Supplemental Notes for Number Theory Andrew V. Sills Georgia Southern University January 29, 2012 i Contents Contents ii 1 Generating Functions 1 1.1 Definition and Simple Examples . 1 1.2 Polygonal Numbers . 2 1.3 The Fibonacci Numbers . 4 1.4 Exercises . 6 2 Integer Partitions 9 2.1 Introduction . 9 2.2 Partitions and Generating Functions . 12 2.3 Partitions with restrictions . 14 2.4 Elementary partition identities . 15 2.5 Graphical and combinatorial methods . 17 2.6 Euler’s pentagonal number theorem . 20 2.7 Euler’s recurrence . 23 2.8 Partition enumeration formulas . 24 2.9 Naudé’s question and two variable generating functions . 27 2.10 Exercises . 28 2.11 Notes . 30 3 Compositions 31 3.1 Definition and Elementary Results . 31 3.2 Exercises . 34 ii Chapter 1 Generating Functions 1.1 Definition and Simple Examples f g1 Definition 1.1. Let an n=0 be a sequence of numbers. Then the gener- ating function of fang is given by X1 n G(an; x) = anx : n=0 In other words, the generating function of a sequence fang encodes the terms of the sequence as coefficients of a power series. Building a power series out of a sequence of numbers is a powerful technique which is often used in combinatorics. (The convergence set of the power series will not concern us here as we will not be substituting particular values in for x.) Example 1.2. Find the generating function for the sequence of all ones: f g1 1 n=0 . Solution: 1 X 1 G(1; x) = xn = : 1 − x n=0 f g1 Example 1.3. Find the generating function for the sequence n n=0. Solution: We place the terms of the sequence (0; 1; 2; 3;:::; ) as coeffi- cients a power series to get x + 2x2 + 3x3 + 4x4 + ::: 1 2 CHAPTER 1. GENERATING FUNCTIONS Notice from the previous example that the geometric series is 1 1 + x + x2 + x3 + x4 + ··· = : (1.1) 1 − x Differentiate both sides of (1.1) to obtain 1 0 + 1 + 2x + 3x2 + 4x3 + ··· = : (1.2) (1 − x)2 Now multiply both sides of (1.2) by x to obtain x x + 2x2 + 3x3 + 4x4 + ··· = : (1 − x)2 Thus, x G(n; x) = : (1 − x)2 Many examples of Maclaurin series are studied in elementary calcu- lus. These Maclaurin series may be thought of as generating functions for particular sequences. For example, G 1 x • ( n! ; x) = e : • G(an; x) = sin x, where { − (n−1)/2 ( 1) if n is odd a = n! : n 0 if n is even (Advice: Write out the details! ) 1.2 Polygonal Numbers Triangular numbers The sequence (1; 1 + 2; 1 + 2 + 3; 1 + 2 + 3 + 4;::: ) = (1; 3; 6; 10; 15;::: ) is called the sequence of triangular numbers. Let ∆n denote the n-th triangular number. It is well known, and easily proved by induction, that Xn n(n + 1) ∆ = 1 + 2 + 3 + 4 + ··· + n = j = : n 2 j=1 We also define ∆0 = 0. 1.2. POLYGONAL NUMBERS 3 Proposition 1.4. The generating function for the triangular numbers is x G(∆ ; x) = n (1 − x)3 Proof. Twice differentiate the geometric series d2 ( ) d2 1 1 + x + x2 + x3 + x4 + ··· = dx2 dx2 (1 − x) to obtain 1 X 2 n(n − 1)xn−2 = : (1 − x)3 n=2 Multiply both sides by x/2, and reindex: 1 X n(n + 1) x xn = : 2 (1 − x)3 n=0 The extended triangular numbers are given by setting n to any integer value, positive, negative, or zero in ∆n = n(n+1)/2. Notice, however, that ∆−n = (−n)(−n + 1)/2 = n(n − 1)/2 = ∆n−1. So, each triangular number of negative index is the same as a certain triangular number of positive index. Square numbers The sequence (02; 12; 22; 32; 42;::: ) = (0; 1; 4; 9; 16; 25;::: ) is called the se- quence of square numbers or perfect squares. Let n denote the n-th square number. Clearly, 2 n = n : Proposition 1.5. The generating function of the square numbers is given by x(1 + x) G( ; x) = : n (1 − x)3 2 We may define the extended square numbers n = n for all integers 2 2 n, but −n = (−n) = n = n: 4 CHAPTER 1. GENERATING FUNCTIONS Pentagonal numbers The sequence (0; 1; 5; 12; 22;::: ) is called the sequence fPng of pentagonal 3 2 − 1 numbers. The n-th pentagonal number is given by 2 n 2 n: The extended pentagonal numbers of negative index are different from those of positive index. We have (:::; 26; 15; 7; 2; 0; 1; 5; 12; 22;::: ) as n runs from −∞ to 1. Hexagonal numbers The n-th hexagonal number is 2n2 − n. The extended hexagonal numbers coincide with the triangular numbers. 1.3 The Fibonacci Numbers Generating function Recall that the Fibonacci numbers may be defined recursively by F0 = 0;F1 = 1;Fn+1 = Fn + Fn−1 for n ≥ 1: (1.3) Let us denote by F(x) the generating function for the Fibonacci se- quence: X1 n F(x) := G(Fn; x) = Fnx : n=0 Multiply (1.3) by xn and sum over n ≥ 1 to obtain X1 X1 X1 n n n Fn+1x = Fnx + Fn−1x : n=1 n=1 n=1 This is equivalent to 1 ( ) F(x) − F − F x = F(x) + xF(x): x 0 1 Thus, F(x) − x = F(x) + xF(x) x 1.3. THE FIBONACCI NUMBERS 5 =)F(x) − x = xF(x) + x2F(x) =)F(x) − xF(x) − x2F(x) = x =) (1 − x − x2)F(x) = x x =)F(x) = : (1.4) 1 − x − x2 A Direct Formula for the nth Fibonacci number As an application, we will now derive a direct (i.e., non-recursive) formula for Fn. First recall that the golden ratio ϕ is defined as p 1 + 5 ϕ := : 2 Let us denote the conjugate of the golden ratio by p 1 − 5 ϕ¯ := : 2 Note that ϕϕ¯ = −1: (1.5) So, X1 n Fnx = F(x) n=0 x = 1 − x − x2 x = − x2 + x − 1 x = − (x + ϕ)(x + ϕ¯) p2 − ϕ p2 + ϕ¯ 5 − 5 = ¯ (by partial fractions) x(+ ϕ )x + ϕ( ) ϕ¯ p2 − ϕ ϕ p2 + ϕ¯ 5 5 = − ϕ¯(x + ϕ) ϕ(x + ϕ¯) p2 ϕ¯ + 1 p2 ϕ − 1 = 5 − 5 xϕ¯ − 1 xϕ − 1 6 CHAPTER 1. GENERATING FUNCTIONS p2 ϕ − 1 p2 ϕ¯ + 1 = 5 − 5 1 − xϕ 1 − xϕ¯ p1 p1 = 5 − 5 − − ¯ 1 (xϕ 1 xϕ ) 1 1 1 = p − 5 1 − ϕx 1 − ϕx¯ 1 1 1 X X = p (ϕx)n − (ϕx¯ )n (by (1.1)) 5 n=0 n=0 1 X ϕn − ϕ¯n = p xn: n=0 5 Compare coefficients of xn in the extremes to find that ( ) ( ) p n p n 1+ 5 1− 5 n n − ϕ − ϕ¯ 2 2 Fn = p = p : 5 5 Thus we see that knowing a recursive formula for a sequence may al- low us to derive its generating function. Then the generating function, together with some techniques from elementary calculus, allowed us to derive a direct formula for the terms of the original sequence. 1.4 Exercises 1. Find a closed form representation for each of the following generat- ing functions. a) G((−1)n; x). cos( πn ) b) G( 2 ; x). n! c) G(an; x), where { 0 if n = 0 an = 1 : n if n > 0 ( ) m(m − 1)(m − 2) ··· (m − k + 1) d) G ; x ; where m is a fixed real k! number. 1.4. EXERCISES 7 ( ) 1 − (−1)n e) G ; x : 2(n!) 2. Express the generating function for the pentagonal numbers as a rational function of x. 3. For n ≥ 3, let Ps(n) denote the nth s-gonal number, so that P3(n) = ∆n, P4(n) = n, etc. NoticeP that for the cases already considered, n − − − i.e. for s = 3; 4; 5; 6, Ps(n) = j=1(s 2)j (s 3), so let us define Xn Ps(n) = (s − 2)j − (s − 3); j=1 for all s = 3. Find a closed form formula for Ps(n). P 1 n 4. Express n=0 Ps(n)x as a rational function of x. f g1 5. The Pell numbers Pn n=0 may be defined recursively by P0 = 0;P1 = 1;Pn+1 = 2Pn + Pn−1 for n ≥ 1: a) Express G(Pn; x) as a rational function of x. b) Find a direct formula for Pn. 6. The Lucas numbers may be defined recursively as L0 = 2;L1 = 1;Ln+1 = Ln + Ln−1 for n ≥ 1: a) Express G(Ln; x) as a rational function of x. b) Find a direct formula for Ln. Chapter 2 Integer Partitions 2.1 Introduction In a letter dated September 4, 1740, Phillip Naudé asked Euler if he could figure out how many ways 50 could be be written as a sum of seven differ- ent positive integers.1 Apparently, Euler got very interested in this ques- tion and related questions, and thus was born the theory of integer parti- tions. Definition 2.1. A partition of an integer n is a representation of n as a sum of positive integers, where the order of the summands is considered irrelevant. Thus 2 + 3 and 3 + 2 are considered to be the same partition of 5. Definition 2.2. Each summand in a partition is called a part of the par- tition. Since the order of the parts is irrelevant, it is often convenient to im- pose a canonical ordering on the parts.