Upper Bounds for Stern s Diatomic Sequence and Related Sequences

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1 Upper Bounds for Stern s Diatomic Sequence and Related Sequences Colin Defant Department of Mathematics University of Florida, U.S.A. cdefant@ufl.edu Sumitted: Jun 18, 01; Accepted: Oct, 016; Pulished: Oct 14, 016 Mathematics Suject Classifications: 0A0; 11B37 Astract Let (s (n)) n=0 denote Stern s diatomic sequence. For n, we may view s (n) as the numer of partitions of n 1 into powers of with each part occurring at most twice. More generally, for integers, n, let s (n) denote the numer of partitions of n 1 into powers of with each part occurring at most times. Using this cominatorial interpretation of the sequences s (n), we use the transfer-matrix method to develop a means of calculating s (n) for certain values of n. This then allows us to derive upper ounds for s (n). In the special case =, our ounds improve upon the current upper ounds for the Stern sequence. We then show that lim sup n s (n) n log φ = ( 1) log φ. Keywords: Stern sequence; Fionacci numer; Lucas numer; over-expansion; transfer-matrix method. 1 Introduction Throughout this paper, F m and L m will denote the Fionacci and the Lucas numers. We have F m+ = F m+1 +F m and L m+ = L m+1 +L m for all integers m (including negative integers). We convene to use the initial values F 1 = F = 1, L 1 = 1, and L = 3. We also let φ = 1 + and φ = 1 = 1. The symol N will denote the set of positive φ integers. Prolem B1 of the 014 William Lowell Putnam Competition defines a ase 10 overexpansion of a positive integer N to e an expression of the form N = d k 10 k + d k 1 10 k d the electronic journal of cominatorics 3(4) (016), #P4.8 1

2 with d k 0 and d i {0, 1,,..., 10} for all i. We may generalize (and slightly modify) this notion to otain the following definition. Definition 1. Let e an integer. A ase over-expansion of a positive integer N is k a word d k d k 1 d 0 over the alphaet {0, 1,..., } such that d k 0 and d i i = N. We refer to the letter d i as the i th digit of the expansion. It is well-known that each positive integer N has a unique ase over-expansion that does not contain the letter (or digit) ; we refer to this expansion as the ordinary ase expansion of N. The Stern-Brocot sequence, also known as Stern s diatomic sequence or simply Stern s sequence, is defined y the simple recurrence relations s(n) = s(n) and s(n + 1) = s(n) + s(n + 1) for all nonnegative integers n, where s(0) = 0. This sequence has found numerous applications in numer theory and cominatorics, and it has several interesting properties which relate it to the Fionacci sequence. For n, it is well-known that s(n) is the numer of ase over-expansions (also known as hyperinary expansions) of n 1 [3]. To generalize Stern s sequence, let s (n) denote the numer of ase over-expansions of n 1. Equivalently, one may wish to think of s (n) as the numer of partitions of n 1 into powers of with each part occurring at most times. We convene to let s (0) = 0 and s (1) = 1. The sequence s (n) satisfies the recurrence relations s (n) = s (n), s (n + 1) = s (n) + s (n + 1), and s (n + i) = s (n + 1) for i {, 3,..., 1}. Equivalently, ( s n ), if n 0 (mod ); ( s (n) = s n 1 ) ( + n 1 s + 1 ), if n 1 (mod ); ( s n i + 1 ) (1), if n i (mod ) and i <. Using (1), one may easily prove the following lemma. Lemma. Let n e a positive integer. If n 1 (mod ), then ( ) ( ) n 1 n + 1 s (n) = s + s. If n + 1 (mod ), then s (n) = s ( n 1 Calkin and Wilf [] determined that ) ( ) n s = 3log φ lim sup n s(n) 1 + φ φ+ n log = , the electronic journal of cominatorics 3(4) (016), #P4.8

3 and they asked for the exact value of lim sup s(n). Here, we give upper ounds for the n log φ values of s (n) for any integer, from which we will deduce that lim sup n s (n) n log φ = ( 1) log φ. φ s(n) In particular, lim sup n n = 3log. While preparing this manuscript, the author discovered that Coons and Tyler [4] had already determined this value of the supremum log φ s(n) limit of ; in the same paper, they mention that this prolem actually dates ack n log φ to Berlekamp, Conway, and Guy in 198 [1]. However, the ounds we derive apply to the more general family of sequences s (n) and are stronger than those given in [4]. In addition, our methods of proof are quite different from those used in [4]. Coons and Tyler use clever analytic estimates to prove their results from the recursive definition of s(n). By contrast, we will make heavy use of the interpretation of s (n) as the numer of ase over-expansions of n 1 in order to prove several of our most important results. In particular, we will comine this cominatorial interpretation of s (n) with the transfer-matrix method in order to prove Theorem 1. In turn, Theorem 1 will allow us to determine the maximum values of s (n) when n is restricted to certain intervals. For each fixed, the sequence s (n) has a somewhat periodic nature. Our upper ounds are local estimates within the periods; they refine the more gloal estimates given in [4] when =. Determining Maximum Values Throughout this section, fix an integer. Our goal is to derive upper ounds for the numers s (n), particularly those values of n that are slightly larger than a power of (we will make this precise soon). To do so, we will make use of the following sequence. Definition 3. Let h 1 = h = 1. For m 3, let h m = 1 + m 3 m i. Alternatively, we may calculate h m using the recurrence relation { h m 1 + 1, if m; h m = h m 1 + 1, if m () along with the initial value h 1 = 1. For example, h 3 = + 1, h 4 = + 1, h = , and h 6 = It is important to note that h m 1 (mod ) for all m N. We state the following lemma for easy reference, although we omit the proof ecause it is fairly straightforward. the electronic journal of cominatorics 3(4) (016), #P4.8 3

4 Lemma 4. For any positive integer m, h m+1 h m = m + ( 1) m. + 1 The following lemma lists three simple ut useful oservations aout the numers s (n). We omit the proof ecause it follows easily from (1). Lemma. Let n and k e nonnegative integers with k n k+1. i. If n = j k for some j {1,,..., 1}, then s (n) = 1. ii. If k = 1, then s (n), where equality holds if and only if n 1 (mod ). iii. If k 1 and n 1 (mod ), then s (n) = s (n ) for some integer n with k 1 n k. Proposition 6. Let k and n e nonnegative integers with k n < k+1. We may write n uniquely in the form n = j k + t, where j {1,,..., 1} and t {0, 1,..., k 1}. We have s (n) F k+, where equality holds if and only if t = h k or t = h k+1. Proof. That we may write n uniquely in the form n = j k + t is trivial. The proof of the rest of the proposition is y induction on k. The case k = 0 is immediate from the first part of Lemma. The case k = 1 follows from the second part of the same lemma. Now, assume k > 1. We divide the proof into three cases. Case 1: In this case, suppose t = h k or t = h k+1. We will assume that t = h k and that k is even; a similar argument holds if k is odd or t = h k+1. Thus, n = j k + h k. Because k is even, we may use () to write h k = h k and h k 1 = h k + 1. Furthermore, h k 1 (mod ) ecause k is even. Since n h k 1 (mod ), we have y Lemma that s (n) = s ( n 1 = s ( j k + h k 1 1 ) ( ) ( n s = s j k + h ) ( k 1 + s j k 1 + h ) k ) ( ) ( ) ( ) + s j k 1 + h k 1 = s j k + h k + s j k 1 + h k 1. ( ) ( ) By induction on k, s j k + h k = Fk and s j k 1 + h k 1 = Fk+1. Thus, s (n) = F k + F k+1 = F k+, as desired. Case : In this case, suppose n 1 (mod ). By Lemma, there exists an integer n with k 1 n k such that s (n) = s (n ). If n = k, then, using Lemma again, s (n) = 1 < F k+. If n < k, then it follows from induction on k that s (n) = s (n ) F k+1 < F k+. Case 3: In this final case, suppose n 1 (mod ) and that t h k and t h k+1. Suppose, y way of contradiction, that s (n) F k+. Since t n 1 (mod ), we may write t = t + 1 for some integer t < k 1. Using (1), we have s (n) = s ( n 1 ) + s ( n 1 ) + 1 = s (j k 1 + t ) + s (j k 1 + t + 1). (3) the electronic journal of cominatorics 3(4) (016), #P4.8 4

5 If t 0, 1 (mod ), then we know from Lemma and the fact that k 1 j k 1 + t < j k 1 + t + 1 k that s (j k 1 + t ) = s (v 1 ) and s (j k 1 + t + 1) = s (v ) for some integers v 1, v [ k, k 1 ]. By (3) and induction on k, it follows that if t 0, 1 (mod ), then s (n) = s (v 1 ) + s (v ) = F k + F k < F k+. This is a contradiction, so t 0, 1 (mod ). We will assume that t 0 (mod ); a similar argument may e used to derive a contradiction in the case t 1 (mod ). Write t = t. Because t < k 1 and j 1, j k 1 + t + 1 k. We know that j k 1 + t + 1 k ecause t 0 (mod ). Therefore, we have the inequalities k j k + t < k 1 < j k 1 + t + 1 < k. We see y (3) and induction that s (n) = s (j k 1 +t )+s (j k 1 +t +1) = s (j k +t )+s (j k 1 +t +1) F k +F k+1 =F k+, where the equality s (j k 1 +t ) = s (j k +t ) is immediate from (1). This last inequality must e an equality since we are assuming s (n) F k+, so we must have s (j k +t ) = F k and s (j k 1 +t +1) = F k+1. The induction hypothesis states that this is only possile if t {h k, h k 1 } and t +1 {h k 1, h k }. Suppose first that t = h k and t +1 = h k 1. We have h k 1 = h k + 1, so it follows from () that k must e even. Using () again, we see that h k = h k = (t + 1) + 1 = t, which contradicts our assumption that t h k. Similarly, if t = h k 1 and t + 1 = h k, then we may derive the contradiction t = h k+1. It is clearly impossile to have t = h k 1 and t + 1 = h k 1 since t = t. Therefore, we are left to conclude that t = h k and t + 1 = h k. If k is even, then we may use () to write t = t = h k 1 = h k 1 = (h k + 1) = h k = t, which is impossile. This means that k must e odd, so h k = h k y (). Since h k = t + 1 = t + 1 = h k + 1, we conclude that h k 1 = h k. It is easy to see from Definition 3 that this is only possile if k = 3. Hence, t = h 1 = 1, t = h 3 1 =, and t = t + 1 = + 1. However, this means that t = h 4 = h k+1, which is our final contradiction ecause we assumed t {h k, h k+1 }. Now that we know the maximum values of s (n) for k n < k+1, we may easily derive the following result. Corollary 7. We have lim sup n s (n) n log φ ( 1) log φ. Proof. We will need Binet s formula for the Fionacci numers, which states that F m = φm ( φ) m for all m Z. For each positive integer k, let u k = k + h k. Let i k = ( 1)h k k, and oserve that { 1, if k; i k = 1, if k. the electronic journal of cominatorics 3(4) (016), #P4.8

6 We have φ k+ ( φ) (k+) 1 φ (k+) ( φ) (k+) lim = lim k ( k+ + i k ) log φ k φ (k+) ( k+ + i k ) = lim 1 φ (k+) ( φ) (k+) = 1. log φ k (1 + i k / k+ ) log φ By Proposition 6 and Binet s formula, s (u k ) u log φ k = F k+ ( k + h k ) log φ = φk+ ( φ) (k+) ( k + h k ) log φ = φk+ ( φ) (k+) ( k + k +i k 1 )log φ so lim sup n = ( 1) log φ s (n) s (u k ) lim n log φ k u log φ k φ k+ ( φ) (k+) ( k+ + i k ) log φ, = ( 1) log φ. s (n) We now wish to show that lim sup n n ( 1) log φ. If k + h log φ k < n < k+1 for some positive integer k, then we know from Proposition 6 that We saw in the proof of Corollary 7 that s (n) n log φ F k+ n log φ = s ( k + h k ) n log φ < s ( k + h k ) ( k + h k ) log φ. lim k s ( k + h k ) ( k + h k ) log φ = ( 1) log φ. Hence, we need only find a sufficiently strong upper ound for s (n) when k n k +h k for some positive integer k. For this purpose, we make the following definitions. Definition 8. For nonnegative integers k, r, y, let { I(k, r, y) = n N: k < n k + } y r i, and µ(k, r, y) = max{s (n): n I(k, r, y)}, ν(k, r, y) = min{n I(k, r, y): s (n) = µ(k, r, y)}. Our goal is to calculate µ(k, r, y) for any given nonnegative integers k, r, y that satisfy y < r < k 1. This will allow us to to derive tight upper ounds for all integers n that satisfy k n k + h k for some k y choosing appropriate values of r and y. One might think that a simple inductive argument ased on the recurrence relation (1) should e ale to derive our upper ounds quite effortlessly. However, the author has found that attempts to prove upper ounds for s (n) using induction often fail or ecome the electronic journal of cominatorics 3(4) (016), #P4.8 6

7 incredily convoluted. Indeed, the reader may wish to look at [4] in order to appreciate the surprising amount of ingenuity that Coons and Tyler need for the derivation of their upper ounds (which are weaker than ours) in the specific case =. Therefore, we shall prove a sequence of lemmas in order to develop more cominatorial means of calculating µ(k, r, y) and ν(k, r, y). Lemma 9. Let a t a t 1 a 0 e the ordinary ase expansion of a positive integer n. Let c l c l 1 c 0 e a ase over-expansion of n. Set c i = 0 for all integers i with l < i t. For any m {0, 1,..., t}, {0, 1, }, if a m = 0; c m {0, 1, }, if a m = 1; {a m 1, a m }, otherwise. Proof. We know that t a i i = t c i i ecause a t a t 1 a 0 and c l c l 1 c 0 are ase over-expansions of the same numer. Therefore, for any j {0, 1,..., t}, j a i i j c i i (mod j+1 ). (4) j j If a i i > c i i for some j {0, 1,..., t}, then it follows from (4) and the fact that a i {0, 1,..., 1} for all i that j c i i j+1 + j a i i j+1 + j ( 1) i = 1, which is impossile. Hence, j a i i j c i i for all j {0, 1,..., t}. Choose some m {0, 1,..., t}. Since c i for all i, we have m c i i c m m + It follows that m 1 a m m i = (c m + 1) m + m a i i m m c i i < (c m + ) m, i+1 < (c m + ) m. () so c m a m 1. To complete the proof, we simply need to show that if c m > a m, then either a m = 0 and m { 1, } or a m = 1 and c m =. Suppose c m > a m. We have m a i i = a m m + m 1 a i i a m m + m 1 c i i < m c i i the electronic journal of cominatorics 3(4) (016), #P4.8 7

8 and m a i i (y (4)), so we know from () that m a i i m+1 + m c i i (mod m+1 ) m c i i < m+1 + (c m + ) m. This implies that a m m < m+1 + c m m + m, so c m a m >. This is impossile if a m {0, 1} ecause c m. If a m = 0, we must have either c m = 1 or c m =, and if a m = 1, then we must have c m =. Lemma 10. Let k, r, y e nonnegative integers with y < r < k 1. If a t a t 1 a 0 is the ordinary ase expansion of ν(k, r, y) 1, then t = k, a k = 1, a j = a 0 = 0 for all j {r + 1,..., k 1}, and a i {0, 1} for all i {0, 1,..., k}. Proof. Let ν = ν(k, r, y), and let a t a t 1 a 0 e the ordinary ase expansion of ν 1. It follows from the fact that ν I(k, r, y) that t = k, a k = 1, and a j = 0 for all j {r + 1,..., k 1}. We still need to show that a 0 = 0 and a i {0, 1} for all i {0, 1,..., k}. Suppose, for the sake of finding a contradiction, that a m {, 3,..., 1} for some m {0, 1,..., k}. Because a k = 1, we know that m {0, 1,..., k 1}. Let ν = ν (a m 1) m. The ordinary ase expansion of ν 1 is simply the word otained from a k a k 1 a 0 y replacing the m th digit (which is a m ) with 1. Oserve that ν < ν and ν I(k, r, y) since k < k + m = k + a m m (a m 1) m ν (a m 1) m = ν. Hence, y the definition of ν in Definition 8, s (ν ) < s (ν). Choose some ase over-expansion c l c l 1 c 0 of ν 1, and let f(c l c l 1 c 0 ) e the word otained from c l c l 1 c 0 y replacing the m th digit (which is c m ) with c m a m + 1. Lemma 9 implies that c m a m + 1 {0, 1} ecause c m {a m 1, a m }. Therefore, f(c l c l 1 c 0 ) is a ase over-expansion of ν 1. We see that f is an injection from the set of ase over-expansions of ν 1 to the set of ase over-expansions of ν 1. This contradicts the fact that s (ν ) < s (ν), so we conclude that a i {0, 1} for all i {0, 1,..., k}. We are left with the task of showing that a 0 = 0. Suppose a 0 0. By the preceding paragraph, we must have a 0 = 1. This means that ν (mod ), so ( ) ( ) ( ) ν ν ν s (ν) = s + 1 < s + s + 1 = s (ν 1) y (1). Since ν (mod ) and ν > k y definition, we know that ν 1 > k. Therefore, ν 1 I(k, r, y) and s (ν) < s (ν 1), which contradicts the definition of ν. It follows from this contradiction that a 0 = 0. the electronic journal of cominatorics 3(4) (016), #P4.8 8

9 Lemma 10 hints that it is of interest to enumerate the ase over-expansions of positive integers whose ordinary ase expansions use only the digits 0 and 1. Let e 0, e 1,..., e l e nonnegative integers with e 0 < e 1 < < e l, and let n = e 0 + e e l. Let Q i denote the operation that changes one ase over-expansion of n into another y increasing the i th digit of an expansion y and decreasing the (i + 1) th digit of the expansion y 1. The operation Q i can only e used if the i th digit of the expansion under consideration is a 0. If we use the operation Q i when the (i + 1) th digit of the expansion is a 0, then this digit is immediately converted to a 1 and the (i + ) th digit is reduced y 1. If the (i + ) th digit is also 0, then it is immediately converted to a 1 and the (i + 3) th digit is reduced y 1. This process continues until a nonzero digit is reduced y 1. After using the operation Q i, any leading 0 s are erased from the expansion. The transformation of 0 s to ( 1) s is analogous to the transformation of 0 s to 9 s that occurs when the numer 1 is sutracted from the numer 10000, resulting in As an example, only the operations Q, Q 4, Q, Q 6, and Q 8 may e used to transform the expansion into a new expansion. The operation Q 4 changes the expansion into the expansion 100( 1)( 1)1011. The operation Q 8 first changes the expansion into ; the leading 0 is then erased, yielding We will transform the ordinary ase expansion of n into a ase over-expansion c t c t 1 c 0 of n y repeated use of the operations just descried. At each step, we will choose the value of c ei for some i {0, 1,..., l} while simultaneously deciding the values of c j for all j {e i 1 +1, e i 1 +,..., e i 1} (or all j < e i in the case i = 0). We proceed y permanently deciding the values of c 0, c 1,..., c e0, then permanently deciding the values of c e0 +1, c e0 +,..., c e1, and so on until we decide the values of c el 1 +1, c el 1 +,..., c el. We omit c el from the word c t c t 1 c 0 if c el = 0; in this case, t = e l 1. For the sake of providing a concrete example of this process, we will suppose that {e 0, e 1,..., e l } = {, 4,, 9, 11} and = 7. Here, the ordinary ase 7 expansion of n is Since e 0 =, we first choose the value of c (while simultaneously deciding the values of c 0 and c 1 ). There is only one way to set c = 1; namely, we keep the same expansion If we want to have c = 0, then we could either perform the operation Q 0 to get the expansion or perform the operation Q 1 to get the expansion Similarly, if we want to have c = 7, then we could either perform the operations Q 0 and then Q to otain or perform the operations Q 1 and then Q to get That is, there are e 0 = ways to set c = 0 and e 0 = ways to set c = 7. By Lemma 9, 0, 1, and 7 are the only possile values of c. For this example, we will assume we chose to use the operations Q 1 and Q to get the expansion Next, we choose the value of c 4 ecause e 1 = 4. Because we used the operation Q, the 4 th digit of the expansion was temporarily converted into a 0. This means that there is no way to set c 4 = 1 using the operations Q i. There is one way to set c 4 = 0; just keep the expansion Similarly, there is one way to set c 4 = 7; just use the operation Q 4 to otain We will assume that we make the former choice and keep the expansion Next, we choose the value of c since e =. The only way to set c = 1 is to keep the expansion the same. Since we have already determined c 0, c 1, c, c 3, c 4, we cannot perform any of the operations Q 0, Q 1, Q, Q 3, Q 4 in order to the electronic journal of cominatorics 3(4) (016), #P4.8 9

10 u 0 u 1 u u 3 u 4 u v 0 v 1 v v 3 v 4 v w 0 w 1 w w w 4 w 0 Figure 1: The edge-weighted digraph G for the given example. Note that each of the vertices u 0, v 0, w 0 is drawn twice. temporarily reduce the th digit of the expansion y 1. In other words, c is stuck with the value 1. We now have the expansion , and we wish to choose the value of c 9 (since e 3 = 9) while simultaneously determining the values of c 6, c 7, c 8. Again, the only way to make c 9 = 1 is to keep the expansion To make c 9 = 0, we could use Q 6 to get , use Q 7 to get , or use Q 8 to get To set c 9 = 7, we could use Q 6 and then Q 9 to otain , use Q 7 and then Q 9 to get , or use Q 8 and then Q 9 to get Hence, there are e 3 e 1 = 3 ways to set c 9 = 0 and e 3 e 1 = 3 ways to set c 9 = 7. We will assume that we choose to use Q 7 (and not Q 9 ) to otain the expansion All that is left to do is decide the value of c 11 (while simultaneously determining c 10 ). We cannot set c 11 = 7 ecause there is no nonzero digit to the left of the 11 th digit from which to orrow. Alternatively, one could oserve that if c 11 = 7, then c 11 c 10 c 0 would e a ase 7 over-expansion for a numer strictly larger than n. We can, however, keep the expansion if we wish to set c 11 = 1. There is one way (ecause e 4 e 3 1 = 1) to set c 11 = 0; simply use the operation Q 10 to otain the expansion Figure 1 depicts an edge-weighted digraph G which encodes all the possile choices that we could have made in this example. The graph has only fifteen vertices, ut we have drawn each of the vertices u 0, v 0, w 0 twice in order to improve the aesthetics of the image. The vertex u i corresponds to choosing to set c ei = 0. The vertex v i corresponds to setting c ei = 1. The vertex w i corresponds to setting c ei = 7. The weights of the edges correspond to the numer of choices possile. For example, if we have chosen to let c = 0 (corresponding to the vertex u ), then there are three ways to set c 9 = 7 (corresponding to the vertex w 3 ). Thus, there is an edge of weight 3 from u to w 3. After setting c 9 = 7, it is impossile to set c 11 = 1, so there is an edge of weight 0 from w 3 to v 4. The reader might ask why there are edges from u 4, v 4, w 4 to u 0, v 0, w 0. We include these edges ecause the electronic journal of cominatorics 3(4) (016), #P4.8 10

11 we wish to interpret ase 7 over-expansion of n in terms of closed walks in the graph G. There are edges of weight from u 4, v 4, w 4 to u 0 ecause there are two ways to set c = 0, regardless of the value of c 11 (recall that we choose c efore choosing c 11 ). Similarly, there are edges of weight 1 from u 4, v 4, w 4 to v 0 and edges of weight from u 4, v 4, w 4 to w 0. Suppose we want to construct a ase 7 over-expansion of n in which c = 0, c 4 = 1, c = 1, c 9 = 7, and c 11 = 0. This choice of the values of c ei for i {0, 1,, 3, 4} corresponds to the closed walk (u 0, v 1, v, w 3, u 4, u 0 ) in G. The weight of this walk (which we calculate as the product of the weights of its edges) is 6, so there are 6 ase 7 over-expansions of n with these specific values of c, c 4, c, c 9, c 11. As another example, there are no ase 7 over-expansions of n in which c = 7, c 4 = 7, c = 1, c 9 = 0, and c 11 = 1 ecause the weight of the closed walk (w 0, w 1, v, u 3, v 4, w 0 ) is 0. We are finally in a position to enumerate the ase 7 over-expansions of n. To do so, we will need the following definition. Definition 11. For any real t, let t 1 t M t = t 1 t and N t = t 1 t t 1 t If we put the vertices of G in the order u 0, v 0, w 0, u 1, v 1, w 1,..., u 4, v 4, w 4, then the adjacency matrix of G (written as a lock matrix) is O M d1 O O O O O M d O O A = O O O M d3 O O O O O B, C O O O O where d i = e i e i 1 1, e l e l B = e l e l = , C =. e 0 1 e 0 e 0 1 e 0 e 0 1 e 0 = and O denotes the 3 3 zero matrix. The total numer of ase 7 over-expansions of n is equal to the sum of the weights of the closed walks of length in G that start at u 0, v 0, or w 0. This, in turn, is equal to the sum of the first three diagonal entries of A. Using elementary linear algera, we see that the first three rows and the first three columns of A intersect in a 3 3 lock given y M 1 M M 3 BC. Therefore, the sum of the first three diagonal entries of A is Tr(M 1 M M 3 BC), where Tr denotes the trace of a matrix. One easily calculates this value to e 18. We now state this result in greater generality (and in a slightly different form). Sketching the proof of the following theorem, we trust the reader to see that the method used in the preceding example is representative of the method used in general., the electronic journal of cominatorics 3(4) (016), #P4.8 11

12 Theorem 1. Let e 0, e 1,..., e l e nonnegative integers with e 0 < e 1 < < e l (where l 1). Let d i = e i e i 1 1 for all i {1,,..., l}. The numer of ase over-expansions of the numer e 0 + e e l is given y Tr(N e0 M d1 M d M dl ). Proof. Let n = e 0 + e e l. Let O M d1 O O O O O O M d O O O O O O O O O A = , O O O O M dl 1 O O O O O O B C O O O O O where B = d l 1 0 d l and C = e 0 1 e 0 e 0 1 e 0 e 0 1 e 0 Let G e the edge-weighted digraph with vertex set {u 0, v 0, w 0, u 1, v 1, w 1,..., u l, v l, w l } and adjacency matrix A. Suppose we wish to construct a ase over-expansion c t c t 1 c 0 of n. As in the example aove, specifying the values of c e0, c e1,..., c el (each such value must e 0, 1, or y Lemma 9) corresponds to choosing a closed walk of length l + 1 in G starting at u 0, v 0, or w 0. The weight of this walk is the numer of ase overexpansions c el c el 1 c 0 of n that have the specified values of c e0, c e1,..., c el. Therefore, the total numer of ase over-expansions of n is equal to the sum of the first three diagonal entries of A l+1. The first three rows and the first three columns of A l+1 intersect in the 3 3 lock M d1 M d M dl 1 BC, so the numer of ase over-expansions of n is Tr(M d1 M d M dl 1 BC). Now, one may easily calculate that BC = M dl N e0. Therefore, using the fact that Tr(XY ) = Tr(Y X) for any square matrices X, Y of the same size, we conclude that Tr(M d1 M d M dl 1 BC) = Tr(M d1 M d M dl 1 M dl N e0 ) = Tr(N e0 M d1 M d M dl ). x 1 z x 1 Definition 13. We define Ξ = x 1 z x 1 : x 1, x, x 3, z 0, z 0 and Ξ = x x 3 x Ξ {I 3 }, where I 3 is the 3 3 identity matrix. We omit the proofs of the following three lemmas ecause they are fairly straightforward. Lemma 14. The set Ξ is closed under matrix multiplication, and Ξ is a monoid under matrix multiplication. For any nonnegative real numer u, M u, N u Ξ.. the electronic journal of cominatorics 3(4) (016), #P4.8 1

13 Lemma 1. Let X e a 3 3 matrix with nonnegative real entries, and let Y Ξ. If the second diagonal entry (the entry in the second row and the second column) of X is positive, then Tr(XY ) > 0. Lemma 16. For any positive integer t, F t F t 1 F t M1 t = F t F t 1 F t. F t 1 F t F t 1 The following lemma attempts to gain information aout the ordinary ase expansion of ν(k, r, y) 1 when y < r < k 1. By Lemma 10, all of the digits in that expansion are 0 s and 1 s, so we may write ν(k, r, y) 1 = e 0 + e e l for some nonnegative integers e 0, e 1,..., e l that satisfy e 0 < e 1 < < e l = k. Because ν(k, r, y) I(k, r, y), we have y k ν(k, r, y) 1 < k + r i. It follows that e l 1 j r j for all j {0, 1,..., y}, where equality cannot hold for all j. Hence, we may consider the smallest nonnegative integer λ such that e l 1 λ r λ 1. Lemma 17. Let k, r, y e nonnegative integers with y < r < k 1. Let a k a k 1 a 0 e the ordinary ase expansion of ν(k, r, y) 1, and write ν(k, r, y) 1 = e 0 + e e l, where e 0, e 1,..., e l are nonnegative integers that satisfy e 0 < e 1 < < e l = k. Let λ e the smallest nonnegative integer such that e l 1 λ r λ 1. There does not exist i {1,,..., e l 1 λ } such that a i = a i+1. Proof. Let ν = ν(k, r, y). Note that it follows from the preceding paragraph that λ y. Because ν 1 = e 0 + e e l, we have { 1, if i = e j for some j; a i = (6) 0, otherwise. for all i {0, 1,..., k}. In particular, e 0 > 0 ecause a 0 = 0 y Lemma 10. We will let d j = e j e j 1 1 for all j {1,,..., l}. Suppose, y way of contradiction, that a i = a i+1 for some i {1,,..., e l 1 λ }, and let m e the smallest such index i. We have five cases to consider. First, assume a 1 = 1 and a m = a m+1 = 0. By the minimality of m, we see that a i = 0 for all nonnegative even integers i < m and a j = 1 for all positive odd integers j < m. In addition, m must e even. Therefore, setting t = m, we have e j = j + 1 for all j {0, 1,..., t}. This means that d j = 1 for all j {1,,..., t}. It follows from (6) and the assumption that a m = a m+1 = 0 that e t = m 1 < m < m + 1 < e t+1 < e t+ < < e l = k. (7) the electronic journal of cominatorics 3(4) (016), #P4.8 13

14 Note that t < l 1 λ ecause e t < m {1,,..., e l 1 λ }. Let us write ν 1 = 1 + t ei+1 +. We have i=t+1 ν 1 = 1 + t ei+1 + i=t m n + n=1 i=t e t+1 n=1 n + i=t+1 < e t i=t+1 = e t l 1 λ i=t+1 + i=l λ Recall from the paragraph immediately preceding this theorem that e l 1 j r j for all j {0, 1,..., y}. Since λ is the smallest nonnegative integer such that e l 1 λ r λ 1, e l 1 i = r i for all nonnegative integers i < λ. Furthermore, e l = k, so we find that λ 1 = k + r i. Therefore, i=l λ ν 1 < e t l 1 λ i=t+1. λ 1 λ 1 + k + r i < el 1 λ+1 + k + λ 1 k + r λ + r i k + y r i, where we have used the inequalities e l 1 λ r λ 1 and λ y. This shows that ν 1 I(k, r, y), so s (ν 1 ) s (ν) y the definition of ν. Our goal is to derive a contradiction y showing that s (ν 1 ) > s (ν). Let B = M dt+ M dt+3 M dl. We saw that t < l 1 λ, so t + 1 < l. This shows that the matrix product defining B is nonempty. Lemma 14 implies that B Ξ. Now, e 0 = (0) + 1 = 1. By Theorem 1, s (ν) = Tr(N 1 M d1 M d M dl ) = Tr(N 1 M t 1M dt+1 B). We may write ν 1 1 = e 0 + e e l, where e j = e j + 1 for all j {0, 1,..., t} and e j = e j for all j {t + 1,..., l}. Setting d j = e j e j 1 for all j {1,,..., l}, we have y Theorem 1. Consequently, s (ν 1 ) = Tr(N e 0 M d 1 M d M d l ) = Tr(N M t 1M dt+1 1B) s (ν 1 ) s (ν) = Tr((N M t 1M dt+1 1 N 1 M t 1M dt+1 )B). We remark that one must take care when considering the case t = 0. In this case, M t 1 is the 3 3 identity matrix. A straightforward calculation invoking Lemma 16 shows that N M t 1M dt+1 1 N 1 M t 1M dt+1 = F t+ N dt+1, r i the electronic journal of cominatorics 3(4) (016), #P4.8 14

15 so s (ν 1 ) s (ν) = F t+ Tr(N dt+1 B). Because d t+1 = e t+1 e t 1 (m + ) (m 1) 1 = y (7), we see that N dt+1 is a matrix with nonnegative entries whose second diagonal entry is positive. By Lemma 1, s (ν 1 ) s (ν) > 0, which is our desired contradiction. Next, we assume a 1 = 1 and a m = a m+1 = 1. The proof is similar to that given in the preceding paragraph. It follows from the minimality of m that a i = 0 for all even nonnegative integers i < m and a j = 1 for all odd positive integers j < m. Also, m must e odd. Let t = m 1. We have e j = j + 1 for all j {0, 1,..., t} and e t = m < m + 1 = e t+1 < e t+ < < e l = k. We know from Lemma 10 that a j = 0 for all j {r +1,..., k 1}, so m < r. This implies t 1 that e t+1 = m + 1 r < k = e l. Setting ν = 1 + ei+1 +, we have t 1 ν 1 = ei+1 + = t ei 1 + i=1 i=t+1 i=t+1 t 1 = ei < ν 1 < k + i=t+1 i=t+1 y r i, so ν I(k, r, y) and ν < ν. It follows from the definition of ν that s (ν ) < s (ν). As in the preceding case, we let B = M dt+ M dt+3 M dl. The matrix product defining B is nonempty ecause e t+1 < e l. In addition, B Ξ, so we may write B = ξ 1 ξ ξ 1 ξ 1 ξ ξ 1 ξ 3 ξ 4 ξ 3 for some nonnegative real ξ 1, ξ, ξ 3, ξ 4 with ξ > 0. Since d t+1 = e t+1 e t 1 = (m + 1) m 1 = 0, we have y Theorem 1 that s (ν) = Tr(N 1 M t 1M 0 B) and s (ν ) = Tr(N M t 1B). As in the previous paragraph, note that M1 t is the 3 3 identity matrix if t = 0. By elementary calculations that make use of Lemma 16, we find that N M1 t N 1 M1M t 0 = F t Thus, s (ν ) s (ν) = Tr((N M t 1 N 1 M t 1M 0 )B) = F t+1 (ξ 3 + ξ 4 ) 0, the electronic journal of cominatorics 3(4) (016), #P4.8 1

16 which is a contradiction. We now assume a 1 = 0, a = 1, and a m = a m+1 = 0. Invoking the minimality of m, we see that a i = 1 for all positive even integers i < m and a j = 1 for all positive odd integers j < m. The index m must e odd. Let t = m 3. We find that e j = j + for all j {0, 1,..., t}, so d j = 1 for all j {1,,..., t}. Coupling (6) with the assumption that a m = a m+1 = 0 shows us that e t = m 1 < m < m + 1 < e t+1 < e t+ < < e l = k. (8) Thus, t < l 1 λ ecause e t < m {1,,..., e l 1 λ }. Let ν 3 = 1++ We find that e t+1 n=1 ν 3 1 = + n + i=t+1 t ei+1 + < e t i=t+1 i=t+1 e t+1 n + n=1 = e t i=t+1 l 1 λ i=t+1 t ei i=l λ Because e l = k and e l 1 i = r i for all nonnegative integers i < λ, we have Consequently, ν 3 1 e t l 1 λ i=t+1 i=l λ λ 1 = k + r i. λ 1 λ 1 + k + r i < el 1 λ+1 + k + λ 1 k + r λ + r i k + y r i,. r i i=t+1 where we have used the inequalities e l 1 λ r λ 1 and λ y. It follows that ν 3 I(k, r, y), so s (ν 3 ) s (ν). Let B = M dt+ M dt+3 M dl. As in the previous two cases, the matrix product defining B is nonempty (ecause t + 1 < l λ l) and B Ξ. Note that e 0 = (0) + =. We have s (ν) = Tr(N M1M t dt+1 B) and s (ν 3 ) = Tr(N 1 M1 t+1 M dt+1 1B), so s (ν 3 ) s (ν) = Tr((N 1 M t+1 1 M dt+1 1 N M t 1M dt+1 )B). Using Lemma 16, one may easily show that N 1 M t+1 1 M dt+1 1 N M t 1M dt+1 = F t+3 N dt+1,. the electronic journal of cominatorics 3(4) (016), #P4.8 16

17 so s (ν 3 ) s (ν) = F t+3 Tr(N dt+1 B). Because d t+1 = e t+1 e t 1 (m + ) (m 1) 1 = y (8), there follows that N dt+1 is a matrix with nonnegative entries whose second diagonal entry is positive. By Lemma 1, s (ν 3 ) s (ν) > 0, a contradiction. The fourth case we consider is that in which a 1 = 0, a = 1, and a m = a m+1 = 1. By now the reader is proaly familiar with the general pattern of the proof. We use the minimality of m to say that a i = 1 for all positive even integers i < m and a j = 1 for all positive odd integers j < m. The index m must e even. Let t = m, and oserve that e j = j + for all j {0, 1,..., t} and that e t = m < m + 1 = e t+1 < e t+ < < e l = k. We know that m < r ecause Lemma 10 tells us that a j = 0 for all j {r + 1,..., k 1}. t 1 Therefore, e t+1 r < k = e l. Let ν 4 = ei+1 +. We have = + t 1 ν 4 1 = + ei+1 + t ei 1 + i=1 i=t+1 = i=t+1 t ei 1 + i=t+1 t 1 = + ei i=t+1 i=t+1 < ν 1 < k + y r i, so ν 4 I(k, r, y) and ν 4 < ν. This implies that s (ν 4 ) < s (ν). Let B =M dt+ M dt+3 M dl. The matrix product defining B is nonempty ecause e t+1 < e l, and B Ξ. Consequently, ξ 1 ξ ξ 1 B = ξ 1 ξ ξ 1 ξ 3 ξ 4 ξ 3 for some nonnegative real ξ 1, ξ, ξ 3, ξ 4 with ξ > 0. Because d t+1 = e t+1 e t 1 = (m + 1) m 1 = 0, we may use Theorem 1 to deduce that One may show that so s (ν) = Tr(N M t 1M 0 B) and s (ν 4 ) = Tr(N 1 M t+1 1 B). N 1 M1 t+1 N M1M t 0 = F t s (ν 4 ) s (ν) = Tr((N 1 M t+1 1 N M t 1M 0 )B) = F t+ (ξ 3 + ξ 4 ) 0., the electronic journal of cominatorics 3(4) (016), #P4.8 17

18 This is our desired contradiction. Finally, we consider the case in which a 1 = a = 0. In this case, ν 1 (mod 3 ). y y Because ν 1 < k + r i and ν 1 k + r i 0 (mod ), it follows that ν 1 k + y r i. This means that ν + I(k, r, y), so s (ν + ) s (ν) y the definition of ν. As ν (mod 3 ), we find y (1) and Lemma that ( ) ( ) (ν + ) 1 (ν + ) + 1 s (ν + ) = s + s and ( ) (( ν 1 ν ) ) 1 s + 1 = s (( ν ) ) 1 > s (( ν ) ) 1 + s + 1 ( ) ( ) ν 1 ν 1 = s = s 3. Therefore, ( ) ( ) (ν + ) 1 (ν + ) + 1 s (ν + ) = s + s ( ) ( ) ( ) ( ) ν 1 ν 1 ν 1 ν 1 = s s + 1 > s s = s (ν). With this contradiction, the proof is complete. Lemma 18. Preserving the notation from Lemma 17, we have e l 1 λ = r λ 1. Proof. The proof is very similar to those of the first and third cases considered in the proof of Lemma 17. Suppose, y way of contradiction, that e l 1 λ r λ, and let l 1 λ ν = 1 + ei+1 + and ν = ν +. We have ν 1 < ν 1 = + < r λ + i=l λ i=l λ l 1 λ +1 + i=l λ e l 1 λ +1 n=1 n + λ 1 = r λ + k + r i = k + i=l λ r λ 1 n=1 λ r i k + n + y r i, i=l λ λ 1 where we convene to let r i = 0 if λ = 0. Therefore, ν, ν I(k, r, y). This means that s (ν ), s (ν ) s (ν), so we will derive a contradiction y showing that s (ν ) > s (ν) or s (ν ) > s (ν). the electronic journal of cominatorics 3(4) (016), #P4.8 18

19 Let B = M dl λ+1 M dl λ+ M dl, where we define B to e the 3 3 identity matrix if λ = 0. By Lemma 14, B Ξ. To ease our notation, let t = l 1 λ and q = d l λ. We have s (ν) = Tr(N e0 M1M t q B), s (ν ) = Tr(N e0 +1M1M t q 1 B), and s (ν ) = Tr(N 1 M e0 1M1M t q 1 B). It is straightforward to show that N M1M t q 1 N 1 M1M t q = F t+ N q and N 1 M1 t+1 M q 1 N M1M t q = F t+3 N q. If λ = 0, then it follows from our assumption that e l 1 λ r λ that q = e l λ e l 1 λ 1 = e l e l 1 1 = k e l 1 1 k (r ) 1 3. If λ > 0, then q = e l λ e l 1 λ 1 = e l 1 (λ 1) e l 1 λ 1 = (r (λ 1)) e l 1 λ 1 (r (λ 1)) (r λ ) 1 = 3. In either case, q 3, so oth F t+ N q and F t+3 N q are 3 3 matrices with nonnegative real entries whose second diagonal entries are positive. It follows from Lemma 1 that if e 0 = 1, then s (ν ) s (ν) = Tr((N M t 1M q 1 N 1 M t 1M q )B) = Tr(F t+ N q B) > 0. Similarly, if e 0 =, then s (ν ) s (ν) = Tr((N 1 M t+1 1 M q 1 N M t 1M q )B) = Tr(F t+3 N q B) > 0. This is a contradiction, so the proof is complete. For nonnegative integers k, r, y with y < r < k 1, let w = a k a k 1 a 0 e the ordinary ase expansion of ν(k, r, y) 1. We summarize here the information we have gained aout this expansion. From Lemma 10, we know that a i {0, 1} for all i. We know from the same lemma that a k = 1, a 0 = 0, and a j = 0 for all j {r + 1,..., k 1}. Therefore, w = 10 k r 1 v, where v is a word of length r +1 over the inary alphaet {0, 1} that ends in the letter (digit) 0. From Lemmas 17 and 18 and (6), we find that there exists some λ {0, 1,..., y} such that A. a r i = 1 and a r i 1 = 0 for all i {0, 1,..., λ 1}. B. a r λ = 0. C. a r λ i+1 = 1 and a r λ i = 0 for all i {1,,..., r/ λ}. That is, This means that w = v = This leads us to the following. { (10) λ 0(10) r/ λ 0, if r; (10) λ 0(10) r/ λ, if r. { 10 k r 1 (10) λ 0(10) r/ λ 0, if r; 10 k r 1 (10) λ 0(10) r/ λ, if r. (9) the electronic journal of cominatorics 3(4) (016), #P4.8 19

20 Definition 19. Let k, r, x e nonnegative integers such that x < r < k 1. Define the word w(k, r, x) y { 10 k r 1 (10) x 0(10) r/ x 0, if r; w(k, r, x) = 10 k r 1 (10) x 0(10) r/ x, if r, and let γ(k, r, x) e the positive integer whose ordinary ase expansion is w(k, r, x). Definition 0. For any integers k, r, x, define V (k, r, x) = 1 ((k r)(l r+ L r 4x+1 ) + L r+1 + L r 4x+ ). Lemma 1. Let k, r, x, y e nonnegative integers such that x y < r < k 1. We have γ(k, r, x) + 1 I(k, r, y) and s (γ(k, r, x) + 1) = V (k, r, x). Proof. The ordinary ase expansion of k + y r i is 10 k r 1 (10) y+1 0 r y 1, so one may see from Definition 19 and the hypothesis that x y that γ(k, r, x)+1 k + y r i. It should also e clear from Definition 19 that k < γ(k, r, x)+1, so γ(k, r, x)+1 I(k, r, y). If we choose to write γ(k, r, x) = for nonnegative integers e 0, e 1,..., e l with e 0 < e 1 < e l, then and e 0 = {, if r; 1, if r, e j = e 0 + j for all j {1,,..., t}, e t+j = r x + j = e 0 + t + j + 1 for all j {1,,..., x}, e l = k, where t = r/ x 1. Oserve that the letter (digit) 1 appears exactly r/ + 1 times in the word w(k, r, x), so it follows from (6) that l = r/ (implying that t + 1 = l x). Let us first assume x = 0. Setting d j = e j e j 1 1 for all j {1,,..., l}, we have { 1, if j {1,,..., l 1}; d j = k r, if j = l. We may invoke Theorem 1 to see that s (γ(k, r, 0) + 1) = Tr(N e0 M d1 M d M dl ) = Tr(N e0 M l 1 1 M k r ). (10) the electronic journal of cominatorics 3(4) (016), #P4.8 0

21 Let T = If r, we may use Lemma 16 to find that Similarly, if r, then F r F r 1 F r F r F r 1 F r N e0 M l 1 1 = N 1 M 1 (r ) 1 = T. N e0 M l 1 1 = N M 1 (r 3) 1 = T. Therefore, regardless of the parity of r, we have s (γ(k, r, 0) + 1) = Tr(T M k r ) y (10). Thus, s (γ(k, r, 0) + 1) = Tr(T M k r ) = (k r)f r+1 + F r+, where the last equality is otained y simply calculating the matrix T M k r y hand. One may easily show that F r+1 = 1 (L r+ L r+1 ) and F r+ = 1 (L r+1 + L r+ ), so s (γ(k, r, 0) + 1) = 1 ((k r)(l r+ L r+1 ) + L r+1 + L r+ ) = V (k, r, 0). Let us now assume x > 0. Again, we set d j = e j e j 1 1 for all j {1,,..., l}, and we see that 1, if j {1,,..., l 1} \ {t + 1}; d j =, if j = t + 1; k r 1, if j = l. By Theorem 1, s (γ(k, r, x) + 1) = Tr(N e0 M d1 M d M dl ) = Tr(N e0 M t 1M M l 1 (t+1) 1 M k r 1 ) = Tr(N e0 M t 1M M x 1 1 M k r 1 ). If r, then Lemma 16 allows us to find that Similarly, if r, then N e0 M t 1 = N 1 M r/ x 1 1 = N e0 M t 1 = N M (r 1)/ x 1 1 = F r x F r x 1 F r x F r x F r x 1 F r x F r x F r x 1 F r x F r x F r x 1 F r x the electronic journal of cominatorics 3(4) (016), #P4.8 1

22 Therefore, no matter the parity of r, we have N e0 M t 1M M x 1 1 M k r 1 = F r x F r x 1 F r x F r x F r x 1 F r x Using Lemma 16 again, we find that F r x F r x 1 F r x F r x F r x 1 F r x M M1 x 1 M k r 1 = where and Therefore, M M x 1 1 M k r 1. χ 1 χ χ 1 χ 1 χ χ χ 1 = (k r)(f r x+3 F x + F r x+1 F x 1 ) F r x+3 F x F r x+1 F x 3 χ = F r x+3 F x + F r x+1 F x 1. s (γ(k, r, x) + 1) = Tr(N e0 M t 1M M x 1 1 M k r 1 ) = χ 1 + χ = (k r)(f r x+3 F x + F r x+1 F x 1 ) + F r x+3 F x 1 + F r x+1 F x. We now make use of the well-known identity to otain F m F n = 1 (L m+n ( 1) n L m n ) F r x+3 F x + F r x+1 F x 1 = 1 (L r+3 L r 4x+3 + L r + L r 4x+ ), = 1 (L r+ + L r+1 + L r L r 4x+1 ) = 1 (L r+ L r 4x+1 ) and F r x+3 F x 1 + F r x+1 F x = 1 (L r+ + L r 4x+4 + L r 1 L r 4x+3 ) Consequently, s (γ(k, r, x) + 1) = (k r) = 1 (L r+1 + L r + L r 1 + L r 4x+ ) = 1 (L r+1 + L r 4x+ ). ( ) 1 (L r+ L r 4x+1 ) + 1 (L r+1 + L r 4x+ ) = V (k, r, x). Before proceeding to our main result, we need one final easy lemma. Lemma. For any integers k, r, x, V (k, r, x + 1) V (k, r, x) = 1 ((k r)(l r 4x+1 L r 4x 3 ) + L r 4x L r 4x+ ). the electronic journal of cominatorics 3(4) (016), #P4.8

23 Proof. Referring to Definition 0, we have V (k, r, x + 1) = A + 1 ((k r)( L r 4(x+1)+1) + L r 4(x+1)+ ) and V (k, r, x) = A + 1 ((k r)( L r 4x+1) + L r 4x+ ), where A = 1 ((k r)(l r+) + L r+1 ). Thus, V (k, r, x + 1) V (k, r, x) = 1 ((k r)(l r 4x+1 L r 4x 3 ) + L r 4x L r 4x+ ). We are finally in a position to prove our main result. Theorem 3. Let k, r, y e nonnegatve integers such that y < r < k 1. If r is even, then ν(k, r, y) = γ(k, r, y) + 1 and µ(k, r, y) = V (k, r, y). If k = r 3 (mod 4) and y r + 1 4, then Otherwise, ν(k, r, y) = γ(k, r, (r 3)/4) + 1 and µ(k, r, y) = V (k, r, (r 3)/4). ν(k, r, y) = γ(k, r, δ(r, y)) + 1 and µ(k, r, y) = V (k, r, δ(r, y)), where δ(r, y) = min { } r 1 4, y. Proof. We saw in the discussion immediately preceding Definition 19 that ν(k, r, y) = γ(k, r, λ) + 1 for some λ {0, 1,..., y}; we simply need to determine the value of λ. By Definition 8, λ is the element of {0, 1,..., y} such that s (γ(k, r, λ) + 1) is maximized and γ(k, r, λ) + 1 I(k, r, y). Lemma 1 tells us that γ(k, r, x) + 1 I(k, r, y) and s (γ(k, r, x) + 1) = V (k, r, x) for all x {0, 1,..., y}, so µ(k, r, y) = max{v (k, r, x): x {0, 1,..., y}}. That is, λ must e the element of the set {0, 1,..., y} that maximizes V (k, r, λ) (if there are multiple such elements, we choose the smallest in accordance with the definition of ν(k, r, y) in Definition 8). We first assume r is even. We wish to show that λ = y. This is immediate if y = 0, so we will assume y > 0. Choose some x {0, 1,..., y 1}. We make use of an easily-proven fact aout Lucas numers that states that if m, n are odd integers (not necessarily positive) with m < n, then L m < L n. In this case, we use the assumption that r is even to see that r 4x 3 and r 4x + 1 are odd integers with r 4x 3 < r 4x + 1. Therefore, L r 4x+1 L r 4x 3 > 0. Because k r y hypothesis, we may use Lemma to find that V (k, r, x + 1) V (k, r, x) 1 ((L r 4x+1 L r 4x 3 ) + L r 4x L r 4x+ ) the electronic journal of cominatorics 3(4) (016), #P4.8 3

24 = 1 (L r 4x 1 L r 4x ). As r 4x and r 4x 1 are odd integers with r 4x < r 4x 1, we see that Because x was aritrary, this shows that so λ = y. V (k, r, x + 1) V (k, r, x) > 0. V (k, r, 0) < V (k, r, 1) < < V (k, r, y), We now assume r is odd. Let x 1, x e integers such that x 1 r 4 < r 1 4 x. We will make use of the fact that L m = L m for any even integer m. Because x 1 r 4, we otain the inequalities r 4x 1 +1 > r 4x 1 3 > 0. Therefore, L r 4x1 +1 L r 4x1 3 > 0. Because k r, we see from Lemma that V (k, r, x 1 + 1) V (k, r, x 1 ) 1 ((L r 4x 1 +1 L r 4x1 3) + L r 4x1 L r 4x1 +) = 1 (L r 4x 1 1 L r 4x1 ). We know that L r 4x1 1 L r 4x1 > 0 ecause r 4x 1 0. Thus, V (k, r, x 1 + 1) > V (k, r, x 1 ). (11) If y < r 1 r 4, then every element of the set {0, 1,..., y 1} is less than, so (11) 4 shows that This shows that if y < r 1 4, then V (k, r, 0) < V (k, r, 1) < < V (k, r, y). λ = y = δ(r, y), which agrees with the statement of the theorem. Therefore, we will henceforth assume that y r 1 4 so that δ(r, y) = r 1 4. It follows from the inequality x r 1 that 4 4x r + 3 > 4x r 1. If 4x r 1 0, then If 4x r 1 =, then L 4x r 1 L 4x r+3 < 0. L 4x r 1 L 4x r+3 = L L = 0. Hence, L 4x r 1 L 4x r+3 0. the electronic journal of cominatorics 3(4) (016), #P4.8 4

25 Because r 4x + 1 and r 4x 3 are even, L r 4x +1 L r 4x 3 = L 4x r 1 L 4x r+3 0. Using Lemma and the fact that k r, we have V (k, r, x + 1) V (k, r, x ) 1 ((L r 4x +1 L r 4x 3) + L r 4x L r 4x +) = 1 (L r 4x 1 L r 4x ). Because 4x r + > 4x r and r 4x 1, r 4x are even, we have This shows that L r 4x 1 L r 4x = L 4x r+1 L 4x r+ < 0. V (k, r, x + 1) < V (k, r, x ). (1) We now have three short cases to consider. In the first case, assume r 1 (mod 4). We have shown through (11) and (1) that and V (k, r, m) > V (k, r, m 1) > V (k, r, m ) > V (k, r, m) > V (k, r, m + 1) > V (k, r, m + ) >, where m = r 1 r 1. This implies that 4 4 is the unique value of x that maximizes r 1, so {0, 1,..., y}. Conse- 4 V (k, r, x). Recall that we have assumed that y r 1 4 quently, λ = r 1 = δ(r, y). 4 For the second case, we suppose that r 3 (mod 4) and r < k. We need to show that λ = δ(r, y). We may use (11) and (1) to see that and where n = r 3. By Lemma, 4 V (k, r, n) > V (k, r, n 1) > V (k, r, n ) > V (k, r, n + 1) > V (k, r, n + ) > V (k, r, n + 3) >, V (k, r, n + 1) V (k, r, n) = 1 ((k r)(l 4 L 0 ) + L 1 L ) = 1 ((k r)(7 ) ) = k r > 0, the electronic journal of cominatorics 3(4) (016), #P4.8

26 so V (k, r, n) < V (k, r, n + 1). This shows that n + 1 = r+1 is the unique value of x 4 that maximizes V (k, r, x). Because y r 1 4 = r+1 r+1, it follows that {0, 1,..., y}. 4 4 Therefore, λ = r+1 = r = δ(r, y). For the third and final case, we assume k = r 3 (mod 4). We wish to show that λ = r 3. Using (11) and (1), we otain 4 and V (k, r, n) > V (k, r, n 1) > V (k, r, n ) > V (k, r, n + 1) > V (k, r, n + ) > V (k, r, n + 3) >, where n = r 3. Lemma tells us that 4 V (k, r, n + 1) V (k, r, n) = 1 ((k r)(l 4 L 0 ) + L 1 L ) = 1 ((k r)(7 ) ) = k r = 0, so V (k, r, n) = V (k, r, n + 1). Therefore, V (k, r, x) otains its maximum when x = n and when x = n+1. This means that µ(k, r, y) = V (k, r, n). Either ν(k, r, y) = γ(k, r, n)+1 or γ(k, r, n+1)+1. Note that γ(k, r, n)+1 < γ(k, r, n+1)+1 y Definition 19. Since ν(k, r, y) is defined to e the smallest element of I(k, r, y) such that s (ν(k, r, y) + 1) = µ(k, r, y), it follows that ν(k, r, y) = γ(k, r, n) + 1. That is, λ = n = r 3 4. We end this section with a definition and a theorem that will prove useful in the next section. Definition 4. For any integers k and m with m k, let G k (m) = k + h m and H k (m) = F m+ + (k m)f m. Although it is possile to give an easy inductive proof of the following theorem using the simple Fionacci-like recurrence we prefer a proof ased on Theorem 1. H k (m) = H k 1 (m 1) + H k (m ), (13) Theorem. For any integers k and m with m k, we have s (G k (m)) = H k (m). Proof. Referring to Definition 3, we see that we may write G k (m) 1 = k + m 3 m i =, the electronic journal of cominatorics 3(4) (016), #P4.8 6

27 where e 0 = m m 3 {, if m; = 1, if m, l = m 1, el = k, and e j = e 0 + j for all j {0, 1,..., l 1}. For example, G 1 (7) 1 = i = Defining d i as in Theorem 1, we have d j = 1 for all j {1,,..., l 1} and d l = e l e l 1 1 = k (e 0 + (l 1)) 1 = k (( m ) ( m 3 + m 1 )) 1 1 = k m + 1. By Theorem 1, s (G k (m)) = Tr(N e0 M1 l 1 M k m+1 ). Using Lemma 16, one may show that ρ 1 ρ ρ 1 N e0 M1 l 1 M k m+1 = ρ 1 ρ ρ 1, where Therefore, ρ 1 = (k m 1)(e 0 F l + F l 1 ) + e 0 F l+ + F l+1 and ρ = e 0 F l + F l 1. s (G k (m)) = ρ 1 + ρ = (k m)(e 0 F l + F l 1 ) + e 0 F l+ + F l+1. If m is even, then e 0 = and l = m, so s (G k (m)) = (k m)(f m + F m 3 ) + F m + F m 1 = (k m)f m + F m+ = H k (m). If m is odd, then e 0 = 1 and l = m 1, so s (G k (m)) = (k m)(f m 1 + F m ) + F m+1 + F m = (k m)f m + F m+ = H k (m). 3 More Manageale Bounds In the previous section, we derived fairly strong upper ounds (relative to those previously known) for the values of s (n) for those integers n satisfying k n k + h k for some integer k 3. Unfortunately, these ounds are somewhat cumersome. In this section, we will weaken them in order to make them cleaner and more easily applicale. Suppose k, m, and n are integers such that m < k and G k (m) < n < G k (m + 1). We wish to use Theorem 3 to show that the point (n, s (n)) lies elow the line segment connecting the points (G k (m), H k (m)) and (G k (m + 1), H k (m + 1)). This leads us to the following definition. the electronic journal of cominatorics 3(4) (016), #P4.8 7

28 Definition 6. For any integers k and m with m < k, define a function J k,m : R R y J k,m (x) = H k(m + 1) H k (m) G k (m + 1) G k (m) (x G k(m)) + H k (m). The graph of J k,m (x) is the line passing through (G k (m), H k (m)) and (G k (m + 1), H k (m + 1)). Our aim is to show that s (n) < J k,m (n) for all n {G k (m) + 1,..., G k (m + 1) 1}. Oserve that we may rewrite J k,m (x) as J k,m (x) = F m 1(k m) (x G k (m)) + H k (m) h m+1 h m ecause H k (m+1) H k (m) = (k m 1)F m+1 +F m+3 (k m)f m F m+ = F m 1 (k m) and G k (m + 1) G k (m) = h m+1 h m. Lemma 7. If m and x are positive integers with m 4 and x < G m 1 (m 3), then s (x) F m. Proof. Let m and x e as in the statement of the lemma. If x < m 1, then it follows from Proposition 6 that s (x) F m. If x = m 1, then s (x) = 1 < F m y Lemma. Therefore, we will assume m 1 < x. Note that the inequalities m 1 < x < G m 1 (m 3) force m 6. Since m 1 < x G m 1 (m 3) 1 = m 1 + m 6 m i, it follows from Definition 8 that x I ( m 1, m, ) m 6. Referring to Definition 8 again, we find that s (x) µ ( m 1, m, ) m 6. Therefore, we simply need to show that µ ( m 1, m, ) m 6 Fm. (14) Suppose m is odd. It follows from Theorem 3 and Definition 0 that µ ( m 1, m, ) ( ) ( ) m 6 = µ m 1, m, m 7 = V m 1, m, m 7 = 1 (4(L m 3 L 10 m ) + L m 4 + L 11 m ). Because m is odd, L 10 m = L m 10 and L 11 m = L m 11. Therefore, For any integer t, let µ ( m 1, m, ) m 6 1 = (4(L m 3 + L m 10 ) + L m 4 + L m 11 ). η t = 1 (4(L t 3 + L t 10 ) + L t 4 + L t 11 ) and κ t = 11F t. the electronic journal of cominatorics 3(4) (016), #P4.8 8

29 It is easy to see that the recurrence relations η t+4 = 3η t+ η t and κ t+4 = 3κ t+ κ t hold for all integers t. In addition, one may verify that κ = η = 0 and κ 7 = η 7 = 11. This implies that κ t = η t for all odd integers t. In particular, µ ( m 1, m, m 6 ) = ηm = κ m = 11F m < F m, where the inequality 11F m < F m is easily-proven for all odd m > 6 (for example, y showing that 11F m = F m F m 8 < 1F m F m 8 = F m ). This proves (14) when m is odd. Next, assume m is even. By Theorem 3, where µ ( m 1, m, m 6 δ ( m, m 6 If m 0 (mod 4), then m 6 4 = m 4, so 4 µ ( m 1, m, m 6 ) = V ( m 1, m, δ ( m, m 6 ) { = min (m ) 1 4, m 6 } = m 6 4. )), ) ( ) = V m 1, m, m = (4(L m 3 L 0 ) + L m 4 + L 1 ) = 1 (8L m 3 + L m 4 7) < 1 (8L m 3 + L m 4 ). Similarly, if m (mod 4), then m 6 4 = m 6, so 4 µ ( m 1, m, m 6 ) ( ) = V m 1, m, m = (4(L m 3 L ) + L m 4 + L 3 ) = 1 (8L m 3 + L m 4 8) < 1 (8L m 3 + L m 4 ). In either case, we may use the identity F u = 1 (L u+1 + L u 1 ) to see that µ ( m 1, m, ) m 6 1 < (8L m 3 +L m 4 ) = 1 (6L m 3 +L m ) = 1 (4L m 3 +L m 1 ) = 1 (L m 3 + L m 1 ) + 1 (L m 3 + L m 4 + L m ) = F m + 1 (L m + L m ) < F m + 1 (L m + L m 4 ) = F m + F m 3 = F m. This proves (14) in the case when m is even, so the proof is complete. Throughout the proofs of the following three lemmas, we assume =. For example, since G k (4) = k + h 4 = k + + 1, it will e understood that G k (4) = k +. Lemma 8. For any positive integers t and x with t log x +, we have s ( t + x) s (x)(t + 1 log x ). the electronic journal of cominatorics 3(4) (016), #P4.8 9

30 Proof. The proof is y induction on t. The inequality t log x + forces t. If t =, then the inequality t log x + forces x = 1 so that s ( t + x) = s () = 3 = s (1)( + 1 log 1 ) = s (x)(t + 1 log x ). Now, suppose t 3. If x is even, then it follows from (1) and induction on t that s ( t + x) = s ( t 1 + x/) s (x/)((t 1) + 1 log (x/) ) = s (x)(t + 1 log x ). Therefore, we may assume x is odd. By induction on t, we have and s ( t 1 + (x 1)/) s ((x 1)/)((t 1) + 1 log ((x 1)/) ) = s ((x 1)/)(t + 1 log x ) s ( t 1 + (x + 1)/) s ((x + 1)/)((t 1) + 1 log ((x + 1)/) ) Therefore, it follows from (1) that s ((x + 1)/)(t + 1 log x ). s ( t + x) = s ( t 1 + (x 1)/ ) + s ( t 1 + (x + 1)/ ) (s ((x 1)/) + s ((x + 1)/)) (t + 1 log x ) = s (x)(t + 1 log x ). Lemma 9. If k, m, and n are integers such that 4 m < k and then s (n) < H k (m). k + m 1 n < k + G m 1 (m 3), Proof. Let k, m, and n e as stated in the lemma. Let x = n k, and note that log x = m 1. We may comine Lemmas 7 and 8 to get s (n) = s ( k + x) s (x)(k + 1 log x ) = s (x)(k m + ) F m (k m + ) < F m (k m) + F m+ = H k (m). Lemma 30. If k, m, and n are integers such that m {, 7}, m < k, and n = k + G m 1 (m 3), then s (n) < J k,m (n). Proof. The proof is very straightforward. We make use of the fact that s (17) = and s (69) = 14. Suppose m =. Then k 6 and n = k + G 4 () = k h = k + 17 y hypothesis. Using Lemma 8 and the remark following Definition 6, we have s (n) = s ( k + 17) s (17)(k + 1 log 17 ) = (k 3) < 34 (k ) + 13 = 3(k ) 1 11 ((k + 17) ( k + 11)) (k ) the electronic journal of cominatorics 3(4) (016), #P4.8 30

31 = F 4(k ) (n G k ()) + F 7 + F (k ) = J k, (n). h 6 h Next, suppose m = 7. Then k 8 and n = k + G 6 (4) = k = k Using Lemma 8 and the paragraph following Definition 6 once again yields s (n) = s ( k + 69) s (69)(k + 1 log 69 ) = 14(k ) < 104 (k 7) (k 7) 1 8(k 7) = 8 43 ((k + 69) ( k + 43)) (k 7) = F 6(k 7) (n G k (7)) + F 9 + F 7 (k 7) = J k,7 (n). h 8 h 7 We now return to our ase assumption that may represent any integer larger than 1. Lemma 31. Let m and y e integers such that m 4 and 1 y m. We have L m 4y 1 L m and L m 4y L m 6. Proof. First, suppose m is even. We use the fact that L u < L v for any odd integers u, v with u < v to see that the maximum possile value of L m 4y 1 is otained when y = 1. That is, L m 4y 1 L m. It is easy to see that L u for all even integers u, so L m 4y > L m 6. Next, suppose m is odd. We use the fact that L u = L u < L v = L v for any nonnegative even integers u, v with u < v to find that the ilateral sequence..., L 4, L, L 0, L, L 4,... is strictly convex. Therefore, the maximum possile value of L m 4y 1 occurs when y = 1 or when y = m = m 3. Because L m 4(1) 1 = L m = L m = L m 4((m 3)/) 1, the maximum possile value of L m 4y 1 is L m. As L u < L v for any odd integers u, v with u < v, the smallest possile value of L m 4y occurs when y = m 3. That is, L m 4y L m 4((m 3)/) = L 6 m. Now, L u = L u for all odd integers u, so L m 4y L 6 m = L m 6. Lemma 3. If m and y are integers such that m 4 and 1 y m, then L m + L m 4y >. F m 1 Furthermore, if m 4 and m {, 7}, then L m + L m 4 F m Proof. We will use the closed-form expressions F n = φn φ n and L n = φ n + φ n for the Fionacci and Lucas numers. We also make use of the fact that L u for any even integer u. For any integer j, we have L m j F m 1 1 φ j 1 = φ m j m j + φ (φ m 1 φ m 1) 1 φ j 1 = 1 φ (φ m j j 1 + φ j 1) ( φ j 1 φ m 1 φ m 1) the electronic journal of cominatorics 3(4) (016), #P4.8 31

32 = 1 φ m j Lj 1 φ j 1 φ m 1 φ m 1 = L j 1 φ m j = ( 1) m j L j 1, φ j 1 F m 1 φ m 1 F m 1 where the last equality follows from the identity φ = 1 φ. Therefore, L m j F m 1 = L j 1 1 φ j 1 + ( 1)m j. (1) φ m 1 F m 1 If m is even, then we may set j = in (1) and invoke the inequality L m 4y (since m 4y is even) to find that L m + L m 4y F m 1 L m + F m 1 > L m F m 1 = 1 φ + ( 1)m L 1 φ m 1 F m 1 = 1 φ + 1 > 1 φ m 1 F m 1 φ >. Let us now suppose m is odd. By Lemma 31, L m + L m 4y F m 1 L m L m 6 F m 1 Using (1), once with j = 3 and once with j =, yields = L m 3 + L m F m 1. L m 3 + L m F m 1 = 1 φ + ( 1)m 3 L φ m 1 F m φ 4 + ( 1)m L 4 φ m 1 F m 1 > 1 φ + 1 φ 4 = ecause ( 1) m 3 L and ( 1) m L 4 are positive. This completes the φ m 1 F m 1 φ m 1 F m 1 proof of the first inequality in the statement of the lemma. Assume, now, that m 4 and m {, 7}. We may use (1), once with j = and again with j = 4, to see that L m + L m 4 F m 1 = 1 φ + ( 1)m L 1 φ m 1 F m φ 3 + ( 1)m 4 L 3 φ m 1 F m 1 = 1 φ + 1 φ 3 + ( 1)m (L 1 + L 3 ) = 1 φ m 1 F m 1 φ + 1 φ 3 + ( 1)m. φ m 1 F m 1 For m 4 and m {, 7}, it is easy to see that ( 1) m φ m 1 F m 1 attains its minimum when m = 9. Thus, L m + L m 4 1 F m 1 φ + 1 φ 3 + ( 1)9 = 8 φ 8 F 8 1, where this last equality is easily verified. Alternatively, we could have proven the second inequality stated in the lemma using the fact that L m + L m 4 F m 1 = F m 3 F m 1. the electronic journal of cominatorics 3(4) (016), #P4.8 3

33 Lemma 33. Let k and m e integers such that k and m k 1. Define a function Θ k,m : R R y Θ k,m (x) = J k,m( k + x) ( k + x) log φ. For any x 0, d dx Θ k,m( k + x) 0. Proof. We egin with the easily-proven fact that F m F m 1 3 this, we have for all integers m 4. From H k (m) (k m)f m 1 = (k m)f m + F m+ (k m)f m 1 (k m)f m + (k m)f m+ (k m)f m 1 = F m + F m+ F m 1 One may easily show that = 4F m 1 + 3F m F m log φ 6 log φ = F m F m 1 6. for any choice of an integer (if 3, this follows from the oservation that > 6 log φ). In addition, using Definition 3 and Lemma 4, we have h m m and Therefore, h m+1 h m m m + m = m 1. k + h m log φ k + m log φ = m 1 ( k m log φ) m 1 ( + 1 log φ) 6 m 1 log φ 6(h m+1 h m ) log φ Let T k (m) = (k m)f m 1 h m+1 h m x 0. We have H k (m) (k m)f m 1 (h m+1 h m ) log φ. so that we otain k +h m log φ H k(m) T k (m) log φ. Choose some xt k (m)(1 log φ) 0 T k (m) [ ] Hk (m) T k (m) log φ h m log φ k = [H k (m) T k (m)h m ] log φ k T k (m). Adding xt k (m) log φ + k T k (m) to each side of this last inequality yields which we may rewrite as ( k + x)t k (m) log φ (xt k (m) + [H k (m) T k (m)h m ]), ( k + x)t k (m) log φ [H k (m) + T k (m)(x h m )] 0. (16) the electronic journal of cominatorics 3(4) (016), #P4.8 33

34 Now, referring to the paragraph immediately following Definition 6, we see that J k,m ( k +x) = T k (m)(x h m )+H k (m). Therefore, (16) ecomes ( k +x)t k (m) J k,m ( k + d x) log φ 0. In addition, dx J k,m( k + x) = T k (m). Consequently, d J k,m ( k + x) dx ( k + x) = (k + x) log φ T k (m) J k,m ( k + x) log φ( k + x) log φ 1 log φ ( k + x) log φ = (k + x)t k (m) J k,m ( k + x) log φ ( k + x) log φ+1 0. We are finally ready to prove the main results of this section. Theorem 34. If k, m, and n are integers such that m < k and G k (m) < n < G k (m + 1), then s (n) < J k,m (n). Proof. By Definition 3 and Definition 4, n G k (m + 1) 1 = k + m m 1 i. Therefore, we may let y e the smallest element of the set { 0, 1,..., } m such that y n k + m 1 i. Referring to Definition 8, we find that s (n) µ(k, m 1, y) ecause n I(k, m 1, y). Hence, it suffices to prove that µ(k, m 1, y) < J k,m (n). (17) It follows from Theorem 3, Definition 0, Definition 4, and the identity F u = 1 (L u+1 + L u 1 ) that µ(k, m 1, 0) = V (k, m 1, 0) = 1 ((k m + 1)(L m+1 L m ) + L m + L m+1 ) ( ) 1 = (k m) (L m+1 L m ) + 1 (L m+ + L m 1 ) ( ) 1 = (k m) (L m+1 + L m 1 ) + 1 (L m+3 + L m+1 ) = (k m)f m + F m+ = H k (m). This shows that if y = 0, then s (n) µ(k, m 1, 0) = H k (m) = J k,m (G k (m)) < J k,m (n), the electronic journal of cominatorics 3(4) (016), #P4.8 34

35 where we have used the trivial fact that J k,m is an increasing function. Therefore, we may y 1 assume y > 0. Because of the way we chose y, this means that n > k + m 1 i k + m 1. Note that this also forces m 4 ecause y m. By a similar token, it follows from Lemma 9 that s (n) = s (n) H k (m) = J k,m (G k (m)) < J k,m (n) if = and n < k + G m 1 (m 3). Therefore, we may assume n k + G m 1 (m 3) if =. If =, m {, 7}, and n = k + G m 1 (m 3), then the desired result is simply Lemma 30. This means that if = and m {, 7}, then we may assume n k + G m 1 (m 3) + 1. We have two cases to consider. Case 1: In this case, assume that µ(k, m 1, y) = V (k, m 1, y). Recall Definition 0 to see that V (k, m 1, y) = 1 ((k m + 1)(L m+1 L m 4y ) + L m + L m 4y+1 ) = 1 ((k m)(l m+1 L m 4y ) + L m+ + L m 4y 1 ). We now use the fact that L m 4y 1 < L m 1 (which follows immediately from Lemma 31) as well as the identity 1 (L u+1 + L u 1 ) = F u to see that V (k, m 1, y) < 1 ((k m)(l m+1 L m 4y ) + L m+ + L m 1 ) = 1 (k m)(l m+1 L m 4y ) + 1 (L m+3 + L m+1 ) = 1 (k m)(l m+1 L m 4y ) + F m+. Because we simply need to show that J k,m (n) = F m 1(k m) h m+1 h m (n G k (m)) + H k (m) = F m 1(k m) h m+1 h m ( n k h m ) + Fm (k m) + F m+, 1 (k m)(l m+1 L m 4y ) F m 1(k m) ( ) n k h m + Fm (k m) h m+1 h m in order to otain (17). After dividing each side y k m and using the identity 1 (L m+1 L m 4y ) = F m+1 1 (L m + L m 4y ), we find that this last inequality ecomes F m+1 1 ( (L m + L m 4y ) F m 1 1 h ) m+1 (n k ) h m+1 h m + F m, the electronic journal of cominatorics 3(4) (016), #P4.8 3

36 which we may rewrite as h m+1 (n k ) h m+1 h m L m + L m 4y F m 1 (18) after sutracting F m+1 from each side and then rearranging terms. We will prove (18) in each of the following two sucases. Sucase 1: In this sucase, assume = and y = 1. Recall that we mentioned at the eginning of the proof that we may assume n k + G m 1 (m 3) when =. Furthermore, we may assume n k + G m 1 (m 3) + 1 if m {, 7}. If m =, then If m = 7, then h m+1 (n k ) h m+1 h m h m+1 (G m 1 (m 3) + 1) h m+1 h m = h 6 (G 4 () + 1) h 6 h = ( ) ( ) ( ) ( ) = 3 10 < 1 3 = L 3 + L 1 F 4 = L m + L m 4y F m 1. h m+1 (n k ) h m+1 h m h m+1 (G m 1 (m 3) + 1) h m+1 h m = h 8 (G 6 (4) + 1) h 8 h 7 = ( ) ( ) ( ) ( ) = 14 < 3 8 = L + L 3 F 6 = L m + L m 4y F m 1. This proves (18) if m {, 7}, so we will assume m {, 7}. By Definition 3, h m+1 = 1 + = 1 + m 1 + m 3 + m m 6 m 1 i = 1 + m 1 + m 3 + m m 1 i m i = m 3 + m 1 + h m 3 = m 3 + G m 1 (m 3). Therefore, using Lemma 4 to write h m+1 h m = m + ( 1) m, we have 3 h m+1 (n k ) h m+1 h m h m+1 G m 1 (m 3) h m+1 h m = If m is even, then m 3 i= m 3 = 3 h m+1 h m m + ( 1) m. If m is odd, then m 9, so h m+1 (n k ) h m+1 h m 3 m 3 m + < 3m 3 m = 3 8 < 8 1. h m+1 (n k ) h m+1 h m 3 m 3 m = = 4 m < 8 1. the electronic journal of cominatorics 3(4) (016), #P4.8 36

37 Either way, h m+1 (n k ) < 8 h m+1 h m 1. Lemma 3 tells us that 8 1 L m + L m 4, so we F m 1 otain (18). Sucase : In this sucase, suppose either y > 1 or. It follows from our choice of y y 1 that n > k + m 1 i. Using Definition 3 and Lemma 4, we see that h m+1 (n k ) = 1 + m m 1 i (n k ) < 1 + m 1 i (n k ) and Therefore, < 1 + y 1 m 1 i m 1 i = 1 + i=y h m+1 h m m + 1. m 1 i = 1 + m+1 y 1 h m+1 (n k ) < 1 + m+1 y /( 1) h m+1 h m ( m )/( + 1) = + m+1 y 1 ( m )( 1). (19) If we treat m as a continuous real variale, then ( ) + m+1 y 1 = 1 ( ) + m+1 y 1 m ( m )( 1) 1 m m = 1 ( m ) m+1 y log ( + m+1 y 1) m log 1 ( m ) = This shows that + m+1 y 1 ( m )( 1) log ( 1)( m ) ( m+ y m+ + m ) < 0. is decreasing in m. Suppose y. This forces m 6 ecause y m decreasing in m, we see from (19) that. Because + m+1 y 1 ( m )( 1) is h m+1 (n k ) h m+1 h m < + m+1 y 1 ( m )( 1) + 7 y 1 ( 6 )( 1) ( 6 )( 1). If =, then h m+1 (n k ) h m+1 h m < ( 6 )( 1) = 11 6 <. the electronic journal of cominatorics 3(4) (016), #P4.8 37

38 If 3, then h m+1 (n k ) < h m+1 h m ( 6 )( 1) < ( 6 )( 1) = ( 1)( 1) No matter the value of, we find from Lemma 3 that = 1 ( 1) 1 3 (3 1) = 1 36 <. h m+1 (n k ) h m+1 h m < < L m + L m 4y F m 1, which is (18). We have proven (18) when y, so assume y = 1. We assumed either y > 1 or, so we must have 3. Since + m+1 y 1 is decreasing in m and m 4, it follows ( m )( 1) from (19) that If = 3, then h m+1 (n k ) h m+1 h m < + m+1 y 1 ( m )( 1) + y 1 ( 4 )( 1) = ( 4 )( 1). h m+1 (n k ) h m+1 h m < ( 4 )( 1) = 3 16 <, which proves (18) with the help of Lemma 3. If 4, then h m+1 (n k ) h m+1 h m < ( 4 )( 1) < ( 4 )( 1) = ( 3 1)( 1) = 1 ( 1) 1 (4 1) = 1 9 <, which proves (18) once again. Case : Here, assume µ(k, m 1, y) V (k, m 1, y). Referring to Theorem 3, we see that ν(k, m 1, y) = γ(k, m 1, x)+1 and µ(k, m 1, y) = V (k, m 1, x) for some x < y. x 1 One finds from Definition 19 that γ(k, m 1, x) + 1 > k + m 1 i. By Lemma 1, γ(k, m 1, x) + 1 I(k, m 1, x). In other words, if we let y e the smallest element of the set { 0, 1,..., y } m such that γ(k, m 1, x) + 1 k + m 1 i, then y = x. By definition, µ(k, m 1, x) is the maximum value of s (j) as j ranges over all elements of I(k, m 1, x). Because γ(k, m 1, x) + 1 I(k, m 1, x), this means that µ(k, m 1, x) s (γ(k, m 1, x) + 1) = s (ν(k, m 1, y)) = µ(k, m 1, y), the electronic journal of cominatorics 3(4) (016), #P4.8 38

39 where we have used Definition 8 to deduce the last equality. Similarly, µ(k, m 1, y) is the maximum value of s (j) as j ranges over all elements of I(k, m 1, y). Since x < y, I(k, m 1, x) I(k, m 1, y). Therefore, µ(k, m 1, x) µ(k, m 1, y). This shows that µ(k, m 1, x) = µ(k, m 1, y) = V (k, m 1, x). ( ] x 1 x Now, choose some integer n k + m 1 i, k + m 1 i (note that γ(k, m 1, x) + 1 is in this interval). { 0, 1,..., m } satisfying n k + Because x is the smallest element of the set x m 1 i and µ(k, m 1, x) = V (k, m 1, x), it follows from Case 1 that s (n ) < J k,m (n ). We may set n = γ(k, m 1, x) + 1, so s (γ(k, m 1, x) + 1) < J k,m (γ(k, m 1, x) + 1). Using the fact that γ(k, m 1, x) + 1 y 1 x k + m 1 i k + m 1 i < n, we have µ(k, m 1, y) = s (γ(k, m 1, x) + 1) < J k,m (γ(k, m 1, x) + 1) < J k,m (n). This proves (17) and completes the proof of the theorem. Corollary 3. We have Proof. Corollary 7 states that lim sup n lim sup n s (n) n log φ = ( 1) log φ. s (n) n log φ ( 1) log φ, so we will now prove the reverse inequality. For each integer k 3, let u k = G k (k) = k + h k. Let θ(x) = log (x) 1 for each x > 0. Recall that we showed in the proof of Corollary 7 that We will show that s (u k ) lim k u log φ k It will then follow from (0) and (1) that lim sup n s (n) lim sup n log φ n = ( 1) log φ. (0) s (n) n s (u θ(n) ) for all n >. (1) log φ (u θ(n) ) log φ s (u θ(n) ) (u θ(n) ) log φ = lim sup k s (u k ) u log φ k = ( 1) log φ, which will complete the proof. In order to derive (1), let us choose some integer n >. Let θ = θ(n), and note that θ < n θ+1. It follows from Proposition 6 that s (n) F θ+ = s ( θ + h θ ) = s (u θ ) the electronic journal of cominatorics 3(4) (016), #P4.8 39

40 (if n = θ+1, then Proposition 6 does not apply, ut the inequality still holds ecause s (n) = 1 < F θ+ ). If n u θ, then this shows that s (n) n s (u θ ) log φ n s (u θ ) log φ u log φ θ which is the inequality we seek to prove. Therefore, we will assume n < u θ. For any t {, 3,..., θ 1}, we may use Lemma 33 to see that Θ θ,t (G θ (t)) = Θ θ,t ( θ + h t ) Θ θ,t ( θ + h t+1 ) = Θ θ,t (G θ (t + 1)), where we have preserved the notation from that lemma. Furthermore, with the help of Definition 6, we find that Θ θ,t (G θ (t+1)) = J θ,t(g θ (t + 1)) G θ (t + 1) = H θ(t + 1) log φ G θ (t + 1) = J θ,t+1(g θ (t + 1)) log φ G θ (t + 1) log φ for all t {, 3,..., θ 1}. Therefore, for all t {, 3,..., θ 1}. That is, Θ θ,t (G θ (t)) Θ θ,t+1 (G θ (t + 1)), = Θ θ,t+1 (G θ (t+1)) () Θ θ, (G θ ()) Θ θ,3 (G θ (3)) Θ θ,θ (G θ (θ)). (3) Because θ < n < u θ = G θ (θ), there exists some m {, 3,..., θ 1} such that G θ (m) n < G θ (m + 1). In other words, we may write n = θ + x for some x {h m, h m + 1,..., h m+1 1}. By Lemma 33 and (), Θ θ,m (n) = Θ θ,m ( θ + x) Θ θ,m ( θ + h m+1 ) = Θ θ,m (G θ (m + 1)) = Θ θ,m+1 (G θ (m + 1)). It follows from (3) that Θ θ,m+1 (G θ (m + 1)) Θ θ,θ (G θ (θ)), so Θ θ,m (n) Θ θ,θ (G θ (θ)). (4) If n = G θ (m), then it follows easily from Theorem and Definition 6, that s (n) = s (G θ (m)) = H θ (m) = J θ,t (G θ (m)) = J θ,m (n). If n G θ (m), then G θ (m) < n < G θ (m + 1), so Theorem 34 shows that s (n) < J θ,m (n). Either way, s (n) J θ,m (n). Hence, using (4), we otain s (n) n J θ,m(n) log φ n log φ = Θ θ,m(n) Θ θ,θ (G θ (θ)) = J θ,θ(g θ (θ)) G θ (θ) log φ = s (G θ (θ)) G θ (θ) = s (u θ ), log φ u log φ θ which proves (1). the electronic journal of cominatorics 3(4) (016), #P4.8 40

41 Theorem 36. For any real x 0 and any integer k 3, let β(x) = log (( 1)x + 1) and f k (x) = 1 [ (k β(x))(φ β(x) + φ β(x) ) + φ β(x)+ + φ β(x) ]. If k and n are positive integers such that k 3 and k < n k + h k, then s (n) < f k (n k ). Proof. Let k 3 e an integer. Let α = k + 1, and note that h k α. We 1 will show that f k is increasing and concave down on the open interval (1, α). The desired inequality will then follow quite easily. Oserve that we may write f k (x) = g k (β(x)), where g k (x) = 1 [ (k x)(φ x + φ x ) + φ x+ + φ x ]. If x log ( k + ), then ( k x k log ( k + ) = log ) k 1 1 k 1 log 1 log 1 4 log, where we have used the inequality log (1 + u) u that holds for all u > 1 (as well log as the inequalities k 3 and ). Therefore, if x log ( k + ), then where and g k(x) = 1 [ (k x)(φ x φ x ) log φ (φ x + φ x ) + (φ x+ φ x ) log φ ] 1 [ 1 ] 4 log (φx φ x ) log φ (φ x + φ x ) + (φ x+ φ x ) log φ ) ( )] + φ x = 1 [ φ x ( φ log φ log φ 4 log 1 = 1 [ C1 φ x + C φ x], C 1 = φ log φ log φ 4 log φ log φ + log φ 4 log 1 C = φ log φ + log φ 4 log If x, then C 1 φ x C 1 φ and C φ x C φ. This means that g k(x) 1 [ C1 φ x + C φ x] C 1 φ + C φ > 0 the electronic journal of cominatorics 3(4) (016), #P4.8 41

42 whenever x log ( k + ). Now, if 1 x α, then β(x) log ( k + ) and β (x) > 0. Consequently, f k(x) = d dx g k(β(x)) = g k(β(x))β (x) > 0 for all x [1, α]. This shows that f k is increasing on the open interval (1, α). We now wish to show that f k is concave down on the interval (1, α). We first calculate and to otain β (x) = 1 1 log ( 1)x + 1 β (x) = 1 ( ) 1 = (β (x)) log log ( 1)x + 1 f k (x) = d d x g k(β(x)) = d dx (g k(β(x))β (x)) = g k(β(x))β (x) + (β (x)) g k(β(x)) = (β (x)) (g k(β(x)) g k(β(x)) log ). Therefore, to show that f k is concave down on the interval (1, α), we just need to show that g k (β(x)) log > g k (β(x)) for all x (1, α). To do so, it suffices to show that log log φ g k(x) > log φ g k(x) () for all x (, log ( k + )). Suppose x (, log ( k + )). We have so (φ x + φ x ) + (φ x+ φ x ) log φ = φ x (φ log φ 1) φ x (φ log φ + 1) φ (φ log φ 1) φ x (φ log φ + 1) φ (φ log φ 1) φ (φ log φ + 1) > 0, g k(x) = 1 [ (k x)(φ x φ x ) log φ (φ x + φ x ) + (φ x+ φ x ) log φ ] In other words, Using the inequality log (1 + u) > 1 [ (k x)(φ x φ x ) log φ ]. log log φ g k(x) > (k x)(φ x φ x ) log. (6) u, which holds for all u > 1, we find that log ) k x > k log ( k + ) = log ( k 1 1 k 1 log, the electronic journal of cominatorics 3(4) (016), #P4.8 4

43 so it follows from the assumption that k 3 that In addition, since x >, k x > 1 log. (7) φ x (log log φ) φ x (log + log φ) > φ (log log φ) φ (log + log φ) > 0. (8) Comining (6), (7), and (8) with the fact that log φ g k(x) = log φ [(k x)(φ x + φ x ) φx φ x log φ yields log log φ g k(x) log φ g k(x) > ] + (φ x+ + φ x ) (k x)(φ x φ x ) log (k x)(φ x + φ x ) log φ + (φ x φ x ) (φ x+ + φ x ) log φ = (k x)(φ x (log log φ) φ x (log + log φ)) + φ x ( φ log φ) φ x ( + φ log φ) > 1 log (φx (log log φ) φ x (log + log φ)) + φ x ( φ log φ) φ x ( + φ log φ) = A 1 φ x A φ x, where A 1 = φ log φ 1 (log log φ) log and A = + φ log φ 1 (log + log φ). log Because, it is easy to see that A 1 > φ log φ 1 φ log φ 1 4 > 0.49 and A < + φ log φ <.19. Since x >, log log φ g k(x) log φ g k(x) > A 1 φ x A φ x > 0.49φ x.19φ x > 0.49φ.19φ > 0, which proves (). We have shown that f k is increasing and concave down on the interval (1, α). Now, suppose m is odd and 3 m k. Because m is odd, one may easily show that h m = 1 + m 3 m i = m + 1. Since 1 G k (m) k = h m = m > m 1 1, the electronic journal of cominatorics 3(4) (016), #P4.8 43

44 we may use the fact that f k is increasing on (1, α) to see that ( ) ( ( )) f k (G k (m) k m 1 m 1 ) > f k = g k β = g 1 k (m) 1 = 1 [ (k m)(φ m + φ m ) + φ m+ + φ m ] = (k m) φm ( 1/φ) m + φm+ ( 1/φ) m+ Similarly, if m is even and m k, then f k (G k (m) k ) = f k (h m ) = f k ( m 1 1 = (k m)f m + F m+ = H k (m). ) = g k ( β = 1 [ (k m)(φ m + φ m ) + φ m+ + φ m ] > 1 [ (k m)(φ m φ m ) + φ m+ φ m ] = (k m) φm ( 1/φ) m + φm+ ( 1/φ) m+ ( )) m 1 = g k (m) 1 = (k m)f m + F m+ = H k (m). Hence, f k (G k (m) k ) > H k (m) for all m {, 3,..., k}. We may now prove that s (n) < f k (n k ) for all n { k + 1, k +,..., k + h k }. Choose such an integer n. If n = G k (m) for some m {, 3,..., k}, then it follows from Theorem and the preceding discussion that s (n) = H k (m) < f k (n k ). Therefore, we will assume n G k (m) for all m {, 3,..., k}. Note that there exists some m {, 3,..., k 1} such that G k (m) < n < G k (m + 1). Let C e the curve {(x, f k (x k )): G k (m) x G k (m + 1)}, and let L = {J k,m (x): G k (m) x G k (m + 1)} e the line segment connecting the points (G k (m), H k (m)) and (G k (m+1), H k (m+1)). Because f k (x) is concave down on the interval (1, α), the curve C is concave down. Since f k (G k (m) k ) > H k (m) and f k (G k (m + 1) k ) > H k (m + 1), the curve C must lie aove the line segment L. In particular, J k,m (n) < f k (n k ). By Theorem 34, s (n) < J k,m (n) < f k (n k ), as desired. 4 Concluding Remarks We wish to acknowledge some of the potential uses and extensions of results derived in this paper. First, we note that Theorem 1 allows us to derive explicit formulas for the values of s (n) for integers n whose ordinary ase expansions have certain forms. For example, we were ale to invoke Theorem 1 in the proof of Theorem in order to show the electronic journal of cominatorics 3(4) (016), #P4.8 44

45 Figure : Two plots of s (n) for 1 n 10. The top plot uses the value =, while the ottom uses = 3. For each k {3, 4,..., 9}, the graph of f k (x k ) for k x k + h k is shown in green. the electronic journal of cominatorics 3(4) (016), #P4.8 4

46 Figure 3: A plot of s 3 (n) for 3 9 n h 9 (where h 9 is defined with = 3). The graph of f k (x 3 9 ) is colored green. The upper ound from Theorem 34 is the polygonal path colored purple. The points (G 9 (m), H 9 (m)) for m {, 3,..., 9} are colored lue. that s (G k (m)) = H k (m). As another example, it is possile to use Theorem 1 to show that s (1 + x 1 + x 1+x + x 1+x +x 3 ) = x 1 x x 3 + x 1 x + x 1 x 3 + x x 3 + x 1 (9) for any positive integers x 1, x, x 3. The equation (9) appears with several similar identities (many of which can e deduced from Theorem 1) in [3]. Second, arguments ased on symmetry and periodicity may e used to extend the upper ounds given y Theorems 3, 34, and 36. For example, referring to the top image in Figure, one will see that the plot of s (n) forms several mound shapes. However, only the left sides of the mounds are ounded aove y the green curves. It is known [, page ] that if k N, then s ( k + x) = s ( k+1 x) for all x {1,,..., k }. (30) This allows us to otain upper ounds over the right sides of the mounds for free. More precisely, since we know from Theorem 36 that s ( k +x) < f k (x) for all x {1,,..., h k }, it follows from (30) that s ( k+1 x) < f k (x) for all such x. Using identities similar to (30) for aritrary values of, one may extend our upper ounds for s (n) to a larger range of values of n. the electronic journal of cominatorics 3(4) (016), #P4.8 46