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### Finding Catalan Numbers

```
Date: 12/15/1999 at 21:34:28
Subject: Catalan Numbers

Dear Dr. Math,

For a research paper, I must study Catalan numbers. Could you possibly
explain them (how to find them, applications, etc.)?

Thanks a million.
```

```
Date: 12/16/1999 at 10:36:52
From: Doctor Anthony
Subject: Re: Catalan Numbers

A physical interpretation is to consider an n x n square grid of
streets and to find the number of routes from one corner of the square
to the corner diagonally opposite that do not cross the diagonal. That
is, while making the journey, the number of blocks north is always
equal to or ahead of the number of blocks east - and of course vice
versa.

Catalan Numbers
---------------
The number of sequences a(1), a(2), a(3), ..., a(2n) of 2n terms that
can be formed by using n +1's and n -1's whose partial sums satisfy

a(1) + a(2) + ... + a(k) >= 0  (where k = 1, 2, ..., 2n)  ....[1]

equals the nth Catalan number

C_n = 1/(n+1) C(2n,n)   (n >= 0)

We call a sequence of n +1's and n -1's acceptable if it satisfies [1]
and unacceptable otherwise. [This is equivalent to the number of
routes consisting of n northerly blocks (+1) and n easterly blocks
(-1) that never cross the diagonal.]

Let A(n) denote the number of acceptable sequences of n +1's and n
-1's and let U(n) denote the number of unacceptable ones. The TOTAL
number of sequences of n +1's and n -1's is

(2n)!/(n!n!) = C(2n,n)

So we must have

A(n) + U(n) = C(2n,n)

and we evaluate A(n) by first evaluating U(n) and subtracting from
C(2n,n).

Consider an unacceptable sequence of n +1's and n -1's. Because the
sequence is unacceptable there is a smallest k such that the partial
sum:

a(1) + a(2) + ... + a(k) < 0

Because k is smallest there is an equal number of +1's and -1's
preceding a(k) and we have:

a(1) + a(2) + ... + a(k-1) = 0   and   a(k) = -1

In particular, k is an ODD integer. We now reverse the signs of each
of the first k terms; that is we replace a(i) by -a(i) for each i = 1,
2, 3, ..., k and leave unchanged the remaining terms. The resulting
sequence a'(1), a'(2), ..., a'(2n) is a sequence of (n+1) +1's and
(n-1) -1's.

This process is reversible:

Given a sequence of (n+1) +1's and (n-1) -1's there is a first
instance when the number of +1's exceeds the number of -1's (since
there are more +1's than -1's). Reversing the +1's and -1's up to that
point results in an unacceptable sequence of n +1's and n -1's. Thus
there are as many unacceptable sequences as there are sequences of
(n+1) +1's and (n-1) -1's. The number of sequences of (n+1) +1's and
(n-1) -1's is the number (2n)!/[(n+1)!(n-1)!] = C(2n,n-1).

Therefore

(2n)!
U(n) = ------------
(n+1)!(n-1)!

and thus
(2n)!      (2n)!
A(n) = ----- - ------------
n! n!   (n+1)!(n-1)!

(2n)!      1      1
= -------- [--- - -----]
n!(n-1)!   n    (n+1)

(2n)!      1
= -------- [------]
n!(n-1)!  n(n+1)

1    (2n)!
= ----- [-----]
(n+1)  n! n!

=  1/(n+1) C(2n,n)

There are many different interpretations of this theorem. The
following are two examples.

Example (1)
------------
There are 2n people in line to get into a theatre. Admission is 50
cents. Of the 2n people, n have 50-cent pieces and n have a \$1 bill.
The box office rather foolishly begins with an empty cash register. In
how many ways can the people line up so whenever a person with a \$1
bill buys a ticket, the box office has a 50-cent piece in order to
give change?

If we regard the people as indistinguishable and identify a 50-cent
piece with a +1 and a dollar bill with -1, then the answer is the
number

C_n = 1/(n+1) C(2n,n)

of acceptable sequences. If the people are regarded as distinguishable

(n!)(n!) 1/(n+1) C(2n,n) = (2n)!/(n+1)

Example (2)
-----------
A city lawyer works n blocks north and n blocks east of his place of
residence. Every day he walks 2n blocks to work. How many routes are
possible if he never crosses (but may touch) the diagonal line from
home to office?

Each acceptable route is either above the diagonal or below the
diagonal. We find the number of acceptable routes above the diagonal
and multiply by 2. Each route is a sequence of n northerly blocks and
n easterly blocks. We identify north with +1 and east with -1. Thus
each route corresponds to a sequence

a(1), a(2) , a(3), ..., a(2n)

of n +1's and n -1's and in order that the route not dip below the
diagonal we must have

a(1) + a(2) + a(3) + ... + a(k) >= 0

for k = 0, 1, 2, ..., 2n

Hence by theorem proved above the total number of acceptable routes is

2.C_n = 2/(n+1) C(2n,n)

Recurrence Relation for Catalan Numbers
----------------------------------------
We have

C_n = 1/(n+1) C(2n,n)

= 1/(n+1) (2n)!/(n!n!)
and
C_(n-1) = (1/n) (2n-2)!/[(n-1)!(n-1)!]

dividing we obtain

C_n     (4n-2)
------- = ------
C_(n-1)   (n+1)

Therefore the Catalan sequence is determined by the following
recurrence relation and initial condition:

4n-2
C_n = ------  C_(n-1)   and   C(1) = 1
n+1

Parenthesizing an Expression
----------------------------
If we have to find all the ways to parenthesize the letters abcd we
could have:

(((ab)c)d)    (((abc     111000
((a(bc))d)    ((a(bc     110100
((ab)(cd))    ((ab(c     110010
(a((bc)d))    (a((bc     101100
(a(b(cd)))    (a(b(c     101010

The connection with Catalan numbers is that the information about
parenthesizing does not require the letter d and we replace each left
parenthesis '(' with '1' and each letter with '0'. Then an acceptable
sequence must have the number of 1's equal to or exceeding the number
of 0's.

In general one can parenthesize the product of n+1 letters in C_n
ways.

So for the letters abcd we can parentesize in  C_3 ways

C(3) = 1/4 C(6,3)

= (1/4) 20

=  5 ways

Catalan Numbers

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Number Theory
High School Number Theory
High School Permutations and Combinations
High School Sequences, Series

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