Combinatorial Enumeration
Fall 2004
Professor D.M. Jackson
CHRIS ALMOST
Contents
1 Introduction
1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Ordinary generating series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Types of decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
2
3
2 Plane planted trees
3
3 Implicit functions
4
4 Partitions of Integers
9
4.1 Elementary properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.2 Haruspication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.3 Ferrers Graphs/Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5 Exponential Generating Series
5.1 A classical problem . . . . . . . . . . . . . . . . .
5.2 An architect’s problem . . . . . . . . . . . . . . .
5.3 Permutations and cycles . . . . . . . . . . . . . .
5.4 The preimage decomposition . . . . . . . . . . .
5.5 Branch decomposition for labelled rooted trees
5.6 Derivative of Sets . . . . . . . . . . . . . . . . . .
5.7 Another combinatorial operation . . . . . . . .
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6 Sign-Reversing Involutions
11
11
13
14
15
15
16
16
17
7 The Pattern Algebra
22
7.1 A brief return to {0, 1}-strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
7.2 Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
8 Pólya Theory
8.1 Colouring Polyhedra . .
8.2 Painting the 2 × 2 board
8.3 Pólya’s Theorem . . . . .
8.4 Rooted Trees . . . . . . .
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26
26
27
27
29
2
COMBINATORIAL ENUMERATION
1
Introduction
Midterm: Wednesday, October 27, 2004 from 5:00 until 7:00 in MC 2054
1.1
Motivation
Let S be a set of combinatorial objects. Let ω : SP→ N be a weight
function. Find cn = |{s ∈ S | ω(s) = n}|.
P
Equivalently, we want to construct a series F (x) = s∈S x ω(s) = n∈N cn x n . This is called the ordinary generating
series for the problem (S, ω). Then cn = [x n ]F (x). We want to get hold of properties of F . We also wish to
discover more algebraic and combinatorial information about S or even F .
Concerning the 4-colour theorem, we deal with planar diagrams (maps, where a map is an embedding of a
graph in an orientable surface). Are there any planar algebras? e.g. consider the Tempoley-Lie algebra generated
by symbols 1, e1 , . . . , en−1 and the following rules.
1. ei2 = ei for all i
2. ei e j = e j ei for all |i − j| > 1
3. ei ei±1 ei = ei for all i
See diagrams for interpretation. This gives an algebraic approach to the proof of the 4-colour theorem.
1.2
Ordinary generating series
We have sets A and B of combinatorial objects and if we have a bijection Ω : A −→
˜ B : a 7→ b, then |A | = |B|.
The reason we want a bijection is that B may be easier to count than A . The problem is to find such an Ω, but
such constructions are sporatic and not easy find. Sometimes enumerative arguments point to the existence of
bijections. We would prefer Ω to be a natural bijection.
A weight function α on A is any function α : A → N. A counting problem: count A with respect to α.
Denoted by (A , α).
Computational engine for working with formal power series: A formal power series is an element of C[[x]],
where x is an indeterminant. Note that x −1 6∈ C[[x]]. C[[x]] is a ring with the obvious sum and product given
by



!
X
X
X X

bj x j =
ai b j  x n
ai x i 
i≥0
j≥0
n≥0
i+ j=n
Note that this is well defined as each coefficient is just a finite sum of complex numbers.
Given a formal power series f , f −1 exists if and only if f (0) 6= 0. For example, (1− x)−1 = 1+ x + x 2 + x 3 +· · · .
P
PROOF: P
(Sketch) Let f (x) ∈ C[[x]], f (x) = ai x i . Suppose that g(x) ∈ C[[x]]
isP
such that g(x) f (x) = 1, say
P
g(x) = b j x j . Then by applying the coefficient operator to the equation ( ai x i )( b j x j ) = 1, we get that the
coefficient of x n is 1 if n = 0 and 0 otherwise. In general,
X
ai b j = δ0,n
i, j≥0
i+ j=n
In particular, a0 b0 = 1, so neither of them can be zero. The coefficients of g(x) can be computed from the
coefficients of f (x) by working your way up. The general formula is
X
bn = −a0−1
ai b j
i>0,b≥0
i+ j=n
As one can easily see, f (x) is invertible in C[[x]] if and only if f (0) 6= 0.
ƒ
PLANE
3
PLANTED TREES
Sometimes we wish to work in the ring of formal Laurent series, C((x)), where the series may have finitely
many terms with a negative exponent on x.
The differential operator is defined formally by D : x n 7→ nx n−1 and extended linearly to all of C[[x]]. The
sum rule and the product rule for the derivative work as one would expect.
1.1 Lemma. (Sum Lemma) Let S be a set of combinatorial objects and let A, B ⊆ S with A ∩ B = ∅. Let ω be a
wight function on S. Then
a
[(A
B, ω)]o = [(A, ω)]o + [(B, ω)]o
1.2 Lemma. (Product Lemma) Let A, B be sets of combinatorial objects. Let α be a weight function on A and β
be a wight function on B. Let ω be the weight function on A × B such that ω : (a, b) 7→ α(a) + β(b). Then
[(A × B, ω)]o = [(A, α)]o · [(B, β)]o
1.3
Types of decompositions
1.3 Example. Say we want to find the generating series for S, given that we already know the generating series
for A and B.
1. Ω : S −→
˜ A × B is called a direct decomposition for S
2. Ω : A −→
˜ S × B is called an indirect decomposition for S. Here S is embedded into a bigger (possibly easier)
problem by adjoining B. Under appropriate weight conditions, S = A · B −1
`
3. S −→
˜ {"} S × S is called a recursive decomposition for S. It leads in C[[x]] under appropriate weight
conditions to S = x + S 2 . This is a functional equation. In more complicated cases we need Lagrange’s
Theorem for Implicit formal power series to solve this equation
2
Plane planted trees
A plane planted tree is a tree with a unique monovalent root vertex that is embedded in the plane. Let cn be
the number of plane planted trees on n vertices. Find cn , assuming that trees that are reflections of each other
are different trees. We may also wish to count with respect to the number of non-root vertices. Consider the
collection of all plane planted trees on 5 vertices (4 non-root vertices). Draw them. There are 5 different trees.
We need a decomposition for the set of all plane planted trees. A simple idea for finding elementary decompositions is to delete the vertex adjacent to the root. This fragments the tree into a single root vertex and a
number of trees isomorphic to plane planted trees (i.e. we can identify a monovalent root vertex). This gives us
a bijection
a
Ω : P −→
˜ {"} ×
Pm
m≥0
This is a recursive construction. Let ω(t) be the number of non-root vertices in t ∈ P . Is Ω ω-preserving? Yes,
and additively! Note that this property is very important in the product lemma.
Let P = [(P , ω)]o . Then by the above bijection and the sum and product lemmas, P(x) satisfies the following
identity
X
x
P(x) = x
P(x)m =
1 − P(x)
m≥0
4
COMBINATORIAL ENUMERATION
This uniquely defines P. Can we do anything more with this?
x
P=
1−P
P − P2 = x
P = x + P2
Now solve the inverse problem, to finds a set C and a weight function θ on C such that C = x + C 2 , where
C = [(C , θ )]o . Consider
a
C −→
˜ {"}
C2
with θ (t) the number of non-root monovalent vertices in t.
3
Implicit functions
3.1 Theorem. (Lagrange’s Implicit Function Theorem for C[[x]]) Let ϕ ∈ C[[t]] be invertible. Then
1. The functional equation w = tϕ(w) for w = w(t) ∈ C[[t]] has a unique solution in C[[t]]. This is called
a functional equation of Langrangian type.
2. Moreover, if f (x) ∈ C[[x]], then
f (w) = f (0) +
X xn
n≥1
n
[λn−1 ]( f 0 (λ)ϕ n (λ))
3. If cn := [λn ]g(λ)ϕ n (λ), where g ∈ C[[λ]], then
X
cn t n =
n≥0
g(w)
1 − tϕ 0 (w)
where w = w(t) is the unique series defined in (1)
PROOF: See notes for a combinatorial proof.
Prileminaries. We work in C((x)), the ring of formal Laurent series. We consider the operator [x −1 ]. Notice
that [x n ] = [x −1 ]x −n−1 , so we only need to know how to be able to work out [x −1 ] and we get the rest for
free. For f ∈ C((x)), [x −1 ] f 0 = 0 since for any n ≥ 1 [x −1 ]D x x n = [x −1 ]nx n−1 = 0 and [x −1 ] is linear. For
f , g ∈ C((x)) we have [x −1 ] f 0 g = −[x −1 ] f g 0 by the product rule. We call [x −1 ] the formal residue operator.
For f ∈ C((x)) let
¨
k if f = x k g, where g ∈ C[[x]] and g(0) 6= 0
val( f ) =
0 otherwise
This is called the valuation of f . For any f , g ∈ C((x)) we have val( f g) = val( f ) + val(g), so val behaves like
deg. If f = x k g, where g ∈ C[[x]] and g(0) 6= 0 then f −1 = x −k g −1 ∈ C((x)).
Claim. (Residue Composition Theorem) Let f , r ∈ C((x)) and let va(r) = k > 0. Then
k[x −1 ] f (x) = [z −1 ] f (r(z))r 0 (z)
IMPLICIT
5
FUNCTIONS
We use the monomial principle to prove this claim. Since k > 0 the composition f (r(z)) exists. Suppose that
f (x) = x n for some n ∈ Z. If n 6= −1 then
[z −1 ]r n r 0 =
1
n+1
Dz r n+1 = 0 = k[x −1 ]x n
If n = −1 then suppose r = z k h for some h ∈ C[[x]] with h(0) 6= 0.
[z −1 ]r −1 r 0 = [z −1 ]z −k h−1 (kz k−1 h + z k h0 ) = [z −1 ](kz −1 +
h0
h
) = k + [z −1 ]Dz log h = k = k[x −1 ]x −1
Extending this linearly proves the claim.
Now to prove Langrange’s theorem. Let Φ(w) = wϕ −1 (w). val(Φ) = 1 > 0 since ϕ has valuation 0 (it is
invertible in C[[x]]). Thus the compositional inverse Φ[−1] of Φ exists. Φ(w) = t, so w(t) = Φ[−1] (t) is the
unique solution of the functional equation. Let n ∈ Z, then
[t n ] f (w) = [t −1 ]t −(n+1) f (w) = [t −1 ]t −(n+1) f (Φ[−1] (t)) = [w −1 ]Φ−(n+1) (w) f (w)Φ0 (w)
by the Residue Composition Theorem. If n 6= 0 then
[t n ] f (w) = [w −1 ]
1
−n
(Dw Φ−n (w)) f (w)
1
[w −1 ] f 0 (w)Φ−n (w)
ϕ(w) n
−1
0
= [w ] f (w)
n
w
1
= [w n−1 ] f 0 (w)ϕ n (w)
n
=
by the Complementation Lemma
n
1
so the result holds in this case. If n = 0 then
[t n ] f (w) = f (w)
t=0
= f (w)
w=0
= f (0)
Thus the second part of the theorem is proven.
For the third part, differentiate with respect to t in the functional equation to get
dw
dt
Therefore
f 0 (w)ϕ(w)
1 − tϕ 0 (w)
=
X
= ϕ(w) + tϕ 0 (w)
t n−1 [λn−1 ] f 0 (λ)ϕ n (λ) =
n≥1
dw
dt
X
t n [λn ] f 0 (λ)ϕ n+1 (λ)
n≥0
Let g(λ) = f (λ)ϕ(λ) and the result follows. (Check this.)
0
ƒ
1
x
3.2 Example. P = 1−P
. Here taking w = P, t = x, and ϕ(λ) = 1−λ
, we see that this equation is in Lagrangian
form, and hence the theorem applies. We want to solve for P, so we should take f (λ) = λ. Thus
P(x) =
X xn
n≥1
n
[λn−1 ]
1
(1 − λ)n
6
COMBINATORIAL ENUMERATION
by Lagrange’s theorem. We still need to actually determine the coefficients.
1
n
[λ
n−1
]
1
(1 − λ)n
1
−n
(−1)n−1
n n−1
1 n + (n − 1) − 1
=
=
n
n−1
1 2n − 2
so cn := [x n ]P(x) =
n
n−1
is the number of plane planted trees on n non-root vertices. Thus
P(x) =
X
xn
n≥1
1 2n − 2
n
n−1
3.3 Example. Find the expansion of P 6 (x).
Solution 1: This is a mess, worst in the cosmos.
X
6
P (x) =
x
n
n≥1
!6
1 2n − 2
n
n−1
Solution 2: Here f (λ) = λ6 . Then
P 6 (x) =
X xn
[λn−1 ]
n
X xn
n≥1
=6
6λ5
(1 − λ)n
1
[λn−6 ]
(1 − λ)n
X 6 2n − 7
xn
=
n n−6
n≥1
n≥1
n
where the coeffients are determined by doctoring the calculation above. This is the number of plane planted
trees on n + 1 non-root vertices with vertex of degree 7 adjacent to the root vertex.
3.4 Example. Solve T = x e T in C[[x]]. By Lagrange’s theorem
T (x) =
X xn
n≥1
n
[λn−1 ]e nλ =
X xn
n≥1
nn−1
n (n − 1)!
=
X
n≥1
xn
nn−1
n!
3.5 Example. Let α, x be indeterminants. Express eαx as a power series in x e−x . (As an aside, if f ∈ C[[x]],
the valuation of f is defined to be the degree of the term in f of lowest degree. We denote this by val( f ). If
f , g ∈ C[[x]]
then val( f g) = val( f ) + val(g). This is an analogue of degree of a polynomialP
for power series. So
P
why is m≥0 cm (x e−x )m ∈ C[[x]]? It is because the valuation of (x e−x )m is m, and so [x k ] m≥0 cm (x e−x )m is a
finite sum, for all k.) Let y = x e−x . We want to express eαx as a power series in y. Note that x = y e x , and this is
a functional equation in Lagrangian form. Then by Lagrange’s theorem, this has a unique solution x = x( y). We
IMPLICIT
7
FUNCTIONS
want eαx rather than x( y).
eαx = 1 +
X yn
[λn−1 ]αeαλ e nλ
n
X yn
=1+α
[λn−1 ]αe(α+n)λ
n
n≥1
n≥1
=1+α
X y n (α + n)n−1
=1+α
Xy
n≥1
=α
X yn
n≥0
n!
(n − 1)!
n
n≥1
n
n!
(α + n)n−1
(α + n)n−1
by working in C[α, α−1 ][[ y]]. Note that this series lies in C[[x]].
3.6 Example. Let a, b, x be indeterminants. Then e(a+b)x = e ax · e bx , and so
(a + b)
X xk
k≥0
k!
(a + b + k)k−1 = a
X xi
i≥0
i!
(a + i)i−1 + b
X xj
j≥0
j!
(b + j) j−1
Applying [x n ] throughout yields
(a + b)
n!
(a + b + n)n−1 = ab
X 1 1
i, j≥0
i+ j=n
(a + b)(a + b + n)n−1 = ab
i! j!
n X
n
i=1
i
(a + i)i−1 (b + j) j−1
(a + i)i−1 (b + n − i)n−i−1
This is called Abel’s extension to the Binomial theorem.
3.7 Theorem. (Taylor’s Theorem for C[[x]]) Let f (x) ∈ C[[x]]. Then f (x) =
C[[x]] → C : f (x) → f (0).
P
xk
k≥0 k!
k
L0 ddx k f (x), where L0 :
PROOF: Consider [x n ]x k = δk,n . On the other hand,
1
k!
L0
dk
d xk
xn =
1
k!
L0 n(n − 1) · · · (n − k + 1)x n−k = δk,n
since if n > k then 0n−k = 0, if n < k then n(n − 1) · · · (n − k + 1) = 0, and finally if n = k then 00 = 1 and
n
1
L0 ddx n x k . Extend linearly to the whole of C[[x]], since {1, x, x 2 , . . .}
n(n−1) · · · (n−k+1) = k!. Thus [x n ]x k = k!
is a basis (?) of C[[x]]. The result follows.
ƒ
Examples of the use of Lagrange’s Theorem
1. Find the number cn of plane planted trees with n non-root vertices and such that each vertex has even up
degree (except the root). Let P be the set of all plane planted trees. then by the universal decomposition
for plane planted trees (the branch decomposition) we have
a
Φ : P −→
˜ {"}
Pk
k≥0
8
COMBINATORIAL ENUMERATION
where k is the up degree of a generic vertex. Let ω be the obvious weight function on P . Then Φ is
additively ω-preserving. Let E be the desired set of trees. Then E ⊂ P . Now consider
a
E 2i
Φ|E : E −→
˜ {"}
i≥0
Then Φ|E is additively ω- preserving. Therefore we use the sum and product lemmas to get
E = x(1 + E 2 + E 4 + · · · ) =
x
1 − E2
where E(x) := [(E , ω)]o . This is a functional equation in Lagrangian form. Solving for E
[x ]E =
n
1
n
[λ
n−1
2 −n
](1 + λ )
=
1
n
[λ
n−1
]
n X
n+k−1
k=0
k
λ2k
For a non-zero contribution, we need n − 1 = 2k for some k ≥ 0, so n is odd. Let n = 2m + 1.
X 2m + k
1
1
3m
2m+1
2m+1
2k
[x
]E = [λ
]
λ =
2m + 1 k≥0
k
2m + 1 m
and is 0 if n is even.
2. Find the average number of non-root vertices of degree one in the set all plane planted trees on n non-root
vertices.
a
Φ : P −→
˜ {"}
Pk
k≥0
with weight function ω(t) = (ω1 (t), ω2 (t)) ∈ N , where ω1 (t) is the number of non-root monovalent
vertices of t and ω2 (t) is the total number of non-root vertices of t. We call ω the refinement of ω1 by ω2
on P . It is also denoted ω1 ⊗ ω2 . Let
X
P(u, x) = [(P , ω)]o :=
uω1 (t) x ω2 (t) ∈ C[u][[x]]
2
t∈P
Then we have
P −→
˜ {"} × (" q P q P 2 q · · · )
and so
P = x(u + P + P 2 + · · · ) = x
1
1−P
+ (u − 1)
This is a functional equation for P as a series for x in Lagrangian form. Using Lagrange’s Theorem in
C[u][x]]
n
X xn
1
n−1
P=
[λ ]
+ (u − 1)
n
1−λ
n≥0
!
n X xn
X
n (u − 1)n−i
n−1
=
[λ ]
n
i (1 − λ)i
n≥0
i=0
!
n X xn X
n
i+n−2
n−i
=
(u − 1)
n
i
n−1
n≥0
i=0
PARTITIONS
OF I NTEGERS
9
Now to find the average number of non-root vertices of degree
one that a plane planted tree with n non2n
1
root vertices has, we must take the average. There are n+1
plane
planted trees on n non-root vertices.
n
If cn,k = [x n uk ]P, then the number we want is thus
Pn
k=0 kcn,k
2n
1
n+1 n
Notice that the numerator is equal to
Lu=1 Du [x ]P =
n
and hence the average number of trees is
4
1
n
n n−1
2n − 3
n−1
n(n+1)(n−2)(n−3)
.
4(2n−1)
Partitions of Integers
Let n, k be a positive integers. We say that (i1 , . . . , ik ) is an integer partition of n if
1. i1 ≥ · · · ≥ ik ≥ 1
2. i1 + · · · + ik = n
Denote that α is a partition of n by α ` n. The number of partitions of n is denoted by p(n) in most books. It has
amazing properties.
4.1 Example. p(5): (5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1) so p(5) = 7.
Is there a closed form expression for p(n)? No! There is an asymptotic expression for p(n) such that the
integer part is exactly p(n). We are not going to do this.
Can we find a generating series for {p(n) | n ≥ 1}? Can we restrict this to other classes of partitions? Can
we show that certain subsets of partitions are equicardinal? If so, can we find natural bijections? Can we find
non-trivial identities based on these enumerations? The answer to all of these questions is yes!
4.1
Elementary properties
Let P be the set of all partitions. The universal decomposition is
P −→
˜ ×i≥1 {i}∗
Let ω(π) = sum of the parts of π. We want to gain information about P = [(P , ω)]o . The bijection is clearly
additively ω- preserving, so we can use the sum and product lemmas.
Y
Y 1
P(x) =
[({i}∗ , ω)]o (x) =
1 − xi
i≥1
i≥1
Caveat: This has a barrier of essential singularities on the unit circle.
Let E ⊆ P be the set of all partitions with only even parts. Then clearly
Y
1
[(E , ω)]o (x) =
1 − x 2i
i≥1
Let OD ⊆ P be the set of all partitions with distinct odd parts. Then clearly
Y
[(OD , ω)]o =
(1 + x 2i−1 )
i≥1
10
4.2
COMBINATORIAL ENUMERATION
Haruspication
Conjecture (on flimsy evidence): The number of partitions of n with only odd parts equals the number of partitions of n with distinct parts. It is enough to show that the generating series are equal, no coefficient extraction
required.
Let A be the set of all partitions with only odd parts. Then
A(x) =
Y
1
i≥1
1 − x 2i−1
=
Y
1
i≥1
=Q
·
1 − x 2i
1 − x 2i−1 1 − x 2i
Q
2i
i≥1 (1 − x )
Qi≥1
i≥1
= Q
(1 − x 2i−1 )(1 − x 2i )
(1 − x 2i )
(1 − x i )
Yi≥1
=
(1 + x i )
i≥1
This is exactly the generating series for D, the set of all partitions with distinct parts. So the conjecture is true!
Even though we cannot convienently extract the coefficients, we can still tell many things about the relationships
between the numbers of partitions. Since the sets are equicardinal, there is a bijection between them. There
should be a natural bijection. Find one.
Consider that 1 + x 2 + x 3 + x 5 = (1 + x 2 )(1 + x 3 ). Using this fact,
Y
Y
(1 + x 2i + x 3i + x 5i ) =
(1 + x 2i )(1 + x 3i )
i≥1
i≥1
=
Y 1 − x 4i 1 − x 6i
i≥1
=
1 − x 2i 1 − x 3i
1
Y
i≥1
(1 −
x 12i−10 )(1 −
x 12i−9 )(1 −
x 12i−6 )(1 − x 12i−3 )(1 − x 12i−2 )
Since
Y
Y
(1 − x 6i ) =
(1 − x 12i−6 )(1 − x 12i )
i≥1
i≥1
Y
Y
(1 − x 4i ) =
(1 − x 12i−8 )(1 − x 12i−4 )(1 − x 12i )
i≥1
i≥1
Y
Y
(1 − x 3i ) =
(1 − x 12i−9 )(1 − x 12i−6 )(1 − x 12i−3 )(1 − x 12i )
i≥1
i≥1
Y
Y
(1 − x 2i ) =
(1 − x 12i−10 )(1 − x 12i−8 )(1 − x 12i−6 )(1 − x 12i−4 )(1 − x 12i−2 )(1 − x 12i )
i≥1
i≥1
We see thusly that the set of partitions in which each part occurs 0, 2, 3, or 5 times is in bijection with the set of
partitions in which each part congruent to 2, 3, 6, 9, or 10 modulo 12 only.
EXPONENTIAL GENERATING SERIES
4.3
11
Ferrers Graphs/Diagrams
See the notes for an explanation and some examples.
Let α ` n. The conjugate of α is the partition of n obtained by reflecting the Ferrers diagram for α about the
˜α) = α and the number of parts in α is equal to the largest part of α̃. Let
diagonal, and it is denoted α̃. Note that ( ˜
Mk ⊆ P be the set of all partitions with at most k parts and let Lk ⊆ P be the set of all partitions with largest
part at most k. Clearly
k
Y
[(Lk , ω)]o =
(1 − x i )−1
i=1
To find [(Mk , ω)]o we note that Mk −→
˜ Lk by conjugation. Thus [(Mk , ω)]o = [(Lk , ω)]o .
Given partition, if we delete the maximal upper left corner square (Durfee square), say it is k × k, then we are
left with a partition from Lk (on the bottom) and a partition from Mk (to the right). This gives us an additively
ω-preserving bijection
[
{Dk } × Mk × Lk
Ω : P −→
˜
k≥0
Taking generating series we get
Y
1
i≥1
1 − xi
=
xk
X
k≥0
Qk
2
i=1 (1 −
x i )2
This is due to one of Euler, Gauß, Cauchy.
Can we preserve more combinatorial information? For example, can we preserve the number of parts of α,
λ(α)?
[(P , ω ⊗ λ)]o (x, u) −→
˜ ×i≥1 {", i, ii, . . . }
So we have
P(x, u) =
Y
X
x ω(α) uλ(α)
i≥1 α∈{",i,... }
YX
=
(x i u) j
i≥1 j≥0
=
Y
1
i≥1
1 − ux i
And thus
Y
i≥1
2
uk x k
=
1
+
Q
Qk
k
i
i
1 − ux i
k≥1
i≥1 (1 − x )
i≥1 (1 − ux )
1
X
which is in C[u][[x]]. Setting u = −1, we get
5
5.1
Y
1
i≥1
1 + xi
2
=1+
(−1)k x k
Qk
2i
k≥1
i≥1 (1 − x )
X
Exponential Generating Series
A classical problem
The derangement problem is a classical problem. A derangement is a rearrangment of a set of things such that
no thing is mapped to itself. Let dn be the number of derangements of n distinct objects. For examples, when
12
COMBINATORIAL ENUMERATION
n = 3, the complete set of derangements is {(231), (312)}, and so d3 = 2. Let Pn be the set of all permuations of
{1, 2, . . . , n}. Then a derangement is a permutation with no fixed points. Encoding P3 in terms of derangements
123
123
123
7→ 123
7→
123
123 {1,2,3}
123
1
12
1 23
7→ 1 32
7→
132
1 {1} 21 {2,3}
123
1
12
3
12
7→ 3 21
7→
213
1 {3} 21 {1,2}
123
123
123
7→ 231
7→
231
231 {1,2,3}
123
1
12
2 13
7→ 2 31
7→
321
1 {2} 21 {1,3}
123
123
123
7→
7→ 312
312 {1,2,3}
312
S∞
Clearly P := n=1 Pn −→
˜ I ∗ D, where I = {", (1), (12), (123), . . . } is the set of all identity permutations
and D = {", (21), (231), (312), . . . } is the set of all derangements. The ∗ denotes that we take the bijection with
all possible labelings of the permutations on symbols {1, . . . , n}, where n is the sum of the size of the derangement
and size of the identity permutation.
We now have a new combinatorial product of sets, called the ∗-product, defined by
[ [
[
A ∗ B :=
((a)α , (b)β )
n≥1 a∈A α⊆{1,...,n}
b∈B β={1,...,n}\α
where the third union is taken over all α, β that partition {1, . . . , n} and |α| = #(labels of a), |β| = #(labels of b).
Can we get a product lemma for the ∗-product? If C = A ∗ B then in the generating series for these problems
we will have
n X
n
cn =
ak bn−k
k
k=0
cn
n!
=
n
X
a
k=0
k
bn−k
k! (n − k)!
n
since there are k ways of choosing the labels for ak (after which the labels are fixed for bn−k ). Thus if we define
P
n
C(x) = k≥0 ci xn! and A(x), B(x) similarily, we have C(x) = A(x)B(x). We call A(x) the exponential generating
series for the problem (A , ω) and is denoted [(A , ω)]e . We have proved
5.1 Lemma. (∗-Product Lemma) [(A ∗ B, ω)]e (x) = [(A , ω)]e (x)[(B, ω)]e (x)
Now to solve the problem. P −→
˜ I ∗ D, and the generating series for P is given by
[(P , ω)]e (x) =
X x ω(π)
π∈P
The generating series for I is
[(I , ω)]e (x) =
ω(π)!
=
X x ω(π)
π∈I
ω(π)!
X
k!
k≥0
=
xk
k!
X xk
k≥0
k!
=
1
1− x
= ex
EXPONENTIAL GENERATING SERIES
13
Thus we can derive the generating series for D implicitly from P(x) = I(x)D(x) to get D(x) =
dn =
5.2
xn
n!
D(x) =
xn
n!
e−x
= n!
1− x
e−x
.
1−x
Thus
n
X
(−1)k
k!
k=0
An architect’s problem
Find the number of m × n {0, 1}-matrices with
1. no rows of 0’s
2. no columns of 0’
3. exactly k 1’s
The idea is to bubble row’s of 0’s to the top of the matrix, taking care of the original row label to the row
so that the original matrix can be recovered. Do a similar thing with the columns (“bubble” them to the left).
From a general {0, 1}-matrix we get (in the bottom right corner) a smaller {0, 1}-matrix with no rows or columns
of 0’s. We can decompose this as a zero matrix and a matrix with no rows or columns of 0’s, where they are
both row and column label disjoint. Let M be the set of all combinatorial {0, 1}-matrices. Let A be the set of
combinatorial matrices with no rows or columns of 0’s and Z be the set of all combinatorial zero matrices. Then
Ω : M −→
˜ Z ∗ r,c A
For indeterminants, take x to be exponential and mark rows, y to be exponential and mark columns, and z to be
ordinary and mark the number of 1’s. Let
ω1 (M ) = number of rows of M
ω2 (M ) = number of columns of M
λ(M ) = number of 1’s in M
Then Ω is additively λ-preserving. Let ω = ω1 ⊗ ω2 ⊗ λ. Let A(x, y; u) =
lemma, we have
m
P
m,n≥0
am,n,k xm!
yn k
u .
n!
[(M , ω)](x, y; u) = [(Z , ω)](x, y; u)[(A , ω)](x, y; u)
X
X xm yn
xm yn
(1 + u)mn
= A(x, y; u)
m! n!
m! n!
m,n≥0
m,n≥0
X
(1 + u)mn
m,n≥0
xm yn
m! n!
= A(x, y; u)e x+ y
where the series for M is obtained by labeling the matrix in a boostrophedon manner. Finally,
–
a M ,N ,K =
xM yN
M! N!
which you can go off and work out yourself.
™
u
K
e−x− y
X
(1 + u)mn
m,n≥0
xm yn
m! n!
By the ∗-product
14
COMBINATORIAL ENUMERATION
5.3
Permutations and cycles
Motivation: A derangement is a permutation with no cycles.
π=
1 2 3 4 5 6 7 8 9 10
3 6 9 1 2 5 8 7 4 10
=
1394
265
78
10
3941
652
87
10
This is called the disjoint cycle decomposition. We may relabel this as
π=
1234
2341
{1,3,9,4}
123
231
12
21
{2,6,5}
1
{7,8}
1
{10}
A permutation is a disjoint union of cycles. Let U be the set of all canonical sets,
U = {∅, {1}, {1, 2}, {1, 2, 3} . . .}
Let C be the set of all canonical cycles,
C = {(1), (1 2), (1 2 3), (1 3 2), . . .}
We build the set P of all permutations from these by a new operation ?, ∗-composition.
P −→
˜ U ?C
5.2 Lemma. (∗-composition) [(A ? B, ωs )]e = [(A , ωs )]e ◦ [(B, ωs )]e where B does not contain the null
structure.
Using this (without proof), letting C(x) = [(C , ωs )]e (x), we have
[(P , ωs )]e (x) = [(U ? C , ωs )]e (x)
[(P , ωs )]e = [(U , ωs )]e ◦ [(C , ωs )]e (x)
X x n X C(x)n
n!
=
n!
n!
n≥0
n≥0
1
= e C(x)
1− x
C(x) = log(1 − x)−1
:= x +
x2
+
x3
+
x4
+ ···
2
3
4
” n—
Thus the number of cycles of length n on n symbols is xn! C(x) = (n − 1)!. For derrangements we have
D −→
˜ U ? (C \ {1})
since a derrangement has no fixed points (no one-cycles). This is a direct decomposition for D. Then
dn =
xn
n!
D(x) =
xn
n!
e C(x)−x =
xn
n!
exp(log(1 − x)−1 − x)) =
xn
n!
e−x
1− x
In a sense we have managed to solve the set equation P −→
˜ I ∗ D for D. We get D −→
˜ U ? (C \ {1}).
EXPONENTIAL GENERATING SERIES
5.4
15
The preimage decomposition
We look as { f : Nn → Nm } = NNn m where Nn = {1, . . . , n}. How many functions/injections/surjections/bijections
are there?
Notice that, given a function f : Nn → Nm , the set { f −1 (i) | i ∈ Nm } of preimages of f partition Nn . Notice
that some of these sets may be empty. Given any ordered
S partition of Nn into exactly m sets (some of which may
be empty) defines a function f : Nn → Nm . Let Fm = n≥0 NNmn . Then we have a bijection
Fm −→
˜ U ?m
Let A be a set of labeled combinatorial structures. Let [Nn ]A denote the set of all members of A with label set
{1, . . . , n}.
1. How many functions? [Nn ]Fm −→
˜ [Nn ]U ?m
|[Nn ]Fm | = |[Nn ]U ?m | =
xn
n!
[(U ?m , ω)]e (x) =
xn
n!
[(U , ω)]m
e (x) =
xn
n!
e mx = mn
2. How many surjections? [Nn ]Sm −→
˜ [Nn ](U \ {∅})?m
?m
|[Nn ]Fm | = |[Nn ](U \ {∅}) | =
=
xn
n!
(e x − 1)m =
xn
n!
xn
?m
[((U \ {∅}) , ω)]e (x) =
X
n m
n!
k=0
k
(−1)m−k e kx =
xn
n!
n X
m
k
k=0
[((U \ {∅}), ω)]m
e (x)
(−1)m−k k n
3. How many injections? [Nn ]Im −→
˜ [Nn ]{∅, {1}}?m
?m
|[Nn ]Fm | = |[Nn ]{∅, {1}} | =
=
xn
n!
xn
n!
?m
[({∅, {1}} , ω)]e (x) =
xn
n!
[({∅, {1}}, ω)]m
e (x)
(1 + x)m = m(m − 1)(m − 2) · · · (m − n + 1) := (m)n , the falling factorial
4. How many bijections? [Nn ]Im −→
˜ [Nn ]{{1}}?m
?m
|[Nn ]Fm | = |[Nn ]{{1}} | =
5.5
xn
n!
?m
[({{1}} , ω)]e (x) =
xn
n!
[({{1}}, ω)]m
e (x)
=
xn
n!
x m = n!δm,n
Branch decomposition for labelled rooted trees
Let TR be the set of all labelled rooted trees. Then by a variation of the bijection above we have that
TR −→
˜ {"} ∗ (U ? TR )
Let T (x) = [(TR , ω0 )]e (x) where w0 (t) is the number of vertices of t. Then T = x e T and this is a functional
equation in Lagrangian form, so
xn
n!
1
nn−1
T = n! [λn−1 ]e nλ = n!
= nn−1
n
n(n − 1)!
16
COMBINATORIAL ENUMERATION
5.6
Derivative of Sets
Let A be a set of labelled
structures. We refer to the elements that carry lables as s-objects. Let
Pcombinatorial
n
A(x) = [(A , ωs )]e (x) = n≥0 an xn! . Let B be the set obtained from A by deleting a canonical s-ojbect in each
a ∈ A (for example, the largest label). Assume that every a ∈ A has a canonical s-object.
B(x) := [(B, ωs )]e (x) =
X
an
n≥0
x n−1
(n − 1)!
=
d
dx
A(x)
We denote B by ∂∂s A . Then [(B, ωs )]e (x) = ddx [(A , ωs )]e (x) where ωs (a) denotes the number of s-objects in
a ∈ A (i.e. the number of labels used in the labelling of a ∈ A ). For example
a
∂B
∂
∂A
B
A∗
(A ∗ B) −→
˜
∂s
∂s
∂s
5.3 Example. Alternating permutations are permutations such that the numbers in the even positions are greater
the the numbers that they follow and the numbers in the odd positions are less than the numbers that they follow.
For example, (2 4 3 5 1) is an alternating permutation on 5 symbols. Let Q (P ) be the set of all alternating
permutations of even (odd) length. The minimal element of P is (1) and the minimal element of Q is ". Consider
∂
P . Let π ∈ P . We delete the greatest label in π, giving us two alternating permutations of odd length (if
∂s
π 6= (1), otherwise it gives us "). Thus
∂
(P \ {(1)}) −→
˜ P ∗P
∂s
Similarily, if π ∈ Q then deleting the greatest label (if π 6= ") gives us an alternating permutation in P and one
in Q. Thus
∂
(Q \ {"}) −→
˜ P ∗Q
∂s
Now, can we solve this simulataneous set of differential equations?
dP
dx
= 1 + P 2 and
dQ
dx
= PQ
This called a pair of Riccati equations. It is also called a matrix Riccati equation. Let P = tan(R). Then
sec (R) and 1 + P = sec (R). Thus d x =
2
2
2
dP
1+P 2
−
sec (R)
dR
sec2 (R)
2
=
= dR, and so R = x + c, where c is constant. Therefore
P = tan(x + c). But P(0) = 0, so c = 0 and P = tan(x). Now
5.7
dP
dR
dQ
Q
= P d x = tan(x)d x, so (solving) Q = sec(x).
Another combinatorial operation
P
P
n
n
Let A(x) be the exponential generating series for A . Let A(X ) = n≥0 an xn! . Then x ddx A(x) = n≥0 nan xn! . We
interpret this replication by distinguishing an s-object in all possible ways. Picture this as “painting a subobject
red”. Denote this operation by s∂
. We have produced a set s∂
A . We have
∂s
∂s
[(
Notice that
s∂
∂s
=s∗
∂
∂s
s∂
∂s
A , ωs )]e (x) =
d
dx
[(A , ωs )]e (x)
as operators.
5.4 Example. Let F be the set of all functions f : Nn → Nn . Let [Nn ]A denote the set of all a ∈ A with n labels
{1, . . . , n}. Let TR be the set of all labelled rooted trees. Then
v∂
∂v
TR −→
˜ Path ? TR −→
˜ (U ? C ) ? TR −→
˜ U ? (C ? TR ) −→
˜
a
n≥0
NNn n −→
˜ F
SIGN-REVERSING INVOLUTIONS
17
It follows that [Nn ] ∂v∂v TR −→
˜ NNn n , and so
|[Nn ]
PROOF:
v∂
T
∂v R
v∂
∂v
TR | = |NNn n | so nt n = nn and hence t n = nn−1
is the set of all labelled rooted with exactly 1 vertex in each distinguished by a spot of red paint
(for sake of argument). That is, they are bi-rooted trees. Let t ∈ ∂v∂v TR . Then there is a unique path between the
root and the red vertex. Then t can be regarded as a set of trees attached to the path by identifying the rootss of
these trees with a vertex of the path. Then
v∂
∂v
TR −→
˜ Path ? TR
Next consider a path. We may write it uniquely as a non-null permutation, thus
Path ↔ P −→
˜ U ?C
by the disjoint cycle decomposition. ? is associative, so
v∂
∂v
TR −→
˜ U ? (C ? TR )
What is this as a combinatorial structure? It is some number of cycles with trees hanging off of the vertices of
cycle. Direct the tree edges towards a vertex on the cycle (this can be done in a unique way), and notice that the
edges of the cycle already have a direction. What do we obtain? A map! Each label in the structure has a unique
arrow eminating from it, pointing to another label. This is a function in NNn n . Conversely, given any function from
Nn to Nn then we may write down a directed graph corresponding to the function. The graph must be in the
form of a disjoint union of cycles with trees hanging off, with the arrows in the trees pointing towards the cycle.
(Think about it). Thus
a
v∂
TR −→
˜
NNn n
∂v
n≥0
and furthermore [Nn ] ∂v∂v TR −→
˜ NNn n .
ƒ
This suggests a combinatorial proof of Lagrange’s theorem, which has been given in the course notes. It works
by asking how refined a weight function is preserved by the string of bijections.
6
Sign-Reversing Involutions
6.1 Example. Given P1 , P2 , Q 1 , Q 2 points in the plane with integer coordinates. We are interested in the number
of pairs of paths (π1 , π2 ) such that
1. π1 is a path from P1 to Q 1 (P1 is the origin and Q 1 is the terminus) and π2 is a path from P2 to Q 2
2. the only steps allowed are ↑: (i, j) 7→ (i, j + 1) and →: (i, j) 7→ (i + 1, j)
3. π1 and π2 have no common points
6.2 Definition. We call pair of paths that satisfies (1) and (2) an ordered 2-path origins (P1 , p2 ) to terminii
(Q 1 , Q 2 ). An ordered 2-path satisfying (3) is called non-intersecting and one not satisfying (3) is called intersecting.
Counting non-intersection 2-paths is hard. However, counting all ordered 2-paths is not: the number of such
a +b a +b paths is 1a 1 2a 2 , where ai is the difference of the x-coordinates of Pi and Q i and bi is the difference of the
1
2
y-coordinates.
18
COMBINATORIAL ENUMERATION
Notation. Let T = T (P1 , P2 ; Q 1 , Q 2 ) denote the set of all ordered 2-paths from (P1 , P2 ) to (Q 1 , Q 2 ). Let T× be the
subset of intersecting ordered 2-paths and T= be the subset of non-intersecting 2-paths.
b (P1 , P2 ; Q 1 , Q 2 ) = T (P1 , P2 ; Q 1 , Q 2 ) ∪ T (P1 , P2 ; Q 2 , Q 1 ) and define T
b× and T
b= similarily. We say that
Let T
(P1 , P2 ; Q 1 , Q 2 ) is proper if T= (P1 , P2 ; Q 2 , Q 1 ) = ∅.
b× (P1 , P2 ; Q 1 , Q 2 ). Let A be the first intersection point encounted upon following π1
Suppose that (π1 , π2 ) ∈ T
starting at P1 . Let π01 be the path obtained by following π1 from P1 to A and then following π2 , and let π02 be the
path obtained by following π2 from P2 to A and then following π1 . If we let
b× → T
b× : (π1 , π2 ) 7→ (π0 , π0 )
φ:T
1
2
b× (i.e. φ 2 = id Tb ). Notice that φ maps elements of T× (P1 , P2 ; Q 1 , Q 2 ) to elements of
then φ is an involution on T
×
b → Z by
T× (P1 , P2 ; Q 2 , Q 1 ), and visa versa. Define ω : T
¨
ω(π1 , π2 ) =
1
−1
if (π1 , π2 ) ∈ T (P1 , P2 ; Q 1 , Q 2 )
otherwise
Then ωφ = −ω. So φ is a sign-reversing involution. Notice that
X
X
ω(π) = −
ω(φ(π))
b×
π∈ T
φ is sign-reversing
b×
π∈ T
=−
X
ω(φ(π))
φ is injective
b×
φ(π)∈ T
=−
X
ω(π)
φ is surjective
b×
π∈ T
Thus
P
b×
π∈ T
ω(π) = 0. Hence, when (P1 , P2 ; Q 1 , Q 2 ) is proper,
|T= (P1 , P2 ; Q 1 , Q 2 )| =
X
ω(π)
π∈T=
=
X
ω(π) +
X
ω(π)
b×
π∈ T
b=
π∈ T
=
X
ω(π)
b
π∈ T
=
X
π∈T (P1 ,P2 ;Q 1 ,Q 2 )
ω(π) +
X
ω(π)
π∈T (P1 ,P2 ;Q 2 ,Q 1 )
= |T (P1 , P2 ; Q 1 , Q 2 )| − |T (P1 , P2 ; Q 2 , Q 1 )|
= |T (P1 ; Q 1 )| · |T (P2 ; Q 2 )| − |T (P1 ; Q 2 )| · |T (P2 ; Q 1 )|
|T (P ; Q )| |T (P ; Q )|
1
1
1
2 =
|T (P2 ; Q 1 )| |T (P2 ; Q 2 )|
Where T (P; Q) has the obvious meaning.
We can say more. Note that |T= | = [(T= , 0)]o (the zero weight). The same reasoning applies to any weight
funtion ω such that
1. ω = ψ ⊗ ψ where ψ is the weight function for a single path
2. ω(π1 , π2 ) cannot depend on the membership of steps in specific paths
SIGN-REVERSING INVOLUTIONS
3. [(T= , ω)]o =
P
π∈T=
19
ω(π)
where
¨
ω(π1 , π2 ) :=
x ω(π1 ,π2 )
−x ω(π1 ,π2 )
if (π1 , π2 ) ∈ T (P1 , P2 ; Q 1 , Q 2 )
otherwise
Then we get that
[(T (P ; Q ), ψ)]
1
1
o
[(T= , ω)]o = [(T (P2 ; Q 1 ), ψ)]o
[(T (P1 ; Q 2 ), ψ)]o [(T (P2 ; Q 2 ), ψ)]o 6.3 Theorem. Involution Theorem Let E ⊆ S where S is a set of combinatorial objects. Let ω : S → R, where
R is a commutative ring. If φ : S \ E → S \ E is a sign-reversing involution (which is to say φ 2 = id and
ωφ = −ω), then
X
X
ω(σ) =
ω(σ)
σ∈E
σ∈S
PROOF:
X
ω(σ) = −
σ∈S \E
X
ω(φ(σ))
φ is sign-reversing
σ∈S \E
=−
X
ω(φ(σ))
φ is one to one
φ(σ)∈S \E
=−
X
ω(σ)
φ is onto
σ∈S \E
Therefore
P
σ∈S \E
ω(σ) = 0 and the result follows as in the example above.
ƒ
Here’s how we apply the theorem. Suppose that we have a combinatorial problem (E , ω). If we can find
• a superset S ⊇ E
• a function ω : S → R
• a function S \ E → S \ E such that
1. φ 2 = id
2. ωφ = −ωP
3. [(E )]o = σ∈E ω(σ)
P
Then [(E )]o = σ∈S ω(σ), where hopefully the latter series is easier to compute.
6.4 Example. A polyomino is a region in the plane formed by joining unit squares at the edges in such a way
the the region formed is simply connected. (Connected in the sense that there is a walk along the integer lattice,
no diagonal motions allowed.) We say that two polyominoes are equivalent if one can be translated to the other,
with no rotations allowed. How many polyonminoes (up to equivalence) are there? The problem of counting
general polyominoes is very hard. However, we can count certain subclasses of polyonminoes.
6.5 Definition. A convex polyomino is a polyomino for which the intersection with any horizontal or vertical line
either has one connected (in the usual sense) component or none at all.
20
COMBINATORIAL ENUMERATION
We wish to count the number c(m) of convex polyominoes with perimeter 2m+4. We will do this by decomposing
it into ordered paths.
Given a convex polyomino P, translate it so that the bottom left corner of its smallest bounding rectangle is
the origin O = (0, 0). Let O0 be the opposite corner of the rectangle. Suppose that O0 = (k + 1, m − k + 1), where
k + 1 is the width of P. Pick a and d to be the points on the boundry of P closest to O on the y and x axes,
respectively. Pick c and b on the boundry of P closest to O0 on the lines parallel to the y and x axes, respectively.
Mark the points on the boundry of P closest to a, b, c, d by . Let ((π1 , π2 ), (π3 , π4 )) be the paths from a to d, b
to c, d to c, and a to b, respectively. (Not really, but from the closest to these points in the correct directions.)
Notice that π1 cannot possibly intersect π3 or π4 , and neither can π2 . Applying our result for non-intersecting
paths, if a = (0, j) and b = (k + 1 − r, m − k + 1) and c = (k
+ 1, m − k + 1 − s) and d = (i, 0) then the number of
−2
non-intersecting 2-paths (π1 , π2 ) is something (where −1 := 1 for today).
m
X
c(m) =
k=0
X
blah
0≤i,r≤k
0≤ j,s≤m−k
which simplifies to
2m−4
c(m) = (2m + 7)2
2m − 4
− 4(2m − 3)
m−2
6.6 Theorem. (Euler’s Pentagonal Number Theorem)
∞
X
(−1)k x
k(3k−1)
2
=
Y
(1 − x i )
i≥1
k=−∞
What is this theorem really saying?
Q Notice that the right hand side looks similar to the generating series for
D, the partitions with distinct parts, i≥1 (1 + x i ). We define
ω : D → Z[[x]] : α 7→ (−1)`(α)) x |α|
Then we have
X
=
α∈D
Y
(1 − x i )
i≥1
n
Now consider the left hand side. The coefficient of x is zero for “most” n, and when it is non-zero the coefficient
is ±1. So if we believe the theorem, it says that for most n the number of partitions into an even number of
k(3k−1)
distinct parts is equal to the number of partitions of n into an odd number of distinct parts. For n =
, these
2
numbers differ by 1. Why is this? Define
C : D → N : α 7→ smallest part of α
R : D → N : α 7→ largest k such that the sucessive diffs of the first k parts are 1
In the Ferrers diagram Fα for α ∈ D, replace the ∗’s counted by C(α) with −’s. Replace the rightmost ∗ in rows
counted by R(α) with +’s. Use ± when necessary. Notice that for any α, there is at most one ±. Let n± (α) be the
number of ±’s in the modified Ferrers diagram for α. Define two operations on D:
• Raising (ρ): remove the −’s from Fα and append a − to the end of the first C(α) rows of Fα (treat ± as a
−)
• Lowering (λ): remove the +’s from Fα and add a new row of R(α) +’s to the bottom of Fα (treat ± as a +)
SIGN-REVERSING INVOLUTIONS
21
We would like to apply λ and ρ only when the result is in D and furthermore, we want each − in ρ Fα to be to
the right of a +.
ρ can be applied when
• C(α) < R(α)
• C(α) = R(α) and n± (α) = 0
λ can be applied when
• C(α) > R(α) + 1
• C(α) = R(α) + 1 and n± (α) = 0
Note that at most one of ρ and λ can be applied to any given α. Let E ⊆ D the the subset of partitions with
distinct parts for which neither ρ nor λ can be applied. Then E consists of precisely the α ∈ D for which either
• C(α) = R(α) and n± (α) = 1, or
• C(α) = R(α) + 1 and n± (α) = 1
But what is E really? When C(α) = R(α) = k, |α| is a pentagonal number, |α| =
(−1)k x
k(3k−1)
2
. Similarily, when C(α) = R(α) + 1 = k + 1 we have |α| =
X
ω(α) =
α∈E
k(3k+1)
,
2
k(3k−1)
.
2
k
Whence ω(α) =
so ω(α) = (−1) x
k(3k+1)
2
. Thus
∞
X
X
X
k(3k−1)
k(3k+1)
k(3k−1)
(−1)k x 2 +
(−1)k x 2 =
(−1)k x 2
k≥0
k≥0
k=−∞
Recall the to prove the theorem we need only prove that
X
X
ω(α) =
ω(α)
α∈D
α∈E
Define a sign-reversing involution φ : D \ E → D \ E by

ρ(α) if C(α) < R(α)


ρ(α) if C(α) = R(α) and n (α) = 0
±
φ(α) =

λ(α) if C(α) > R(α) + 1


λ(α) if C(α) = R(α) + 1 and n± (α) = 0
Let P = {α ∈ D \ E | φ(α) = ρ(α)} and Q = {α ∈ D \ E | φ(α) = λ(α)}. Check that ρ : P → Q and λ : Q → P ,
and that λρ = idQ and ρλ = idP . It follows from these observations that φ 2 = φ, so φ is an involution. Also,
ω(φ(α)) = (−1)`(φ(α)) x |φ(α)|
= (−1)`(α)±1 x |α|
= −(−1)`(α) x |α|
= −ω(α)
Therefore φ is sign-reversing as well. Therefore
∞
X
k=−∞
Whence the theorem is proved.
(−1)k x
k(3k−1)
2
=
X
α∈E
ω(α) =
X
α∈D
ω(α) =
Y
(1 − x i )
i≥1
22
COMBINATORIAL ENUMERATION
7
The Pattern Algebra
This section is under heavy construction. See the notes for a more sane presentation.
7.1
A brief return to {0, 1}-strings
Let a, b be non-commuting indeterminants and let Q〈〈a, b〉〉 be the ring of all formal series in a and b. Encoding
a ↔ 0 and b ↔ 1, the set of all {0, 1}-stings is generated by
(1 − (a + b))−1 = 1 + (a + b) + (a + b)2 + · · · ↔ " q {0, 1} q {02 , 01, 10, 12 } q · · ·
We call (1 − (a + b))−1 the generator for the set {0, 1}∗ What about the well known decomposition {0, 1} =
1∗ (01∗ )? Can we deduce this algebraically?
1 − (a + b) = 1 − b − a = (1 − a(1 − b)−1 )(1 − b)
Thus
−1
(1 − (a + b))−1 = (1 − a(1 − b)−1 )(1 − b)
= (1 − b)−1 (1 − a(1 − b)−1 )−1
which is exactly what we want. Notice also that
1 − (a + b) = 1 − a − b + a b − a b = (1 − a)((1 − b) − (1 − a)−1 ab) = (1 − a)(1 − (1 − a)−1 ab(1 − b)−1 )(1 − b)
Therefore
([1 − (a + b))−1 = (1 − b)−1 (1 − (1 − a)−1 ab(1 − b)−1 )−1 (1 − a)−1
This gives us the decomposition {0, 1}∗ = 1∗ (0∗ 010∗ )0∗ .
Now let’s generalize this to a larger alphabet {1, 2, 3 . . . }. We look for for complex substructures than just the
occurrences of 1’s, 2’s, etc. We might want:
1. length of maximal strictly increasing substrings
2. length of maximal blocks of symbols
3. restriction to permutations with these conditions (i.e. distinct symbols)
7.2
Patterns
Let σ = 4 5 9 9 10 7 6 5 5 8. We call uudduddddu the pattern of σ, where u ↔< and d ↔≥. Denote this
by patt(σ) = u2 d 2 ud 4 u. Let xi mark the occurrence of i ∈ {1, 2, 3 . . . }. Let fi mark the occurrence of a maximal
<-substring. Then
σ ↔ x4 x5 x29 x8 x10 x7 x6 x25 x8 and the maximal <-substrings are f3 f1 f2 f13 f2
The contribution to the generating series that counts strings is
(f3 f1 f2 f13 f2 )(x4 x5 x29 x8 x10 x7 x6 x25 x8 ) ∈ Q〈f1 , f2 , . . . 〉〈〈x1 , x2 , . . . 〉〉
We may generalize this further to any relation on N and it’s complement. Let π1 ⊆ N2 and let π2 = N2 \ π1 .
For example, < and ≥, ≤ and >, = and 6=.
THE PATTERN ALGEBRA
23
7.1 Example. Find all sequences of length 3 and pattern ud over the alphabet {1, 2, 3} where u ↔< and d ↔≥.
121, 131, 132, 133, 231, 232, 233
These correspond to
x1 x2 x1 + x1 x3 x1 + x1 x3 x2 + x1 x3 x3 + x2 x3 x1 + x2 x3 x2 + x2 x3 x3
Now, trivially, the set of all strings with pattern ud is equal to the set of all strings that start with u minus all
strings of the form uu. That is, the above expression is equal to
<
<
(x1 x2 + x2 x3 + x1 x3 )(x1 + x2 + x3 ) − (x 1 x2 x3 ) =: γ<
2 γ1 − γ3
7.2 Example. Alternating sequences of odd length have patterm (ud)k for some k. The generator for such
sequences is 1 + ud + (ud)2 + · · · = (1 − ud)−1 . Let [(1 − ud)−1 ]o denote the ordinary generating series for
sequences with these patterns.
We need an algebraic device to look after the repeated application of the combinatorial construction. Let
ϕ : Q〈〈u, d, w〉〉/u + d = w → Q〈f 〉〈〈x〉〉 such that
1. ϕ(λa + µb) = λϕ(a) + µϕ(b) for all linear combinations of patterns a, b and scalars λ, µ (i.e. ϕ is linear)
2. ϕ(aw b) = ϕ(a)ϕ(b) where w = u+d (so w ↔ π1 qπ2 = N2 , which is to say that w represents no relation)
π
π
3. ϕ(uk−1 ) = γk 1 (x1 , x2 , . . .), where γk 1 represents the collection of all strings of length k that have pattern
k−1
u .
Then if a is a linear combination of patterns then [a]o = ϕ(a).
7.3 Example. Compute [ud]o . We want to calculate ϕ(ud) using the axioms.
ϕ(ud) = ϕ(u(w − u))
= ϕ(uw − u2 )
= ϕ(uw) − ϕ(u2 )
= ϕ(uw1) − ϕ(u2 )
= ϕ(u)ϕ(1) − ϕ(u2 )
<
<
= γ<
2 γ1 − γ3
π
k
Now what if we want to restrict this to permutations only? Let ∆ : γk 1 7→ xk! and extend as a homomorphism.
h 3i
Then, for example, compute x3! ∆[ud]0 . This gives an an answer of 2 for the above problem, which is correct.
Why does it work?
7.4 Example. Compute [(1 − ud)−1 ]o =: [h−1 ]o . We need to compute ϕ(h−1 ). Note first that h = 1 − ud =
1 + u2 − uw =: q − uw. But q−1 = q−1 hh−1 = q−1 (q − uw)h−1 = h−1 − q−1 uwh−1 and thus
ϕ(q−1 ) = ϕ(h−1 − q−1 uwh−1 ) = ϕ(h−1 ) − ϕ(q−1 uwh−1 )
= ϕ(h−1 ) − ϕ(q−1 u)ϕ(h−1 )
= (1 − ϕ(q−1 u))ϕ(h−1 )
P
Whence ϕ(h−1 ) = (1 − ϕ(q−1 u))−1 ϕ(q−1 ). But q−1 = (1 + u2 )−1 = k≥0 (−1)k u2k , so
X
X
X
π1
ϕ(q−1 ) = ϕ( (−1)k u2k ) =
(−1)k ϕ(u2k ) =
(−1)k γ2k+1
k≥0
k≥0
k≥0
24
COMBINATORIAL ENUMERATION
and
1 − ϕ(q−1 u) = 1 −
X
X
X
π1
π
(−1)k ϕ(u2k+1 ) = 1 −
(−1)k γ2k+2
=
(−1)k γ2k1
k≥0
k≥0
k≥0
Thus
ϕ(h−1 ) =
X
π
(−1)k γ2k1
!−1
k≥0
X
π1
(−1)k γ2k+1
!
k≥0
If π1 =< and we wish to find all permutations for which this pattern holds then we must further apply ∆ to
get
−1
∆ϕ(h ) =
X
x 2k
(−1)k
(2k)!
k≥0
!−1
X
(−1)k
k≥0
x 2k+1
(2k + 1)!
!
=
sin x
cos x
= tan x
We wish to list all sequences with the order in which the maximal π1 substrings occur.
7.5 Theorem. (Maximal Decomposition Theorem) Let fi mark a maximal π1 -string of length i. Let xi mark the
occurrence of i as a symbol in the sequence. Let
π
π
π
f (x) = 1 + f1 x + f2 x 2 + · · · and γπ1 = (γ1 1 (x1 , x2 , . . . ), γ2 1 , γ3 1 , . . . )
Then the enumerator for all such sequences is F , where
1 + F = (f −1 ◦ γπ1 (x1 , x2 , . . . ))−1
π
and x k ◦ γπ1 := γk 1 extended linearly.
PROOF: The generator for all patterns is
(1 − w)−1 = (1 − (u + d))−1
= ((1 − u) − d)−1
= ((1 − u)(1 − (1 − u)−1 d))−1
= (1 − (1 − u)−1 d)−1 (1 − u)−1
= (1 − (1 + u + u2 + · · · )d)−1 (1 + u + u2 + · · · )
Now mark maximal π1 -substrings of vertex length k by fk . The generator becomes
(1 − (f1 + f2 u + f3 u2 + · · · )d)−1 (f1 + f2 u + f3 u2 + · · · )
The generating series F that we want is ϕ applied to this. Let g(u) = f1 + f2 u + f3 u2 + · · · . Then, noting that
THE PATTERN ALGEBRA
25
1 + x g(x) = f (x)
F = ϕ (1 − g(u)d)−1 g(u)
= ϕ (1 − g(u)(w − u))−1 g(u)
= ϕ (1 + ug(u) − g(u)w)−1 g(u)
= ϕ (1 − (1 + ug(u))−1 g(u)w)−1 (1 + ug(u))−1 g(u)
= ϕ (1 − f −1 (u)g(u)w)−1 f −1 (u)g(u)
!
X
( f −1 (u)g(u)w)k f −1 (u)g(u)
1+F =1+ϕ
k≥0
=1+
X
ϕ( f −1 (u)g(u))k ϕ( f −1 (u)g(u))
k≥0
=
X
ϕ( f −1 (u)g(u))k
k≥0
−1
= 1 − ϕ( f −1 (u)g(u))
π
1
:= x k+1 ◦ γπ1 = x(x k ) ◦ γπ1 , so if h(x) is a power series in x then ϕ(h(u)) = (xh(x)) ◦ γπ1 .
Now ϕ(uk ) = γk+1
Then
−1
1 + F = 1 − (x f −1 (x)g(x)) ◦ γπ1
But 1 − x f −1 (x)g(x) = 1 − f −1 (x)( f (x) − 1) = f −1 (x), so
1 + F = f −1 (x) ◦ γπ1
−1
ƒ
7.6 Example. Find all <-alternating sequences of even length starting with a <. We can characterize the pattern
of such sequences by the lengths of maximal <-substrings. These have length 2 only, so set f1 = f3 = f4 = · · · = 0
and f2 = 1. Then in the maximal decomposition theorem f (x) = 1 + x 2 , so
!
X
X
X
f −1 ◦ γ< =
(−1)k x 2k γ< =
(−1)k x 2k ◦ γ< =
(−1)k γ<
2k
k≥0
So F =
€P
k≥0 (−1)
k <
γ2k
Š−1
k≥0
. For permutations we have ∆F =
k≥0
P
k x 2k
k≥0 (−1) (2k)!
−1
= sec x.
<
<
7.7 Lemma. (Permutation Lemma) Let Φ be a formal power series in γ<
1 , γ2 , . . . and let ∆ : γk 7→
<
as a homomorphism (we can do this since the γi ’s are algebraically independent). Then
n
x
[x 1 . . . x n ]χΦ =
∆Φ
n!
xk
k!
be extended
where χ denotes the commutative image.
PROOF: See notes.
ƒ
7.8 Example. Find the number of permutations of odd length with no maxima or minima in even positions. We
first need the pattern generator. The patterns of all sequences of odd length are generated by {uu, dd, du, ud}∗ .
The last two are not allowed, so the generator is (1 − (u2 + d 2 ))−1 =: h−1 . We want ∆[h−1 ]o . Now
h = 1 − u2 − d 2 = 1 − u2 − dw + du = 1 − 2u2 − dw + wu
26
COMBINATORIAL ENUMERATION
We have put h in the form h = s +
Pk
j=1 l j w r j
= s + LwR T where L = [l1 , . . . , l k ] and R = [r1 , . . . , rk ]. Then
s−1 = s−1 hh−1 = s−1 (s + LwR T )h−1 = h−1 + s−1 LwR T h−1
Hence
R T s−1 = R T h−1 + R T s−1 LwR T h−1
so
ϕ(R T s−1 ) = ϕ(R T h−1 ) + ϕ(R T s−1 L)ϕ(R T h−1 ) = (1 + ϕ(R T s−1 L))ϕ(R T h−1 )
Thus
ϕ(R T h−1 ) = (1 + ϕ(R T s−1 L))−1 ϕ(R T s−1 )
Now substituting back for our problem yields
‚
1 −1
1 0
1 −1 ”
−d
h
=
+
s
u
0 1
u
o
1
Œ−1 —
1
o
u
s
−1
o
Now take the commutative image. By Cramer’s Rule, the generating series we now seek is therefore
[s−1 ]
[s−1 ]o −1 o
[us d]o 1 + [us−1 ]o ∆[s−1 ]o
=
G = ∆
−1
−1
1 − [s−1 d]
1 + ∆[us−1 ]
[s
]
∆[s
]
o
o
o
o
−[us−1 d]o 1 + [us−1 ]o x + ∆[u2 s−1 ]o 1 + ∆[us−1 ]o This simplifies as G =
8
p
p
2 tanh( 2x)
p
.
1
1− p2 x tanh( 2x)
Pólya Theory
We will now study counting combinatorial structures under the action of a group.
8.1
Colouring Polyhedra
How many different ways are there of painting the faces of a cube with two different colours? 1 + 1 + 2 + 2 + 2 +
1 + 1 = 10 (What about for more complicated solids?)
·
·
·
·
·
·
·
·
The automorphism group of the cube (presented with respect to faces) is a subgroup of S6 with 24 elements.
The cycle index polynomial of Aut(cube), acting of the faces is
ZAut(cube,faces) (x 1 , . . . , x 6 ) =
1 €
24
x 16 + 3x 12 x 22 + 6x 12 x 4 + 6x 23 + 8x 32
Š
PÓLYA THEORY
27
Suppose we cover the cube with 2 colours white and black. Now replace x i with w i + b i . The number of colours
of the face cube with i white faces and j black faces is
[w i b j ]ZAut(cube,faces) 〈w + b〉 = [w i b j ]ZAut(cube,faces) (x 1 , . . . , x 6 ) x 7→w i +bi
i
i=1,...,6
€
= [w i b j ] w 6 + w 5 b + 2w 4 b2 + 2w 3 b3 + 2w 2 b4 + w b5 + b6
8.2
Š
Painting the 2 × 2 board
Number the squares of the board with the numbers {1, 2, 3, 4} in a boostrophedon manner. Fix coordinates around
the
outside
of the
board.
(These are intended to stay fixed as we rotate the board.) Consider the colourings
W B
B W
and
. These colourings are the same. We may encode them as functions {1, 2, 3, 4} → {W, B},
B W
W B
1234
1 2 3 4
the first being W BW B and the second being BW BW . These functions are different. We need a definition of the
equivalence of such functions to reflect the notion of “combinatorially equivalent” colourings.
8.3
Pólya’s Theorem
8.1 Definition. Let G be finte group acting on a finite set X . For g ∈ G, let τi (g) be the number of i-cycles in g
(presented as an element of a symmetric group). The cycle index polynomial of the permutation group of G on
X is
1 X τ1 (g) τ1 (g)
x
x1
···
ZG (x 1 , x 2 , . . . ) :=
|G| g∈G 1
8.2 Definition. Let F (x 1 , . . . , x n ) and f ( y1 , . . . , y r ) be formal power series. The Pólya substitution of f into F is
defined to be
F 〈 f ( y1 , . . . , y r )〉 := F (x 1 , . . . , x n ) x = f ( y ,..., y )
i
1
r
i=1,...,n
8.3 Definition. Recall from Group Theory
1. We say that G has an action on X if
(a) g x ∈ X for all g ∈ G and x ∈ X
(b) (hg)x = h(g x) for all g, h ∈ G and x ∈ X
(c) 1G x = x for all x ∈ X
2. x, y ∈ X are said to be G-equivalent if there is g ∈ G such that g x = y. We write x ∼G y. ∼G is an
equivalence relation on X .
3. We want to determine |X / ∼G |, the number of equivalence classes in X under G-equivalence.
4. (Counting Lemma) Let θ (x) be the size of the unique equivalence class in X / ∼G containing x ∈ X (i.e. θ
is a class function (constant on classes)). Then trivially,
|X / ∼G | =
5. H is said to be a subgroup of G if
(a) H is closed under product in G
(b) H is closed under taking inverses in G
X
1
x∈X
θ (x)
28
COMBINATORIAL ENUMERATION
We write H < G
6. For g ∈ G, the R-coset of H in G containing g is a set of the form
H g := {hg | h ∈ H}
These partition G. We denote the number of distinct R-cosets of H in G by [G : H].
7. (Lagrange’s Theorem) Let H < G. Then
|G|
[G : H] =
|H|
8. (a) For x ∈ X , the orbit of x with respect to G is
G x := {g x | g ∈ G}
(b) For x ∈ X , the stabilizer of x with respect to G is
G x := {g ∈ G | g x = x}
9. From Lagrange’s Theorem, |G| = |G x||G x | for any x ∈ X . This is called the Orbit-Stabilizer Theorem.
10. For g ∈ G, the fix of g in X is the set of fixed points of g,
FixX (g) := {x ∈ X | g x = x}
8.4 Lemma. (Burnside’s Lemma) Let G be a finite group acting on a finite set X . Then
|X / ∼G | =
1 X
|G|
g∈G
X
1
|FixX (g)|
PROOF:
|X / ∼G | =
=
θ (x)
x∈X
X 1
|G x|
1 X
=
|G x |
|G| x∈X
1 XX
=
1
|G| x∈X g∈G
x∈X
g x=g
=
=
1 XX
|G|
1 X
|G|
1
g∈G x∈X
g x=g
|FixX (g)|
ƒ
g∈G
8.5 Theorem. (Pólya’s Theorem) Let G be a finite group acting on a finite set A . Let B = {b1 , . . . , bn } be a set
and let B A denote the set of all functions from A to B.
PÓLYA THEORY
29
1. We claim that gϕ = ϕ g −1 is an action of G on B A .
2. ϕ, ψ ∈ B A are said to be (induced) G-equivalent if there is g ∈ G such that ϕ = gψ. We write ϕ ∼0 ψ
(induced G-equivalence). Then
|B A / ∼0 | =
3. Let ρ be a weight function on B that records the number of occurrences of each bi .
[(B A / ∼, ρ)] = ZG 〈b1 + · · · + bn 〉
PROOF:
1. Let h, g ∈ G. Two of the conditions for G-action are trivial. The remaining condition is proved by
noting
(hg)ϕ = ϕ(hg)−1 = ϕ g −1 h−1 = h(ϕ g −1 ) = h(gϕ)
|B A / ∼0 | =
1 X
|G|
[(FixB A (g), ρ)]o
by Burnside’s Lemma
g∈G
<++>
[(ψ, ρ)]o =
n
Y
−1
(bi )|
[({bi }, ρ)]|ψ
o
i=1
If φ ∈ Fix(g) then gφ = φ, so
φ(a) = (gφ)(a) = φ(g −1 a)
by definition of the action of G on φ. Thus φ(g k a) = φ(g k−1 a), whence φ(g k a) = φ(a) for all k ≥ 1, so φ is
constant on cycles of g ∈ G. Thus the theorem is proved? Write this up better. =/
ƒ
For extensions of this theorem, look up deBruijn’s Theorem. It can be used for counting self complimenting
graphs (a graph Γ is self-complimenting when Γ = ΓC ). Also see the Redfield-Read Superposition Theorem.
8.4
Rooted Trees
Let n be the number of vertices. When n = 1 we have a single vertex, when n = 2 we have a single stalk.
When n = 3 there are two rooted trees, one beanpole and one vee. Notice that since these are not plane planted
trees, reflections of the same tree count as the same tree. Thus there are only four rooted trees on four vertices
(whereas there are five plane planted trees). How many trees are there on n vertices for general n?
We use the branch decomposition. Since the order doesn’t matter, we must “mod out” by the symmetric
group. Hence
a
N
TR −→
˜ {} ×
TR k /Sk
k≥0
so by Pólya’s Theorem
T (x) = x
X
ZSk 〈T (x)〉
k≥0
The cycle index polynomial of the symmetric group is the generating series for all permutations in Sk with
respect to cycle type. Now P −→
˜ U ? C , so we get
‚
Œ
X
X uk x
k
k
u ZSk (x 1 , . . . , x k ) = exp
k
k≥1
k≥0
30
COMBINATORIAL ENUMERATION
Let T (x) =
P
i≥1 t i x
i
, where t i is the number of rooted trees with n non-root vertices. Then
X
‚
t i x = x exp
i
i≥1
X1X
k≥1
‚
= x exp
X
k
Œ
ti x
ik
i≥1
Œ
t i log(1 − x )
i≥1
=x
Y
exp log(1 − x i )−t i
i≥1
Y
=x
(1 − x i )−t i
i≥1
Thus t k = [x k−1 ]
Q
i≥1 (1 −
x i )−t i . Check a few examples.
i −1