=17, i_7=19),(j_1=1,j_2=5,j_3=11,j_4=13)$ has the
 representation shown in Fig.~3.
The configuration contains
$(9+8+5+4+3+2)+(6+5+2)+4=48$ bullets and three crosses.

If $k\ge 0$, the configuration has the same shape as the
configuration described in Fig.~1. If $k\le -1$ the
configurations in ${\cal B}'(k,n)$ can be deduced from the
configurations in
${\cal B}'(-k,n)$ by making a rotation about the axis
$x+y=0$. Assume $k\ge 0$. Each  configuration may be partitioned
into five components $A$, $B$, $C$, $D$, $E$ whose {\it
dimensions} have the same constraints as in the case of Fig.~1.
Only their {\it contents} are different. The staircase~$E$
exactly has $k$ crosses along its diagonal and $1+2+\cdots+
(k-1)=(k-1)k/2$ bullets. As each bullet stands for two
crosses, the staircase~$E$ contains the equivalent of $k+2\times
(k-1)k/2 =k^2$ crosses. In particular $n\ge k^2$. 

We can apply the bijection $\pi\mapsto F$ described in Fig.~2. If
$\pi$ is in ${\cal B}'(k,n)$  $(k\ge 0)$, then $F$ is a Ferrers
diagramme containing
$n/2-(l+m)/2+\min(l,m)-(k-1)k/2=n/2-k^2/2$ bullets. Thus
$\pi\mapsto F$ is a bijection of ${\cal B}'(k,n)$ onto
${\cal P}(n/2-k^2/2)$.\qed

\medskip
It follows from Proposition 2 that
$$\eqalign{
d=\sum_{n\ge 0} d(n)\,q^n
&=\sum_{k\in {\bboard Z},\, n\ge k^2}
(-1)^k\,x^{2k}\, p((n-k^2)/2)\,q^n\cr
&=\sum_{n\ge 0,\,k\in {\bboard Z}}
(-1)^k\,x^{2k}\, p(2n)\,q^{2n+k^2}\cr
&=\sum_{n\ge 0}p(2n)\,q^{2n}
\sum_{k\in {\bboard Z}}
(-1)^k\,x^{2k}\,q^{k^2},\cr}
$$
which is precisely identity (1.2).

\medskip
The bijection $\pi\mapsto F$ may also serve to proving the finite
versions (1.4) and (1.5). We illustrate the method for the latter identity. 
Let ${\cal P}_{L,M}(n)$ be the set of partitions of~$n$ whose greatest
part is at most equal to~$L$ and number of parts at most equal to~$M$.
Also let $p_{L,M}(n)$ denote the cardinality of ${\cal P}_{L,M}(n)$. We
make use of the classical identity (easy to derive)
$$
\sum_{n\ge 0} q^n\, p_{L,M}(n)={L+M\brack L}_q.
$$
On the other hand, let ${\cal B}'_{L,M}(k,n)$ be the subset of pairs
of sequences
$(i_1,\ldots,i_l)$, $(j_1,\ldots,j_m)$ in
${\cal B}'(k,n)$ with the further condition that $l\le L$ and $m\le M$
and let $b'_{L,M}(k,n)$ be the cardinality of ${\cal B}'_{L,M}(k,n)$. Then
$$
d_{L,M}(n)=\sum_k(-1)^kx^{2k}\,b'_{L,M}(k,n).
$$
Now it is readily seen that the above bijection $\pi\mapsto F$ maps
${\cal B}'_{L,M}(k,n)$ onto ${\cal P}_{L-k,M+k}((n-k^2)/2)$. As
$(x^{2};q^2)_L\,(x^{-2};q^2)_M=\sum\limits_{n\ge 0} q^n\,d_{L,M}(n)$,
we obtain
$$\eqalignno{
(x^{2};q^2)_L\,(x^{-2};q^2)_M
&=\sum_{n\ge 0}q^n\sum_k(-1)^kx^{2k}\,b'_{L,M}(k,n)\cr
&=\sum_{n\ge 0}q^n\sum_k(-1)^kx^{2k}\,p_{L-k,M+k}((n-k^2)/2)\cr
&=\sum_k (-1)^kx^{2k}
\sum_{n\ge 0}q^n\,p_{L-k,M+k}((n-k^2)/2)\cr
&=\sum_k (-1)^kx^{2k}q^{k^2}
\sum_{n\ge 0}q^n\,p_{L-k,M+k}(2n)q^{2n}\cr
&=\sum_k (-1)^kx^{2k}q^{k^2}{L+M\brack L-k}_{q^2},\cr}
$$
which is identity (1.5) with the substitutions
$n\leftarrow M$ and $m\leftarrow L$.

\goodbreak
\section 5. The quintuple product identity|
To derive the quintuple product identity (1.3) it suffices to prove
$$\displaylines{\quad
\prod_{i=1}^\infty
{1\over 1-q^i}
\sum_{k\in {\bboard Z}} (-1)^k x^k
q^{k(k+1)/2}\times
\prod_{i=1}^\infty {1\over 1-q^{2i}}
\sum_{k\in {\bboard Z}} (-1)^k x^{2k} q^{k^2}\hfill\cr
\hfill{}
=\prod_{i=1}^\infty
{1\over 1-q^i}
\sum_{k\in {\bboard Z}} q^{(3k^2+k)/2}
(x^{3k}-x^{3k-1}),\quad\cr
\noalign{\hbox{or by using the Euler pentagonal number
identity}}
\prod_{i\ge 1} (1-q^i)=\sum_{k\in {\bboard Z}}
(-1)^k\,q^{(3k^2-k)/2},\cr
\noalign{\hbox{to prove the identity}}
\sum_{k\in {\bboard Z}} (-1)^k x^k
q^{k(k+1)/2}\times
\sum_{l\in {\bboard Z}} (-1)^l x^{2l} q^{l^2}\hfill\cr
\hfill{}
=\sum_{n\in {\bboard Z}} (-1)^n\,q^{3n^2-n}\times
\sum_{m\in {\bboard Z}} q^{(3m^2+m)/2}
(x^{3m}-x^{-3m-1}).\quad\cr}
$$
Write the product of the two series of the left-hand side as the
sum of three series denoted by
$S_0$, $S_1$, $S_2$:
$$\eqalignno{
\sum_{k,l}(-1)^{k+l}\,x^{k+2l}q^{k(k+1)/2+l^2}
&=\sum_mx^{3m}\sum_{k+2l=3m}(-1)^{k+l}\,q^{k(k+1)/2+l^2}\cr
&\quad{}+\sum_mx^{3m-1}
\sum_{k+2l=3m-1}(-1)^{k+l}\,q^{k(k+1)/2+l^2}\cr
&\quad{}+\sum_mx^{3m-2}\sum_{k+2l=3m+2}
(-1)^{k+l}\,q^{k(k+1)/2+l^2}\cr
&=S_0+S_1 +S_2.\cr
}
$$
For $S_0$ notice that $k+2l=3m$ and $l-m=n$ imply: $k+l=2m-n$
and $k(k+1)/2+l^2=(3m^2+m)/2+3n^2-n$. Hence
$$
S_0=\sum_m q^{(3m^2+m)/2}\,x^{3m}
\sum_n(-1)^n\, q^{3n^2-n}.
$$
For $S_1$ the change of indices $k+2l=3m-1$ et $l-m=n$
imply:
$k+l=2m-n-1$ and
$k(k+1)/2+l^2=(3m^2-m)/2+3n^2+n$. Hence
$$\eqalign{
S_1&=-\sum_m q^{(3m^2-m)/2}\,x^{3m-1}
\sum_n(-1)^n\, q^{3n^2+n}\cr
&=-\sum_m  q^{(3m^2+m)/2}\,x^{-3m-1}\,
\sum_n(-1)^n\, q^{3n^2-n}.\cr}
$$
Finally, for $S_2$ make the change of indices
$k+2l=3m-2$ and $l-m=n$, so that
$k+l=2m-n-2$ et
$k(k+1)/2+l^2=(3m^2-3m+2)/2+3n^2+3n$. Hence
$$
S_2=\sum_m x^{3m-2}\,q^{(3m^2-3m+2)/2}
\sum_n(-1)^n\, q^{3n^2+3n}.
$$
But
$\sum\limits_{n\in {\bboard Z}}(-1)^n\, (q^3)^{n(n+1)}=0$,
and $S_2=0$. The sum $S_0+S_1$ is exactly the right-hand
side of the quintuple product identity~(1.3).\qed


\vfill\eject
\vglue 2cm

\centerline{\bf References}

{\eightpoint
\vskip.3cm
\livre Ad85|B. Adiga, B. C. Berndt, S. Bhargava and G .N.
Watson|Ramanujan's second notebook: Theta-functions and
$q$-series {\rm Chap. 16}|Amer. Math. Soc., Prodidence
{\oldstyle 1985} ({\sl Amer. Math. Soc. Memoirs}, vol.~{\bf 315})|


\article An65|G. E. Andrews|A simple proof of Jacobi's triple
product identity|Proc. Amer. Math. Soc.|16|1965|333-334|

\article An74|G. E. Andrews|Applications of basic hypergeometric
functions|SIAM Rev.|16|1974|441--484|

\livre An76|G. E. Andrews|The Theory of
Partitions|Addison-Wesley, Reading, {\oldstyle 1976}
({\sl Encyclopedia of Math. and Its Appl.,} vol.~{\bf 2})|

\livre An84|G. E. Andrews|Generalized Frobenius partitions|Amer.
Math. Soc., Providence, {\oldstyle 1984}
({\sl Amer. Math. Soc. Memoirs}, vol.~{\bf 49}), pp.~4-5|

\divers An98|G. E. Andrews|Private communication, {\oldstyle 1998}|

\article At54|A. O. L. Atkin and P. Swinnerton-Dyer|Some
properties of partitions|Proc. London. Math. Soc.|4|1954|84--106|

\article Ba51|W. N. Bailey|On the simplification of some
identities of the Rogers-Ramanujan type|Proc. London Math.
Soc.|1|1951|217--221|


\divers Br97|D. Bressoud|Proofs and Confirmations. The story of
the alternating sign matrix conjecture. Preprint, Macalester
College, Saint Paul, {\oldstyle 1997}|

\article Ca72|L. Carlitz and M. V. Subbarao|A simple proof of the
quintuple product identity|Proc. Amer. Math. Soc.|32|1972|42--44|


\article Che64|M. S. Cheema|Vector partitions and combinatorial
identities|Math. Comp.|18|1964|414--420|

\article Ew81|J. A. Ewell|An easy proof of the triple-product
identity|Amer. Math. Monthly|88|1981|270--272|

\divers Ga86|F. G. Garvan|Generalizations of Dyson's Rank.
Ph. D Thesis, Pennsylvania State University, May {\oldstyle 1986}, 128~p|

\livre Ga90|G. Gasper and M. Rahman|Basic Hypergeometric
Series|Cambridge Univ. Press, Cambridge and New York,
{\oldstyle 1990} (Encyclopedia of Math. and its Appl., vol.~{\bf
35}), p. 134|

\article Go61|B. Gordon|Some identities in combinatorial
analysis|Quart. J. Math. (Oxford)|12|1961|285--290|

\livre Gu80|H. Gupta|Selected Topics in Number Theory|Abacus
Press, Tunbridge Wells, {\oldstyle 1980}, pp. 213-215|

\article Gu87|R. A. Gustafson|Multilateral summation theorems for
ordinary and basic hypergeometric series in $U(n)$|SIAM J. Math.
Anal.|18|1987|1576--1596|

\livre Ha38|G. H. Hardy and E. M. Wright|An Introduction to the
Theory of Numbers|Clarendon Press, Oxford, {\oldstyle 1938}.
Reprinted with corrections, {\oldstyle 1985}|

\article Hi88|M. D. Hirschhorn|A generalization of the quintuple
product identity|J. Austral. Math. Soc.|A 44|1988|42--45|

\article Jo89|T. T. Joichi and D. Stanton|An involution for
Jacobi's identity|Discrete Math.|73|1989|261--272|

\article Ka78|V. G. Kac|Infinite-dimensional algebras, Dedekind's
$\eta$-function, classical M\"obius function and the very
strange formula|Advances in Math.|30|1978|85--136|


\livre Ka85|V. G. Kac|Infinite Dimensional Lie Algebras {\rm 2nd.
Edition}|Cambridge Univ. Press, Cambridge {\oldstyle 1985}|

\vfill\eject
\divers Kr99|Ch. Krattenthaler|Private communication, 
{\oldstyle 1999}|

\article Le78|J. Lepowsky and S. Milne|Lie algebraic approaches
to classical partition identities|Advances in
Math.|29|1978|15--59|

\article Le84|R. P. Lewis|A combinatorial proof of the triple
product identity|Amer. Math. Monthly|91|1984|420--423|

\livre Ma15|(Major) P. A. MacMahon|Combinatory Analysis {\rm vol.
I, II}|Cambridge Univ. Press, {\oldstyle 1915}. Reprinted by
Chelsea, New York, {\oldstyle 1960}|

\article Ma82|I. G. Macdonald|Some conjectures for root
systems|SIAM J. Math. Anal.|13|1982|988--1007|

\article Me65|P. K. Menon|On Ramanujan's continued fraction and
related identities|J. London Math. Soc.|40|1965|49--54|

\article Mi85|S. Milne|An elementary proof of the Macdonald
identities for $A_l^{(1)}$|Adv.  in Math.|57|1985|34--70|

\article Se52|D. B. Sears|Two identities of Bailey|J. London Math.
Soc.|27|1952|510--511|

\article Su70|M. V. Subbarao and M. Vidyasagar|On Watson's
quintuple product identity|Proc. Amer. Math. Soc.|26|1970|23--27|


\article Su66|C. Sudler|Two enumerative proofs of an identity of
Jacobi|Proc. Edinburgh Math. Soc.|15|1966|67--71|

\article Sy82|J. J. Sylvester|A constructive theory of partitions
in three acts, an interact and an exodion|Amer. J.
Math.|5|1882|251--330 (or pp.~1-83 of the Coll. Math. Papers of
J.J. Sylvester, vol.~4, Cambridge Univ. Press, London and New
York, 1912; reprinted by Chelsea, New York, 1974)|

\article Wa29|G. N. Watson|Theorems stated by Ramanujan. VII:
Theorems on continued fractions|J. London Math.
Soc.|4|1929|39--48|

\article Wr65|E. M. Wright|An enumerative proof of an identity of
Jacobi|J. London Math. Soc.|40|1965|55--57|

\divers Ze98|J. Zeng|Private communication, {\oldstyle 1998}|

\article Zo79|J. Zolnowsky|A direct combinatorial proof of the
Jacobi identity|Discrete Math.|9|1974|293--298|

}
\vskip 1cm
\line{\quad\vtop{\halign{#\hfil\cr
Dominique Foata\cr
D\'epartement de math\'ematique\cr
Universit\'e Louis Pasteur\cr
7, rue Ren\'e-Descartes\cr
F-67084 Strasbourg\cr
{\tt foata@math.u-strasbg.fr}\cr}}\hfil
\vtop{\halign{#\hfil\cr
Guo-Niu Han\cr
{\sevenrm I.R.M.A.} et {\sevenrm C.N.R.S.}\cr
Universit\'e Louis Pasteur\cr
7, rue Ren\'e-Descartes\cr
F-67084 Strasbourg\cr
{\tt guoniu@math.u-strasbg.fr}\cr}}\quad}


\bye
