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\topmatter
\title A DETERMINANT EVALUATION AND SOME ENUMERATION RESULTS FOR PLANE 
PARTITIONS
\endtitle
\rightheadtext{A determinant evaluation}
\author Christian Krattenthaler \endauthor
\affil Institut f\"ur Mathematik\\Universit\"at Wien\\Strudlhofgasse 
4\\A-1090 Vienna\\AUSTRIA\endaffil
\keywords Plane partitions, generating functions, hook lengths, 
contents.\endkeywords
\subjclass Primary 11P57; Secondary 05A10, 05A15, 05A17, 11B65, 11P68, 
15A15. \endsubjclass
\abstract{For fixed integers $\alpha
$ and $\beta
$, planar arrays of integers of 
a given shape, in which the entries decrease at least by $\alpha
$ along 
rows and at least by $\beta
$ along columns, are considered. The number of 
these $(\alpha
,\beta
)$--plane partitions in which the first part and the last 
part in the $i$'th row are bounded by some integers $a\sb
i$ and $b\sb
i$, 
respectively, can be given by a determinant. We give a new 
evaluation of this determinant for some special choices of the row 
bounds $a\sb
i$ and $b\sb
i$. }\endabstract
\endtopmatter
\subheading{1.~Introduction   }When enumerating plane partitions (we 
give all relevant definitions in \S2 below), one is lead to 
certain determinantal expressions, which one succeeds to simplify in 
some special cases. The origins of this kind of work may be found in 
the work of MacMahon \cite{10}. Since then, many authors have 
contributed important advances in the theory of plane partitions. 
Roughly speaking, there are three different kinds of enumeration 
problems. Firstly, counting plane partitions {\sl of\/} a given shape 
subject to some restrictions (some references are 
\cite{{4,}{5,}{6,}{8,}{10,}15}), secondly, counting 
plane partitions {\sl contained\/} in a given shape subject to some 
restrictions, and, thirdly, counting {\sl symmetry classes\/} of plane 
partitions.

In this paper we deal with the first type of enumeration problems. 
Recently, Gessel and Viennot \cite{6} and the author \cite{8} proved 
determinant formulas for various generating functions of rather 
general classes of plane partitions of a given shape. As mentioned 
above, there are several special cases where those determinants may 
be evaluated in order to result in closed expressions. Important 
examples are the hook length and hook--content formulae of Stanley 
\cite{13} and 
Gansner \cite{{4,}5}. 

It was discovered in \cite{8} that a single, very simple determinant 
lemma \cite{8, Lem\-ma~2.2} serves to do all known evaluations for 
type one--problems. 
Besides, it could be applied to obtain some new evaluations. (The 
situation is quite different for the second and third type of 
enumeration problems, which are related in some sense. Here one has to 
encounter very intricate evaluations of Gaussian binomial determinants. 
The above cited determinant lemma does not apply in this case. But 
nevertheless there are beautiful results and beautiful conjectured 
results \cite{{1,}{2,}{3,}{7,}{9,}{11,}{12,}{13,}14}, see also the 
surveys in 
\cite{{16,}17}).

Gessel's and Viennot's paper also contains three very nice, new 
evaluations of binomial deteminants \cite{6, Theorems~21,\,22,\,23}. It 
turned out 
that our deteminant lemma may be applied to derive these results, too.
This was the motivation to write this paper. In Theorem~8 below, we are able 
to give a 
common generalization of Gessel's and Viennot's evaluations. 
This theorem concerns the enumeration of {\sl reverse\/} plane 
partitions. There is also a counterpart for the enumeration of plane 
partitions (Theorem~5). 
It 
generalizes the well--known hook--content formula for the number of 
column--strict plane partitions, given in this form by Stanley
\cite{15, Theorem~15.3 for $x=1$}. This fact becomes clear in 
Theorem~7.
A special case (Corollary~10) of Theorem~5
gives the number of ordinary plane partitions of a (what we call) 
``staircase" shape with  parts  not exceeding a given 
integer. This result also seems to be new. Moreover, the ``staircase" shape 
and the rectangular shape (up to now) are the only ones for which 
MacMahon's determinant (\cite{10, sec.494}, for a clearer 
presentation of this formula see \cite{8, Theorem~6.1 for $\alpha
=0$, $\beta
=1$, 
$b\sb
i=1$, 
$a\sb
i=a$,  $\mu=0$, and $x=1$}) could be evaluated. (For rectangular 
shapes even the norm generating function has a closed expression 
(\cite{10, sec.429}, see also \cite{15, Theorem~18.1}).)  

Another unsolved problem is to give a closed expression for the number 
of row and column--strict plane partitions of a given shape with parts 
not exceeding a given integer. A solution for staircase shapes is 
given in Corollary~11, which also is an implication of Theorem~5.

\bigpagebreak
\subheading{2.~Definitions and Assumptions   }This section contains 
all relevant definitions. Besides, we recapitulate the determinant 
lemma (Lemma~4 below) and the enumeration results (Propositions~1,2 
below) of 
\cite{8} which we shall need in the sequel.

An $r$--tupel 
$\lambda
=(\lambda
\sb
1,\lambda
\sb
2,\dots,\lambda
\sb
r)$ of positive integers satisfying
$\lambda
\sb
1\ge\lambda
\sb
2\ge\dots\ge\lambda
\sb
r$ is called a {\sl partition\/}. For 
a partition $\lambda
$ we write 
$\vert\lambda
\vert$ for $\lambda
\sb
1+\lambda
\sb
2+\dots+\lambda
\sb
r$.

Let $\alpha
,\beta
$ be arbitrary integers and $\lambda
$ a partition. We 
call an array $\pi$ of integers of the form
$$\matrix \pi\sb
{11}
&\pi\sb
{12}&\innerhdotsfor3\after\quad &\pi\sb
{1,\lambda
\sb
1} \\
\pi\sb
{21}&\pi\sb
{22}&\innerhdotsfor2\after\quad &\pi\sb
{2,\lambda
\sb
2}\\
\vdots& & &\mathinner{\mskip1mu\raise1pt\vbox{\hbox{.}}\mskip2mu
\raise4pt\hbox{.}\mskip2mu\raise7pt\vbox{\kern7pt\hbox{.}}\mskip1mu}\\
\pi\sb
{r1}&\innerhdotsfor1\after\quad &\pi\sb
{r,\lambda
\sb
r}& & &
\endmatrix\tag2.1$$
an $(\alpha
,\beta
)$--{\sl plane partition of shape\/} $\lambda
$ if
$$\align\pi\sb
{ij}\ge\pi\sb
{i,j+1}+\alpha
\quad\quad &\text{for}\quad 1\le i\le r,\ 
0<j<\lambda
\sb
i\tag2.2)(a\\
\intertext{and\nopagebreak}
\pi\sb
{ij}\ge\pi\sb
{i+1,j}+\beta
\quad \quad &\text{for}\quad 1\le i\le r,\ 
0<j\le\lambda
\sb
{i+1}\ .\tag2.2)(b\endalign$$
To give an example, the array in Figure~1 below is a $(0,2)$--plane 
partition of shape (5,3,3,1).
$$\matrix \format &\quad \r\\7&6&6&3&1\\
5&4&3\\2&1&1\\-1
\endmatrix$$\centerline{\eightpoint Figure 1}

\medpagebreak
The entries of $\pi$ are called {\sl parts\/} of $\pi$. A 
$(0,0)$--plane partition containing only positive parts is an 
ordinary plane partition, a $(0,1)$--plane partition containing only 
positive parts is a column--strict plane partition, a $(1,0)$--plane 
partition is a row--strict plane partition, and a $(1,1)$--plane 
partition is a row and column--strict plane partition.

An $(\alpha
,\beta
)$--{\sl reverse plane partition of shape\/} $\lambda
$ is an 
array $\pi$ of the form $(2.1)$ satisfying
$$\align\pi\sb
{ij}\le\pi\sb
{i,j+1}+\alpha
\quad\quad &\text{for}\quad 1\le i\le r,\ 
0<j<\lambda
\sb
i\tag2.3)(a\\
\intertext{and\nopagebreak}
\pi\sb
{ij}\le\pi\sb
{i+1,j}+\beta
\quad \quad &\text{for}\quad 1\le i\le r,\ 
0<j\le\lambda
\sb
{i+1}\ .\tag2.3)(b\endalign$$

If $\lambda
\sb
i'$ is the number of parts in the $i$'th column of an array 
$\pi$ of the form $(2.1)$ (i.e. of shape $\lambda
$), we call 
$\lambda
'=(\lambda
\sb
1',\lambda
\sb
2',\dots,\lambda
'\sb
{\lambda
\sb
1})$ the {\sl conjugate partition} 
to $\lambda
$.

For a partition $\lambda
$ with $\vert\lambda
\vert=p$ define the {\sl hook 
lengths} of $\lambda
$ to be the integers $\lambda
\sb
i+\lambda
'\sb
j-i-j+1$, for $1\le 
i\le r$ and $1\le j \le \lambda
\sb
i$. (With other words, only pairs $(i,j)$ 
are allowed for which $\pi\sb
{ij}$ is a part in an array $\pi$ of shape 
$\lambda
$.) We write $h\sb
1,h\sb
2,\dots,h\sb
p$ for the hook lengths of $\lambda
$ (in 
some order). Similarly, we define the {\sl contents\/} of $\lambda
$ to be 
the integers $j-i$, for $1\le i\le r$ and $1\le j\le \lambda
\sb
i$. We write 
$c\sb
1,c\sb
2,\dots,c\sb
p$ for the contents of $\lambda
$ (in some order). For 
instance, for the shape $(5,3,3,1)$ of the array in Figure~1, the hook 
lengths and contents are as follows:
$$\matrix 8&6&5&2&1\\5&3&2\\4&2&1\\1\\\hbox to 0pt{hook 
lengths\hss}\endmatrix\quad \quad \quad 
\matrix\format&\,\, 
\r\\0&1&\hphantom{-}2&\hphantom{-}3&\hphantom{-}4\\-1&0&1\\-2&-1&0\\-3\\ 
\hbox to 
0pt{contents\hss}\endmatrix$$

We refer the reader to  \cite{15} 
for a detailed exposition of the terminology concerning plane partitions.

\bigpagebreak
Next we come to enumeration results for $(\alpha
,\beta
)$--plane partitions 
of shape $\lambda
$ subject to certain part restrictions. The first 
proposition is the special case $\mu=0$, $x=1$ of \cite{8, 
Theorem~6.1}.

\proclaim{  Proposition~1}Let $\lambda
$ be a partition and $a,b$ be 
$r$--tupels of integers satisfying
$$a\sb
i+(1-\beta
)\ge a\sb
{i+1} \quad \text{ and }\quad 
b\sb
i+\alpha
(\lambda
\sb
i-\lambda
\sb
{i+1})+(1-\beta
)\ge b\sb
{i+1}\ .\tag2.4 $$
The number of $(\alpha
,\beta
)$--plane partitions of shape $\lambda
$ in which 
the first part in row $i$ is at most $a\sb
i$ and the last part in row 
$i$ is at least $b\sb
i$ is 
$$\det\sb
{1\le s,t\le 
r}\left(\binom{(1-\alpha
)\lambda
\sb
s-\beta
(s-t)+a\sb
t-b\sb
s+\alpha
}{\lambda
\sb
s-s+t}\right)\ 
.\qed\tag2.5$$
\endproclaim

\medpagebreak
The analogue for $(\alpha
,\beta
)$--reverse plane partitions is the special 
case $\mu=0$, $x=1$ of \cite{8, Corollary~6.4}.
\proclaim{  Proposition~2}
Let $\lambda
$ be a partition and $a,b$ be 
$r$--tupels of integers satisfying
$$a\sb
i+\alpha
(\lambda
\sb
{i-1}-\lambda
\sb
{i})+(1-\beta
)\ge a\sb
{i-1} \quad \text{ and }\quad 
b\sb
i+(1-\beta
)\ge b\sb
{i-1}\ .\tag2.6 $$
The number of $(\alpha
,\beta
)$--reverse plane partitions of shape $\lambda
$ in which 
the first part in row $i$ is at least $b\sb
i$ and the last part in row 
$i$ is at most $a\sb
i$ is 
$$\det\sb
{1\le s,t\le 
r}\left(\binom{(1-\alpha
)\lambda
\sb
s-\beta
(s-t)+a\sb
s-b\sb
t+\alpha
}{\lambda
\sb
s-s+t}\right)\ 
.\qed\tag2.7$$
\endproclaim

\bigpagebreak
We are interested in finding closed forms for these determinants. Of 
course, in general this will not be possible. Therefore, we have to 
restrict ourselves to suitable choices of the row bounds $a\sb
i$ and 
$b\sb
i$. In \cite{8} some work in this direction was done for the even 
more general problem of evaluating the determinants for the {\sl generating 
functions\/} of those classes of $(\alpha
,\beta
)$--plane partitions. In this 
paper we consider only determinant evaluations for the {\sl number\/} 
of $(\alpha
,\beta
)$--plane partitions, restricted by some special row 
bounds, because it does not seem to be possible to obtain closed 
expressions for the corresponding generating functions.

To get a flavour of how such results look like, we cite the celebrated 
hook--content formula for the number of column--strict plane 
partitions \cite{15, Theorem~15.3 for $x=1$}.
\proclaim{Proposition~3} Let $\lambda
$ be a partition with $\vert 
\lambda
\vert=p$ and $a$ a positive integer. The number of column--strict 
plane partitions of shape $\lambda
$ with parts $\le a$ is
$$\frac {(a+c\sb
1)(a+c\sb
2)\cdots(a+c\sb
p)}{h\sb
1\cdot h\sb
2\cdots h\sb
p}\ 
,\tag2.8$$
where the $h\sb
i$'s and $c\sb
i$'s are the hook lengths and contents of 
$\lambda
\sb
i$, respectively.\qed
\endproclaim
The determinant which has to be evaluated to obtain $(2.8)$ is the 
determinant in $(2.5)$ for $\alpha
=0$, $\beta
=1$, $a\sb
i=a$, $b\sb
i=1$. This 
evaluation may be performed using the following determinant lemma, 
which also serves for the evaluation which we give \linebreak in \S3.
\proclaim{  Lemma~4}{\rm(\cite{8, Lemma~2.2})} Let 
$X\sb
1,\dots,X\sb
r,A\sb
2,\dots,A\sb
r,B\sb
2,\dots,B\sb
r$ be indeterminates. Then
$$\multline\det\sb
{1\le s,t\le 
r}\Bigl((X\sb
s+A\sb
r)\cdots(X\sb
s+A\sb
{t+1})(X\sb
s+B\sb
t)\cdots(X\sb
s+B\sb
2)\Bigr)=\\
=\prod\sb
{1\le i<j\le r}(X\sb
i-X\sb
j)\prod\sb
{2\le i\le j\le r}(B\sb
i-A\sb
j)\ .
\endmultline$$
with the convention that empty products (like $
(X\sb
s+A\sb
r)\cdots(X\sb
s+A\sb
{t+1})$ for $t=r$) are set equal to one.\qed
\endproclaim

\bigpagebreak
\subheading{3.~The Determinant evaluation   }
\proclaim{  Theorem~5}Let $\lambda
$ be a partition 
and  $B$ be a positive real number such that \linebreak
$b+(B-\alpha
)\lambda
\sb
i+(1-\beta
-B)i$ is an integer for $i=1,2,\dots,r$. The number of 
$(\alpha
,\beta
)$--plane partitions $\pi$ of shape $\lambda
$ in which the first 
part in row $i$ is at most $a-\beta
 i$ (or equivalently $\pi\sb
{11}\le 
a-\beta
 $) 
and the last part in row $i$ is at least $b+(B-\alpha
)\lambda
\sb
i+(1-\beta
-B)i$ 
is 
$$\aligned\dsize\prod\sb
{i=1}\sp
r\frac 
{(a+\alpha
-b-(B-1)(\lambda
\sb
i-i))!}{(a+\alpha
-b-1-B(\lambda
\sb
i-i))!}&\cdot 
\dsize\prod\sb
{i=1}\sp
r (a+\alpha
-b+(B-1)i+1)\sb
{i-1} \\ 
&\hskip-1cm\times\frac 
{\dsize\prod\sb
{1\le i<j\le 
r}(\lambda
\sb
i-i-\lambda
\sb
j+j)}{\dsize\prod\sb
{i=1}\sp
r(\lambda
\sb
i-i+r)!\ 
}\ ,\endaligned\tag3.1$$
where the upper factorial $(A)\sb
n$ is defined by 
$A(A+1)\cdots(A+n-1)$.
\endproclaim
\demo{Proof}Because of $B\ge0$, the row bounds satisfy $(2.4)$. Hence 
we may apply Proposition~1 with $a\sb
i=a-\beta
 i$ and $b\sb
i=
b+(B-\alpha
)\lambda
\sb
i+(1-\beta
-B)i$ to obtain for the desired number the 
expression
$$\det\sb
{1\le s,t\le 
r}\left(\binom{(1-B)(\lambda
\sb
s-s)+a+\alpha
-b}{\lambda
\sb
s-s+t}\right)\ .$$
Taking some factors out of the determinant leads to
$$\align\det\sb
{1\le s,t\le 
r}\biggl(&\binom{(1-B)(\lambda
\sb
s-s)+a+\alpha
-b}{\lambda
\sb
s-s+t}\biggr)=\\
&\hskip-0.6cm=\frac 
{\dsize\prod\sb
{i=1}\sp
r((1-B)(\lambda
\sb
i-i)+a+\alpha
-b)!}{\dsize\prod\sb
{i=1}\sp
r 
((-B)(\lambda
\sb
i-i)+a-\alpha
-b-1)!\,\dsize\prod\sb
{i=1}\sp
r(\lambda
\sb
i-i+r)!}\\ 
&\hskip-0.2cm\times\det\sb
{1\le 
s,t\le 
r}\Bigl((\lambda
\sb
i-i+r)\cdots(\lambda
\sb
i-i+t+1)\\ 
 &\hphantom{\times }\cdot((-B)(\lambda
\sb
i-i)+a+\alpha
-b-t+1)\cdots
((-B)(\lambda
\sb
i-i)+a+\alpha
-b-1)\Bigr)\ .\endalign$$
If the  determinant in this expression is rewritten in the form
$$\align(-B)\sp
{\binom r2}\det\sb
{1\le s,t\le r}\Bigl(
(\lambda
\sb
i-i+&r)\cdots(\lambda
\sb
i-i+t+1)\cdot\\
&\hskip-1cm\cdot(\lambda
\sb
i-i-\frac {a+\alpha
-b-t+1}{B})\cdots(\lambda
\sb
i-i-\frac 
{a+\alpha
-b-1}{B})
\Bigr)\ ,\endalign$$
Lemma~4 with $X\sb
i=\lambda
\sb
i-i$, $A\sb
i=i$ and $B\sb
i=-(a+\alpha
-b-i+1)\bigl/B$ can be 
applied. Thus we get
$$\align\det\sb
{1\le s,t\le 
r}\biggl(&\binom{(1-B)(\lambda
\sb
s-s)+a+\alpha
-b}{\lambda
\sb
s-s+t}\biggr)=\\
&=\frac 
{\dsize\prod\sb
{i=1}\sp
r((1-B)(\lambda
\sb
i-i)+a+\alpha
-b)!}{\dsize\prod\sb
{i=1}\sp
r 
((-B)(\lambda
\sb
i-i)+a-\alpha
-b-1)!}
\frac 
{\dsize\prod\sb
{1\le i<j\le 
r}(\lambda
\sb
i-i-\lambda
\sb
j+j)}{\dsize\prod\sb
{i=1}\sp
r(\lambda
\sb
i-i+r)!}
\\ &\quad\quad \times\dsize\prod\sb
{2\le i\le j\le r}\left(-\frac 
{a+\alpha
-b-i+1}{B}-j\right)
\cdot(-B)\sp
{\binom r2}\ .
\endalign$$
Finally, simple manipulation turns the right--hand side of this 
identity into the expression $(3.1)$.\qed
\enddemo


There is also a ``hook--content formulation" for this theorem. The 
hook lengths are the same, but we have to generalize the contents of a 
partition $\lambda
=(\lambda
\sb
1,\dots,\lambda
\sb
r)$ with $\vert\lambda
\vert=p$. If $(i,j)$ is 
a pair of 
integers such that $1\le i\le r$ and $1\le j\le \lambda
\sb
i$ we define a 
``generalized content" $d(i,j)$ as follows:
$$d(i,j)=\cases -B\lambda
\sb
i+(B-1)i+j&\text{if }\ i\le 
j\\\hphantom{-}B\lambda
\sb
j'-i-(B-1)j+B&\text{if }\ i> j\ .\endcases\tag3.2$$
We write $d\sb
1,d\sb
2,\dots,d\sb
p$ for these generalized contents of $\lambda
$ 
(in some order). Obviously, for $B=0$, the $d\sb
i$'s are identical with 
the ordinary  contents $c\sb
i$.

\medpagebreak
For the proof of the hook--content form of Theorem~5 we need the 
following lemma:
\proclaim{Lemma~6   }Let $\lambda
=(\lambda
\sb
1,\lambda
\sb
2,\dots,\lambda
\sb
r)$ be a 
partition. Let $s$ be the largest integer for which $\lambda
\sb
s-s\ge0$. 
Then the $r$ numbers 
$$r+1+\lambda
\sb
i-i\quad (s+1\le i\le r),\quad \quad r+j-\lambda
\sb
j'\quad (1\le 
j\le s)\tag3.3$$
are a permutation of $\{1,2,\dots,r\}$.
\endproclaim
\demo{Proof}First we have to show that these numbers actually are 
elements of \linebreak 
$\{1,2,\dots,r\}$. This is clear for $r+1+\lambda
\sb
i-i$, since 
for $s+1\le i\le r$ we assumed $\lambda
\sb
i-i<0$. Considering the 
numbers $r+j-\lambda
\sb
j'$ observe that $j-\lambda
\sb
j'$ is increasing for 
$j=1,2,\dots,s$. But since $\lambda
\sb
s\ge s$, also by assumption, obviously 
also $\lambda
\sb
s'\ge s$ holds. Therefore we have $j-\lambda
\sb
j'\le s-\lambda
\sb
s'\le 
0$ for $j=1,2,\dots,s$.

The fact that the numbers in $(3.3)$ are pairwise distinct is an easy 
consequence of \cite{9, p.3, $(1.7)$} with $n=r+1$ and $m=r$.\qed
\enddemo

\proclaim{  Theorem~7}
Let $\lambda
$ be a partition with $\vert\lambda
\vert=p$ 
and  $B$ be a positive real number such that 
$b+(B-\alpha
)\lambda
\sb
i+(1-\beta
-B)i$ is an integer for $i=1,2,\dots,r$. The number of 
$(\alpha
,\beta
)$--plane partitions $\pi$ of shape $\lambda
$ in which the first 
part in row $i$ is at most $a-\beta
 i$ (or equivalently $\pi\sb
{11}\le 
a-\beta
 $) 
and the last part in row $i$ is at least $b+(B-\alpha
)\lambda
\sb
i+(1-\beta
-B)i$ 
is 
$$\frac {(a+\alpha
-b+d\sb
1)\cdots(a+\alpha
-b+d\sb
p)}{h\sb
1\cdots h\sb
p}\ ,\tag3.3$$
where the $h\sb
i$'s are the hook lengths of $\lambda
$ and the $d\sb
i$'s are 
the generalized contents of $\lambda
$, which were introduced in 
$(3.2)$.\endproclaim
\demo{Proof}It is a simple matter of fact that
$$
\frac 
{\dsize\prod\sb
{1\le i<j\le 
r}(\lambda
\sb
i-i-\lambda
\sb
j+j)}{\dsize\prod\sb
{i=1}\sp
r(\lambda
\sb
i-i+r)!}=
\frac {1}{h\sb
1\cdots h\sb
p}
$$
\cite{9, p.9, Example~1(4)}. Leaves to transform the expression in 
the first line of $(3.1)$ into the numerator of $(3.3)$. Such as in 
Lemma~6, let $s$ be the largest integer for which $\lambda
\sb
s-s\ge0$. 
Breaking the left product of that expression into two factors, we are 
able to perform some cancellations:
$$\align 
\dsize\prod\sb
{i=1}\sp
r\frac 
{(a+\alpha
-b-(B-1)(\lambda
\sb
i-i))!}{(a+\alpha
-b-1-B(\lambda
\sb
i-i))!}&\cdot 
\dsize\prod\sb
{i=1}\sp
r (a+\alpha
-b+(B-1)i+1)\sb
{i-1}=  \\&\hskip-4.5cm 
=\prod\sb
{i=1}\sp
s
\frac 
{(a+\alpha
-b-(B-1)(\lambda
\sb
i-i))!}{(a+\alpha
-b-1-B(\lambda
\sb
i-i))!} \prod\sb
{i=s+1}\sp
r
\frac 
{(a+\alpha
-b-(B-1)(\lambda
\sb
i-i))!}{(a+\alpha
-b-1-B(\lambda
\sb
i-i))!}\\
&\hskip-3cm \times \dsize\prod\sb
{i=1}\sp
r (a+\alpha
-b+(B-1)i+1)\sb
{i-1}\\
&\hskip-4.5cm=\prod\sb
{i=1}\sp
s(a+\alpha
-b-B(\lambda
\sb
i-i))\sb
{\lambda
\sb
i-i+1}\\
\vspace{-4pt}
&\hskip-3cm\times\frac {\dsize\prod\sb
{i=1}\sp
r (a+\alpha
-b+(B-1)i+1)\sb
{i-1}}{
\dsize\prod\sb
{i=s+1}\sp
r(a+\alpha
-b+(B-1)(i-\lambda
\sb
i)+1)\sb
{i-\lambda
\sb
i-1}}\\
&\hskip-4.5cm=\prod\sb
{i=1}\sp
s(a+\alpha
-b-B(\lambda
\sb
i-i))\sb
{\lambda
\sb
i-i+1} 
\underset \text{for all }k \to{\underset j\ne k-\lambda
\sb
k 
\to{\prod\sb
{j=1}\sp
r}} (a+\alpha
-b+(B-1)j+1)\sb
{j-1}
\endalign$$
The expression in the last line may be rewritten in the form
$$\prod\sb
{1\le i\le j\le \lambda
\sb
i}\hskip-1pt(a+\alpha
-b-B(\lambda
\sb
i-i)+j-i) \underset 
\text{for all }k\to{\underset
 j\ne r+1+\lambda
\sb
k-k \to{\prod\sb
{1\le j\le i< 
r}}} \hskip-3pt(a+\alpha
-b+(B-1)(r+1-j)+r-i)\ .$$
Because of Lemma~6, the index $j$ of the second product only attains 
the values $r+l-\lambda
\sb
l'$, $l=1,2,\dots,s$. Therefore the last 
expression is equal to
$$\prod\sb
{1\le i\le j\le \lambda
\sb
i}\hskip-1pt(a+\alpha
-b-B(\lambda
\sb
i-i)+j-i) \underset 
r+j-\lambda
\sb
{j}' \le i<r \to{\prod\sb
{1\le j\le 
s}}\hskip-3pt(a+\alpha
-b+(B-1)(\lambda
\sb
j'-j+1)+r-i)\ .$$
Replacing $i$  by $i+r-\lambda
\sb
j'-1$ in the second product, we obtain 
after some additional manipulation
$$\multline\dsize\prod\sb
{i=1}\sp
r\frac 
{(a+\alpha
-b-(B-1)(\lambda
\sb
i-i))!}{(a+\alpha
-b-1-B(\lambda
\sb
i-i))!}\cdot 
\dsize\prod\sb
{i=1}\sp
r (a+\alpha
-b+(B-1)i+1)\sb
{i-1} =\\=
\prod\sb
{1\le i\le j\le \lambda
\sb
i}(a+\alpha
-b-B\lambda
\sb
i+(B-1)i+j) \prod\sb
{ 1\le 
j<i\le\lambda
\sb
j'} (a+\alpha
-b+B\lambda
\sb
j'-i-(B-1)j+B)\ , \endmultline$$
which is equivalent with our claim.\qed
\enddemo
Proposition~3 is the special case $\alpha
=0$, $\beta
=1$, $B=0$, $b=1$, 
$a\to a+1$ of the above theorem.

The counterparts of Theorems~5 and~7 for $(\alpha
,\beta
)$--reverse 
plane partitions 
read as follows:
\proclaim{  Theorem~8}
Let $\lambda
$ be a partition with $\vert\lambda
\vert=p$ 
and  $B$ be a positive real number such that 
$a-(B-\alpha
)\lambda
\sb
i-(1-\beta
-B)i$ is an integer for $i=1,2,\dots,r$. The number of 
$(\alpha
,\beta
)$--reverse plane partitions $\pi$ of 
shape $\lambda
$ in which the first 
part in row $i$ is at least $b+\beta
 i$ (or equivalently $\pi\sb
{11}\ge 
b+\beta
$) 
and the last part in row $i$ is at most $a-(B-\alpha
)\lambda
\sb
i-(1-\beta
-B)i$ 
is 
$$\aligned\dsize\prod\sb
{i=1}\sp
r\frac 
{(a+\alpha
-b-(B-1)(\lambda
\sb
i-i))!}{(a+\alpha
-b-1-B(\lambda
\sb
i-i))!}&\cdot 
\dsize\prod\sb
{i=1}\sp
r (a+\alpha
-b+(B-1)i+1)\sb
{i-1} \\ 
&\hskip-1cm\times\frac 
{\dsize\prod\sb
{1\le i<j\le 
r}(\lambda
\sb
i-i-\lambda
\sb
j+j)}{\dsize\prod\sb
{i=1}\sp
r(\lambda
\sb
i-i+r)!\ 
}\ ,\endaligned\tag3.5$$
or, in terms of hooks and contents,
$$\frac {(a+\alpha
-b+d\sb
1)\cdots(a+\alpha
-b+d\sb
p)}{h\sb
1\cdots h\sb
p}\ ,\tag3.6$$
where the $h\sb
i$'s are the hook lengths of $\lambda
$ and the $d\sb
i$'s are 
the generalized contents of $\lambda
$, which were introduced in 
$(3.2)$.
\endproclaim
\demo{Proof}Because of $B\ge0$, the row bounds satisfy $(2.6)$. Hence 
we may apply Proposition~2 with $a\sb
i=a-(B-\alpha
)\lambda
\sb
i-(1-\beta
-B)i$ and 
$b\sb
i=b+\beta
 i$ to obtain for the desired number the expression
$$\det\sb
{1\le s,t\le 
r}\left(\binom{(1-B)(\lambda
\sb
s-s)+a+\alpha
-b}{\lambda
\sb
s-s+t}\right)\ .$$
But this expression has already been evaluated in the proof of 
Theorem~5. The hook--content formulation follows from the proof of 
Theorem~7.\qed
\enddemo

Gessel's and Viennot's theorems, which were mentioned in the 
Introduction, are special cases of Theorem~8. To obtain \cite{6, 
Theorem~21} we have to take $\alpha
=0$, $\beta
=1$, $B=2$, $a=r-1$ and 
$b=0$. Indeed, our generalized contents for $B=2$ differ from those 
given by Gessel and Viennot, but this is not an essential difference. 
If we consider the array $\big(d(i,j)\big)$ of shape $\lambda
$\quad ($
1\le i\le r$, $1\le j\le \lambda
\sb
i $), where $d(i,j)$ are the 
$B=2$ --numbers of 
$(3.2)$, then, to obtain Gessel's and Viennot's contents 
from ours, above the main diagonal we only have to reverse the order 
of elements in each row, and below we have to reverse the order of 
elements in each column. More precisely, for $i\ge j$ let 
$D(i,j):=d(i,\lambda
\sb
i+i-j)$, and for $i<j$ let $D(i,j):=d(\lambda
\sb
j'+j-i+1,j)$. 
Then $D(i,j)$ is identical with Gessel's and Viennot's definition of 
contents (save to a constant, which is simply 1). We have chosen the 
form $(3.2)$ for our definition, because for $B=0$ our $d(i,j)$'s 
reduce to ordinary contents.

Setting $\alpha
=1$, $\beta
=0$, $B=2$, $b=0$, $a=2k$ or $2k-1$ in 
$(3.5)$, respectively, we get Theorems~22 and~23 of \cite{6}, though 
in ``conjugate" formulation. We remark that Gessel and Viennot used a 
different determinant to obtain these theorems. Not surprisingly, this 
determinant can also be computed by Lemma~4.
The hook--content formula $(3.6)$ in this special 
case is also new.



\bigpagebreak
Finally we turn our attention to $(\alpha
,\beta
)$--plane partitions of some 
particular shape, which we name ``staircase" shape. Let $L$ be a 
nonnegative integer and $\lambda
\sb
i=\Lambda
-Li$,  $i=1,2,\dots,r$, $\Lambda
$ 
sufficiently large, such that $\lambda
\sb
i$  is positive for 
$i=1,2,\dots,r$. The shape $\lambda
$ given by the partition 
$\lambda
=(\lambda
\sb
1,\lambda
\sb
2,\dots,\lambda
\sb
r)=(\Lambda
-L, \Lambda
-2L,\dots,\Lambda
-Lr)$ we call a 
{\sl staircase shape\/}. The reason is that a typical staircase shape 
looks like a reversed staircase:\hskip0.4cm
$$\gather\vbox{\hrule height 0pt depth1pt\hbox{\vrule width1pt\vrule
 height1pt width0.5cm \vrule 
height0.5cm width1pt \raise0.5cm \hbox{\vrule height0pt depth1pt 
width1cm \vrule 
height0.5cm width1pt 
\raise0.5cm \hbox{\vrule height0pt depth1pt width1cm \vrule 
height0.5cm width1pt 
\raise0.5cm \hbox{\vrule height0pt depth1pt width1cm \vrule 
height0.5cm width1pt 
\raise0.5cm \hbox{\vrule height0pt depth1pt width1cm \vrule 
height0.5cm width1pt}%
}}}}}\\\eightpoint\text{Figure 2}\endgather$$
(Here we chose $\Lambda
=11$, $L=2$, and $r=5$.)

In Theorem~5 for $\lambda
=(\Lambda
-L,\dots,\Lambda
-Lr)$ take $B=\frac 
{1-\beta
+L\alpha
}{L+1} $ and $b=1-\frac {1-\alpha
-\beta
}{L+1}\Lambda
$ in order to 
get $b\sb
i=1$ for all $i$. If we also replace $a$ by $a+\beta
$ we obtain
\proclaim{  Corollary~9}Let $\lambda
$ be a staircase shape, 
$\lambda
=(\Lambda
-L,\Lambda
-2L,\dots,\Lambda
-Lr)$, with $L$ satisfying $Lr<\Lambda
$ and 
$L\alpha
-\beta
+1\ge0$. The number of $(\alpha
,\beta
)$--plane partitions $\pi$ 
of shape $\lambda
$ in which $\pi\sb
{11}\le a$ and the last part in each row 
is positive is
$$\multline\prod\sb
{i=1}\sp
r\frac {(a+\alpha
+\beta
-1+\Lambda
(1-\alpha
) -(\beta
-L\alpha
+L)i)!} 
{(a+\alpha
+\beta
-2-\alpha
\Lambda
-(\beta
-L\alpha
-1)i)!}\times \\\times\prod\sb
{i=1}\sp
r\frac 
{\left(a+\alpha
+\beta
+\frac {1-\alpha
-\beta
}{L+1}\Lambda
-\frac {\beta
-L\alpha
+L}{L+1}i\right)\sb
{i-1}}
{(r+\Lambda
-i(L+1))!}\cdot (L+1)\sp
\binom r2 2\sp
{r-2}3\sp
{r-3}\cdots (r-1)\sp
1\ 
.\qed
\endmultline\tag3.7$$
\endproclaim

\bigpagebreak
For $\alpha
=\beta
=0$ this result gives the number of plane partitions of 
staircase shape with parts $\le a$.
\proclaim{  Corollary~10}
Let $\lambda
$ be a staircase shape, 
$\lambda
=(\Lambda
-L,\Lambda
-2L,\dots,\Lambda
-Lr)$, with $L$ satisfying $Lr<\Lambda
$.
The number of plane partitions 
of shape $\lambda
$ with largest part $\le a$ is
$$\prod\sb
{i=1}\sp
r\frac {(a+\Lambda
-Li-1)!}{(a+i-2)!} \prod\sb
{i=1}\sp
r\frac 
{\left(a+\frac {\Lambda
-Li}{L+1}\right)\sb
{i-1}}{(r+\Lambda
-i(L+1))!}\cdot 
(L+1)\sp
\binom r2 2\sp
{r-2}3\sp
{r-3}\cdots (r-1)\sp
1\ .\qed\tag3.8
$$\endproclaim

\bigpagebreak
Another interesting special case of Corollary~9 is the case 
$\alpha
=\beta
=1$.
\proclaim{ Corollary~11}
Let $\lambda
$ be a staircase shape, 
$\lambda
=(\Lambda
-L,\Lambda
-2L,\dots,\Lambda
-Lr)$, with $L$ satisfying $Lr<\Lambda
$.
The number of row and column--strict plane partitions 
of shape $\lambda
$ with largest part $\le a$ is
$$\prod\sb
{i=1}\sp
r\frac {(a+1-i)!}{(a-\Lambda
+Li)!}\prod\sb
{i=1}\sp
r\frac 
{\left(a+2-\frac {\Lambda
+i}{L+1}\right)\sb
{i-1}}{(r+\Lambda
-i(L+1))!}\cdot
(L+1)\sp
\binom r2 2\sp
{r-2}3\sp
{r-3}\cdots (r-1)\sp
1\ .\qed\tag3.9
$$\endproclaim

\bigpagebreak
Clearly, there are also analogues of these corollaries for reverse 
plane partitions. We omit them here for sake of brevity.

\Refs
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conjecture\yr 1978
\pages 131--150 \inbook Studies in Foundations and Combinatorics\publ 
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\ref\no2\bysame \paper Plane partitions (II): The equivalence of the 
Bender--Knuth and the MacMahon conjectures\jour Pacific J. Math.\vol 
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\ref \no3\bysame \paper Plane partitions (III): The weak Macdonald 
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\moreref \jour Erratum \vol44 \yr1987\page310 \endref


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