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\centerline {\bf THE NUMBER OF ADMISSIBLE SEQUENCES FOR} 
\centerline {\bf  INDECOMPOSABLE SERIAL RINGS} 
\centerline {\bf }
\centerline {Joshua O. Hanes}
\centerline {University of Mississippi, Oxford, MS, 38677}
\centerline { Darren D. Wick}
\centerline {Ashland University, Ashland, OH, 44805}
\next
\centerline {\bf THE NUMBER OF ADMISSIBLE SEQUENCES FOR} 
\centerline {\bf  INDECOMPOSABLE SERIAL RINGS} 
\centerline {\bf }
\rm
\vskip15pt\noindent
ABSTRACT: We give a formula for the number of admissible sequences for indecomposable 
serial rings with $n$ indecomposable projective modules whose minimum composition length is
less than or equal to
$m$. In particular, if
$n=m$ is prime, we show that the number of such admissible sequences is  
$${{2n-1}\choose n} +{1\over n} \bigg[ (n-1)^2 - {{2n-2}\choose {n}}
\bigg]  $$
\vskip25pt\noindent
%\endinsert 
%%
%
%
\centerline {\bf {1. INTRODUCTION}}
\noi
To each indecomposable serial ring (with unity) $R$ there is associated a set 
$\lbrace e_1, \cdots ,e_n \rbrace$ of basic 
primitive idempotents and a set 
$\{ Re_1 ,Re_2 ,\cdots ,Re_n \}$ of
pairwise  non-isomorphic indecomposable 
projective 
left $R$-modules, each having a unique composition series. 
Denoting the composition lengths of $Re_i$ by $c_i =c(Re_i )$,
then the sequence
$c_1 ,c_2 ,\cdots ,c_n $ is called an {\it admissible sequence} for $R$, and
satisfies the following inequalities [1]:
$$ 2\le c_l \le c_{l-1} +1 \ \ {\rm for } \ l=2,\cdots , n \leqno (\bf 1.1) $$
$$c_1 \le c_n +1 \leqno (\bf 1.2)$$
The admissible sequence for a serial ring is unique, except for cyclic permutation of the indices.
We have shown that if $R$ has a simple projective module, then the number of possible admissible
sequences of length $n$ is $b_{n-1}$, the $n-1^{st}$ Catalan number [2].  In this paper
we count the number of admissible sequences for all serial rings. We adopt the convention
that
$\ds {s\choose t} =0$ for all integers
$s<t$ and for all
$s<0$ or
$t<0$. Furthermore, we will refer to various combinatorial identities and have collected them in
the appendix.
\next
\cl {\bf 2. COUNTING ADMISSIBLE SEQUENCES}
\noi
{\bf DEFINITION 2.1}: Let $i,j,k$ be positive integers and $\delta$ denote the Kronecker delta.
Define $a_{i,j}^k$ as follows:
\item {} a) For all $i,k \geq 1$, $a_{i,1}^k=0$
\item {} b) For all $k\geq 1$ and $j\geq 2$, $a_{1,j}^k =\delta_{j,k}$ 
\item {} c) For all $k\geq 1$ and $i, j \geq 2$, $\ds a_{i,j}^k = \sum_{l\geq j-1} a_{i-1,l}^k $
\noi
We note that each of the sums in part c) above is finite by the following lemma.
\noi
{\bf LEMMA 2.2}:
For all $i,k\geq 1$ and for all $j\geq 0$, $a_{i,i+k+j}^k =0$. 
\noi
{\bf Proof}: We induct on $i$. For $i=1$, we have $$a_{1,k+j+1}^k=\delta_{k+j+1,k} = 0 \ \ \ \forall
j\geq 0$$
Let $i>1$. Then
$$a_{i,i+k+j}^k =\sum_{l\geq (i-1)+k+j} a_{i-1,l}^k $$
and by the induction hypothesis, each term in this summation is zero. 
\qed
{\bf LEMMA 2.3}: For all $i \geq 1$ and $j,k\geq 2$, $a_{i,j}^k$ equals the number of 
sequences
$c_1 ,c_2 ,\cdots ,c_i $ with $c_1 =k$ and $c_i =j$ that satisfy (1.1). 
\noi
{\bf Proof}: Fix $c_1 =k$ and induct on $i$. The case $i=1$ is trivial, as $c_1=k$ is the only
such sequence and $a_{1,j}^k=\delta_{j,k}$.
Let $i>1$. Suppose $c_1 ,c_2 ,\cdots ,c_i $ satisfies (1.1) with $c_i =j$. Then clearly $c_1 ,c_2
,\cdots ,c_{i-1} $ also satisfies (1.1). Conversely, if $c_1 ,c_2 ,\cdots ,c_{i-1} $ satisfies
(1.1), then so does $c_1 ,c_2 ,\cdots ,c_i =j $ iff $2\leq j\leq c_{i-1}+1$. Thus every sequence
of length $i$ satisfying (1.1) is obtained from a sequence of length $i-1$ satisfying (1.1). In
particular, the number of  sequences
$c_1 ,c_2 ,\cdots ,c_i =j $ satisfying (1.1) is equal to the number of sequences $c_1 ,c_2 ,\cdots
,c_{i-1}$ satisfying (1.1) with $c_{i-1}\geq j-1$. By induction, the latter number is 
$\ds \sum_{l\geq j-1} a_{i-1,l}^k $ which by definition is $a_{i,j}^k$.
\qed
 We next
give a closed form of $a_{i,j}^k$.
\noi
{\bf LEMMA 2.4}: For all $i,j,k \geq 2$, we have 
$$a_{i,j}^k = {{2i-j+k-3} \choose {i-j+k-1}}-{{2i-j+k-3} \choose {i+k-2}}$$
{\bf Proof}: We induct on $i$. Let $i=2$. Then
$$a_{2,j}^k =\sum_{l\geq j-1} a_{1,l}^k =\sum_{l\geq j-1} \delta _{l,k} = \cases {1 & if $j\leq k+1$
\cr 0 & else \cr}$$
On the other hand, 
$${{2i-j+k-3} \choose {i-j+k-1}}-{{2i-j+k-3} \choose {i+k-2}}
={{k-j+1} \choose {k-j+1}}-{{k-j+1} \choose {k}} $$ 
But since $j\geq 2$, we have that $k-j+1\leq k-1<l$ so that $\ds {{k-j+1 } \choose {k}}=0$.
Furthermore, $\ds {{k-j+1}\choose {k-j+1}}=1$ iff $j\leq k+1$, and is zero otherwise.
\noi
Let $i>2$. Then by the induction hypothesis, 
$$\eqalign {a_{i,j}^k =\sum_{l\geq j-1} a_{i-1,l}^k 
& = \sum_{l\geq j-1} {{2(i-1)-l+k-3}\choose {i-1 -l+k-1}}- 
{{2(i-1)-l+k-3}\choose {i-1+k-2}} \cr
& =\sum_{l\geq j-1} {{2i+k -l-5}\choose {i+k-l-2}}- 
\sum_{l\geq j-1} {{2i+k-l-5}\choose {i+k-3}} \cr}$$
Let $A$ denote the first sum immediately above, and $B$ denote the second sum.
Since $i\geq 3$ we have that $2i+k-l-5\geq i+k-l-2$. Furthermore, the last non-zero term in the
summation $A$ is when $i+k-l-2=0$ or $l=k+i-2$. We then apply identity (A.1) with
$s=2i+k-(j-1)-5$, $t=i+k-(j-1)-2$ and $p=l-(j-1)$ to get that 
$$\eqalign {A= & \sum_{l=j-1}^{i+k-2} {{2i+k-l-5}\choose {i+k-l-2}} \cr
= & \sum_{p=0}^{i+k-(j-1)-2} {{2i+k-5-(j-1)-p}\choose {i+k-2-(j-1)-p}} \cr
= & {{2i+k-(j-1)-5+1}\choose {i+k-(j-1)-2}}  = {{2i-j+k-3}\choose
{i-j+k-1}} \cr}$$
Next we note that the only non-zero terms in the summation $B$ occur when $2i+k-l-5 \geq i+k-3$
or $k\leq i-2$, so that 
$$B= \sum_{l=j-1}^{i-2} {{2i+k-l-5}\choose {i+k-3}}$$
We first consider  when $i\leq j$, in which case the summation above is empty. But if
$i\leq j$ we also have that $2i-j+k-3 <i+k-2$ and thus $\ds {{2i-j+k-3}\choose {i+k-2}}=0$. 
\noi
Next consider the case when $i>j$. We then apply identity (A.2) to $B$ with 
$s=2i+k-(j-1)-5$, $t=i+k-3$ and $p=l-(j-1)$. We thus have
$$\eqalign { B= & \sum_{l=j-1}^{i-2} {{2i+k-l-5}\choose {i+k-3}} \cr
= & \sum_{p=0}^{i-j-1} {{2i+k-5-(j-1)-p}\choose {i+k-3}} \cr
= & {{2i+k-(j-1)-5+1}\choose {i+k-3+1}} = {{2i-j+k-3}\choose {i+k-2}} \cr}$$
Thus we see that $A-B$ gives the desired result.
\qed
The next lemma will be used in section 3.
\noi
{\bf LEMMA 2.5}: Let $i\geq 2$. Then for all $r, \ 1\leq r\leq i-1$, $$
\sum_{l=2}^i a_{i-r,i}^l={{2i-2r-2}\choose {i-r-1}}$$
\noi
{\bf Proof}: First consider the case when $r=i-1$. Then $\ds \sum_{l=2}^i a_{i-r,i}^l
=\sum_{l=2}^i a_{1,i}^l = a_{1,i}^i=1$. Conversely, $\ds {{2i-2r-2}\choose
{i-r-1}}={{2i-2(i-1)-2}\choose {i-(i-1)-1}}={0\choose 0}=1$.
\noi
Let $1\leq r\leq i-2$. By Lemma 2.4 we have that 
$$ \eqalign {\sum_{l=2}^i a_{i-r,i}^l = & \sum_{l=2}^i \bigg[ {{2i-2r-i+l-3}\choose {i-r-i+l-1}}
-{{2i-2r-i+l-3}\choose {i-r+l-2}} \bigg] \cr
= & \sum_{l=2}^i \bigg[ {{i-2r+l-3}\choose {l-r-1}}
-{{i-2r+l-3}\choose {i-r+l-2}} \bigg] \cr}$$
But since $r\geq 1$, we have $i-2r+l-3\leq i-r+l-4\leq i-r+l-2$ so that all of the terms above of
the form $\ds {{i-2r+l-3}\choose {i-r+l-2}}$ are zero. Furthermore, if $l<r+1$, then $l-r-1<0$
in which case all terms above of the form  $ \ds {{i-2r+l-3}\choose {l-r-1}}$ are zero. Therefore, 
letting $p=l-r-1$,
$$\sum_{l=2}^i a_{i-r,i}^l = \sum_{l=r+1}^i  {{i-2r+l-3}\choose {l-r-1}}
= \sum_{p=0}^{i-r-1}  {{i-r-2+p}\choose {p}}$$
Finally we apply identity (A.1) with $s=2i-2r-3$ and $t=i-r-1$ to get the desired result
$$\sum_{l=2}^i a_{i-r,i}^l =  \sum_{p=0}^{i-r-1}  {{i-r-2+p}\choose {p}}
={{2i-2r-2}\choose {i-r-1}}$$ 
\qed
\noi
For each $k\geq 2$ and $n\geq 1$, let the set of all admissible sequences $c_1 ,c_2 ,\cdots ,c_n$ with $c_1
=k$ be denoted by 
$\sigma_n^k$. In order to satisfy (1.2) we must have that $c_n \geq c_1 -1 =k-1$. Thus 
$$| \sigma_n^k |=\sum_{j\geq k-1} a_{n,j}^k=a_{n+1,k}^k$$
%Note that in the sum above the largest possible value of $c_i$ is $k+i-1$ and comes
%from the sequence $k,k+1,k+2,\cdots ,k+i-1$. Thus we have that
%$$| \sigma_i^k |=\sum_{j\geq k-1} a_{i,j}^k = \sum_{j= k-1}^{k+i-1} a_{i,j}^k =a_{i+1,k}^k$$
%\noi
For each $m\geq 2$ and $n\geq 1$, let $T_{n,m}$ denote the set of all admissible sequences $c_1 ,c_2
,\cdots , c_n$ with 
$2\leq c_1 \leq m$. Then 
$$T_{n,m}=\bigcup_{l=2}^m \sigma_n^l$$
and 
$$|T_{n,m} |=|\bigcup_{l=2}^m \sigma_n^l|=\sum_{l=2}^m a_{n+1,l}^l$$
\noi
{\bf COROLLARY 2.6}: Let $m\geq 2$ and $n\geq 1$. Then
$$|T_{n,m} |=(m-1){{2n-1} \choose n}-\sum_{l=2}^m
{{2n-1}\choose {n+l-1}}
$$ 
\noi
{\bf Proof}:
Applying
Lemma 2.4 we get that
$$\eqalign {|T_{n,m} |=\sum_{l=2}^m a_{n+1,l}^l = &\sum_{l=2}^m \bigg[ {{2(n+1)-l+(l-3)}\choose
{n+1-l+l-1}} -{{2(n+1)-l+(l-3)} \choose {n+1+(l-2)}}\bigg] \cr
= & \sum_{l=2}^m \bigg[ {{2n-1} \choose n}-{{2n-1}\choose {n+l-1}}\bigg] \cr
= & (m-1){{2n-1} \choose n}-\sum_{l=2}^m {{2n-1}\choose {n+l-1}} \cr}$$
\qed
{\bf COROLLARY 2.7}: Let $m\geq 2$ and $n\geq 1$. Then for $m\geq n$
$$|T_{n,m} |=m{{2n-1} \choose n}- 2^{2n-2}  $$
\noi
{\bf Proof}:
Whenever  $l>n$ we have that 
$n+l-1>2n-1$ so that $\ds {{2n-1}\choose {n+l-1}}=0$. We then apply identity (A.3) to Corollary 2.6 to
obtain 
$$\eqalign {|T_{n,m} |= & (m-1){{2n-1} \choose n}-\sum_{l=2}^m {{2n-1}\choose {n+l-1}}
=(m-1){{2n-1} \choose n}-\sum_{l=2}^n {{2n-1}\choose {n+l-1}} \cr
= & (m-1){{2n-1} \choose n}-\Big[  2^{2n-2}-{{2n-1}\choose {n}}\Big]
=m{{2n-1} \choose n}- 2^{2n-2} \cr}$$
\qed
\noi
\cl {\bf 3. CYCLIC PERMUTATIONS}
\noi
Since the admissible sequence for a serial ring is unique only up to cyclic permutation, 
we next consider the equivalence classes of the elements of $T_{n,m}$ under cyclic permutation.
Notice that $T_{n,m}$ does not contain all cyclic permutations of its own elements (e.g. 
$a=m,m+1, \cdots ,m+n-1 \in
T_{n,m}$ but every non-trivial cyclic permutation of $a$ is not in $T_{n,m}$). Thus we enlarge
$T_{n,m}$ to include these elements as follows: 
\noi
\item {} 
For $m\geq 2$ and $n\geq 1$, let $C_n=\{\alpha_0 ,\alpha_1 ,\cdots ,\alpha_{n-1}\}$ denote the cyclic group
of order
$n$ acting on the set $T_{n,m}$ as follows:
Let $a=c_1 ,c_2 , \cdots c_n \in T_{n,m}$, then $\alpha_0 (a)=a$ and for $1\leq i\leq
n-1$, $\alpha_i (a)=c_{i+1}, \cdots c_n , c_1 , \cdots c_{i}$. Let $S_{n,m}=\{\alpha_i (a):
a\in T_{n,m}, 0\leq i\leq n-1\}$. Thus $S_{n,m}$ is the set of all cyclic permutations of the
elements of $T_{n,m}$ and $T_{n,m}\subset S_{n,m}$. 
\noi
Regarding two admissible sequences as equivalent if one is a cyclic permutation of the
other, we see that the number of equivalence classes of admissible sequences is just the number
of orbits of $C_n$ on $S_{n,m}$. We denote this number of equivalence classes by $O_{n,m}$. We first
consider the size of the set $S_{n,m}$. Also note that if $a=c_1 ,\cdots c_n \in S_{n,m}$, then 
$min\{ c_1, \cdots c_n \} \leq m$. 
\noi
{\bf LEMMA 3.1}: Let $m\geq 2$ and $n\geq 1$. Then 
$$|S_{n,m} \setminus T_{n,m}|=\sum_{k=1}^{n-1} \bigg[
k|T_{k,2} |
\Big(
\sum_{i=2}^m a_{n-k,m}^i \Big) \bigg]$$
{\bf Proof}: 
Let $a=c_1 ,\cdots ,c_n \in T_{n,m}$. Then all cyclic permutations of $a$ are in $T_{n,m}$ iff $c_i \leq m
$ for all $i, \ 1\leq i\leq n$. Thus there exists a cyclic permutation of
$a$ that is not in
$T_{n,m}$  iff there exists $j$ with $1\leq j \leq n-1$ such that 
\item {} i) $c_{j} =m$
\item {} ii) $c_{j+1} =m+1 $
\item {} iii) There exists $k$ with $  1\leq k\leq n-j$ so that $c_i >m$ for all $i$ with $ j+1 \leq i \leq
j+k$ 
\item {} iv) $c_{[j+k+1]} \leq m$ (where $[p]$ denotes the least positive residue of $p$ modulo $n$)
\vskip5pt\noindent
Without loss of generality, we permute $a$ and consider $a' =\alpha_{j} (a)$, where
%=a_1 ,a_2 \in T_{n,m}$
%with
%$a_1 =d_1 ,\cdots ,d_{n-k} \in T_{n-k,m}$ and $d_{n-k} =m$ and $a_2 =d_{n-k+1} ,\cdots ,d_n$ with
%$d_{n-k+1}=m+1$ and $d_i >m$ for all $i, \ n-k+1 \leq i \leq n$.
% Note also that any other cyclic permutations of $a$ that are
%not in $T_{n,m}$ can be identified as above.  given above, where 
\item {} i) $a' =a_1 ,a_2 $ with $a_1$ a subsequence of length $k$ and $a_2$ a subsequence of
length $n-k$
\item {} ii) $a_1 =d_1 ,\cdots , d_k $ with all $d_i > m$ and $d_1 =m+1$
\item {} iii) $a_2 =d_{k+1}, \cdots d_n \in T_{n-k,m}$ and $d_n =m$ 
\item {} iv) $1\leq k \leq n-1$
\vskip5pt\noindent
Clearly there exist $k$ permutations of $a'$ (and hence of $a$) that are not in $T_{n,m}$, namely
$\alpha_i (a')$ for $0\leq i \leq k-1$. Moreover, every element of $S_{n,m} \setminus T_{n,m}$ is  of the
form $\alpha_i (a')$ for some $i$,  $0\leq i \leq k-1$, and for some $a'$ as above. Thus, the number of
sequences in
$S_{n,m}
\setminus T_{n,m}$  is found by multiplying  the number of sequences of the form $a'$ by $k$.
\noi
The number of sequences of the form $a_2 $ above is $\ds \sum_{i=2}^m a_{n-k,m}^i$. Let $U_k $ be
the set of sequences of the form $a_1 $ above. Then there is a bijection $f:T_{k,2}\rightarrow U_k $
where $f(d_1 , \cdots d_k )=d_1 +m-1 , \cdots ,d_k +m-1$. Thus $|U_k |=|T_{k,2}|$. Multiplying by the $k$
cyclic permutations of $a'$ that are not in $T_{n,m}$ and summing over all $k$, $1\leq k \leq n-1$, gives
the desired result.
\qed
We note that by Corollary 2.6 we have $$|T_{k,2}|={{2k-1}\choose {k}}-{{2k-1}\choose {k+1}} 
={{1}\over {k+1}} {{2k}\choose {k}}$$ which is the $k^{th}$ Catalan number. Applying this to Lemma 3.1 and
Corollary 2.6 we have
\noi
{\bf COROLLARY 3.2}: Let $m\geq 2$ and $n\geq 1$. Then 
$$ \eqalign {|S_{n,m}|=
& |T_{n,m}|+|S_{n,m} \setminus T_{n,m}| \cr 
= & (m-1){{2n-1}\choose n} - \sum_{k=2}^m {{2n-1}\choose {n+k-1}} +\sum_{k=1}^{n-1} \bigg[ {k\over {k+1}}
{{2k}\choose {k}} \Bigg(
\sum_{i=2}^m a_{n-k,m}^i \Bigg) \bigg] \cr}$$
\qed
We consider the special case when $n=m\geq 2$. Applying Corollary 3.2, Lemma 2.5, and Corollary 2.7, we
have 
\noi
{\bf COROLLARY 3.3}:  Let $n\geq 2$. Then
$$|S_{n,n}|=
n{{2n-1}\choose n} - 2^{2n-2}  +\sum_{k=1}^{n-1}{k\over {k+1}} {{2k}\choose {k}} {{2n-2k-2}\choose
{n-k-1}}  $$
\qed
{\bf LEMMA 3.4}:  Let $n\geq 2$. Then
$$|S_{n,n}|=
(n-1){{2n-1}\choose n} $$
{\bf Proof}:
We first note that $${k\over {k+1}} {{2k}\choose k} = {{2k}\choose {k-1}}$$
and that
$${{2k}\choose {k}} - {{2k} \choose {k-1}} = {1\over {k+1}} {{2k}\choose k}$$
We then apply (A.4) and (A.5) to Corollary 3.3 to obtain
$$ \eqalign {|S_{n,n}|= &
n{{2n-1}\choose n} - 2^{2n-2}  +\sum_{k=1}^{n-1}{k\over {k+1}} {{2k}\choose {k}} {{2n-2k-2}\choose
{n-k-1}} \cr
= &
n{{2n-1}\choose n} - \sum_{k=1}^{n-1} {{2k}\choose k} {{2(n-k-1)}\choose {n-k-1}} 
+\sum_{k=1}^{n-1} {{2k}\choose {k-1}} {{2(n-k-1)}\choose {n-k-1}} \cr
= & n{{2n-1}\choose n} - \sum_{k=0}^{n-1} \bigg[{{2k}\choose k}-{{2k}\choose {k-1}} \bigg]
{{2(n-k-1)}\choose {n-k-1}} \cr
= & n{{2n-1}\choose n}-\sum_{k=0}^{n-1} {1\over {k+1}}{{2k}\choose k}  {{2(n-k-1)}\choose {n-k-1}} 
\cr
= & n{{2n-1}\choose n}-{{2(n-1)+1}\choose {n-1+1}} \cr
= & (n-1){{2n-1}\choose n} \cr}$$
\qed
\noi
\cl {\bf 4. ORBITS OF $C_n$ ON $S_{n,m}$}
\noi
For each $\alpha \in C_n$, let $fix(\alpha )=\{ a\in S_{n,m} : \alpha (a) =a\}$. We note that for
the identity $\alpha_0$, $fix(\alpha_0 )=S_{n,m}$. For $n\geq 1$ and $m\geq 2$, we apply
Burnsides Theorem to get
$$O_{n,m} ={1\over {|C_n|}} \sum_{\alpha \in C_n } fix(\alpha ) 
={1\over n} \bigg[|S_{n,m} | + \sum_{\alpha \in C_n \setminus \{\alpha_0 \} } |fix(\alpha)|
\bigg]  \leqno (\bf 4.1) $$ 
Note that for $m= 1$, we must include the number of admissible sequences of length $n$ with $c_1 =1$. In
this case, since the rings are indecomposable with a simple projective module, no cyclic permutations are
needed, and the number of such sequences is the $n-1^{st}$ Catalan number,  $\ds b_{n-1} ={1\over n}
{{2n-2}\choose {n-1}}$ [2].  Thus
for
$n\geq 1$ and
$m\geq 2$, the total number of equivalence classes of admissible sequences is
$$b_{n-1}+O_{n,m} =  
{1\over n} \bigg[ {{2n-2}\choose {n-1}}+|S_{n,m} | + \sum_{\alpha \in C_n \setminus \{\alpha_0
\} } |fix(\alpha)|
\bigg]  \leqno (\bf 4.2) $$
%\cl {\bf SPECIAL CASE n=m IS PRIME}
We consider the special case when $n=m$ is prime. Then for each $\alpha \in C_n \setminus \{
\alpha_0
\}$, the only elements of 
$fix(\alpha ) $ are the sequences $k,k,\cdots k$ where $2\leq k\leq n$, so that $|fix(\alpha
)|=n-1$.  Applying this together with the Lemma 3.4 and (4.2), we have that the number of
equivalence classes of admissible sequences is
$$ b_{n-1}+O_{n,m} =  {1\over n} \bigg[{{2n-2}\choose {n-1}}+ (n-1){{2n-1}\choose n} + (n-1)^2
\bigg]  \leqno {(\bf 4.3)} $$
Simplifying $(4.3)$, we have
\noi
{\bf THEOREM 4.4}: Let $n$ be prime. Then the number of equivalence classes of admissible sequences for
indecomposable serial rings with $n$ indecomposable projective modules whose minimum composition length is
less than or equal to $n$ is
$${{2n-1}\choose n} +{1\over n} \bigg[ (n-1)^2 - {{2n-2}\choose {n}}
\bigg]   $$
\qed
\noi
\next
\cl {\bf APPENDIX}
$$ \sum_{k=0}^{t} {{s-k}\choose {t-k}}  =\sum_{k=0}^{t} {{s-t+k}\choose {k}} =
{{s+1}\choose {t}}
\ \ \ \ \  \ {\rm for \ s\geq t\geq 0
\ \ \ \ \ \ \ \ \ \ \ \ 
([3] \ pg. 7)}
\leqno (\bf A.1)  $$
$$ \sum_{k=0}^{s-t} {{s-k}\choose {t}} = {{s+1}\choose {t+1}} \ \ \ \ \ \ {\rm for \ s \geq t\geq 0
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ 
([3] \ pg. 7)}
\leqno (\bf A.2) $$
$$ \sum_{k=1}^{s} {{2s-1}\choose {s+k-1}} = 2^{2s-2} \ \ \ \ \ \  {\rm for \ s\geq 1
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \  ([3] \ pg. 34)}
\leqno (\bf A.3)
$$ 
$$ \sum_{k=0}^{s} {{2k}\choose {k}} {{2(s-k)}\choose {s-k}} = 2^{2s} \ \ \ \ \ \  {\rm for \ s\geq 0
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ ([3] \ pg. 130)}
\leqno (\bf A.4)
$$ 
$$ \sum_{k=0}^{s} {1\over {k+1}} {{2k}\choose k}  {{2(s-k)}\choose {s-k}} = {{2s+1}\choose {s+1}} \ \ \ \ \
\  {\rm for
\ s\geq 0
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
([3] \ pg. 120)}
\leqno (\bf A.5)
$$ 
\noi
\cl {\bf REFERENCES}
\noi
\item {1.} F.W. Anderson and K.R. Fuller, {\it {Rings and Categories of
Modules}}, Springer-Verlag,  New York 
and Berlin, 2nd ed., 1992.
\noi
\item {2.} J.O. Hanes and D.D. Wick, "The Number of Admissible Sequences for Indecomposable Serial
Rings with a Simple Projective Module," {\it Proceedings of the Louisiana-Mississippi Section of the
Mathematical Association of America}, Spring 2000.
\noi
\item {3.} J. Riordan, {\it {Combinatorial Identities}}, John Wiley and Sons, Inc., New York, London and
Sydney, 1968.
\bye