c 2010 Melissa Ann Dennison
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1 c 2010 Melissa Ann Dennison
2 A SEQUENCE RELATED TO THE STERN SEQUENCE BY MELISSA ANN DENNISON DISSERTATION Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate College of the University of Illinois at Urbana-Champaign, 2010 Urbana, Illinois Doctoral Committee: Professor Bruce C. Berndt, Chair Professor Bruce Reznick, Director of Research Professor Kenneth B. Stolarsky Professor Alexandru Zaharescu
3 Abstract In this dissertation we define and study a two-parameter family of recursive sequences which we call the bow sequences. The general bow sequence is defined similarly to the Stern sequence, and many of the properties of the bow sequences are related to known properties of the Stern sequence. In particular, we derive the generating function for the general bow sequence, and give interpretations of the generating function for two basic cases. We also determine properties of the bow sequences modulo 2 and 3, and give conjectures for the behavior of the bow sequences modulo d for d 4. Finally, we discuss ideas for future research. ii
4 For my parents, for always believing in me. iii
5 Acknowledgements I would like to thank my advisor, Bruce Reznick, for his guidance, enthusiasm, and patience. I m also grateful to my committee members, Bruce Berndt, Kenneth Stolarsky, and Alexandru Zaharescu, for their advice and support. I would also like to thank my husband Brad for his love and encouragement. And a special thanks to my daughter Kajsa, for brightening my days. iv
6 Table of Contents Chapter 1 Introduction History Questions of Interest Chapter 2 The General Bow Sequence Preliminaries Triatomic Arrays Greatest Common Divisors and Maxima Formulas for b α,β (2 r n + k) Properties of the Summatory Function Chapter 3 Generating Functions The Generating Function for the Stern Sequence The Generating Function for the General Bow Sequence A Relative of the Bow Sequences Chapter 4 The General Bow Sequence Modulo The Behavior of b α,β (n) Modulo The Behavior of b 1 (n) and b 0 (n) Modulo Chapter 5 Higher Moduli Background Distributions Modulo d Chapter 6 Another Representation of b 0 (n) Preliminaries A New Formula for b 0 (n) Chapter 7 Directions for Future Research Non-Standard Binary Representations Bow Sequences Modulo d More Formulas for b α,β (2 r + k) Primitive Values Chapter 8 References v
7 Appendix A Appendix B Appendix C vi
8 Chapter 1 Introduction In this dissertation we define and study a family of recursive sequences which we call the bow sequences. The general bow sequence is defined similarly to the Stern sequence, and many of the properties of the bow sequences are related to known properties of the Stern sequence. We begin by discussing the main properties of the Stern sequence, and follow with the corresponding properties for the bow sequences as well as properties for some special cases. 1.1 History In 1858, the mathematician M. A. Stern [17] studied the so-called diatomic array of integers, which was motivated by a function studied by Eisenstein. The diatomic array can be completely understood by consideration of the Stern sequence. The Stern sequence is defined as follows: s(0) = 0, s(1) = 1; (1.1) s(2n) =s(n), for n 1; (1.2) s(2n + 1) = s(n)+s(n + 1), for n 1. (1.3) The terms in the Stern sequence can be written in an array, where the r th row consists of s(n) for 2 r n 2 r+1 for r 0. The even entries in each row are copied from the previous row, and the odd entries are found by adding adjacent entries in the row above; it is a Pascal triangle with memory. The first five rows of the array are given in Table 1.1. De Rham [5] was the first to consider the Stern sequence as defined above. He attributed the name to Bachmann [2], who considered only the diatomic array. The related Stern-Brocot array [7] was used in defining Minkowski s?-function [11], and the Stern sequence has recently been used to understand 2-regular sequences [1] and the Tower of Hanoi graph [9]. The following are 1
9 Table 1.1: The first five rows of the diatomic array for the Stern sequence a few of the important properties of the Stern sequence: Proposition [4], [10], [14], [17] For n 0, every pair of consecutive terms of the Stern sequence is relatively prime, or gcd(s(n),s(n + 1)) = 1. Moreover, given a, b 1 such that gcd(a, b) =1, there is exactly one n such that s(n) =a and s(n + 1) = b. Proposition implies the following proposition. Proposition [10], [14] There are exactly φ(m) odd integers n 1 with the property that s(n) =m. The reason is that, since ( s(2k),s(2k + 1) ) = ( s(k),s(k)+s(k + 1) ), (1.4) it follows that in every pair (a, b) =(s(2k),s(2k + 1)) we have a < b and gcd(a, b) = 1. In particular, for a fixed m, there are exactly φ(m) possible values for s(2k) when s(2k + 1) = m. Next we will show how we can quickly determine all such odd n. shown in [10], continued fractions can be used to determine a formula for the Stern sequence. Let t(n) be defined as follows: t(n) := As s(n) s(n + 1). (1.5) Then we can see by Proposition that (t(n)) provides an enumeration of the positive rationals. Moreover, one can recover s(n) and s(n+1) from t(n). By considering binary representations, we can find a simple closed formula for t(n). This formula gives rise to a direct formula for s(n). Suppose n is odd. Then the binary representation of n consists of a 1 1 s, followed by a 2 0 s, a 3 1 s, and so on, ending with a 2v 0 s and a 2v+1 1 s. We shall write n [a 1,..., a 2v+1 ]. Then we have the following proposition. 2
10 Proposition [10] If n is odd and n [a 1,..., a 2v+1 ], then t(n) = s(n) s(n + 1) = a 1 2v+1 + a 2v + 1. (1.6) + 1 a 1 This formula can be used to determine specific solutions to s(n) = m for a fixed m. Example 1. Suppose n is odd and s(n) = 10. Since gcd(10,s(n + 1)) = 1 and s(n + 1) < 10, we must have s(n + 1) {1, 3, 7, 9}. We can compute the finite continued fraction in each case as follows: 10 1 = 10, 10 3 = , 10 7 = = Since the second and fourth continued fractions have an even number of denominators, we adjust the last denominator as follows: 10 3 = , = Now, reading these denominators from right to left, we see that s(n) = 10 when [n] 2 [10], [1, 2, 3], [3, 2, 1], or [1, 8, 1]. By converting back from binary, we find that s(n) = 10 for n odd exactly when n = 1023, 39, 57, or 513. For example, 39 = [ ] 2. Next, we consider the terms of the Stern sequence modulo d. While the patterns are more complicated for higher moduli, the terms of the Stern sequence exhibit a very regular property modulo 2. Proposition [13], [14], [15] For n 0, s(n) is even precisely when 3 n., 3
11 The formulas for the terms of the Stern sequence which are congruent to k modulo d are more complex. Formulas for the cases d = 3 and 4 can be found in [14], and [15]. We will, however, state the following result for the distribution of pairs modulo d. Proposition [14] The pair (s(n) (mod d),s(n + 1) (mod d)) is uniformly distributed amongst all possible pairs (i, j) with gcd(i, j, d) = 1. Remark. This proposition implies that the density of terms in the Stern sequence which are divisible by a prime p is 1 p Questions of Interest In this dissertation we define a new two-parameter family of recursive sequences called the bow sequences, which have the flipped recursion from the Stern recursion. The general bow sequence is defined as follows: b α,β (0) = 0, b α,β (1) = α, b α,β (2) = β; (1.7) b α,β (2n) =b α,β (n)+b α,β (n + 1), for n 2; (1.8) b α,β (2n + 1) = b α,β (n), for n 1. (1.9) We shall discuss properties of the bow sequences and how these properties relate to known properties of the Stern sequence. In particular, we would like to answer the following questions: Question 1. What is the greatest common divisor of consecutive terms in the various bow sequences? It is not true, in general, that the greatest common divisor of two consecutive terms in the bow sequences is 1. In fact, b 1,0 (13) = b 1,0 (14) = 2, and b 0,1 (21) = b 0,1 (22) = 2, for example. We can, however, state a property similar to Proposition for three consecutive terms in the general bow sequence. In Chapter 2 we will show that gcd(b α,β (n),b α,β (n + 1),b α,β (n + 2)) = gcd(α, β). 4
12 Question 2. Can we find a formula for b α,β (n) that does not depend on the recursion? Although we have not discovered a binary or continued fraction expression for the general bow sequence which is similar to the formula for the Stern sequence found in Proposition 1.1.3, we have determined a generating function for the general bow sequence and the values of the bow sequence for special values of α, β, and n. These formulas and generating functions can be found in Chapters 2 and 3. We also have work towards a reduced form involving matrices in Chapter 7. Question 3. What is the distribution of terms b α,β (n) modulo d for a given pair (α, β) and an integer d? For the case d = 2 we find that 3 of the terms are even for a given pair 7 {α, β} where gcd(α, β) is odd. Clearly, in the case where gcd(α, β) is even, all of the terms in the bow sequence are even. Theorems detailing exactly where even terms occur for each pair {α, β} can be found in Chapter 4. For higher moduli, we give a conjecture for the fraction of terms which are congruent to k modulo d in Chapter 5. We have proved the conjecture to be true in the case d = 3, and we present numerical evidence for d 10 in Appendix C. Question 4. What is the distribution of triples (b α,β (n) (mod d), b α,β (n + 1) (mod d), b α,β (n + 2) (mod d)) for a given pair {α, β} and integer d? Although pairs of terms are considered in the Stern sequence, we will mainly consider triples for the general bow sequence. Similar to Proposition 1.1.5, in Chapter 5 we show for d = 2 and 3 that if gcd(α, β) = 1, the triple (b α,β (n) (mod d), b α,β (n + 1) (mod d), b α,β (n + 2) (mod d)) is uniformly distributed amongst all triples (i (mod d), j(mod d), k(mod d)) such that gcd(i, j, k, d) =1. 5
13 Remark. This implies that if gcd(α, β) = 1, then the density of terms in the bow sequence which are divisible by a prime p is p2 1. We have proved the p 3 1 formula to be true for p = 2 and 3, and we give numerical evidence to support this conjecture for larger values in Appendix B. We have a more complicated conjecture detailing the density of terms which would be divisible by an integer d in Chapter 5. Question 5. For a given m, how many even integers n 2 are there such that b α,β (n) =m? In Proposition we considered the number of odd integers n 1 with the property that s(n) = m. For the general bow sequence, we shall consider the even terms since (1.9) tells us that if b α,β (n) =m, then b α,β (2 r (n + 1) 1) = m, for r 0. Accordingly, for specific pairs {α, β}, we would like to consider even integers n 2 with the property that b α,β (n) =m. A preliminary discussion of these properties can be found in Chapter 7. 6
14 Chapter 2 The General Bow Sequence We begin by defining the bow sequences recursively as in Chapter 1. The general bow sequence, b α,β (n), for α, β Z is defined by: b α,β (0) = 0, b α,β (1) = α, b α,β (2) = β; (2.1) b α,β (2n) =b α,β (n)+b α,β (n + 1), for n 2; (2.2) b α,β (2n + 1) = b α,β (n), for n 1. (2.3) We define b α,β (0) = 0 to simplify results, although the designation does not affect later terms in the sequence, as b α,β (0) does not enter into the recurrence. Here is a table listing the first 17 values of the general bow sequence: n b α,β (n) α 2 β 3 α 4 α + β 5 β 6 2α + β 7 α 8 α +2β 9 α + β 10 2α +2β 11 β 12 3α + β 13 2α + β 14 2α +2β 15 α 16 2α +3β Table 2.1: The general bow sequence 7
15 Note that the terms of the general bow sequence are not necessarily distinct. In particular, we can see from Table 2.1 that b α,β (10) = b α,β (14). It is also worth noting that (2.2) fails in the case where n = 1, since if α 0, then b α,β (2) b α,β (1) + b α,β (2). 2.1 Preliminaries By considering the recursion, we see that b α,β (n) is linear in α and β. Accordingly, we can write b α,β (n) =αb 1,0 (n)+βb 0,1 (n). (2.4) For simplicity, we shall define b 0 (n) := b 0,1 (n), and b 1 (n) := b 1,0 (n). In this dissertation we mainly consider the sequences b 0 (n) and b 1 (n), as all other cases of the general bow sequence can be determined from these two cases by applying equation (2.4). Remark. Three applications of the recursion to 2n, 2n + 1, and 2n + 3 give the following. For n 2, b α,β (2n) =b α,β (2n + 1) + b α,β (2n + 3). (2.5) For α, β 0, the following inequalities follow directly: b α,β (2n) b α,β (2n + 1), b α,β (2n) b α,β (2n + 3). Notice that equation (2.5) can be rewritten as b α,β (2n + 1) = b α,β (2n 2) b α,β (2n 1). (2.6) 8
16 By applying (1.2) and (1.3) to the Stern sequence, we can construct a property similar to (2.5) for the Stern sequence: for n 1, s(2n + 1) = s(2n)+s(2n + 2). (2.7) Remark. Note that iterating (2.2) and (2.3) also gives the following identities: b α,β (4n) =2b α,β (n)+b α,β (n + 1), (2.8) b α,β (4n + 1) = b α,β (n)+b α,β (n + 1), b α,β (4n + 2) = b α,β (n)+b α,β (n + 1) + b α,β (n + 2), b α,β (4n + 3) = b α,β (n). This means that for α, β 0 we have b α,β (4n + 3) b α,β (4n + 1) b α,β (4n + 2), and b α,β (4n) =b α,β (2n)+b α,β (n). (2.9) 2.2 Triatomic Arrays We can present the values of the general bow sequence as a triatomic array where the first row is b α,β (1), b α,β (2), b α,β (3), and the second row is b α,β (3), b α,β (4), b α,β (5), b α,β (6), b α,β (7). In general the r-th row contains b α,β (2 r 1),..., b α,β (2 r+1 1). The odd entries in each row are copied from the previous row in Table 2.2. The even entries are found by adding the two numbers in the previous row that occur to the right of the entry. For example, the second entry in the third row, b α,β (8) = α +2β, is obtained by adding the second and third entries in the second row of the array, namely, α + β and β. The last even entry in each row is obtained by summing the first two entries in that same row. Here are the first three rows of the array: α β α α α+ β β 2α + β α α α+2β α+ β 2α +2β β 3α + β 2α + β 2α +2β α Table 2.2: First three rows of the triatomic array for the general bow sequence 9
17 The triatomic arrays for b 0 (n) and b 1 (n) are quite different from each other. As we will see in Chapter 5, b 0 (n) is much more regular than b 1 (n), but the two sequences share many properties. Consider the tables below: Table 2.3: First four rows of the triatomic array for b 0 (n) Table 2.4: First four rows of the triatomic array for b 1 (n) 2.3 Greatest Common Divisors and Maxima One interesting property of the bow sequences is that although two consecutive terms may share a nontrivial factor, three consecutive terms can only share factors which divide gcd (α, β). Theorem For all {α, β} and n>0, gcd(b α,β (n),b α,β (n + 1),b α,β (n + 2)) = gcd(α, β). (2.10) Proof. Observe that gcd(b α,β (n),b α,β (n + 1),b α,β (n + 2)) = gcd(α, β) for n = 1. Now assume that this holds true for n < k. 10
18 Case 1: k =2n + 1, with n < k. Then, gcd(b α,β (k),b α,β (k + 1),b α,β (k + 2)) = gcd(b α,β (2n + 1),b α,β (2n + 2),b α,β (2n + 3)) = gcd(b α,β (n),b α,β (n + 1) + b α,β (n + 2),b α,β (n + 2)) = gcd(b α,β (n),b α,β (n + 1),b α,β (n + 2)) = gcd(α, β). Case 2: k =2n, with n < k. Then, gcd(b α,β (k),b α,β (k + 1),b α,β (k + 2)) = gcd(b α,β (2n),b α,β (2n + 1),b α,β (2n + 2)) = gcd(b α,β (n)+b α,β (n + 1),b α,β (n),b α,β (n + 1) + b α,β (n + 2)) = gcd(b α,β (n),b α,β (n + 1),b α,β (n + 2)) = gcd(α, β). Thus gcd(b α,β (n),b α,β (n + 1),b α,β (n + 2)) = gcd(α, β) for all {α, β} and n>0. Surprisingly, Theorem has implications for pairs. Although we have seen that pairs frequently share common factors, when the pairs (b 0 (n),b 0 (n + 1)), and (b 1 (n),b 1 (n + 1)) are taken together, the resulting quadruple does not have a common factor. Theorem For n 0, gcd(b 0 (n),b 0 (n + 1),b 1 (n),b 1 (n + 1)) = 1. Proof. Suppose p is prime, and let p gcd(b 0 (n),b 0 (n + 1),b 1 (n),b 1 (n + 1)). Then, p must necessarily divide b 0 (n) and b 0 (n + 1), so we know by Theorem that p b 0 (n + 2). Similarly, p b 1 (n + 2). So let b 0 (n + 2) r (mod p) and b 1 (n + 2) s (mod p), where 1 r, s p 1. 11
19 Consider b s, r (n) =sb 1 (n) rb 0 (n). Thus p b s, r (n), and p b s, r (n+1). Then, since b s, r (n + 2) sr +( r)s 0 (mod p), we know that p b s, r (n + 2). But this implies that p gcd(r, s), which is a contradiction. Thus gcd(b 0 (n),b 0 (n + 1),b 1 (n),b 1 (n + 1)) = 1. Remark. Note that Theorem fails if we consider only three of the four terms. For example, consider n = Then we have (b 0 (n),b 0 (n + 1),b 1 (n),b 1 (n + 1)) = (2070, 1815, 1430, 1254). When we consider the triples, we find that gcd(b 0 (n),b 0 (n + 1),b 1 (n)) = 5, gcd(b 0 (n),b 0 (n + 1),b 1 (n + 1)) = 3, gcd(b 0 (n),b 1 (n),b 1 (n + 1)) = 2, gcd(b 0 (n + 1),b 1 (n),b 1 (n + 1)) = 11. However, clearly gcd(b 0 (n),b 0 (n + 1),b 1 (n),b 1 (n + 1)) = 1. Similarly, we have the following theorem. Theorem For n 0, gcd(b 0 (n),b 0 (n + 2),b 1 (n),b 1 (n + 2)) = 1. Proof. Suppose p is prime, and let p gcd(b 0 (n),b 0 (n + 2),b 1 (n),b 1 (n + 2)). Then, p must necessarily divide b 0 (n) and b 0 (n + 2), so we know by Theorem that p b 0 (n + 1). Similarly, p b 1 (n + 1). So let b 0 (n + 1) r (mod p) and b 1 (n + 1) s (mod p), where 1 r, s p 1. Consider b s, r (n) =sb 1 (n) rb 0 (n). Thus p b s, r (n), and p b s, r (n+2). Then, since b s, r (n + 1) sr +( r)s 0 (mod p), 12
20 we know that p b s, r (n + 1). But this implies that p gcd(r, s), which is a contradiction. Thus gcd(b 0 (n),b 0 (n + 2),b 1 (n),b 1 (n + 2)) = 1. For the next theorem, we will need the Fibonacci numbers, defined as usual by F 0 =0, F 1 = 1; F n = F n 1 + F n 2, for n 2. Theorem For r 1, b α,β (2 r )=αf r 1 + βf r. (2.11) In particular, for r 1, b 0 (2 r )=F r, b 1 (2 r )=F r 1, and b 1,1 (2 r )=F r+1. Proof. First, note that by applying (2.9) we find b α,β (2 r )=b α,β (2 r 1 )+b α,β (2 r 2 ). Secondly, recall by (2.4) that b α,β (2 r )=αb 1 (2 r )+βb 0 (2 r ). Thus we consider only these two cases. All that remains is to check the initial conditions. We find that (b 0 (2),b 0 (4)) = (0, 1) = (F 0,F 1 ), and (b 1 (2),b 1 (4)) = (1, 1) = (F 1,F 2 ). Thus the initial conditions are satisfied. Next, we estimate the size of b α,β (n). Let I r = (2 r 1, 2 r ] Z. To be 13
21 specific, I r = {2 r 1 +1,, 2 r }. Note that I r = (2I r 1 1) 2I r 1. We have the following upper bound. Theorem For r 1, Moreover, for β α 0, max b α,β (n) max { α, β }F r+1. (2.12) n I r max b α,β (n) = b α,β (2 r )=αf r 1 + βf r. (2.13) n I r Proof. Since b α,β (n) 0 for α, β 0, b α,β (n) b α, β (n), and we may assume that α, β 0. Observe that for α, β 0 and γ = max{α, β}, b α,β (n) b γ,γ (n) =γb 1,1 (n) = γ (b 0 (n)+b 1 (n)). For r =1, 2, it is easy to see that max n Ir b 1,1 (n) is F r+1. Now assume that this holds for all r < k. Let r = k. Then max b 1,1 (n) = max {b 1,1 (2n),b 1,1 (2n 1)} n I k n I k 1 = max n I k 1 {b 1,1 (n)+b 1,1 (n + 1),b 1,1 (n 1)}. 14
22 However, one of n or n + 1 is odd. Thus, if n, n +1 I k 1, then max {b 1,1 (n)+b 1,1 (n + 1),b 1,1 (n 1)} max{f k + F k 1,F k } n I k 1 = max{f k+1,f k } = F k+1. The previous argument fails if n + 1 is not in I k 1. However, then n + 1 = 2 k 1 +1, and b 1,1 (2 k 1 + 1) = b 1,1 (2 k 2 )=F k 1 by Theorem Thus we have shown that max b γ,γ (n) γf k+1. n I k Now we will show that max n Ir b 0 (n) F r. For r =1, 2, it is easy to see that max n Ir b 0 (n) =F r. Now assume that this statement also holds for all r < k. Let r = k. Then max b 0 (n) = max {b 0 (2n),b 0 (2n 1)} n I k n I k 1 = max n I k 1 {b 0 (n)+b 0 (n + 1),b 0 (n 1)}. Similarly, one of n or n+1 is odd, with the same result if n+1 = 2 k Thus, if n, n +1 I k 1, then max {b 0 (n)+b 0 (n + 1),b 0 (n 1)} max{f k 1 + F k 2,F k 1 } n I k 1 = max{f k,f k 1 } = F k. With δ> 0 and b 0,δ (n) =δb 0 (n), we have max n Ir b 0,δ (n) δf r. 15
23 Now we shall show that the maximum occurs at n =2 r. Let β α 0. Then, by setting γ = α and δ = β α in the previous results above, we find b α,β (2 r )=αb 1,1 (2 r )+(β α)b 0 (2 r ) αf r+1 +(β α)f r = αf r 1 + βf r. However, by Theorem we know that b α,β (2 r )=αf r 1 + βf r, thus the maximum occurs at n =2 r. 2.4 Formulas for b α,β (2 r n + k) Here are some results on b α,β (2 r n + k) for fixed values of k. Theorem The following formulas hold for the general bow sequence. (1). b α,β (2 r n 4) = F r 1 b α,β (n)+f r 2 b α,β (n + 1) + 2b α,β (n 1), for r 2, n 2. (2). b α,β (2 r n 3) = F r 1 b α,β (n)+f r 2 b α,β (n + 1) + b α,β (n 1), for r 2, n 2. (3). b α,β (2 r n 2) = F r b α,β (n) +F r 1 b α,β (n + 1) + b α,β (n 1), for r 1, n 2. (4). b α,β (2 r n 1) = b α,β (n 1), for r 0, n 1. (5). b α,β (2 r n)=f r+1 b α,β (n)+f r b α,β (n + 1), for r 0, n 2. (6). b α,β (2 r n + 1) = F r b α,β (n)+f r 1 b α,β (n + 1), for r 1, n 2. (7). b α,β (2 r n+2) = (F r+1 1)b α,β (n)+f r b α,β (n+1)+b α,β (n+2), for r 1, n 2. (8). b α,β (2 r n + 3) = F r 1 b α,β (n)+f r 2 b α,β (n + 1), for r 2, n 2. (9). b α,β (2 r n + 4) = (F r + F r 2 1)b α,β (n) +(F r 1 + F r 3 )b α,β (n + 1) + b α,β (n + 2), for r 3, n 2. 16
24 Proof. To prove (4) we apply the recurrence repeatedly and obtain b α,β (2 r n 1) = b α,β (2 r 1 n 1) = = b α,β (n 1). We show (5) by induction on r. First note by (2.9) that b α,β (2 r+2 n)=b α,β (2 r+1 n)+b α,β (2 r n). So all we need to do is to verify the base cases r = 0, and 1: b α,β (n) =F 1 b α,β (n)+f 0 b α,β (n + 1), b α,β (2n) =F 2 b α,β (n)+f 1 b α,β (n + 1). Since F 0 = 0, and F 1 = F 2 = 1, these are immediate. Property (6) is due to property (5) along with the recursion. We show (7) by induction on r. First note that for r =1, 2 the assertion is that b α,β (2n + 2) = (F 2 1)b α,β (n)+f 1 b α,β (n + 1) + b α,β (n + 2), b α,β (4n + 2) = (F 3 1)b α,β (n)+f 2 b α,β (n + 1) + b α,β (n + 2). Since F 1 = F 2 = 1, and F 3 = 2, these are immediate. Assuming that this holds for r < k, we now let r = k: b α,β (2 k n + 2) = b α,β (2 k 1 n + 1) + b α,β (2 k 1 n + 2) = F k 1 b α,β (n)+f k 2 b α,β (n + 1) + (F k 1)b α,β (n) + F k 1 b α,β (n + 1) + b α,β (n + 2) =(F k+1 1)b α,β (n)+f k b α,β (n + 1) + b α,β (n + 2). Property (8) is due to property (6). Property (9) is due to (7) and (8). Property (3) is due to (4) and (5). Property (2) is due to property (3). Lastly, property (1) is due to properties (3) and (4). We get the following corollary for b 0 (n) and b 1 (n). Corollary For r 1, the following are true for b 0 (n) and b 1 (n): (1). b 0 (3 2 r )=F r, b 0 (5 2 r )=F r+2, b 0 (7 2 r )=2F r, b 0 (9 2 r ) = 2F r +F r+1, b 0 (11 2 r ) = 2F r+2, b 0 (13 2 r )=2F r + F r+1, and b 0 (15 2 r ) = 3F r. 17
25 (2). b 1 (3 2 r )=F r+2, b 1 (5 2 r )=2F r, b 1 (7 2 r )=F r+2, b 1 (9 2 r )=2F r +F r+1, b 1 (11 2 r ) = 3F r, b 1 (13 2 r ) = 2F r+2, and b 1 (15 2 r ) = 2F r + F r+1. Proof. These properties can be verified by letting n =3, 5, 7, 9, 11, 13, and 15 in Theorem 2.4.1(5). Remark. It is interesting to note that the second property in (2) implies that b 1 (5 2 r ) is always even. Then, by the recurrence, we also know that b 1 (5 2 r + 1) is also always even. This gives an infinite number of even pairs {b 1 (5 2 r ),b 1 (5 2 r + 1)}. Also, the third property in (1) implies that b 0 (7 2 r ) is always even. Hence, we have an infinite number of even pairs {b 0 (7 2 r ),b 0 (7 2 r + 1)}. Similarly, we also have an infinite number of pairs {b 0 (15 2 r ),b 0 (15 2 r + 1)} and {b 1 (11 2 r ),b 1 (11 2 r + 1)} which are multiples of three. Remark. Note that b 0 (5 2 r )=F r+2 = b 1 (7 2 r ), and b 1 (5 2 r )=2F r = b 0 (7 2 r ). Additionally, we note that b 0 (9 2 r )=b 1 (9 2 r ), which we will use in Chapter 7. Lemma b 0 (n) = 0 n =2 r 1, for r 0. Proof. Theorem shows that if n =2 r 1, then b 0 (n) = 0. Suppose b 0 (n) = 0. If n is even, then b 0 (2k) = 0 means b 0 (k) +b 0 (k + 1) = 0, and the only time this happens is when k = 0, as we can see from Table 2.3. Otherwise, n must be odd, which means that the other zeroes come from the recursion, and thus n =2 r 1. Lemma b 1 (n) = 0 n =0or n =3 2 r 1, for r 0. Proof. First, we know by Theorem 2.4.1(4) that for r 0, b 1 (3 2 r 1) = b 1 (2) = 0. 18
26 Now, suppose b 1 (n) = 0. If n was even, then b 1 (2k) = 0 means n = 2 or b 1 (k)+b 1 (k + 1) = 0, which never happens, as we can see from Table 2.4. So n must be either 2 or odd. Thus all the other zeroes come from the recursion, and n =3 2 r 1. For r 1, there exist infinitely many n and infinitely many m such that F r gcd(b 0 (n),b 0 (n + 1)), and F r gcd(b 1 (m),b 1 (m + 1)). In fact, if we let P r = {n n = (2 r 1)2 j,j 1} and Q r = {n n = (3 2 r 1)2 j,j 1}, then we have the following theorem. Theorem For r 1, if n P r, then F r gcd(b 0 (n),b 0 (n + 1)), and if r 2 and m Q r 2, then F r gcd(b 1 (m),b 1 (m + 1)). Proof. By Theorem 2.4.1(5) we know that b α,β (2 r (2k + 1)) = F r+1 b α,β (2k + 1) + F r b α,β (2k + 2) = F r+1 b α,β (k)+f r (b α,β (k + 1) + b α,β (k + 2)). By Lemmas and 2.4.4, k =2 r 1 = b 0 (k) = 0, and k =3 2 r 1 = b 1 (k) =0. So if we choose n = (2 r 1)2 j for r 2, j 1, and m = (3 2 r 1)2 j for r, j 1, then F r gcd(b 0 (n),b 0 (n + 1)), and F r+2 gcd(b 1 (m),b 1 (m + 1)). Lemma For any integer r, there exists m such that r F m. Proof. For d 2, let S d denote the set {u, v} of pairs of residue classes modulo d. Then the (invertible) map h on S d defined by h(u, v) = (v, u + v) has the property that h(f n, F n+1 )=(F n+1, F n+2 ). Since h induces a permutation on S d,(0, 1) belongs to a cycle of length r, and it follows that h r (0, 1) = (0, 1). In particular, for every integer k, h kr (0, 1) = (0, 1), so that F kr = 0; that is F kr 0 (mod d) or d F kr. 19
27 Theorem and Lemma imply the following theorem. Theorem For all d > 0, there exist infinitely many n and infinitely many m such that d gcd(b 0 (n),b 0 (n + 1)) and d gcd(b 1 (m),b 1 (m + 1)). Proof. Given d, there exists r so that d F kr for each k, and it follows that b 0 (n) and b 0 (n + 1) are both multiples of d for n P kr. Similarly, b 1 (m) and b 1 (m + 1) are both multiples of d for m Q kr. 2.5 Properties of the Summatory Function Next, we consider the function which is the sum of the r th row of the triatomic array, F α,β (r) := Lemma For r 5, n I r+1 b α,β (n) = 2 r+1 n=2 r +1 b α,β (n). F α,β (r) = 3F α,β (r 1) αf r 5 βf r 4. (2.14) Proof. Starting with the definition of F α,β (r) and applying the recurrence, 20
28 we find F α,β (r) = b α,β (n) n I r+1 = b α,β (2n)+b α,β (2n 1) n I r = b α,β (n)+b α,β (n + 1) + b α,β (n 1) n I r = ( n I r 3b α,β (n) ) + b α,β (2 r + 1) b α,β (2 r 1 + 1) + b α,β (2 r 1 ) b α,β (2 r ) =3F α,β (r 1) + b α,β (2 r 1 ) b α,β (2 r 2 )+b α,β (2 r 1 ) b α,β (2 r ). Then, by applying Theorem and simplifying we get F α,β (r) =3F α,β (r 1) + 2(αF r 2 + βf r 1 ) αf r 3 βf r 2 αf r 1 βf r =3F α,β (r 1) + α (F r 1 3F r 3 )+β (F r 3F r 2 ) =3F α,β (r 1) + α (F r 2 2F r 3 )+β (F r 1 2F r 2 ) =3F α,β (r 1) + α (F r 4 F r 3 )+β (F r 3 F r 2 ) =3F α,β (r 1) αf r 5 βf r 4. This lemma suggests a relationship of the following form. Theorem For r 1, ( 7 F α,β (r) =α 5 3r F r 4 ) ( 2 5 F r 1 + β 5 3r 1 5 F r + 3 ) 5 F r 1. Proof. By considering Table 2.1, we check that this holds for small r. Sup- 21
29 pose that the formula holds for r < k. Then by Lemma 2.5.1, ( 7 F α,β (k) = 3α 5 3k F k 1 4 ) ( 2 5 F k 2 +3β 5 3k F k ) 5 F k 2 + αf k 1 3αF k 3 + βf k 3βF k 2 ( 7 = α 5 3k F k 1 12 ) 5 F k 2 + F k 1 3F k 3 ( 2 + β 5 3k 3 5 F k ) 5 F k 2 + F k 3F k 2 ( 7 = α 5 3k F k 4 ) ( 2 5 F k 1 + β 5 3k 1 5 F k + 3 ) 5 F k 1. Thus the theorem holds for r 1. Remark. Theorem implies that F α,β (r) 3 r = 7α +6β 15 + O (( ) r ) φ = 3 7α +6β 15 + o(1). Remark. Interestingly, the sum of the r th row of the diatomic array for the Stern sequence is 3 r + 1. Remark. Note that F 6,7 (r) = 5F r +9F r 1, which means that the average value of b 6,7 (n) 0 as n. Now suppose we sum all the terms of the general bow sequence up to the N th term. Define a new function as follows: Definition. E α,β (N) is defined as the finite sum of the first N terms of the bow sequence E α,β (N) := N b α,β (k). (2.15) k=1 Theorem For N 1 we have the following E α,β (2N) =3E α,β (N) α b α,β (N)+b α,β (N + 1). (2.16) Proof. One can calculate quickly that this formula holds for N < 4. Let N 4. We start with the definition, separate terms and apply the recursion 22
30 to find E α,β (2N) = 2N k=1 b α,β (k) = b α,β (1) + b α,β (2) + b α,β (3) + = b α,β (1) + b α,β (2) + b α,β (3) + = b α,β (1) + b α,β (2) + b α,β (3) + Using Table 2.1 we see that E α,β (2N) =2α + β + = 2N k=4 b α,β (k) N (b α,β (2k)+b α,β (2k + 1)) b α,β (2N + 1) k=2 N (2b α,β (k)+b α,β (k + 1)) b α,β (2N + 1). k=2 N (2b α,β (k)+b α,β (k + 1)) b α,β (2N + 1) k=2 N 2b α,β (k)+ k=1 N b α,β (k + 1) b α,β (2N + 1). k=1 We reindex the second sum and apply (2.15) to obtain E α,β (2N) = 2E α,β (N)+ N+1 k=2 b α,β (k) b α,β (2N + 1) =2E α,β (N)+E α,β (N) α + b α,β (N + 1) b α,β (2N + 1). Then by applying the recursion we get the desired result E α,β (2N) =3E α,β (N) α b α,β (N)+b α,β (N + 1). Corollary For r 3 and N =2 r we have E α,β (2N) =3E α,β (N) α(f r 3 + 1) βf r 2. (2.17) 23
31 Proof. By Theorem we have E α,β (2N) =3E α,β (N) α b α,β (2 r )+b α,β (2 r + 1). By applying Theorem and simplifying we obtain E α,β (2N) = 3E α,β (N) α (αf r 1 + βf r )+(αf r 2 + βf r 1 ) =3E α,β (N) α αf r 3 βf r 2 =3E α,β (N) α(f r 3 + 1) βf r 2. Corollary suggests the following relationship. Theorem For r 2 we have ( 7 E α,β (2 r+1 )=α 10 3r F r F r ) ( 1 + β 2 5 3r F r 1 ) 5 F r 1. (2.18) Proof. By considering Table 2.1, we check that this holds for small r 2. Suppose that the formula holds for r < k. Then, by Corollary 2.5.4, ( 7 E α,β (2 r ) = 3α 10 3r F r F r ( 1 +3β 5 3r F r 1 1 ) 5 F r 2 α(f r 4 + 1) βf r r F r F r ) 2 F r 4 1 ( 1 + β 5 3r F r 1 3 ) 5 F r 2 F r 3. ( 7 = α Then, by simplifying the Fibonacci terms, we find ( 7 E α,β (2 r )=α 10 3r F r F r 2 + F r ) 2 ( 1 + β 5 3r F r ) 5 F r 2 ( 7 = α 10 3r F r F r ) ( 1 + β 2 5 3r F r 1 ) 5 F r )
32 Thus the theorem holds for r 2. Remark. Theorem implies that E α,β (2 r ) 3 r = 7α +6β 30 + O (( ) r ) φ = 3 7α +6β 30 + o(1). Remark. By comparison, the sum of the first 2 r terms of the Stern sequence is 1 2 (3r + 1). 25
33 Chapter 3 Generating Functions We use the following notation for generating functions: G a 1,a 2,a 3,...,a m (x) := x := (1 + x a 1 2 j + x a 2 2 j + + x am 2j ) (3.1) j=0 c a 1,a 2,a 3,...,a m (n)x n. (3.2) n=1 Combinatorially, this means that c a 1,a 2,a 3,...,a m (n) is the number of ways of writing n 1 as the sum i 0 γ i2 i where γ i {0,a 1,a 2,a 3,..., a m }. First consider 1 x G1 (x) = (1 + x 2j )= 1 1 x = x k. (3.3) n=0 Thus we recover the fact that every integer n>0 has a unique representation k=0 n = γ i 2 i, (3.4) i=0 where γ i {0, 1}. In this chapter we examine G 1,2 (x), G 1,3 (x), and G 2,3 (x). 26
34 3.1 The Generating Function for the Stern Sequence The generating function for the Stern sequence, s(n), is defined by S(x) := s(n)x n. (3.5) Separating into even and odd terms as in [14] and [15] we find S(x) = = = s(2n)x 2n + n=0 s(n)x 2n + n=0 n=0 s(2n + 1)x 2n+1 n=0 s(n)x 2n+1 + n=0 ( 1+x + 1 ) S(x 2 ). x s(n + 1)x 2n+1 n=0 Since s(0) = 0 we can write S(x) =xt (x). T (x) = s(n + 1)x n (3.6) n=0 and xt (x) = ( 1+x + 1 ) x 2 T (x 2 ), x thus T (x) = (1 + x + x 2 )T (x 2 ). (3.7) Since 1 s(n) n, we know that lim n (s(n)) 1/n = 1. Thus S(x) has radius of convergence 1, is analytic on the open unit disk, and S(x r ) is defined for x < 1. 27
35 Iterating (3.7) N times we obtain T (x) = We find that for x < 1, ( N 1 ) (1 + x 2j + x 2j+1 ) T j=0 Thus we find, as in [15], that ( ) lim T x 2N = T (0) = s(1) = 1. N S(x) =xt (x) =x ( x 2N ). (1 + x 2j + x 2 2j ). (3.8) j=0 By equation (3.1) we see that S(x) =G 1,2 (x). Remark. Combinatorially, the generating function tells us that s(n) =c 1,2 (n). Remark. Consider the generating function modulo 2. Proceeding as in [13] we find the following S(x) =x = x x (1 + x 2j + x 2 2j ) j=0 j=0 j=0 1 x 3 2j 1 x 2j 1+x 3 2j 1+x 2j (mod 2) 1 x (1 x) (mod 2) 1 x3 x + x2 1 x 3 (mod 2). (3.9) Thus we can clearly see that s(n) is even precisely when 3 n. We obtain similar results for the bow sequences. 28
36 3.2 The Generating Function for the General Bow Sequence We can now consider the generating function for the general bow sequence. First, let B α,β (x) = b α,β (n)x n = αb 1,0 (x)+βb 0,1 (x). (3.10) n 0 It follows from (2.12) that B 1,0 (x) and B 0,1 (x) have radius of convergence 1. By repeatedly applying the recursion and tweaking the limits, we have B α,β (x) =αx + βx 2 + n 3 = αx + βx 2 + n 1 = αx + βx 2 + n 1 = αx + βx 2 + n 1 b α,β (n)x n b α,β (2n + 1)x 2n+1 + n 2 b α,β (n)x 2n+1 + n 2 b α,β (n)x 2n+1 + n 2 b α,β (2n)x 2n (b α,β (n)+b α,β (n + 1))x 2n b α,β (n)x 2n + n 2 b α,β (n + 1)x 2n = αx + βx 2 + b α,β (n)x 2n+1 + b α,β (n)x 2n + b α,β (n)x 2n 2 n 1 n 2 n 3 = αx + βx 2 + x b α,β (n)x 2n + b α,β (n)x 2n + 1 b x 2 α,β (n)x 2n n 1 n 2 n 3 = αx + βx 2 + x b α,β (n)x 2n + n 1 n b x 2 α,β (n)x 2n α βx 2 n 1 = α + αx αx 2 + ( ) 1 x +1+x B 2 α,β (x 2 ) b α,β (n)x 2n αx 2 = α(1 x + x 2 )+ 1 x 2 (1 + x2 + x 3 )B α,β (x 2 ). Since b α,β (0) = 0 and b α,β (1) = α, we can write B α,β (x) =αx+x 2 C α,β (x), with C α,β (0) = β, so B α,β (x 2 )=αx 2 + x 4 C α,β (x 2 ). Thus we have 29
37 B α,β (x) =αx + x 2 C α,β (x) = 1 x 2 (1 + x2 + x 3 )(αx 2 + x 4 C α,β (x 2 )) α(1 x + x 2 ). Solving for C α,β (x), we find C α,β (x) = 1 ) (α(1 x + x 2 + x 3 )+x 2 C x 2 α,β (x 2 )(1 + x 2 + x 3 ) α(1 x + x 2 ) = αx + C α,β (x 2 )(1 + x 2 + x 3 ). We can feed this identity back into the equation to get C α,β (x) =αx + (1 + x 2 + x 3 )(αx 2 + C α,β (x 4 )(1 + x 4 + x 6 )) = αx + (1 + x 2 + x 3 )αx 2 + (1 + x 2 + x 3 )(1 + x 4 + x 6 )C α,β (x 4 ). After N steps, C α,β (x) =αx + α N x 2 2k k N+1 (1 + x 2 2j + x 3 2j )+ (1 + x 2 2k + x 3 2k )C α,β (x 2N+2 ). k=0 j=0 k=0 Since, C α,β (x) =β+x P (x) for some P (x), C α,β (x 2N+2 ) β = x 2N+2 P (x 2N+2 ). Thus for x < 1, C α,β (x 2N+2 ) β 0 as N ; hence we have C α,β (x) =αx + α x 2 2k Thus we have the following theorem. k (1 + x 2 2j + x 3 2j )+β (1 + x 2 2k + x 3 2k ). k=0 j=0 k=0 30
38 Theorem The generating function for the general bow sequence is: B α,β (x) =αx + αx 3 + αx 2 x 2 2k k=0 + βx 2 (1 + x 2 2k + x 3 2k ). k=0 k (1 + x 2 2j + x 3 2j ) (3.11) Since B α,β (x) = αb 1,0 (x) +βb 0,1 (x), we can state two combinatorial implications. First, we have B 0,1 (x) := b 0 (n)x n = x 2 n=0 k=0 j=0 (1 + x 2 2k + x 3 2k )=xg 2,3 (x) (3.12) for the case when α = 0, β = 1. We interpret this statement as: Corollary Combinatorially, b 0 (n) =c 2,3 (n 1) is the number of ways of writing n 2 as the sum i c i2 i where c i {0, 2, 3}. We also have B 1,0 (x) := b 1 (n)x n = x + x 3 + x 2 n=0 x 2 2k k=0 k (1 + x 2 2j + x 3 2j ) (3.13) for the case when α = 1, β = 0. We interpret this statement as: Corollary Combinatorially, b 1 (n) is the number of ways of writing n 2 2 k+1 as the sum k i=0 c i2 i where c i {0, 2, 3} and k N. Remark. By rearranging, we find that b 1 (n) is also the number of ways of writing n 2 as the sum k i=0 c i2 i where k N, c i {0, 2, 3} for i k 1 and c k {2, 4, 5}. Alternatively, since k i=0 2i =2 k+1 1, we get the following formula B 1,0 (x) =x + x 3 + x 3 k=0 j=0 j=0 k (x 1 2j + x 3 2j + x 4 2j ), (3.14) which means that b 1 (n) is the number of ways of writing n 3 as a finite sum k i=0 c i2 i where c i {1, 3, 4}. 31
39 3.3 A Relative of the Bow Sequences After considering the generating functions G 1,2 (x) and G 2,3 (x), one might be curious about the sequence that has the following generating function Y (x) := G 1,3 (x) (3.15) = x (1 + x 2k + x 3 2k )= y(n)x n. (3.16) k=0 n=0 We see that y(n) =c 1,3 (n) is the number of ways of writing n 1 as the sum i 0 c i2 i where c i {0, 1, 3}. We can define y(n) recursively as follows: y(0) = 0, y(1) = 1; (3.17) y(2n + 1) = y(n + 1), for n 1; (3.18) y(2n) =y(n)+y(n 1), for n 1. (3.19) By analogy with b α,β (n), we can define the general sequence by: y α,β (0) = α, y α,β (1) = β; (3.20) y α,β (2n + 1) = y α,β (n + 1), for n 1; (3.21) y α,β (2n) =y α,β (n)+y α,β (n 1), for n 1. (3.22) In particular, y(n) =y 0,1 (n). We can find the generating function of y α,β (n) using the techniques of this chapter. Let the generating function be defined as Y α,β (x) := y α,β (n)x n. n 0 32
40 Then Y α,β (x) =α + βx + y α,β (n)x n n 2 = α + βx + y α,β (2n)x 2n + y α,β (2n + 1)x 2n+1 n 1 n 1 = α + βx + (y α,β (n)+y α,β (n 1))x 2n + y α,β (n + 1)x 2n+1 n 1 n 1 = α + βx + y α,β (n)x 2n α + y α,β (n)x 2n+2 + y α,β (n)x 2n 1 n 0 n 0 n 2 = βx + Y α,β (x 2 )+x 2 Y α,β (x 2 )+ n 0 y α,β (n)x 2n 1 α x βx = α x + 1 x Y α,β(x 2 )+Y α,β (x 2 )+x 2 Y α,β (x 2 ) = α x + 1 x Y α,β(x 2 )(1 + x + x 3 ). Now, let Y α,β (x) =α + xd α,β (x). So Y α,β (x 2 )=α + x 2 D α,β (x 2 ). Substituting this into our equation for Y α,β (x) we get Y α,β (x) = α x + 1 x Y α,β(x 2 )(1 + x + x 3 ) = α x + 1 x (α + x2 D α,β (x 2 ))(1 + x + x 3 ). Then we solve for D α,β (x) as follows xd α,β (x) = α x + 1 x (α + x2 D α,β (x 2 ))(1 + x + x 3 ) = α x α +(α x + xd α,β(x 2 ))(1 + x + x 3 ) = α x α + α x + α + αx2 = xd α,β (x 2 )+x 2 D α,β (x 2 )+x 4 D α,β (x 2 ) = αx 2 + D α,β (x 2 )(x + x 2 + x 4 ). Dividing each side by x we find D α,β (x) =αx + D α,β (x 2 )(1 + x + x 3 ). Next, we shall substitute this equation into itself repeatedly to obtain the 33
41 following equation after N iterations D α,β (x) =αx + α N k=0 x 2 2k k N (1 + x 2j + x 3 2j )+D α,β (x 2N+1 ) (1 + x 2j + x 3 2j ). j=0 Let N. By the definition of D α,β (x) we see that for x < 1, D α,β (x 2N+1 ) β as N. Now we get the following formula D α,β (x) =αx + α k=0 x 2 2k j=0 Thus we obtain the following theorem. Theorem For y α,β (n), Y α,β (x) =α + αx 2 + αx + βx j=0 k (1 + x 2j + x 3 2j )+β (1 + x 2j + x 3 2j ). k=0 x 2 2k (1 + x 2j + x 3 2j ). j=0 j=0 k (1 + x 2j + x 3 2j ) (3.23) Since Y α,β (x) =αy 1,0 (x)+βy 0,1 (x), we can state another combinatorial implication. First, we have Y 1,0 (x) = 1 + x 2 + x k=0 x 2 2k j=0 k (1 + x 2j + x 3 2j ) (3.24) for the case where α = 1, β = 0. We interpret this statement as: Corollary Combinatorially, y 1,0 (n) is the number of ways of writing n 1 2 k+1 as the sum k i=0 c i2 i where k N, c i {0, 1, 3}. Remark. By rearranging, we find that y 1,0 (n) is also the number of ways of writing n 1 as the sum k i=0 c i2 i, where k N, c i {0, 1, 3} for i k 1, and c k {2, 3, 5}. Alternatively, since k i=0 2i =2 k+1 1, we get the following formula Y 1,0 (x) = 1 + x 2 + x 2 k=0 j=0 j=0 k (x 1 2j + x 2 2j + x 4 2j ), (3.25) 34
42 which means that y 1,0 (n) is the number of ways of writing n 2 as a finite sum k i=0 c i2 i, where c i {1, 2, 4}. Let and define Y n (x) := x B n (x) := x 2 n k=0 n k=0 ( 1+x 2k + x 3 2k), (3.26) ( 1+x 2 2k + x 3 2k). (3.27) Then we know that Y n (x) =Y (x) up to x 2n+1, and B n (x) =B 0,1 (x) up to x 2n+2. Consider If we multiply by ( ) 1 Y n = 1 x x n k=0 x 3 ( ). x 2k x 3 2k n k=0 x 3 2k, we find ( ) 1 n x 3 (2n+1 1)+3 Y n = x 2 x k=0 = B n (x). ( ) x 3 2k + x 2 2k +1 Thus, Theorem ( ) 1 x 3 2n+1 Y n = B n (x). x Theorem can be used to find a weak relationship between y(n) and b 0 (m), but first we need the following definition. 35
43 Definition. For an integer N 1, we say that a(n)x n n=0 if a(n) =b(n) for n < N. b(n)x n (mod x N ) n=0 Remark. No information is assumed about a(n) and b(n) for n N. Now, consider the following generating functions: V (x) := W (x) := For an integer r 0, let V r (x) = W r (x) = v(n)x n = n=0 w(n)x n = n=0 v r (n)x n = n=0 w r (n)x n = n=0 k=0 k=0 r k=0 r k=0 ( 1+x 2k + x 3 2k), ( 1+x 2 2k + x 3 2k). ( 1+x 2k + x 3 2k), ( 1+x 2 2k + x 3 2k). As in our work towards Theorem 3.3.3, observe that ( ) 1 r ( W r = ) x x 2 2k x 3 2k k=0 ( r ) r 1 ( ) = x 3 2k + x 2k +1, x 3 2k k=0 so that V r (x) =x 3(2r+1 1) W r ( 1 x ). We can see that V r(x) is a polynomial of degree M r := 3(2 r+1 1). So we have k=0 hence M r M r v r (n)x n = w r (n)x Mr n, n=0 n=0 v r (n) =w r (M r n). (3.28) 36
44 Now we consider V (x) up to x Mr. We see that ( V (x) =V r (x) 1+x 2r+1 + x 3 2r+1)( 1+x 2r+2 + x 3 2r+2) ( V r (x) 1+x 2r+1)( 1+x 2r+2) (mod x Mr ) ( V r (x) 1+x 2r+1 + x 2r+2) (mod x Mr ). Since 3 2 r >M r, it is also true that ( ) V (x) V r (x) 1+x 2r+1 + x 2r+2 + x 3 2r+1 + (mod x Mr ). Thus, V (x) V r(x) 1 x 2r+1 (mod x Mr ), or equivalently, ( 1 x 2r+1) V (x) V r (x) (mod x Mr ). Thus, with the restriction that if n<0, v(n) = 0, for m < M r, we have v r (m) =v(m) v(m 2 r+1 ). (3.29) Similarly, we have ( W (x) =W r (x) 1+x 2 2r+1 + x 3 2r+1)( 1+x 2 2r+2 + x 3 2r+2) ( W r (x) 1+x 2 2r+1) (mod x Mr ). Since 4 2 r+1 >M r, 1 W (x) W r (x) (mod x Mr ), 1 x 2 2r+1 or equivalently, ( 1 x 2 2r+1) W (x) W r (x) (mod x Mr ). 37
45 Thus, with the restriction that if n<0, w(n) = 0, for m<m r, we have w r (m) =w(m) w(m 2 2 r+1 ). (3.30) have: If we now limit ourselves to 0 n M r, by (3.28), (3.29), and (3.30) we v(n) v(n 2 r+1 )=w(3 2 r+1 3 n) w(2 r+1 3 n). (3.31) Recalling that v(n) =y(n + 1) and w(n) =b 0 (n + 2), we substitute these into (3.31) to find y(n + 1) y(n +1 2 r+1 )=b 0 (3 2 r+1 1 n) b 0 (2 r+1 1 n). Setting m = n + 1 we have the following theorem. Theorem For m 3 2 r+1 2, y(m) y(m 2 r+1 )=b 0 (3 2 r+1 m) b 0 (2 r+1 m), (3.32) where y(k) =b 0 (k) =0for k<0. Remark. There is no m such that both m 2 r+1 > 0 and 2 r+1 m>0. Next, consider the generating function G 2,3 (x) =x (1 + x 2 2j + x 3 2j ). (3.33) j=0 We may examine this generating function mod 2. Theorem G 2,3 (x) x+x3 +x 4 +x 5 1 x 7 (mod 2). Proof. We shall show that x k (1 + x 2 2j + x 3 2j ) x + x3 + x 4 + x 5 (3.34) 1 x 7 j=0 (1 + x2 + x 3 + x 4 )x 2k+2 +1 (1 + x 2k+1 ) (mod 2). 1 x 7 38
46 For the case k = 0, (3.34) becomes x(1 + x x 3 20 ) x + x3 + x 4 + x 5 = x5 + x 8 + x 10 + x 11 1 x 7 1 x 7 which is what we want. k+1 x = = (1 + x2 + x 3 + x 4 )x 5 (1 + x 2 ) 1 x 7, Now, assuming that (3.34) holds for k, we see that for k + 1 we have (1 + x 2 2j + x 3 2j ) x + x3 + x 4 + x 5 1 x 7 j=0 ( (1 + x 2 + x 3 + x 4 )x 2k+2 +1 (1 + x 2k+1 ) + x + x3 + x 4 + x 5 1 x 7 1 x 7 x + x3 + x 4 + x 5 1 x 7 = 1+x2 + x 3 + x 4 1 x 7 ) (1 + x 2 2k+1 + x 3 2k+1 ) [( ( x 2k x 2k+1) )( + x 1+x 2 2k+1 + x 3 2k+1) ] x = 1+x2 + x 3 + x 4 1 x 7 [x 2 2k x 3 2k x 4 2k x 5 2k x 2 2k x 5 2k x 6 2k x 3 2k x x ]. Then, considering this equation modulo 2, we remove terms with coefficient 2 to obtain (1 + x 2 2j + x 3 2j ) x + x3 + x 4 + x 5 1 x 7 k+1 x j=0 1+x2 + x 3 + x 4 1 x 7 ( x 4 2k x 6 2k+1 +1 ) (mod 2) (1 + x2 + x 3 + x 4 )(x 2k+3 +1 )(1 + x 2k+2 ) (mod 2). 1 x 7 Thus as k we have x k (1 + x 2 2j + x 3 2j ) x + x3 + x 4 + x 5. (3.35) 1 x 7 j=0 39
47 From Theorem it is easy to see that if G 2,3 (x) = c 2,3 (n)x n, n=0 then c 2,3 (n) is even exactly when n 0, 2, 6 (mod 7). Note that c 2,3 (n) =b 0 (n + 1). Thus we have the following corollary. Corollary b 0 (n) is even precisely when n 0, 1, 3 (mod 7). Next, consider Y (x) =G 1,3 (x) modulo 2. Theorem G 1,3 (x) x+x2 +x 3 +x 5 1 x 7 (mod 2). Proof. We shall show, to be precise, that x k (1 + x 2j + x 3 2j ) x + x2 + x 3 + x 5 (3.36) 1 x 7 j=0 (1 + x + x2 + x 4 )x 2k+1 +1 (1 + x 2k+2 ) (mod 2). 1 x 7 For the case k = 0, (3.36) becomes x(1 + x 20 + x 3 20 ) x + x2 + x 3 + x 5 = x3 + x 4 + x 5 + x 8 + x 9 + x 11 1 x 7 1 x 7 which is what we want. = (1 + x + x2 + x 4 )x 3 (1 + x 4 ) 1 x 7, 40
48 k+1 x = Now, assuming that (3.36) holds for k, we see that for k + 1 we have (1 + x 2j + x 3 2j ) x + x2 + x 3 + x 5 1 x 7 j=0 ( (1 + x + x 2 + x 4 )x 2k+1 +1 (1 + x 2k+2 ) + x + x2 + x 3 + x 5 1 x 7 1 x 7 x + x2 + x 3 + x 5 1 x 7 = 1+x + x2 + x 4 1 x 7 ) (1 + x 2k+1 + x 3 2k+1 ) [( ( x 2k x 2k+2) )( + x 1+x 2k+1 + x 3 2k+1) ] x = 1+x + x2 + x 4 1 x 7 [x 2k x 3 2k x 2 2k x 4 2k x 2 2k x 4 2k x 6 2k x 3 2k x x ]. Then, considering this equation modulo 2, we remove terms with coefficient 2 to obtain (1 + x 2j + x 3 2j ) x + x2 + x 3 + x 5 1 x 7 k+1 x j=0 1+x + x2 + x 4 1 x 7 ( x 2 2k x 6 2k+1 +1 ) (mod 2) (1 + x + x2 + x 4 )(x 2k+2 +1 )(1 + x 2k+3 ) (mod 2). 1 x 7 Thus as k we have x k (1 + x 2j + x 3 2j ) x + x2 + x 3 + x 5. (3.37) 1 x 7 j=0 From Theorem it is easy to see that if G 1,3 (x) = n=0 c1,3 (n)x n, then c 1,3 (n) is even exactly when n 0, 4, 6 (mod 7). Recall that y(n) =c 1,3 (n). We have the following corollary. Corollary y(n) is even precisely when n 0, 4, 6 (mod 7). We also observe the following theorem. Theorem G 1,k 1,k (x) = x. (1 x)(1 x k 1 ) 41
49 Proof. Starting with the definition, we know that G 1,k 1,k (x) =x j=0 ( 1+x 2j + x (k 1) 2j + x k 2j). Then, by factoring and simplifying, we find that G 1,k 1,k (x) =x (1+x 2j) j=0 i=0 1 = x 1 x 1 1 x k 1 x = (1 x)(1 x k 1 ). (1+x (k 1) 2i) Remark. We can write G 1,k 1,k (x) as G 1,k 1,k = x(1 + x + x 2 + )(1 + x k 1 + x 2(k 1) + ) = x(1 + x + + x k 2 +2x k x 2k 3 +3x 2k 2 + ). Thus a 1,k 1,k = n k Corollary Combinatorially, a 1,k 1,k (n) is the number of ways of writing n 1 as the sum i 0 c i2 i where c i {0, 1,k 1,k}. Remark. Alternatively, if we let c i = a i +(k 1)b i, with a i,b i {0, 1}, then a 1,k 1,k (n) is the number of ways of writing n 1=r +(k 1)s, where r = i 0 a i2 i, and s = i 0 b i2 i. 42
50 Chapter 4 The General Bow Sequence Modulo 2 We begin with a few statements about the general bow sequence modulo 2 before proceeding to consider b 0 (n) and b 1 (n) in greater detail. 4.1 The Behavior of b α,β (n) Modulo 2 Let A α,β (m) denote the following expression: A α,β (m) =b α,β (m 1) + b α,β (m)+b α,β (m + 2). (4.1) Then we have the following theorem. Theorem For m 2, A α,β (m) is always even, except that for k 0, we have A α,β (3 2 k ) α (mod 2). Proof. For small m and k, we use Table 2.1 to note the following: A α,β (2) = b α,β (1) + b α,β (2) + b α,β (4) = 2α +2β 0 (mod 2), A α,β (3) = b α,β (2) + b α,β (3) + b α,β (5) = α +2β α (mod 2). Then, by applying (2.5) for m 2, we see that A α,β (2m + 1) = b α,β (2m)+b α,β (2m + 1) + b α,β (2m + 3) 0 (mod 2). Thus we know that A α,β (3) α (mod 2), and A α,β (2m + 1) 0 (mod 2) for m 2. 43
51 For even terms, by applying the recurrence, we find A α,β (2m) =b α,β (2m 1) + b α,β (2m)+b α,β (2m + 2) = b α,β (m 1) + b α,β (m)+2b α,β (m + 1) + b α,β (m + 2) A α,β (m) (mod 2). Putting these together, we find that for k 0 and m 2, A α,β (3 2 k ) α (mod 2), and otherwise A α,β (m) 0 (mod 2). Remark. Note that this implies that for m 1, b 0 (m 1) + b 0 (m)+b 0 (m + 2) 0 (mod 2). (4.2) b 1 (n). We use the following theorem to prove several results about b 0 (n) and Theorem If (x n ) Z is a sequence of integers, and for all n 0 x n+3 + x n+1 + x n 0 (mod 2), then for all n, x n+7 x n (mod 2). Proof. We have x n+3 x n + x n+1 (mod 2), so x 3 x 0 + x 1 (mod 2), x 4 x 1 + x 2 (mod 2), x 5 x 2 + x 3 x 0 + x 1 + x 2 (mod 2), x 6 x 3 + x 4 x 0 + x 2 (mod 2), x 7 x 4 + x 5 x 0 +2x 1 +2x 2 x 0 (mod 2), x 8 x 5 + x 6 2x 0 + x 1 +2x 2 x 1 (mod 2), x 9 x 6 + x 7 2x 0 + x 2 x 2 (mod 2). So we can see that the lemma holds for small n. Suppose the lemma holds 44
52 for n<k. Similarly, by the induction hypothesis we have x k x k 3 + x k 2 (mod 2) x k 10 + x k 9 (mod 2) x k 7 (mod 2). Thus for all n, x n+7 x n (mod 2). 4.2 The Behavior of b 1 (n) and b 0 (n) Modulo 2 First, define v α,β (n) as follows: v α,β (n) := (b α,β (7n),b α,β (7n + 1),b α,β (7n + 2),..., b α,β (7n + 6)) (mod 2). There is only one possibility for v 0,1 (n). We have the following theorem. Theorem For b 0 (n) and all n 0, v 0,1 (n) v (0) (n) := (0, 0, 1, 0, 1, 1, 1) (mod 2). (4.3) Proof. First, we can see from Table 2.1 that the theorem holds for small n. Then, by Theorem 4.1.1, for n 1, b 0 (n 1) + b 0 (n)+b 0 (n + 2) 0 (mod 2). So by Theorem 4.1.2, for all n, v 0,1 (n) v (0) (n) := (0, 0, 1, 0, 1, 1, 1) (mod 2). Remark. This theorem also follows immediately from Corollary Corollary For n 0, 2 gcd(b 0 (n),b 0 (n + 1)) 7 n. (4.4) 45
53 Proof. The statement is an immediate consequence of Theorem Next, consider b 1 (n). By Theorem we know that unless m =3 2 k, b 1 (m + 2) b 1 (m 1) + b 1 (m) (mod 2), and for m =3 2 k we have b 1 (m + 2) b 1 (m 1) + b 1 (m) + 1 (mod 2). Since 3 2 k 3, 5, 6 (mod 7), by Theorem we should have three main cases for v 1,0 (n), and possibly three transition cases. Consider the first 28 cases for v 1,0 (n) given below: v 1,0 (0) (0, 1, 0, 1, 1, 0, 0) (mod 2), v 1,0 (1) (1, 1, 1, 0, 0, 1, 0) (mod 2), v 1,0 (2) (0, 1, 0, 1, 1, 1, 0) (mod 2), v 1,0 (3) (0, 1, 0, 1, 1, 0, 0) (mod 2), v 1,0 (4) v 1,0 (6) (1, 0, 1, 1, 1, 0, 0) (mod 2), v 1,0 (7) v 1,0 (13) (1, 1, 1, 0, 0, 1, 0) (mod 2), v 1,0 (14) v 1,0 (26) (0, 1, 0, 1, 1, 1, 0) (mod 2), v 1,0 (27) (0, 1, 0, 1, 1, 0, 0) (mod 2). It appears that v 1,0 (n) cycles through 4 cases. below: v (1) (n) := (0, 1, 0, 1, 1, 1, 0) v (1 ) (n) := (0, 1, 0, 1, 1, 0, 0) v 1,0 (n) v (2) (n) := (1, 0, 1, 1, 1, 0, 0) v (3) (n) := (1, 1, 1, 0, 0, 1, 0). These cases are listed Remark. This would imply that we have the following 4 cases for v 1,1 (n): v (4) (n) := (0, 1, 1, 1, 0, 0, 1) v (4 ) (n) := (0, 1, 1, 1, 0, 1, 1) v 1,1 (n) v (5) (n) := (1, 0, 0, 1, 0, 1, 1) v (6) (n) := (1, 1, 0, 0, 1, 0, 1). 46
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