Bijective Proofs: You Say You Want an Involution?

Home , Bijective proof, Involution (mathematics)

Bijective Proofs: You say you want an involution?

David “D$” Lonoﬀ and Daniel “DMac” McDonald

Carleton College

March 5, 2009 A B A

Deﬁnition An involution is a bijection from a set to itself which is its own inverse.

Example

Introduction

Deﬁnition A bijection is a one-to-one and onto mapping. Deﬁnition An involution is a bijection from a set to itself which is its own inverse.

Introduction

Deﬁnition A bijection is a one-to-one and onto mapping.

Example A B A

Introduction

Deﬁnition A bijection is a one-to-one and onto mapping.

Deﬁnition An involution is a bijection from a set to itself which is its own inverse.

Example A B Introduction

Deﬁnition A bijection is a one-to-one and onto mapping.

Deﬁnition An involution is a bijection from a set to itself which is its own inverse.

Example A B A Example n X n = 2n k k=0 n n! ( k = k!(n−k)! is the number of unordered subsets of size k from a set of size n)

Example Are there an even or odd number of people in the room right now?

Bijective proof Involutive proof

Introduction

The philosophy of combinatorial proof Example n X n = 2n k k=0 n n! ( k = k!(n−k)! is the number of unordered subsets of size k from a set of size n)

Example Are there an even or odd number of people in the room right now?

Involutive proof

Introduction

The philosophy of combinatorial proof Bijective proof Involutive proof

Example Are there an even or odd number of people in the room right now?

Introduction

The philosophy of combinatorial proof Bijective proof

Example n X n = 2n k k=0 n n! ( k = k!(n−k)! is the number of unordered subsets of size k from a set of size n) Example Are there an even or odd number of people in the room right now?

Introduction

The philosophy of combinatorial proof Bijective proof Involutive proof

Example n X n = 2n k k=0 n n! ( k = k!(n−k)! is the number of unordered subsets of size k from a set of size n) Introduction

The philosophy of combinatorial proof Bijective proof Involutive proof

Example n X n = 2n k k=0 n n! ( k = k!(n−k)! is the number of unordered subsets of size k from a set of size n)

Example Are there an even or odd number of people in the room right now? Example There are 5 partitions of 4, as 4=3+1=2+2=2+1+1=1+1+1+1.

A partition λ of n with parts λ1 ≥ λ2 ≥ ... ≥ λs is either written m1 m2 mk as such or in the form λn1 λn2 . . . λnk , where each λni is a distinct

part in λ with λn1 > λn2 > . . . > λnk , and there are mi copies of

λni . Thus the partition 6 + 4 + 4 + 3 of 17 could be written as 6 ≥ 4 ≥ 4 ≥ 3 or 614231.

Partitions

Deﬁnition A partition of a positive integer n is an expression of n as the sum of a sequence of weakly decreasing positive integers, called the parts of the partition. A partition λ of n with parts λ1 ≥ λ2 ≥ ... ≥ λs is either written m1 m2 mk as such or in the form λn1 λn2 . . . λnk , where each λni is a distinct

part in λ with λn1 > λn2 > . . . > λnk , and there are mi copies of

λni . Thus the partition 6 + 4 + 4 + 3 of 17 could be written as 6 ≥ 4 ≥ 4 ≥ 3 or 614231.

Partitions

Deﬁnition A partition of a positive integer n is an expression of n as the sum of a sequence of weakly decreasing positive integers, called the parts of the partition.

Example There are 5 partitions of 4, as 4=3+1=2+2=2+1+1=1+1+1+1. Thus the partition 6 + 4 + 4 + 3 of 17 could be written as 6 ≥ 4 ≥ 4 ≥ 3 or 614231.

Partitions

Deﬁnition A partition of a positive integer n is an expression of n as the sum of a sequence of weakly decreasing positive integers, called the parts of the partition.

Example There are 5 partitions of 4, as 4=3+1=2+2=2+1+1=1+1+1+1.

A partition λ of n with parts λ1 ≥ λ2 ≥ ... ≥ λs is either written m1 m2 mk as such or in the form λn1 λn2 . . . λnk , where each λni is a distinct

part in λ with λn1 > λn2 > . . . > λnk , and there are mi copies of

λni . Partitions

Deﬁnition A partition of a positive integer n is an expression of n as the sum of a sequence of weakly decreasing positive integers, called the parts of the partition.

Example There are 5 partitions of 4, as 4=3+1=2+2=2+1+1=1+1+1+1.

A partition λ of n with parts λ1 ≥ λ2 ≥ ... ≥ λs is either written m1 m2 mk as such or in the form λn1 λn2 . . . λnk , where each λni is a distinct

part in λ with λn1 > λn2 > . . . > λnk , and there are mi copies of

λni . Thus the partition 6 + 4 + 4 + 3 of 17 could be written as 6 ≥ 4 ≥ 4 ≥ 3 or 614231. Example 7 5 3

Ferrers diagram

Deﬁnition

The Ferrers diagram of a partition λ = λ1 ≥ λ2 ≥ ... ≥ λs is a diagram of left-justiﬁed boxes with λi boxes in the ith row from the top. Ferrers diagram

Deﬁnition

The Ferrers diagram of a partition λ = λ1 ≥ λ2 ≥ ... ≥ λs is a diagram of left-justiﬁed boxes with λi boxes in the ith row from the top.

Example 7 5 3 Example

4 3 3 2 1 2 1

Ferrers diagram

Deﬁnition

The conjugate of a partition λ = λ1 ≥ λ2 ≥ ... ≥ λs is the partition whose Ferrers diagram has λi boxes in the ith column from the left. Ferrers diagram

Deﬁnition

The conjugate of a partition λ = λ1 ≥ λ2 ≥ ... ≥ λs is the partition whose Ferrers diagram has λi boxes in the ith column from the left.

Example

4 3 3 2 1 2 1 Example D(6) : {6, 5 + 1, 4 + 2, 3 + 2 + 1} O(6) : {5 + 1, 3 + 3, 3 + 1 + 1 + 1, 1+1+1+1+1+1}

Euler’s Theorem

Euler’s Theorem Let O(n) denote the set of partitions of n into odd parts, and let D(n) denote the set of partitions of n into distinct parts. Then for all n, |O(n)| = |D(n)|. Euler’s Theorem

Euler’s Theorem Let O(n) denote the set of partitions of n into odd parts, and let D(n) denote the set of partitions of n into distinct parts. Then for all n, |O(n)| = |D(n)|.

Example D(6) : {6, 5 + 1, 4 + 2, 3 + 2 + 1} O(6) : {5 + 1, 3 + 3, 3 + 1 + 1 + 1, 1+1+1+1+1+1} p For λ = λ1 > λ2 > . . . > λs ∈ D(n), let λi = 2 i µi , where µi is p odd. Let φ(λ) = µ, where µ has 2 i parts of µi , so that µ ∈ O(n). m1 m2 mk i1 i2 iq Write µ as µn1 µn2 . . . µnk . Each mi = 2 + 2 + ... + 2 for exactly one sequence of nonnegative integers i1 > i2 > . . . > iq. Thus µ could only be the image under φ of some λ with parts of it the form 2 µi , so φ is one-to-one. Since such a λ must have distinct parts, we have a well-deﬁned inverse mapping φ−1 : O(n) → D(n), so φ is onto.

Proof of Euler’s Theorem

We give a bijection φ : D(n) → O(n) devised by Glaisher (1883). p Let φ(λ) = µ, where µ has 2 i parts of µi , so that µ ∈ O(n). m1 m2 mk i1 i2 iq Write µ as µn1 µn2 . . . µnk . Each mi = 2 + 2 + ... + 2 for exactly one sequence of nonnegative integers i1 > i2 > . . . > iq. Thus µ could only be the image under φ of some λ with parts of it the form 2 µi , so φ is one-to-one. Since such a λ must have distinct parts, we have a well-deﬁned inverse mapping φ−1 : O(n) → D(n), so φ is onto.

Proof of Euler’s Theorem

We give a bijection φ : D(n) → O(n) devised by Glaisher (1883). p For λ = λ1 > λ2 > . . . > λs ∈ D(n), let λi = 2 i µi , where µi is odd. i i iq Each mi = 2 1 + 2 2 + ... + 2 for exactly one sequence of nonnegative integers i1 > i2 > . . . > iq. Thus µ could only be the image under φ of some λ with parts of it the form 2 µi , so φ is one-to-one. Since such a λ must have distinct parts, we have a well-deﬁned inverse mapping φ−1 : O(n) → D(n), so φ is onto.

Proof of Euler’s Theorem

We give a bijection φ : D(n) → O(n) devised by Glaisher (1883). p For λ = λ1 > λ2 > . . . > λs ∈ D(n), let λi = 2 i µi , where µi is p odd. Let φ(λ) = µ, where µ has 2 i parts of µi , so that µ ∈ O(n). m1 m2 mk Write µ as µn1 µn2 . . . µnk . Thus µ could only be the image under φ of some λ with parts of it the form 2 µi , so φ is one-to-one. Since such a λ must have distinct parts, we have a well-deﬁned inverse mapping φ−1 : O(n) → D(n), so φ is onto.

Proof of Euler’s Theorem

We give a bijection φ : D(n) → O(n) devised by Glaisher (1883). p For λ = λ1 > λ2 > . . . > λs ∈ D(n), let λi = 2 i µi , where µi is p odd. Let φ(λ) = µ, where µ has 2 i parts of µi , so that µ ∈ O(n). m1 m2 mk i1 i2 iq Write µ as µn1 µn2 . . . µnk . Each mi = 2 + 2 + ... + 2 for exactly one sequence of nonnegative integers i1 > i2 > . . . > iq. Thus µ could only be the image under φ of some λ with parts of it the form 2 µi , so φ is one-to-one. Since such a λ must have distinct parts, we have a well-deﬁned inverse mapping φ−1 : O(n) → D(n), so φ is onto. Theorem Let Q(n) be the set of partitions of n into distinct parts where the smallest part is odd. Then |Q(n)| is odd if and only if n is a perfect square.

Example Q(6): {5 + 1, 3 + 2 + 1} Q(7): {7, 6 + 1, 4 + 3, 4 + 2 + 1} Q(9): {9, 8 + 1, 6 + 3, 6 + 2 + 1, 5 + 3 + 1} Q(10): {9+1, 7+3, 7 + 2 + 1, 6 + 3 + 1, 5 + 4 + 1, 4 + 3 + 2 + 1}

Fine’s Theorem Example Q(6): {5 + 1, 3 + 2 + 1} Q(7): {7, 6 + 1, 4 + 3, 4 + 2 + 1} Q(9): {9, 8 + 1, 6 + 3, 6 + 2 + 1, 5 + 3 + 1} Q(10): {9+1, 7+3, 7 + 2 + 1, 6 + 3 + 1, 5 + 4 + 1, 4 + 3 + 2 + 1}

Fine’s Theorem

Theorem Let Q(n) be the set of partitions of n into distinct parts where the smallest part is odd. Then |Q(n)| is odd if and only if n is a perfect square. Fine’s Theorem

Theorem Let Q(n) be the set of partitions of n into distinct parts where the smallest part is odd. Then |Q(n)| is odd if and only if n is a perfect square.

Example Q(6): {5 + 1, 3 + 2 + 1} Q(7): {7, 6 + 1, 4 + 3, 4 + 2 + 1} Q(9): {9, 8 + 1, 6 + 3, 6 + 2 + 1, 5 + 3 + 1} Q(10): {9+1, 7+3, 7 + 2 + 1, 6 + 3 + 1, 5 + 4 + 1, 4 + 3 + 2 + 1} blahb ahb bahbahabhahb bhabhhabh abhahbahb bahahbhabhahb ahbhabhahbabhabhabh ahhabhab hahbhabh abhahba abhahbab abhab babjajbjab f ashdafoipa ashd as fd hsasdh asdhfa sdfha asdh dsahfdsa asdfha

Idea of the proof: Define an involution κ : Q(n) → Q(n) which has exactly one fixed point if and only if n is a perfect square (and has no fixed points if and only if n is not a perfect square).

Fine’s Theorem

Theorem Let Q(n) be the set of partitions of n into distinct parts where the smallest part is odd. Then |Q(n)| is odd if and only if n is a perfect square. blahb ahb bahbahabhahb bhabhhabh abhahbahb bahahbhabhahb ahbhabhahbabhabhabh ahhabhab hahbhabh abhahba abhahbab abhab babjajbjab f ashdafoipa ashd as fd hsasdh asdhfa sdfha asdh dsahfdsa asdfha

Fine’s Theorem

Theorem Let Q(n) be the set of partitions of n into distinct parts where the smallest part is odd. Then |Q(n)| is odd if and only if n is a perfect square.

3 If tα = e1 = ∞, then do nothing.

If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. 2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1.

3 If tα = e1 = ∞, then do nothing.

Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. 2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1.

3 If tα = e1 = ∞, then do nothing.

Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. 2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1.

3 If tα = e1 = ∞, then do nothing.

Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. 2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1.

3 If tα = e1 = ∞, then do nothing.

1 If tα < e1, then split sα into 2α − 1 and tα.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case. 2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1.

3 If tα = e1 = ∞, then do nothing.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα. 3 If tα = e1 = ∞, then do nothing.

Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα.

2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1. Proof of Fine’s Theorem

Let λ ∈ Q(n), with odd parts sk > sk−1 > . . . > s1 and smallest even part e1. If there are no even parts, deﬁne e1 = ∞. Deﬁne ti = si − (2i − 1), for i ≤ k, and let tk+1 = ∞. Let α be the smallest i such that ti > 0. Now, we construct κ(λ) by performing one of the following three actions, depending on the case.

1 If tα < e1, then split sα into 2α − 1 and tα.

2 If tα ≥ e1, and e1 < ∞, then combine sα−1 with e1.

3 If tα = e1 = ∞, then do nothing. Proof of Fine’s Theorem Proof of Fine’s Theorem Proof of Fine’s Theorem Proof of Fine’s Theorem Proof of Fine’s Theorem Proof of Fine’s Theorem Proof of Fine’s Theorem Kolberg’s Theorem

Kolberg’s Theorem The partition function p(n) takes inﬁnitely many even and odd values. Then p(n) has the same parity as |SC(n)|.

5 3 2 1 1

Let DO(n) = D(n) ∩ O(n) be the set of partitions of n into distinct odd parts. Then |SC(n)| = |DO(n)|.

5 3 9 2 3 1 1

Thus p(n) has the same parity as |DO(n)|.

Proof of Kolberg’s Theorem

Let SC(n) denote the set of self-conjugate partitions of n. Let DO(n) = D(n) ∩ O(n) be the set of partitions of n into distinct odd parts. Then |SC(n)| = |DO(n)|.

5 3 9 2 3 1 1

Thus p(n) has the same parity as |DO(n)|.

Proof of Kolberg’s Theorem

Let SC(n) denote the set of self-conjugate partitions of n. Then p(n) has the same parity as |SC(n)|.

5 3 2 1 1 Then |SC(n)| = |DO(n)|.

5 3 9 2 3 1 1

Thus p(n) has the same parity as |DO(n)|.

Proof of Kolberg’s Theorem

Let SC(n) denote the set of self-conjugate partitions of n. Then p(n) has the same parity as |SC(n)|.

5 3 2 1 1

Let DO(n) = D(n) ∩ O(n) be the set of partitions of n into distinct odd parts. Thus p(n) has the same parity as |DO(n)|.

Proof of Kolberg’s Theorem

Let SC(n) denote the set of self-conjugate partitions of n. Then p(n) has the same parity as |SC(n)|.

5 3 2 1 1

Let DO(n) = D(n) ∩ O(n) be the set of partitions of n into distinct odd parts. Then |SC(n)| = |DO(n)|.

5 3 9 2 3 1 1 Proof of Kolberg’s Theorem

Let SC(n) denote the set of self-conjugate partitions of n. Then p(n) has the same parity as |SC(n)|.

5 3 2 1 1

Let DO(n) = D(n) ∩ O(n) be the set of partitions of n into distinct odd parts. Then |SC(n)| = |DO(n)|.

5 3 9 2 3 1 1

Thus p(n) has the same parity as |DO(n)|. Let DO1(n) = {λ ∈ DO(n)|λ1 − λ2 = 2, λs > 1}. 7 5 3

Then |DO(n)| = |DO(n + 1)\DO1(n + 1)|. 9 9 1

5 3 7 1 3 Thus |DO(n)| diﬀers in parity from |DO(n + 1)| precisely when |DO1(n + 1)| is odd. Thus to prove the theorem we can show that |DO1(n)| is odd for inﬁnitely many n.

Proof of Kolberg’s Theorem

For any λ ∈ DO(n) denote the parts of λ by λ1 > λ2 > . . . > λs . Then |DO(n)| = |DO(n + 1)\DO1(n + 1)|. 9 9 1

5 3 7 1 3 Thus |DO(n)| diﬀers in parity from |DO(n + 1)| precisely when |DO1(n + 1)| is odd. Thus to prove the theorem we can show that |DO1(n)| is odd for inﬁnitely many n.

Proof of Kolberg’s Theorem

For any λ ∈ DO(n) denote the parts of λ by λ1 > λ2 > . . . > λs . Let DO1(n) = {λ ∈ DO(n)|λ1 − λ2 = 2, λs > 1}. 7 5 3 Thus |DO(n)| diﬀers in parity from |DO(n + 1)| precisely when |DO1(n + 1)| is odd. Thus to prove the theorem we can show that |DO1(n)| is odd for inﬁnitely many n.

Proof of Kolberg’s Theorem

For any λ ∈ DO(n) denote the parts of λ by λ1 > λ2 > . . . > λs . Let DO1(n) = {λ ∈ DO(n)|λ1 − λ2 = 2, λs > 1}. 7 5 3

Then |DO(n)| = |DO(n + 1)\DO1(n + 1)|. 9 9 1

5 3 7 1 3 Thus to prove the theorem we can show that |DO1(n)| is odd for inﬁnitely many n.

Proof of Kolberg’s Theorem

For any λ ∈ DO(n) denote the parts of λ by λ1 > λ2 > . . . > λs . Let DO1(n) = {λ ∈ DO(n)|λ1 − λ2 = 2, λs > 1}. 7 5 3

Then |DO(n)| = |DO(n + 1)\DO1(n + 1)|. 9 9 1

5 3 7 1 3 Thus |DO(n)| diﬀers in parity from |DO(n + 1)| precisely when |DO1(n + 1)| is odd. Proof of Kolberg’s Theorem

For any λ ∈ DO(n) denote the parts of λ by λ1 > λ2 > . . . > λs . Let DO1(n) = {λ ∈ DO(n)|λ1 − λ2 = 2, λs > 1}. 7 5 3

Then |DO(n)| = |DO(n + 1)\DO1(n + 1)|. 9 9 1

5 3 7 1 3 Thus |DO(n)| diﬀers in parity from |DO(n + 1)| precisely when |DO1(n + 1)| is odd. Thus to prove the theorem we can show that |DO1(n)| is odd for inﬁnitely many n. Then for all positive integers j, |DOj (n)| = |DOj (n + 2j + 2)\DOj+1(n + 2j + 2)|. 9 11 7 9 5 7 3 3

φ : DO (24) DO (30) \ DO (30) 2 2 2 3

Proof of Kolberg’s Theorem

Let DOj (n) = {λ ∈ DO(n)|λ1 − λ2 = λ2 − λ3 = ... = λj − λj+1 = 2, λs > 1}. 11 9 7 3 Thus |DOj (n)| diﬀers in parity from |DOj (n + 2j + 2)| precisely when |DOj+1(n + 2j + 2)| is odd.

Proof of Kolberg’s Theorem

Let DOj (n) = {λ ∈ DO(n)|λ1 − λ2 = λ2 − λ3 = ... = λj − λj+1 = 2, λs > 1}. 11 9 7 3

Then for all positive integers j, |DOj (n)| = |DOj (n + 2j + 2)\DOj+1(n + 2j + 2)|. 9 11 7 9 5 7 3 3

φ : DO (24) DO (30) \ DO (30) 2 2 2 3 Proof of Kolberg’s Theorem

Let DOj (n) = {λ ∈ DO(n)|λ1 − λ2 = λ2 − λ3 = ... = λj − λj+1 = 2, λs > 1}. 11 9 7 3

Then for all positive integers j, |DOj (n)| = |DOj (n + 2j + 2)\DOj+1(n + 2j + 2)|. 9 11 7 9 5 7 3 3

φ : DO (24) DO (30) \ DO (30) 2 2 2 3

Thus |DOj (n)| differs in parity from |DOj (n + 2j + 2)| precisely when |DOj+1(n + 2j + 2)| is odd. Note 2 that DOk−2(k − 1) = {(2k − 1) + (2k − 3) + ... + 3} and 2 2 therefore DOk−2(k − 1) = 1. Thus DOk−3(k − 1) and 2 DOk−3(k − 1 − 2(k − 2)) differ in parity, so we can pick 2 2 nk−3 ∈ k − 1, k − 1 − 2(k − 2) such that |DOk−3(nk−3)| is odd. Then |DOk−4(nk−3)| and |DOk−4(nk−3 − 2(k − 3))| differ in parity, so we can pick nk−4 ∈ {nk−3, nk−3 − 2(k − 3)} such that |DOk−4(nk−4)| is odd. We iterate this process to find n1. Then |DO1(n1)| is odd, and 2 n1 ≥ k − 1 − (2(k − 2) + 2(k − 3) + ... + 4) = k2 − 1 − 2(2 + 3 + ... + k − 2) = k2 − 1 − k(k − 3) = 3k − 1 > k.

Proof of Kolberg’s Theorem

Given k ≥ 1, we construct n > k such that |DO1(n)| is odd. 2 Thus DOk−3(k − 1) and 2 DOk−3(k − 1 − 2(k − 2)) differ in parity, so we can pick 2 2 nk−3 ∈ k − 1, k − 1 − 2(k − 2) such that |DOk−3(nk−3)| is odd. Then |DOk−4(nk−3)| and |DOk−4(nk−3 − 2(k − 3))| differ in parity, so we can pick nk−4 ∈ {nk−3, nk−3 − 2(k − 3)} such that |DOk−4(nk−4)| is odd. We iterate this process to find n1. Then |DO1(n1)| is odd, and 2 n1 ≥ k − 1 − (2(k − 2) + 2(k − 3) + ... + 4) = k2 − 1 − 2(2 + 3 + ... + k − 2) = k2 − 1 − k(k − 3) = 3k − 1 > k.

Proof of Kolberg’s Theorem

Given k ≥ 1, we construct n > k such that |DO1(n)| is odd. Note 2 that DOk−2(k − 1) = {(2k − 1) + (2k − 3) + ... + 3} and 2 therefore DOk−2(k − 1) = 1. Then |DOk−4(nk−3)| and |DOk−4(nk−3 − 2(k − 3))| diﬀer in parity, so we can pick nk−4 ∈ {nk−3, nk−3 − 2(k − 3)} such that |DOk−4(nk−4)| is odd. We iterate this process to ﬁnd n1. Then |DO1(n1)| is odd, and 2 n1 ≥ k − 1 − (2(k − 2) + 2(k − 3) + ... + 4) = k2 − 1 − 2(2 + 3 + ... + k − 2) = k2 − 1 − k(k − 3) = 3k − 1 > k.

Proof of Kolberg’s Theorem

Given k ≥ 1, we construct n > k such that |DO1(n)| is odd. Note 2 that DOk−2(k − 1) = {(2k − 1) + (2k − 3) + ... + 3} and 2 2 therefore DOk−2(k − 1) = 1. Thus DOk−3(k − 1) and 2 DOk−3(k − 1 − 2(k − 2)) diﬀer in parity, so we can pick 2 2 nk−3 ∈ k − 1, k − 1 − 2(k − 2) such that |DOk−3(nk−3)| is odd. We iterate this process to ﬁnd n1. Then |DO1(n1)| is odd, and 2 n1 ≥ k − 1 − (2(k − 2) + 2(k − 3) + ... + 4) = k2 − 1 − 2(2 + 3 + ... + k − 2) = k2 − 1 − k(k − 3) = 3k − 1 > k.

Proof of Kolberg’s Theorem

Given k ≥ 1, we construct n > k such that |DO1(n)| is odd. Note 2 that DOk−2(k − 1) = {(2k − 1) + (2k − 3) + ... + 3} and 2 2 therefore DOk−2(k − 1) = 1. Thus DOk−3(k − 1) and 2 DOk−3(k − 1 − 2(k − 2)) diﬀer in parity, so we can pick 2 2 nk−3 ∈ k − 1, k − 1 − 2(k − 2) such that |DOk−3(nk−3)| is odd. Then |DOk−4(nk−3)| and |DOk−4(nk−3 − 2(k − 3))| diﬀer in parity, so we can pick nk−4 ∈ {nk−3, nk−3 − 2(k − 3)} such that |DOk−4(nk−4)| is odd. Proof of Kolberg’s Theorem

Given k ≥ 1, we construct n > k such that |DO1(n)| is odd. Note 2 that DOk−2(k − 1) = {(2k − 1) + (2k − 3) + ... + 3} and 2 2 therefore DOk−2(k − 1) = 1. Thus DOk−3(k − 1) and 2 DOk−3(k − 1 − 2(k − 2)) differ in parity, so we can pick 2 2 nk−3 ∈ k − 1, k − 1 − 2(k − 2) such that |DOk−3(nk−3)| is odd. Then |DOk−4(nk−3)| and |DOk−4(nk−3 − 2(k − 3))| differ in parity, so we can pick nk−4 ∈ {nk−3, nk−3 − 2(k − 3)} such that |DOk−4(nk−4)| is odd. We iterate this process to find n1. Then |DO1(n1)| is odd, and 2 n1 ≥ k − 1 − (2(k − 2) + 2(k − 3) + ... + 4) = k2 − 1 − 2(2 + 3 + ... + k − 2) = k2 − 1 − k(k − 3) = 3k − 1 > k. Special thanks to our advisor Mark Krusemeyer.

Sources and Suggested Reading: G. Andrews and K. Eriksson. Integer Partitions. Cambridge University Press, 2004. C. Bessenrodt, I. Pak. Partition Congruences by Involutions. Europ. J. Combinatorics, vol. 25, 2004, 1139-1149. Pak, Igor. On Fine’s Partition Theorems, Dyson, Andrews, and Missed Opportunities. Math. Intelligencer, vol. 25, no.1, 2003, 10-16.