Delving deeper Thomas Koshy

The Ubiquitous Catalan Numbers

J

ust as Fibonacci and Lucas numbers are a great source of fun and excitement for both amateurs and professionals alike (Askey 2005, Koshy 2002), so are the less well-known Catalan numbers. They too are excellent candidates for mathematical activities such as experimentation, exploration, and conjecture. I was surprised to see that, like Fibonacci and Lucas numbers, Catalan numbers seemed to show up in several problems I had assigned to students over the years. Example 1. Find the number of mountain ranges that can be drawn with n upstrokes and n downstrokes. In other words, find the number of different paths we can choose from the origin to the point (2n, 0) on the xy-plane subject to the following conditions:

4

42

• We can touch the x-axis, but we cannot cross it. • From the point (x, y), we can climb up to the point (x + 1, y + 1) or climb down to the point (x + 1, y – 1). See figure 1.

(4, 2) (1, 1) (3, 1) (4, 2)

2

(0, 0) (1, 1)(2,(3,0)1) –2

(0, 0)

Solution. When we have no clue about the solution, it is often a good technique to solve the prob-

(2, 0)

(5, 1) 5

(6, 0) (5, 1)

5

(6, 0)

–2

Fig. 1 The two possible paths from the point (x, y) This department focuses on mathematics content that appeals to secondary school teachers. It provides a forum that allows classroom teachers to share their mathematics from their work with students, their classroom investigations and projects, and their other experiences. We encourage submissions that pose and solve a novel or interesting mathematics problem, expand on connections among different mathematical topics, present a general method for describing a mathematical notion or solving a class of problems, elaborate on new insights into familiar secondary school mathematics, or leave the reader with a mathematical idea to expand. Send submissions to “Delving Deeper” by accessing mt.msubmit.net. “Delving Deeper” can accept manuscripts in ASCII or Word formats only. Edited by Al Cuoco, [email protected] Center for Mathematics Education, Education Development Center Newton, MA 02458 E. Paul Goldenberg, [email protected] Center for Mathematics Education, Education Development Center Newton, MA 02458

lem in a few simple cases and then look for a pattern, as illustrated in the following steps. Step 1. Collect enough data by performing simple experiments. Figure 2 shows the various possibilities for n = 0, 1, 2, 3, and 4. Notice that some mountain ranges are reflections of others. Step 2. Arrange the results of the five experiments in a table, as in table 1. The numbers in the bottom row that count the number of mountain ranges are called the Catalan numbers, Cn. Step 3. Look for a pattern. (The pattern does not seem to be obvious.) Step 4. If a pattern exists, can you conjecture a formula for the nth Catalan number Cn?

184 Mathematics Teacher | Vol. 100, No. 3 • October 2006 Copyright © 2006 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in any other format without written permission from NCTM.

n=0

Table 1

n=1

Data from Example 1

n=2

n=3

n=4

n

1

2

3

4

5

…

n

Number of mountain ranges

1

2

5

14

?

…

?

n=1

1

–1

n=2

1

1

–1

–1

1

–1

1

–1

1

1

1

–1

–1

–1

1

1

–1

1

–1

–1

1

–1

1

–1

1

–1

1

–1

1

1

–1

–1

1

1

–1

–1

1

–1

n=3

Fig. 2 Mountain ranges for n = 0, 1, 2, 3, and 4

Step 5. If you conjectured a formula, can you establish it? The next example deals with partial sums. If S = a1 + a2 + a3 + L + an is a sum of numbers, each of the sums a1, a1 + a2, a1 + a2 + a3, K a1 + a2 + a3 + L + an is a partial sum of S. Example 2. Arrange n 1s and n –1s in a row in such a way that every partial sum is a nonnegative integer. Find the number of such possible sequences. Solution. Once again, we have no clue about the solution or the answer. So we proceed as in example 1. Step 1. Collect enough data by performing simple experiments. Notice that every partial sum must be nonnegative, so every sequence must begin with a 1. Figure 3 shows the various possible arrangements corresponding to 0 ≤ n ≤ 3. Step 2. Arrange the data in a table, as in table 2. Step 3. Look for a pattern. (The pattern, as before, is not obvious.) Step 4. If a pattern exists, can you conjecture a formula for the nth Catalan number Cn? Step 5. If you conjectured a formula, can you establish it?

← Catalan numbers

Fig. 3 Various arrangements corresponding to 0 ≤ n ≤ 3

Table 2 Data from Example 2 n

1

2

3

4

5

…

n

Number of partial sums

1

2

5

14

?

…

?

← Catalan numbers again

(( )( ))

((( )))

Table 3 Data from Example 4 n

Correctly parenthesized sequences

1

()

2

( )( )

(( ))

3

( )( )( )

(( ))( )

( )(( ))

expressions. For example, for numbers a, b, c, and d, all of these expressions are correctly parenthesized: ((ab)c)d, (a(bc)d, (ab)(cd), (a(bc)d), and a(b(cd)), but this expression is not:

Example 3. Bertrand’s ballot problem (Feller 1968) is another version of the first two examples: Two candidates A and B are running for president. Each person gets n votes. In how many ways can the 2n votes be counted in such a way that at each count the number of votes received by A is greater than or equal to the number of votes received by B? Example 4. This example, originally studied by Catalan, deals with correctly parenthesized algebraic

a)(bc)(d Suppose we are given n pairs of left and right parentheses. Find the number of correctly parenthesized sequences that can be formed. Solution. For convenience, we drop the letters. So “( )” and “(( ))( )” are correctly parenthesized but “)( )(” is not. The results corresponding to 0 ≤ n ≤ 3 are in table 3. It would be a good exercise to list the possible sequences corresponding to n = 4. Readers might try showing that examples Vol. 100, No. 3• October 2006 | Mathematics Teacher 185

T3 =

= 1, T4 = = 2, and T5 = 2! 3! 2 • 6 • L • ( 4n − 2) Cn = , n ≥ 1. (n• L + 1•)!( 4n − 2) 2•6 Cn = , n ≥ 1. (n + 1)! When n = 1, this yields 1. So we 0; •but 1 = •6 • ( 4 n − 6) 4n − C 21 =2 C 10 •CL • = could define CC 0 nto be 1. In other words, we have n4+n1− 2 2 • 6 • 10n•!L • ( 4n − 6) conjectured a recursive for Cn. • Cn 4 =n − formula n +2 1• C n! = n −1 n + 1 4 n − 2 Conjecture. Cn satisfies the •following recurrence: = Cn − 1 n+1 C0 = 1

n=3

n=4

n=5

n=6

Fig. 4 Triangulations of n-gons, where 3 ≤ n ≤ 6

1–4 are isomorphic—they are the same problems wrapped in different contexts. We now turn to a triangulation problem studied by Euler. An n-gon (a polygon with n sides) is convex if every diagonal of the polygon lies inside it. Example 5. Find the number of ways (the interior of) a convex n-gon can be partitioned into triangles by drawing non-intersecting diagonals, where n ≥ 3. Solution. The various possibilities for 3 ≤ n ≤ 6 are shown in figure 4. Using an induction argument, which he characterized as “quite laborious,” Euler showed that the number of triangulations Tn of a convex n-gon is given by

4!

=

C 4 =n1− 2 •C Cn =0 , n≥1 n4+n1− 2 n − 1 •C Cn = , n≥1 n −1 n+1 For example, here is a verification for n = 4: 14 •C C4 = 5 14 3 •C C4 14 = 10 3 •C = 5• 5 14 4 10 2 • • = 14 10 6 C2 = 5• 4• • C1 5 14 4 103 6 • • • = 10 14 6 2 C1 = 5• 4• •3 • C0 5 14 4 103 62 2 • • • •C = = 14 5 4 3 2 0 = 14 1 2•26n•10 • L • ( 4n − 10) This agrees with Tn =data. , n ≥ 3. Cn =our  (n − 1)! n + 11 is n 2only n a conjecture at this The recurrence Cn above =  n  n + 1try point, and readers might proving it by show• 6examples 1 2 any of 2 2 • 6 • 10 ing that it correctly1models the  10 T3 = = 1, T4 = = 2, and T5 = C = through 4. There curious 5 aretwo ! other 3! phenom4! 6 15210  ena, each worthyC5of=a closer look:   6 5  252 = 2=•42 6 •.L • ( 4n − 2) 6= , our n ≥ recur1. • Why should theC=numbers produced by n 252 = 42 (n.+ 1)! rence be integers? 6  2n  theorem about • How might one prove 2n Euler’s Cn =  4−n − 2 2 • 6 • 10 • L • ( 4n − 6) triangulations?C n=2n  n−•21n  Cn =n  n+−1 n!  n   n − 1 4 n − 2 ANOTHER RECURSIVE FORMULA •C = n −1 In 1759, Johann Andreas developed n +von 1 Segner another recursive formula for Cn :

2 • 6 • 10 • L • ( 4n − 10) 2 •T6n •=10 • L • ( 4n − 10) , n ≥ 3. (1) C0 = 1 Tn = (n − 1)! , n ≥ 3. (3) Cn = C0Cn – 1 + C1Cn – 2 + L + Cn – 2C1 + Cn – 1C0 (n − 1)! 4n − 2 For example, = (C0, C1, C . .n .=, Cn – 1) • •(C Cnn −–11,, Cnn ≥– 21, . . . , C0) n+1 2 2•6 2 • 6 • 10 2 T = =212,••T 6• 10 2 •T6 •=10 = • L •=( 42n , and = 5. T3 = =3 1,TT 2,3!and T5 =− 10) ,5 n ≥=345.!. where the dot indicates the dot product of the two 24!== 64 = n 2! 3! 4 ! 14 (n − 1)! vectors. •C C4 = 2 • 6 • 10 • L • ( 4n − 10) 3 T = , n ≥ 3 . 5 After a nfew2more calculations, it looks as if these For example, • 6 •L • ( 4n − 2) ( n − 1 )! • ( 4n − 2) 2 •C6n •=L , n shifted n ≥ 1. by two • • 14 10 areCthe Catalan • .6 2(n numbers 2 6 10 = same n ≥21C +11, ,)! • •C = n T = = T = = 2 , and T = = 5 . ( n + 1 )! 3 4 5 spaces to the right.2That is, it seems as if Cn = Tk + 2.4 ! C5 = (C0, C1, C2, C3,5C4) 4• (C4,2 C3, C2, C1, C0) ! 3! 2 2•6 2 • 6 • 10 • (14, 5, 2, 1, 1) If this were = (1, 1, 2, 5, 14)14 10 6 T3 true, = 4n=−we 12, Twould = have = 2, and T = = 5. • •C = 42 • 6 • 10 • L • ( 4 n −56) 1+ 14(1) = 1(14) + 1(5) +52(2) +• 5(1) 4! • !( 4 n − 6) • • L3 4nC−n = 2 22! • 62 •• 10 4 3 • • 6 L ( 4 n − 2 ) •n + 1 Cn = n ! (2) = 42. , n ≥ 1. 14 10 6 2 n + 1 Cn = n! • • • •C = 4n• 6−•2L • ((n4n+ −1)! 2 2 ) • 4nC− = 2 Cn − 1 5 4 3 2 0 = , n ≥ 1 . • = C Using formula (2), Formulas (2) and (3) can be used to obtain an n n n+− 11(the n + value 1)! of Cn could then be = 14 n+1 4n − 2 2 • 6 • 10 • L • ( 4n − 6) rewritten as follows: explicit formula for Cn (Guy 1990): • Cn = n+1 n! C =1 C0 = 1 0 4n − 2 2 • 6 • 10 • L • ( 4n − 6) 1  2n Cn = 4n −42n •− 2 Cn = (4) • n ! n − 1 4nC−n = 2 n=+ 1n +• C1n − 1C , n≥1 n + 1  n  •n Cn = Cnn+−− 112, n ≥ 1 4 n + 1= •C n −1 n+1 For instance, C14 = 1 1  10 0 C5 =   •C 14C4 = 3 6 5  −2 C4 =| Vol.C•100, C No. 5 34n 186 Mathematics Teacher • October • C 2006 , n≥1 5 0 =3C1n = n −1 n+1 14 10 252 14 10 = 4n •− 2 • C = = 42. = C•n = •5C 4 • Cn2− 1 , n ≥ 1 6

C0 = 1 1  2n Cn =   4n n− +2 1  n  •C Cn = , n≥1 n −1 n+1 1  10 C514 =   6• C 5  C4 = 5 2523 = 10 = 42. 14 • 6 •C = 2 5 4 14 10 6 CATALAN MEETS= PASCAL • 2n• • C21n  Cn a5=fascinating Formula (4) contains Every  n4  −3  n −treasure:  14 10 6 2 1by Catalan number Cn =can be • obtained • • • C dividing the 5 4in Pascal’s 3 2 0triangle by central binomial coefficient 14 n + 1. For example, =consider the Pascal’s triangle in figure 5, where the various central binomial coefficients are boxed. To 2n C4, read the central 1 find Cn =70 in row binomial coefficient 8 (that is, 2n); divide  n + 1  n  it by 5 (that is, n + 1). This yields C4 = 14. Catalan numbers can be obtained from Pascal’s  10 triangle in three other1ways. C5 =   6 5  1. The difference of a central binomial coefficient 252 = in the = 42 . and its adjacent entry same row is a Catalan 6 number:

(5)

 2n  2n  Cn =   −   n   n − 1

Fig. 5 Pascal’s triangle

Fig. 6 Subtract the boxed numbers from the central cofficient.

For example,  6  6 C3 =   −    3  2

 6  6 C3 =   −    3  2

= 20 − 15 = 5.

See figure 6.

 6  6 C3 =   6−2n6  2n + 1  −  CC3ntwice =3= 2 a−2central  2. The difference of coef 3 n 2  nbinomial = 20 − 15 ficient, and the binomial coefficient that lies along = 20 − 15 5. in the row below, is a Catathe same diagonal =and = 5.  6  7 lan number: C3 = 2   −    3  3  2n  2n + 1 6 2 Cn C= 2== 2 (6) 6− 235 n + 1 • n−20  n  Cn3 = 2n3 −− = 5. n  2 n  For example, = 20 − 15  6  7 C3 = 2= 5.26−n 72n  CCn + 1==3 2 −3−  3  3n 3n − 2 = 2 • 20  2n− 35 2n + 1 C=n5=. 2 • 20 − 35  n   n  = 5. 8  8 C5 =   −    4  2 See figure 7.  2n  2n  6  72n  Cn + 1 C= ==2 70 2n−−−28 n −32  Cn +31=n  3− = 42 . 3. Every Catalan number,  n except   n −C2, is the differ-

= 20 − 15 = 5. Fig. 7 Double a central coefficient and subtract its northwest neighbor.  2n  2n + 1 Cn = 2   −   n   n   6  7 C3 = 2   −    3  3 = 2 • 20 − 35 = 5. n  2nof near neighbors Fig. 8 Catalan numbers asa2difference

For example,

 8  8 C5 =   −    4  2

0

= 2 • 20 coefficient − 35 ence of the central binomial and the     8 8 binomial coefficient two spaces away in the same = 5 .  C5 =  28−n  8 row: C5=4 −2   4n  2 = 70 −228 n  2n  = 70 − 28 (7) Cn=+ 142 . n  −  n − 2 = 42.1  2n Cn = n + 1  n   2n    8 8 C5 n= 2 n −    n41  2 if n = 0 

Cn + 1 =   −   n   n − 2

= 70 − 28 = 42.

See figure 8.  n The study of Catalan2numbers raises some inter n  esting questions for discussion. Vol. 100, No.  3•October 2006 | Mathematics Teacher 187

Cn = 1  2n

1 2n n + 1  n 

 n   n − 2  2n2n 2n2n + 1 Cn +C = 2  −−  1 n =  2n   n8n  n −8 C5 =   −    4  2  1. A quick look at figure8  7that shows  every central =570 6−828 C5C= = 2 −  − integer. binomial coefficient3 is 4 an even Is it always   2  = 42 3.  3 true? In other words, is = =702−• 28 20 − 35 = =425..2n  n 

In another direction, we have seen several articles in this department (e.g., Kobayashi 2006) that explain why the sum of the entries in the nth row of Pascal’s triangle is 2n. What about the sum of the squares of the entries in the nth row?

  BIBLIOGRAPHY  2n2n  2n  C = 6 − 6 Cn + 1 =   −  3   always Askey, Richard. “Fibonacci and Lucas Numbers.”  2 integer?  3an even n    6  n 61 2nn− 2 C3that = Cn =−     2. We have found Mathematics Teacher 98 (May 2005): 610–15. = 20 − 15 36 n2+61  n  Feller, W. An Introduction to Probability Theory and C3 =  1−  2n = 5. = 20 − 15 8 Its Applications. Vol. 1, 3rd ed. New York: Wiley, CnC==3 8 2 −   n +4−1115 n 2  1968. ==55.20 if n = 0 1  2n   2n  2n + 1 Gardner, M. “Mathematical Games.” Scientific AmeriCn= =70 − and C = , 4 n – 2 n 28 C = 5. Cn = 2   −  if n > 0 n +234 1  (June n  1976): 120–25. n −1 4.nIs+it1 always is an integer true? In can  n  for n0 ≤ n=2≤42 n1 .  2n + 1if n = 0 1  2n other words, is of to the Editor. Mathematics MagaCnnC=+2=1always  –− 2 a factor and Cn =Guy, R. K.Letters ,  4 n  n n n + 1if  C2nn  2 n n > 0 n zine + 1 61 n (Oct. 1988): 269. − 1  =n2 n+41− n − 2  1 2n − 2  6  71Cn 2 ———. “The Second Law of Small Numbers.” MathC3 = 2   −    =2nn  ?  n  .  3 n3+ 1  n 6n  n7+ 1  n  n − 1  ematics Magazine 63 (Feb. 1990): 3–20. C23n= 2  4 n−− 2 1  2n − 2 Jarvis, F. “Catalan Numbers.” Mathematical Spectrum = 2 • 201− 35 3 3    nfour  =explicit  . C ,  6n +−1 7formulas 3. We=have found for 36 (2003–2004): 9–12. n + 1 n n − 1  C = 2 n     2 n 2 n − 2 4 n − 2 5. 3 2 • 20  Are .there 2n(5). =31 −335 through namely, formulas=C(2) Kobayashi, Yukio. “Relations among Powers of 2,    = = n5n . n + 1n n n − 1  other such formulas? Combinations, and Symbolic Algebra.” Mathemat= 2 • 20 − 35  2 n 2 − 2  2n − (3)  can  2establish 2n one n  = 4nformulas 4. How through (6)? ics Teacher 99 (April 2006): 577–78. . Cn + 1 =   −   = 5.  n   n − 1 to   in  n  other  n −n2 5. Are there ways, formulas Koshy, T. Discrete Mathematics with Applications. Bur2n  1addition 2n  if n = 0  Elsevier/Academic Press, 2004. 1  2nMA: =  Catalan − (5) through C (7), numbers can be and C = lington, n + 1 that  Cnn = 4n– n 2−2n2 n  n  , 2 n C if n > 0 n + 1 extracted ———. Fibonacci and Lucas Numbers with Applica=  triangle?  8 from  8CnPascal’s n−+ n 1 − 2n −1 +1  n (4)from formula (2)? = we obtain −   formula 6. C Can tions. New York: Wiley, 2002. 5  4  2  8  8 Larcombe, P. “The 18th Century Chinese Discovery of C5 2=n  −4n − 2  1  2n − 2 = 70notes: − 28 1 Thomas Editors’ the Catalan Numbers.” Mathematical Spectrum 32 4=8Koshy  2 gives us much . food  nthis  −n+81provides  n  n − glimpses = article n + 1C 1  for thought. (1999–2000): 5–6. ∞ = 42. Indeed, 5= 70   − 28  4  2 of many surprising connections, but it gives us very ==42 . 28 70 few proofs. We would be −interested in submissions  2n 4n − 2  2n − 2  2n = 42 . from readers that provide proofs for some =     . of the  n  n  like n − 1questions  nin  2 nthe  paper, questions that arise 1–6  n  above. Here are some others. 2n There 1areidentities 2n  nand  results embedded in the Cn =that make article no mention of Catalan numbers.   n + 1 n  2n of these properties, 1 both For example, if CCn satisfies = n n +11 n2n C = 1 n if n = 0 n + 1  n  1  2n  Cn =  4n – 2 and Cn = ,  1 if n = 0 n + 1  n   n + 1 Cn − 1 if 1  2n  n>0 Cn =  4n – 2 and Cn = ,  n +11 n2n n1+ 1 Cn − 1 ififnn>=00 Cn =that and Cn = ,  –2  2n it would  2n − 42n  4n − 2be  Cn − 1 if n > 0  1true n + 1  n  1 then = . n + 1  n + 1  n   n 1+ 1 2nn n −41n− 2  1  2n − 2 = . n +11 n2n  n4n+ −1 2n1n2n− −1 2 = .  2n 4n − 2  2 n + 1  nn− 2 . n + 1  n  n − 1  = Multiplying    by n + 1, wewould have n both − 1 4n  n  2nnsides − 2 2n − 2 THOMAS KOSHY, [email protected]  n  = n  n − 1  .  2n 4n − 2  2n − 2 .edu, is a professor of mathematics  n  = n  n − 1  .

One can prove this by algebraic calculation, but is there a combinatorial proof? Speaking of combinatorial proofs, can readers find such proofs that formulas (5)–(7) all produce the same numbers? 188 Mathematics Teacher | Vol. 100, No. 3 • October 2006

at Framingham State College, Framingham, MA 01701. An author of six mathematics books, he loves to share his passion for mathematics and its beauty and power with others.