Download 3. CATALAN NUMBERS Corollary 1. cn = 1

MATH IN MOSCOW
COMBINATORICS, SPRING 2012
3. CATALAN NUMBERS
Denition 1. A Dyck path is a continuous function f : [0; 2n] → [0; +∞) linear with the slope 1 or −1 at every
interval (k; k + 1) for k = 0; : : : ; 2n − 1, and such that f (0) = f (2n) = 0. The number of Dyck paths of length 2n
is called the n-th Catalan number and denoted (usually) by cn .
Theorem 1. The Catalan numbers satisfy the identity cn =
Pn−1
k=0 ck cn−k−1
for any n = 1; 2; : : : .
Proof. If a Dyck path f has no points k = 1; : : : ; 2n − 1 such that f (k) = 0 then the function g(t) def
= f (t + 1) − 1,
0 ≤ t ≤ 2n − 2, is a Dyck path of length 2n − 2. The number of such Dyck paths is cn−1 = c0 cn−1 (c0 = 1 by
denition).
Otherwise, let 2k + 2, 0 ≤ k ≤ n − 1, be the minimum positive integer such that f (2k + 2) = 0 (it is easy to see
that such integer must be even). Then g1 (t) = f (t + 1) − 1 for 1 ≤ t ≤ 2k + 1 is a Dyck path of length 2k, and
g2 (t) = f (t + 2k + 2) for 0 ≤ t ≤ 2n − 2k − 2 is a Dyck path of length 2n − 2k − 2. Thus, the number of Dyck paths
with the rst zero 2k + 2 is equal to ck cn−k−1 .
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Theorem 2. The generating function of Catalan numbers is
P∞
n
n=0 cn t
√
1− 1−4t .
2t
n
by t and
=
Proof. Let C (t) def
= c0 + c1 t + : : : . Multiply
the identity
from Theorem 1
sum it over all n = 1; 2; : : : .
P∞
P∞ Pn−1
n
n
Since c0 = 1, one has C (t) − 1 = n=1 cn t = n=1 k=0 ck cn−k−1 t . Changing the summation variable to
P
P∞
k+m+1 = tC 2 (t). The quadratic equation obtained has two
m def
= n − k − 1 gives C√(t) − 1 = ∞
k=0 m=0 ck cm t √
solutions: C (t) = (1 + 1 − 4t)=(2t) and C (t) = (1 − 1 − 4t)=(2t), but the rst one has a pole at t = 0 and so
cannot be developped into a power series.
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Corollary 1. cn = n+1
¡2n¢
(2n)!
n = n!(n+1)! .
Proof. By Newton's binomial formula for the exponent 1=2 one has
µ ¶
µ
¶ µ
¶
µ
¶
∞
∞
X
X
√
1=2 n n
11
1
3
2n − 3 n n
1 − 4t = 1 + (−1)n
4 t = 1 + (−1)n
· −
· −
· ··· · −
4 t
n
n! 2
2
2
2
n=1
n=1
=1+
√
∞
X
n=1
(−1)2n−1
P
∞
X
1 · 3 · · · · · (2n − 3) n n
(2n − 2)!
2n n
4
t
=
1
−
2n−1 (n − 1)!n! 2 t ;
2n n!
2
n=1
(2n−2)! n−1 . Changing the summation variable n 7→ n + 1 one obtains the
hence, (1 − 1 − 4t)=(2t) = ∞
n=1 n!(n−1)! t
required formula.
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Examples of combinatorial problems solved by Catalan numbers:
Proposition 1. The number of balanced bracket structures of length n (i.e. ways to place correctly n pairs of
brackets) is cn .
For example, for one pair or brackets the only structure is (); for two pairs there are two: ()() and (()), etc.
Proof. To every bracket structure of length n one can associate a Dyck path f of length 2n: if the k-th character
in the structure is ( then the slope of f at the interval (k − 1; k) is 1, otherwise it is −1. An easy induction proves
that so Dyck paths and bracket structures are put into one-to-one correspondence.
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Proposition 2. The number of triangulations of a convex (n + 2)-gon by diagonals is cn .
To count triangulations correctly one should have the vertices of the polygon numbered. For example, for a
square with the vertices 1, 2, 3, 4 there are c2 = 2 triangulations: a diagonal 1; 3 or a diagonal 2; 4. For a pentagon
there are c3 = 5 triangulations; if the pentagon is regular they are mapped one to the other by rotations.
Proof. Let Tn be the number of such triangulations; apparently, T1 = 1, and take T0 = 1 by denition. Number
the vertices of the polygon counterclockwise and consider the vertex 1. If the triangulation contains a triangle
[1; 2; (n+2)], then cut it o obtaining a triangulated (n+1)-gon; the number of such triangulations is Tn−1 = T0 Tn−1 .
If [1; 2; (n + 2)] is not present, then 1 is an endpoint of some diagonals. Take the leftmost of them, let it join 1
with k + 2, 0 ≤ k ≤ n − 1. Then one has a triangulation of the (k + 2)-gon [1; 2; : : : ; (k + 2)] plus a triangulation
of the (n − k + 1)-gon [k + 2; k + 3; : : : ; n + 2; 1]. The total number of such triangulations is Tk Tn−k−1 . Therefore
the sequence Tn satises the formula of Theorem 1, so that Tn = cn .
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Proposition 3. Let a0 ; : : : ; an be variables and ∗ be the symbol of a binary operation. The number of ways to put
brackets correctly in the expression a0 ∗ a1 ∗ · · · ∗ an is cn .
For example, if n = 2 then there are c2 = 2 ways: (a0 ∗ a1 ) ∗ a2 and a0 ∗ (a1 ∗ a2 ). Note that Proposition 3 and
Proposition 1 are dierent!
Proof 1. Denote by dn the answer to the problem. Let the binary operation corresponding to the k-th star be the
last to be performed. To the left from the star there are k letters and k − 1 stars with correctly put brackets, and to
the right, there are n − k − 1 letters and n − k − 2 stars. Thus the number of ways to put brackets so is dk dn−k−1 .
Summing over all possible k one obtains the identity of Theorem 1.
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Proof 2 (a sketch). Consider a triangulation of the (n + 2)-gon, and mark its sides, except one, by the letters
a0 ; a1 ; : : : ; an counterclockwise (the unmarked side is between a0 and an ). Then for every triangle of the triangulation repeatedly do the following. If one side of the triangle is empty (unmarked) and the two remaining sides
are marked by the expressions p and q (looking counterclockwise from the empty side) then mark an empty side
with p ∗ q (putting brackets around p and q unless these are single letters). In all the other cases leave the triangle
intact. An induction proves that after a nite number of steps all the sides and diagonals of the polygon will be
marked, and the marking does not depend on the order of triangles to process. The expression on the side of the
polygon that was initially empty contains all the letters a0 ; : : : ; an in correct order. This puts the expressions in
question into a one-to-one correspondence with the triangulations.
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