Download Toolkit for the Math Olympiad 1. N T N

WORKSHOP ON PROBLEM SOLVING IN MATHEMATICS
(CLASSES 8 & 9)
OFFERED JOINTLY BY
RISHI VALLEY SCHOOL, A.P., & INDIAN INSTITUTE OF SCIENCE, BANGALORE
8–10 AUGUST, 2008
Toolkit for the Math Olympiad
1. N T
NT-1.
Notation: a | b means: “a is a divisor of b”. We read it as: “a divides b”.
Example: 4 | 12, but 4 ∤ 13.
NT-2.
Notation: a ≡ b (mod c) means: “a − b is divisible by c”. We read it as: “a is
congruent to b modulo c”.
Example: 19 ≡ 4 (mod 5).
NT-3.
The congruence relation modulo n for a fixed non-zero integer n is reflexive,
symmetric, and transitive. Thus: if a ≡ b (mod n), and b ≡ c (mod n), then a ≡ c
(mod n).
NT-4.
Notation: ⌊x⌋ (“Floor x”) means: “the largest integer not greater than x”.
Example: ⌊3.2⌋ = 3, ⌊0.75⌋ = 0, ⌊−5.7⌋ = −6.
NT-5.
Notation: (a, b) means: “the greatest common divisor of a and b”.
Example: (10, 15) = 5, (8, 9) = 1.
NT-6.
We say that a, b are coprime if (a, b) = 1.
Example: 15 and 22 are coprime. Two consecutive integers are coprime.
NT-7.
Notation for various sets:
• N = {1, 2, 3, 4, . . .} is the set of all positive integers.
• N0 = {0, 1, 2, 3, 4, . . .} is the set of all non-negative integers.
• P = {2, 3, 5, 7, 11, 13, 17, . . .} is the set of prime numbers. Note that 1 < P. We
call 1 a “unit” — it is neither prime nor composite.
• Z = {. . . , −4, −3, −2, −1, 0, 1, 2, 3, 4, . . .} is the set of all integers.
• Zn = {0, 1, 2, 3, . . . , n − 1}.
• Q is the set of all rational numbers; 23 ∈ Q, − 73 ∈ Q, etc.
• R is the set of all real numbers.
• C is the set of all complex numbers.
NT-8.
Some useful and far reaching results: For any n ∈ Z, we have:
• Either n2 ≡ 0 (mod 3) or n2 ≡ 1 (mod 3). That is, all squares are of the form
3k or 3k + 1; a square cannot be of the form 3k + 2.
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RVS & IISC
• Either n2 ≡ 0 (mod 4) or n2 ≡ 1 (mod 4). That is, all squares are of the form
4k or 4k + 1; a square cannot be of the form 4k + 2 or 4k + 3.
• If p is a prime number, and p | ab, then p | a or p | b (or both).
This claim is not true for composite moduli. That is, if n is composite, and
n | ab, we cannot conclude that n | a or n | b or both.
• If a, b are coprime positive integers, and ab is a square, then both a and b are
squares.
• If a, b are coprime integers, and ab is a cube, then both a and b are cubes.
• Suppose that a, b, c, d are positive integers, and ab = cd. Further, suppose that
a, b are coprime, and c, d are coprime. Then either a = c and b = d, or a = d
and b = c. In any case, {a, b} = {c, d}.
NT-9.
NT-10.
Multiplicative inverse. If p is a prime number, and a is coprime to p, then an integer
b can be found such that ab ≡ 1 (mod p). We call b the multiplicative inverse of p.
Example: Let p = 11. The multiplicative inverses of 2, 3, 4, 5 are 6, 4, 3, 9, respectively.
Little Fermat Theorem. If p is a prime number, and a is coprime to p, then
ap−1 ≡ 1 (mod p).
Example: 26 ≡ 1 (mod 7), and 34 ≡ 1 (mod 5).
NT-11.
Another form of the Little Fermat Theorem. If p is a prime number, and a is any integer,
then ap ≡ a (mod p).
NT-12.
Wilson’s Theorem. If p is a prime number, then
(p − 1)! + 1 ≡ 0 (mod p).
Example: 6! + 1 = 721 ≡ 0 (mod 7).
NT-13.
Notation: For any positive integer n,
ϕ(n) = # positive integers between 1 and n which are coprime to n.
This is the Euler phi function.
Example: ϕ(1) = 1, ϕ(2) = 1, ϕ(3) = 2, ϕ(10) = 4. Note that:
n is a prime number ⇐⇒ ϕ(n) = n − 1,
n is a power of 2 ⇐⇒ ϕ(n) = 12 n.
NT-14.
The Euler phi function is multiplicative. This means that if m, n are coprime, then
ϕ(mn) = ϕ(m) · ϕ(n).
Example: ϕ(12) = ϕ(3) · ϕ(4).
NT-15.
Here is a quick way of computing ϕ(n): List the distinct primes p which divide n, then
p−1
multiply n by the product of p for all such p.
Example: Take n = 20. The distinct primes dividing 20 are 2 and 5, so ϕ(20) =
20 × 12 × 45 = 8.
Example: Take n = 350. The distinct primes dividing 350 are 2, 5, 7, so
ϕ(350) = 350 × 12 × 45 × 67 = 120.
PROBLEM SOLVING WORKSHOP
NT-16.
3
Euler’s generalization of the Little Fermat Theorem. If n is any positive integer, and a
is any positive integer coprime to n, then
aϕ(n) ≡ 1 (mod n).
Example: 34 ≡ 1 (mod 10), 1510 ≡ 1 (mod 22).
NT-17.
Infinitude of primes.
• There are infinitely many prime numbers.
• There are infinitely many prime numbers of each of the types 1 (mod 4) and
3 (mod 4):
1 (mod 4) : {5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, . . .},
3 (mod 4) : {3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, . . .}.
• There are infinitely many prime numbers of each of the types 1 (mod 3) and
2 (mod 3):
1 (mod 3) : {7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, . . .},
2 (mod 3) : {2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, 107, . . .}.
NT-18.
If n ≡ 3 (mod 4), then n has at least one prime factor of the form 3 (mod 4).
NT-19.
If p is a prime number of the type 3 (mod 4), then it cannot be expressed as x2 + y2 ,
where x, y are integers.
NT-20.
If p is a prime number of the type 1 (mod 4), then it can be expressed as x2 + y2 ,
where x, y are integers. Moreover, this representation is unique.
Example: 13 = 22 + 32 , 89 = 52 + 82 .
NT-21.
If a positive integer n can be expressed as x2 + y2 where x, y are integers, then:
• n has at least one prime factor p of the form 1 (mod 4).
• If a positive integer n can be expressed as x2 + y2 where x, y are integers, then
the number of primes p which divide n and which are of the form 3 (mod 4)
is even.
Example: Take n = 2205. It can be expressed as 212 + 422 , and its prime factorization is 2205 = 32 × 5 × 72 . Note that it has a prime factor of the type 1 (mod 4), and
the number of primes p which divide n and which are of the form 3 (mod 4) is 4
(two 3’s and two 5’s).
NT-22.
Pythagorean triples. The equation x2 + y2 = z2 has infinitely many “primitive
solutions” (i.e., with x, y, z coprime). They may be found as follows: Choose any
two positive integers u, v of opposite parity, with u > v. Put
x = u2 − v2 ,
y = 2uv,
z = u2 + v2 .
(We can switch the roles of x and y: put x = 2uv, y = u2 − v2 .) This generates the
entire set of primitive solutions.
Example: Put u = 5, v = 2; we get (x, y, z) = (21, 20, 29).
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2. A
Alg-1.
Difference of two squares. This is of use more often than one would expect:
a2 − b2 = (a − b) · (a + b).
Alg-2.
Two simple and useful factorizations.
xy + x + y + 1 = (x + 1)(y + 1),
Alg-3.
xy − x − y + 1 = (x − 1)(y − 1).
Sophie Germain identity.
a4 + 4b4 = a2 + 2b2 + 2ab · a2 + 2b2 − 2ab .
Alg-4.
Alg-5.
Let a, b, c be real numbers, a , 0. Then the roots of the quadratic equation ax2 +
bx + c = 0 are real if and only if b2 − 4ac ≥ 0.
If the roots of the quadratic equation ax2 + bx + c = 0 (a , 0) are α, β then
b
c
α+β=− ,
αβ = .
a
a
Alg-6. If u, v are given numbers, then the quadratic equation whose roots are u, v is
(x − u)(x − v) = 0.
Alg-7.
If the roots of the cubic equation ax3 + bx2 + cx + d = 0 (a , 0) are α, β, γ then
b
c
d
α+β+γ=− ,
αβ + βγ + γα = ,
αβγ = − .
a
a
a
Alg-8. Remainder theorem. If f (x) is a polynomial in x, and c is any real number, then the
remainder in the division f (x) ÷ (x − c) is f (c).
Alg-9.
The sum of a n-term arithmetic progression a, a + d, a + 2d, . . . , a + (n − 1)d is
n−1
X
(first term + last term)
n(2a + (n − 1)d)
=n×
.
(a + kd) =
2
2
k=0
Example: 1 + 2 + 3 + · · · + n = 21 n(n + 1), 1 + 3 + 5 + · · · + (2n − 1) = n2 .
Alg-10.
The sum of a n-term geometric progression a, ar, ar2 , . . . , arn−1 (r , 1) is
n
r −1
.
a
r−1
Example: 1 + 2 + 22 + · · · + 2n−1 = 2n − 1, 1 + 3 + 32 + · · · + 3n−1 = 12 (3n − 1).
Alg-11.
We have:
• 1 + 2 + 3 + · · · + n = 21 n(n + 1).
• 12 + 22 + 32 + · · · + n2 = 16 n(n + 1)(2n + 1).
• 13 + 23 + 33 + · · · + n3 = 41 n2 (n + 1)2 .
Alg-12.
The harmonic series 1 + 21 + 13 + 14 + · · · does not converge to a finite number.
There is no simple formula for the sum 1 + 21 + · · · + n1 . Rather:
1
1
+ · · · + ≈ ln n + γ,
2
n
where ln n is the “natural logarithm” of n, and γ ≈ 0.577216 is the “EulerMascheroni constant”.
1+
PROBLEM SOLVING WORKSHOP
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3. G
Triangle inequality. Any two sides of a non-degenerate triangle together exceed
the third side.
Geom-1.
Pythagoras’s theorem and its converse. Given any △ABC, with sides a, b, c and angles
∠A, ∠B, ∠C, we have:
Geom-2.
a2 + b2 < c2 ⇐⇒ ∠C is acute,
a2 + b2 = c2 ⇐⇒ ∠C is a right angle,
a2 + b2 > c2 ⇐⇒ ∠C is obtuse.
SAS inequality. In △ABC, let the lengths of sides AB, AC be fixed, but let ∠A vary.
Then the length a of the third side BC is an increasing function of ∠A. That is, the
larger the angle A, the larger the side a, and conversely.
Geom-3.
Geom-4.
Angle bisector theorem and its converse.
• In △ABC, let the internal bisector of ∠A meet the opposite side BC at D. Then
BD : DC = AB : AC.
• If D is a point on side BC of △ABC such that BD : DC = AB : AC, then AD
bisects ∠A.
Geom-5.
Ceva’s theorem and its converse.
• In △ABC, let P be any point, and let the lines AP, BP, CP meet the sidelines
BC, CA, AB in the points D, E, F, respectively. Then the following equality
holds:
BD CE AF
·
·
= 1.
DC EA FB
The lengths here are signed lengths. Thus, if D lies on the extension of BC
BD
is
(rather than on side BC), then BD and DC have opposite sign, so the ratio DC
negative.
• Let D, E, F be points on the sidelines BC, CA, AB of △ABC. Suppose that the
following equality holds:
BD CE AF
·
·
= 1.
DC EA FB
Then the lines AD, BE, CF meet in a point.
Geom-6.
The following concurrences are true for any triangle.
• The perpendicular bisectors of the sides of a triangle concur (at the circumcenter).
• The internal angle bisectors of a triangle concur (at the incenter).
• The medians of a triangle concur (at the centroid).
• The altitudes of a triangle concur (at the orthocenter).
Geom-7.
Menelaus’s theorem and its converse.
• Let a straight line ℓ cut the sidelines BC, CA, AB of △ABC in the points D, E, F,
respectively. Then the following equality holds:
BD CE AF
·
·
= −1.
DC EA FB
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(As earlier, the lengths are signed lengths.)
• Let D, E, F be points on the sidelines BC, CA, AB of △ABC. Suppose that the
following equality holds:
BD CE AF
·
·
= −1.
DC EA FB
Then the points D, E, F lie in a straight line.
Geom-8.
Apollonius’s theorem. Let D be the midpoint of side BC of △ABC. Then:
AB2 + AC2 = 2 AD2 + BD2 .
Stewart’s theorem. Let D be an point on side BC of △ABC, and let BD = x, DC = y,
AD = p. Then:
a p2 + xy = b2 x + c2 y.
Geom-9.
If D is the midpoint of BC, this reduces to Apollonius’s theorem.
Cyclic quadrilateral. A quadrilateral ABCD is cyclic (i.e., it may be inscribed in a
circle) if and only if ∠A + ∠C = 180◦ = ∠B + ∠D.
Geom-10.
Ptolemy’s theorem and its converse. A quadrilateral ABCD is cyclic if and only if
the following equality holds:
Geom-11.
AB · CD + AD · BC = AC · BD.
Geom-12.
Sine rule. Let the radius of the circumcircle of △ABC be R. Then:
a
b
c
=
=
= 2R.
sin A sin B sin C
Geom-13.
Cosine rule. We have:
c2 = a2 + b2 − 2ab cos C,
Geom-14.
c = a cos B + b cos A.
Area of a triangle. There are several formulas for the area of a given △ABC:
• Area (△ABC) = 12 bc sin A = 21 ca sin B = 21 ab sin C;
p
• Area (△ABC) = s(s − a)(s − b)(s − c), where s is the semi-perimeter of the
triangle;
• Area (△ABC) = rs, where r is the radius of the incircle of the triangle;
abc
.
4R
Geom-15. Area of a cyclic quadrilateral (Brahmagupta’s formula). We have:
p
Area (quadrilateral ABCD) = (s − a)(s − b)(s − c)(s − d),
• Area (△ABC) =
where s = 12 (a + b + c + d) is the semi-perimeter of ABCD.
Area of a bicentric quadrilateral. A bicentric quadrilateral is one which has both a
circumcircle and an incircle. If ABCD is such a quadrilateral, then:
√
Area (quadrilateral ABCD) = abcd.
Geom-16.
Gergonne point. Let the incircle of △ABC touch the sides BC, CA, AB at points
P, Q, R, respectively. Then AP, BQ, CR meet in a point K called the Gergonne point.
Geom-17.
PROBLEM SOLVING WORKSHOP
7
Nagell point. Let the excircles of △ABC opposite vertices A, B, C touch the sides
BC, CA, AB at points U, V, W, respectively. Then AU, BV, CW meet in a point J called
the Nagell point.
Geom-18.
Symmedian point. If the median of △ABC through vertex A is reflected in the
bisector of ∠A, the resulting line is called the symmedian through A. There are three
such lines, one through each vertex of the triangle, and they meet in a point called
the symmedian point.
Geom-19.
Power of a point. Let C(O, r) be a circle (the notation means that its center is O,
and its radius is r), and let P be a point. Consider any line ℓ through P. Suppose
it cuts C(O, r) at A, B. Then the product PA · PB does not depend on ℓ, and so is
the same no matter which line is drawn (so it depends only on P, O, r). In fact:
PA · PB = OP2 − r2 .
As in Ceva’s and Menelaus’s theorems, the lengths here are signed lengths.
Hence, if PA and PB are oriented in opposite directions (which will be the case if
P lies within the circle), then PA · PB < 0.
The quantity OP2 − r2 is called the power of P with respect to C.
• If P lies on C, then its power w.r.t C is 0.
• If P lies outside C, then its power w.r.t C is the square of the length of the
tangent from P to C.
• If P lies within C, then its power w.r.t C is negative.
Geom-20.
Radical axis of two circles. Let C1 and C2 be two given circles. Consider the locus
of points P for which the power of P w.r.t. C1 is the same as the power of P w.r.t.
C2 . This locus is a straight line; it is called the radical axis of C1 , C2 .
• If C1 , C2 intersect at points A, B, then the radical axis is line AB.
• If C1 , C2 touch each other at a point P, then the radical axis is the line tangent
to both circles at P.
• If C1 , C2 are concentric, then the locus is empty.
Geom-21.
Circumscribable quadrilateral. A convex quadrilateral ABCD possesses an incircle
C(I, r) if and only if AB + CD = AD + BC.
area of ABCD
.
If this condition is satisfied, then r =
semi-perimeter of ABCD
Geom-22.
Brocard point. Given any triangle ABC, points W, W ′ may be found within it such
that ∠WAB = ∠WBC = ∠WCA, and ∠W ′ BA = ∠W ′ CB = ∠W ′ AC. These are the
Brocard points of △ABC. Let ∠WAB = θ, ∠W ′ BA = θ′ . Then:
• θ = θ′
Geom-23.
• cot θ = cot A + cot B + cot C
• csc2 θ = csc2 A + csc2 B + csc2 C
• sin3 θ = sin(A − θ) · sin(B − θ) · sin(C − θ)
4. I
Ineq-1.
Trivial inequality. If x is any real number, we have: x2 ≥ 0.
This seems “trivial” but is the basis for every other inequality!
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Ineq-2.
Arithmetic Mean-Geometric Mean Inequality. Let a1 , a2 , . . . , an be n non-negative
numbers. The AM-GM inequality states that
√
a1 + a2 + · · · + an
≥ n a1 a2 · · · an ,
n
Q 1/n
1 P
i.e., n ai ≥ ( ai ) , with equality if and only if a1 = a2 = · · · = an .
Example: For the numbers 9, 12, 54, the AM is 25, and the GM is 18.
Ineq-3.
Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality. Let
a, b be positive numbers; the RMS-AM-GM-HM inequality states that:
r
√
2ab
a2 + b2 a + b
≥
≥ ab ≥
,
2
2
a+b
with equality precisely when a = b.
More generally, let a1 , a2 , . . . , an be n positive numbers; then
s
a21 + · · · + a2n a1 + · · · + an
√
n
,
≥
≥ n a1 · · · an ≥ 1
n
n
+ ··· + 1
a1
an
with equality if and only if a1 = a2 = · · · = an .
Ineq-4.
Cauchy-Schwartz inequality. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be any 2n real numbers; then
n
n
n
X
X
2 X
a2i ·
ai bi ≤
b2i ,
i=1
i=1
i=1
with equality precisely when there exist constants µ, λ, not both zero, such that
µai = λbi for all i.
Ineq-5.
Triangle inequality. In any non-degenerate triangle, the sum of any two sides
exceeds the third side. (“Non-degenerate” means that the triangle has positive
area. If the triangle collapses into a line, i.e, if its area vanishes, then the “triangle
inequality” becomes an equality.)
One form of the triangle inequality is: if a, b are real numbers, then
|a + b| ≤ |a| + |b|,
|a − b| ≥ |a| − |b|.
5. C
Comb-1.
Notation.
• |A| is the cardinality of the set A
• Factorials: n! = 1 × 2 × · · · × n; by definition, 0! = 1.
n!
; so n0 = 1 = nn .
• Combinatorial coefficient: nk = k!(n−k)!
By definition, nk = 0 if k is not an integer with 0 ≤ k ≤ n.
One-to-one correspondence. If the elements of two finite sets A, B can be placed into
one-to-one correspondence, then |A| = |B|.
Comb-2.
Comb-3.
Permutations. n distinct objects can be arranged in a line in n! distinct ways.
Combinations. From a set containing n elements, a subset with k elements can be
chosen in nk distinct ways.
Comb-4.
PROBLEM SOLVING WORKSHOP
9
Comb-5.
Two laws of enumeration.
• Law of addition. If A, B are two sets, then |A ∪ B| = |A| + |B| − |A ∩ B|.
• Law of multiplication. If A, B are two sets, then |A × B| = |A| · |B|. Here, A × B is
the Cartesian product of the sets A, B.
Comb-6.
Fundamental identity of the combinatorial coefficients. This is the identity
!
!
!
n
n−1
n−1
=
+
,
k
k
k−1
for n, k ∈ N, 1 ≤ k ≤ n.
Taxicab paths. If we have to walk on the coordinate plane from the initial point
O(0, 0) to the terminal point P(m, n) where m, n ∈ N0 , so that our path consists of
steps one unit “North” or one unit “East”, then the number of possible paths is
m+n
n .
Comb-7.
Principle of inclusion-exclusion (PIE). This is a far reaching generalization of the
law of addition. Let A1 , A2 , A3 , . . . , An be n sets. For n = 2 we have:
Comb-8.
|A1 ∪ A2 | = |A1 | + |A2 | − |A1 ∩ A2 |.
For n = 3:
|A1 ∪ A2 ∪ A3 | = |A1 | + |A2 | + |A3 | − |A1 ∩ A2 | − |A2 ∩ A3 | − |A3 ∩ A1 | + |A1 ∩ A2 ∩ A3 |.
In general:
n
n
X
[
X
X Ai =
Ai ∩ A j ∩ Ak − · · · + (−1)n |A1 ∩ · · · ∩ An |
Ai ∩ A j +
|Ai | −
i=1 i=1
i<j
i<j<k
Pigeon hole principle. If n + 1 objects are distributed over n sets, then at least one
of the sets contains more than one of the given objects. (“If n + 1 pigeons stay
in n pigeon holes, then at least one of the pigeon holes contains more than one
pigeon.”)
Comb-9.
Recursion. Sometimes a sequence is defined recursively. This means that we
compute each element in terms of the elements preceding it, using some fixed
rule. This applies to all elements except for a few initial terms which are fixed
independently.
• Powers of 2. Let an = 2n for n ∈ N. Then: a1 = 2, an = 2an−1 for n > 1.
• Squares. Let an = n2 for n ∈ N. Then: a1 = 1, an = an−1 + 2n − 1 for n > 1.
• Factorials. Let an = n! for n ∈ N. Then: a1 = 1, an = nan−1 for n > 1.
Comb-10.
Compositions. For n ∈ N, let an be the number of ways of writing n as a sum of one
or more positive integers, with order being taken into account (so, 1 + 2 is counted
separately from 2 + 1). These expressions are called the compositions of n. So the
compositions of 2 are 2, 1 + 1, and the compositions of 3 are 3, 2 + 1, 1 + 2, 1 + 1 + 1.
We may show that: an = 2n−1 .
Comb-11.
Comb-12.
Fibonacci numbers. The Fibonacci numbers Fn for n ∈ N0 are defined thus:
F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 for n ∈ N, n ≥ 1.
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RVS & IISC
Here are the first few Fibonacci numbers:
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 · · ·
Fn 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 · · ·
These numbers are ubiquitous. For example:
• The number of compositions of n in which all the summands exceed 1 is a
Fibonacci number; in fact it is Fn−1 .
Example: For n = 6 we get the compositions 6, 4 + 2, 3 + 3, 2 + 4, 2 + 2 + 2.
• The number of compositions of n in which the summands are only 1’s and 2’s
is a Fibonacci number; in fact it is Fn+1 .
Example: For n = 4 we get the compositions 1 + 1 + 1 + 1, 2 + 1 + 1, 1 + 2 +
1, 1 + 1 + 2, 2 + 2.
• The number of subsets of the n-element set {1, 2, . . . , n} in which no two
consecutive numbers occur is a Fibonacci number; in fact it is Fn+2 .
Example: For n = 3 we get the 5 subsets
{} , {1} , {2} , {3} , {1, 3} .
For n = 4 we get the 8 subsets
{} , {1} , {2} , {3} , {4} , {1, 3} , {1, 4} , {2, 4} .
Catalan numbers. If we have to walk on the coordinate plane from the initial
point O(0, 0) to the terminal point P(n, n) where n ∈ N, so that our path consists
of steps one unit “North” or one unit “East” and never goes above the line y = x,
then the number of possible paths is defined to be the nth Catalan number, Cn . We
may show that
!
2n
1
·
.
Cn =
n+1 n+1
They may be recursively defined: C0 = 1, and
n
X
Cn+1 = C0 Cn + C1 Cn−1 + C2 Cn−2 + . . . =
Ck Cn−k .
Comb-13.
k=0
Here are the first few Catalan numbers:
6
7
8
9
10
11
···
n 1 2 3 4 5
Cn 1 2 5 14 42 132 429 1430 4862 16796 58786 · · ·
Like the Fibonacci numbers, the Catalan numbers too are ubiquitous:
• The number of ways a convex n-sided polygon can be triangulated is a Catalan
number; in fact it is Cn−2 .
• The number of correctly matched strings of n pairs of parenthesis is Cn . For
example, n = 3:
((( ))),
( )(( )),
( )( )( ),
(( ))( ),
(( )( )).
The Catalan numbers have many arithmetical properties. For example, Cn is odd
precisely when n is of the form 2k − 1.