Numeration Systems. S. E. Payne General Numeration Systems 2. 3 Combinatorial Numeration Numeration with forbidden substrings 14

Transcription

1 Numeration Systems S. E. Payne 2010 Contents 1 General Numeration Systems 2 2 Greedy Representations 3 3 Combinatorial Numeration 13 4 Numeration with forbidden substrings 14 5 The dominant root of the recurrence relation 16 6 Initial conditions 17 7 Numeration system corresponding to an anti-lyndon word 19 8 The lazy expression Example Example The Non-anti-Lyndon case Fibonacci Nim Just for fun this time! 26 1

2 1 General Numeration Systems Let b be an integer greater than 1. It is well known that each positive integer N has a unique representation of the form N = b 0 + b 1 b + b 2 b b r b r, where 0 b i b 1, 0 i r. (1) The proof amounts to showing that the b r+1 distinct expressions of the form in Eq. 1 give b r+1 distinct numbers, and these numbers range from 0 up through b r+1 1. A second useful numeration system is the following. Define the Fibonacci numbers in the slightly uncommon way: F 0 = 1, F 1 = 2, and F n = F n 1 + F n 2. (This is uncommon because the usual sequence starts with two terms equal to 1.) If d r, d r 1,..., d 1, d 0 is a binary string of length r + 1, put N(d r d r 1 d 0 ) = d r F r + d r 1 F r d 1 F 1 + d 0 F 0. (2) This version of the Fibonacci sequence {F n } n=0 has the following properties. Theorem 1.1. Fibonacci Numbers (i) For 0 i < j Z, F i < F j, so {F n } n=0 is strictly increasing. (ii) 2 F 0 = F 1, but for n 1 we have 2 F n > F n+1. (iii) F n is the number of binary strings of length n not containing 11 as a substring. Proof. F 0 counts the empty string. F 1 counts the two strings 0, 1. In general for n 2, the number of strings of length n having no substring 11 is the number F n 2 of binary strings x 1 x n 2 01 having no substring 11, plus the number F n 1 of binary strings x 1 x n 1 0 having no substring 11. (iv) F n+1 1 is the largest integer N corresponding to a binary string of length at most n + 1 and not having 11 as a subword. 2

3 Proof. It is fairly easy to see that the largest integer corresponding to a binary string of length at most n + 1 that does not have 11 as a subword must have a binary string of the form It is easy to check that F 2 + F 0 = F 3 1 and F 3 + F 1 = F 4 1. Suppose that F 2(k 1) + F 2(k 2) + + F 0 = F 2k 1 1. Then F 2k + F 2(k 1) + + F 0 = F 2k + F 2k 1 1 = F 2k+1 1. It is similarly easy to show by induction that F 2k+1 + F 2k F 1 = F 2(k+1) 1 for k 1. This shows that F n+1 1 = F n + F n 2 + F n 4 + is the largest N corresponding to a binary string of length at most n+1 and not having 11 as a subword. (v) (Zeckendorf 1939) Each positive integer may be represented in a unique way as the sum of nonconsecutive Fibonacci numbers. Proof. Let N be any one of the F n+1 integers satisfying 0 N F n+1 1. If N = 0, N corresponds to the empty binary word, which does not have 11 as a subword. For N > 0, let m be the largest integer such that F m N, i.e., F m N < F m+1. Since 2F m F m 1 + F m = F m+1 > N, it follows that if kf m N, then k = 0 or k = 1. Then F m 1 > N F m. Now choose the largest m such that F m N F m, i.e., F m N < F m +1. We just showed that m m 2. Continuing in this fashion we see that each positive integer N has at least one representation as a sum of nonconsecutive Fibonacci numbers. On the other hand, there are exactly F n+1 1 positive integers less than F n+1, and exactly F n+1 1 nonzero binary sequences of length n + 1 not having a subword equal to 11. It follows that each integer has a unique representation as a sum of nonconsecutive Fibonacci numbers. 2 Greedy Representations In this section we present two results of Fraenkel [1]. To start, let 1 = u 0 < u 1 < u 2 < be a finite or infinite sequence of integers. Let N be any nonnegative integer, and suppose that u n is the largest number in the sequence not exceeding N (except that we let n = 0 if N = 0). Dividing N 3

4 by u n and iterating gives N = d n u n + r n, r n = d n 1 u n 1 + r n 1, r n 1 = d n 2 u n 2 + r n 2,. r i+1 = d i u i + r i,. 0 r n < u n 0 r n 1 < u n 1 0 r n 2 < u n 2. 0 r i < u i. Collecting terms we get r 2 = d 1 u 1 + r 1, 0 r 1 < u 1 r 1 = d 0 u 0. N = d n u n + d n 1 u n d 0 u 0 (d i 0, i 0). This is the greedy representation of N in the numeration system S = {u 0, u 1, u 2,...}. The process above shows that every nonnegative integer has a (unique greedy) representation in the system S. Note that r i+1 = d i u i + d i 1 u i d 0 u 0 < u i+1 (i 0). Conversely, if N = n i=0 d iu i, where d i u i + d i 1 u i d 0 u 0 < u i+1 (i 0), (3) then n i=0 d iu i is the unique greedy representation of N by S, as shown in the following theorem: Theorem 2.1. Let 1 = u 0 < u 1 < u 2 < be any finite or infinite sequence of integers. Each nonnegative integer N has precisely one representation in the system S = {u 0, u 1,...} of the form N = n i=0 d iu i, where the d i are nonnegative integers satisfying Eq. 3. Proof. It remains only to establish uniqueness. representations: Suppose that N has two N = c n u n + + c 0 u 0 = d n u n + d 0 u 0, 4

5 where the digits c j and d j are nonnegative integers and satisfy Eq. 3. Let i be the largest integer such that c i+1 d i+1, say c i+1 > d i+1 (and c j = d j for 0 j i). Then u i+1 (c i+1 d i+1 )u i+1 = (d i c i )u i + + (d 0 c 0 )u 0 d i u i + + d 0 u 0 < u i+1, an impossibility. The argument given above shows that each nonnegative integer N has at least one representation satisfying Eq. 3, hence the proof is complete. Note that Eq. 3 implies 0 d i < u i+1 u i (i 0). (4) Sometimes Eq. 4 implies Eq. 3, and sometimes it does not. When the u i are defined recursively, the situation depends on the length of the recurrence relation. If the recurrence relation contains only one term (viz., u n = b (n) u n 1 ), then Eq. 4 does imply Eq. 3. We see this as follows. Assume that Eq. 4 does hold. Then d 0 u 0 < b (1) u 0 = u 1, so Eq. 3 holds for i = 0. As an induction hypothesis assume that Eq. 4 holds for i = n 1. Then d n u n + d n 1 u n d 0 u 0 (b (n+1) 1)u n + d n 1 u n d 0 u 0 < (b (n+1) 1)u n + u n = b (n+1) u n = u n+1. Hence for many of the standard numeration systems (e.g., b-ary representations for b > 1) the condition in Eq. 4 is a necessary and sufficient condition for unique representation. But for systems in which the recurrence relation contains more than one term, Eq. 4 is only a necessary condition, not a sufficient one. This will become clear from the deeper theorem below. For m 1 let b 1 = b (n) 1 be a function of n, and b 2,..., b m be constant where b 1, b 2,..., b m are integers satisfying 1 b m b 2 b (n) 1. Let u m+1, u m+2,..., u 1 be fixed nonnegative integers (in case m > 1), and let u 0 = 1, u n = b (n) 1 u n 1 + b 2 u n b m u n m (n 1), (5) 5

6 be an increasing sequence of positive integers. Suppose that N = n j=0 d ju j is the unique (greedy) representation satisfying Eq. 3 of an integer N in the system S = {u 0, u 1,...}, where the u i are defined by Eq. 5. Recall that this implies that {u i } n=0 must also satisfy Eq. 4. If m = 1, then Eq. 4 implies d i < u i+1 /u i = b (i+1) 1 (i 0). If m > 1, then d i < (b (i+1) 1 u i + b 2 u i b m u i m+1 )/u i b i+1) 1 + (b (i) 1 u i 1 + b 2 u i b m u i m /u i = b (i+1) (i 1). = d i b (i+1) 1 (if i 1 and m > 1). By Eq. 4, d 0 < u 1, so that by putting n = 1 in Eq. 5 we have d 0 < b (1) 1 + m j=2 b ju 1 j when m > 1. We record this as follows: 0 d i < b (i+1) 1 ( if i 0 and m = 1); 0 d i b (i+1) 1 ( if i 1 and m > 1); 0 d 0 < b (1) 1 + m j=2 b ju 1 j (if m > 1). Note: The condition of Eq. 6 is necessary for uniqueness of the representation N in the system S, but it is not necessarily sufficient. Fraenkel[1] gives the following full explanation of the situation. Theorem 2.2. Let S = {u i } be a sequence of the form in Eq. 5. Each nonnegative integer N has precisely one representation in S of the form N = n i=0 d iu i where the digits d i are nonnegative integers satisfying the following (two-fold) condition: (i) Let k m 1. For any j satisfying 0 j m 2, if (6) (d k, d k 1,..., d k j+1 ) = (b (k+1) 1, b 2,..., b j ), (7) then d k j b j+1 ; and if Eq. 7 holds with j = m 1, then d k m+1 < b m. In case m = 1 and j = 0, this means 0 d k < b (k+1) 1 for k 0. (ii) Let 0 k < m 1. If Eq. 7 holds for some j satisfying 0 j k 1, then d k j b j+1 ; and if Eq. 7 holds with j = k, then d 0 < m i=k+1 b iu k+1 i. 6

7 Before starting into the proof of Theorem 2.2 we look at an example and then consider a little more carefully just what the theorem is saying. Example Consider the numeration system defined by m = 3, u 2 = 0, u 1 = u 0 = 1, u n = 3u n 1 + 2u n 2 + u n 3 (n 1). Thus b 1 = 3, b 2 = 2, b 3 = 1, and S = {1, 5, 18, 65, 236,...}. By Eq. 6 the digit bounds are 0 d i 3 for i 1, and 0 d 0 4. First consider subcondition (i). Let k m 1 = 2 and 0 j m 1 = 2. In fact j = 0 gives a vacuous condition. If j = 1 and k 2, then d k = b 1 = 3 implies d k 1 b 2 = 2. If j = m 1 = 2, then (d k, d k 1 ) = (b 1, b 2 ) = (3, 2) implies that d k 2 < b 3 = 1, so d k 2 = 0. Second, consider subcondition (ii). Suppose 0 k < m 1 = 2. If for some j with 0 j k 1 we have (d k, d k 1,..., d k j+1 ) = (b 1, b 2,..., b j ), then d k j b j+1. If (for j = k) we have (d k,..., d 1 ) = (b 1,..., b k ), then we have d 0 < m i=k+1 b iu k+1 i. The following cases turn up. (k, j) = (0, 0) : = d 0 < b 1 u 0 + b 2 u 1 + b 3 u 2 = 5, so d 0 4, nothing new. (k, j) = (1, 0). This is a vacuous case. (k, j) = (1, 1). Here (d 1 = b 1 = 3 implies d 0 < 3 i=2 b iu 2 i = b 2 u 0 + b 3 u 1 = 3, so d 0 2. So putting this together we have the following for this specific example: d k = 3 fork 1 = d k 1 2; (d k, d k 1 ) = (3, 2) for k 2 = d k 2 = 0. We practice writing some integers in the system S using representations satisfying the conditions of Theorem 2.2. Case 1. N = 18. Clearly we could write 18 = , i.e., = 33 S, but in this case d 1 = 3 would force d 0 < 3. Hence the correct representation is = 100 S. Case 2. N = 36. Clearly 200 S = 36 10, and this representation satisfies the conditions of Theorem 2.2. It is easy to see that we must have d 2 1 (since d 0 4 and d i 3 for i 1). Then d 2 = 1 quickly forces d 1 = d 0 = 3, but d 1 = 3 forces d 0 2. So 200 S is the only acceptable representation. Case 3. N = 65. If we try to get by using only u 0, u 1 and u 2, we are quickly lead to 65 = 321 S. But (d 2, d 1 ) = (3, 2) forces d 0 = 0. so the only acceptable representation is 1000 S. Case 4. N = 236. Clearly = S satisfies the conditions of Theorem 2.2. If we try to get by using only u 0, u 1, u 2 and u 3, using d 0 4 and d i 3 for i 1, we are quickly led to = 3210 S, but (d 3, d 2 ) = (3, 2) should force d 1 = 0. Hence S is the only acceptable representation. 7

8 Case 5. N = 235. Just using the conditions d 0 4 and d i 3 for i 1 we quickly see that d 3 = 3. So d 2 2. Clearly d 2 = 1 is too small for u 1 and u 0 to carry the rest of the weight. So d 2 = 2. Then (d 3, d 2 ) = (3, 2) implies d 1 = 0, leaving d 0 = 4. So the unique representation satisfying the condition of Theorem 2.2 is 235 = 3204 S. Note that the conditions (i) and (ii) of Theorem 2.2 are both concerned with blocks of consecutive digits. They differ only in the location of these blocks: in (i) the right-hand digit d k m+1 of a block of maximal size m is in some position k m Such a block may appear anywhere within the digit sequence. If it is shifted right until its rightmost digit goes beyond position 0, we get a truncated block: the right-hand digit of the truncated block (of smaller maximal size k + 1 < m) occupies position 0. This is the situation considered in (ii). Next we want to see that the condition of Eq. 6 is implied by the condition of the Theorem. For m = 1, the first part of (i) is empty and (ii) is empty. The second part of (i) must be interpreted to say that 0 d k < b (k+1) 1 for k 0, so that for m = 1 the condition of Theorem 2.2 is equivalent to Eq. 6. Now suppose that m > 1 and put j = 0 in the first part of (i). Eq. 7 holds vacuously and the conclusion must be interpreted to say d k b (k+1) 1 for k 1, i.e., the first inequality in eq. 6 holds. Put j = k = 0 in the second part of (ii). Then d 0 < m i=1 b iu 1 i. To motivate further the somewhat curious condition of Theorem 2.2, consider the violation of (i) in the special case when Eq. 7 holds for j = m 1 but d k m+1 b m. Then the integer M = k i=k m+1 d iu i has two representations: and M = b (k+1) 1 u k + k 1 i=k m+2 b k i+1 u i + d k m+1 u k m+1, M = u k+1 + (d k m+1 b m )u k m+1. The latter, which violates Eq. 3, follows from the former representation using the recurrence Eq. 5. (After Lemma 2.3 below, it may be seen that also the other violations of the condition of Theorem 2.2 boil down to the case of a sequence of digits sufficiently large, so that a number has an additional representation, violating Eq. 3, obtained by using the recurrence Eq. 5. Proof. (of Theorem 2.2) We start by showing that a representation satisfying the condition of the Theorem always exists. Let N be any nonnegative 8

9 integer and let N = n i=0 d iu i be its (greedy) representation (so Eqs. 3 and 4 hold). Suppose that (i) does not hold. First suppose that there is some j satisfying 1 j m 2 for which Eq. 7 holds (i.e., (d k, d k 1,..., d k j+1 ) = (b (k+1) 1, b 2,..., b j )) but d k j > b j+1 (so d k j b j = d k j u k j (b j+1 + 1)u k j ). Then k d i u i i=0 j+1 k d i u i b (k+1) 1 u k + b i u k+1 i + u k j i=k j i=2 i=2 j+1 = b (k+1) 1 u k + b i u k+1 i + b (k j) 1 u k j 1 + i=2 m b i u k j i j+1 m j 1 b (k+1) 1 u k + b i u k+1 i + b (k j) 1 u k j 1 + b i u k j i j+1 b (k+1) 1 u k + = b (k+1) 1 u k + i=2 b i u k+1 i + b j+2 u k j 1 + m b i u k+1 i = u k+1, i=2 i=2 i=2 m j 1 i=2 b j+1+i u k j i violating Eq. 3. Secondly, suppose that Eq. 7 holds with j = m 1 but d k m+1 b m. This means (d k, d k 1,..., d k (m 2) ) = (b (k+1) 1, b 2,..., b m 1 ) and d k m+1 b m. Hence we have k d i u i = i=0 m 2 j=0 d k j u k j + k j=m 1 d k j u k j m 2 = b (k+1) 1 u k + b j+1 u k j + d k m+1 u k m+1 + j=1 m 2 b (k+1) 1 u k + b j+1 u k j + b m u k m+1 = b (k+1) 1 u k + j=1 m b j u k+1 j = u k+1. j=2 9 k d k j u k j j=m

10 Then k i=0 d iu i u k+1 contradicts Eq. 3, which must hold for the greedy representation. Note that with m = 1 and j = 0 this violation of (i) means d k u k b (k+1) 1 u k. But k 0 means u k > 0, so this means d k b (k+1) 1, contradicting the top row of Eq. 6. Now suppose that (ii) does not hold. First suppose that 0 k < m 1. Suppose there is a j satisfying 0 j k 1 for which d k j > b j+1, so d k j b j+1 + 1, and (d k, d k 1,..., d k j+1 ) = (b (k+1) 1, b 2,..., b j ). (Note that in this case 0 j k 1 m 3.) u k+1 > d k u k + + d k (j 1) u k (j 1) + d k j u k j + k j 1 i=0 d i u i b (k+1) 1 u k + + b j u k+1 j + b j+1 u k j + u k j = b (k+1) 1 u k + + b j+1 u k j + b (k j) 1 u k j b m (k j) u 2k 2j m + + b m u k j m b (k+1) 1 u k + + b j+1 u k j + b j+2 u k j 1 + b m u k+1 m = u k+1. Here we used b r b r+j+1. Finally, suppose that (ii) fails in the following way: Eq. 7 holds with j = k and d 0 m i=k+1 b iu k+1 i. So (d k, d k 1,..., d 1 ) = (b (k+1) 1, b 2,..., b k ). (Recall that u 0 = 1.)Then u k+1 > d k u k + + d 0 u 0 = b (k+1) 1 u k + b 2 u k b k u 1 + d 0 u 0 b (k+1) 1 u k + b k u 1 + b k+1 u b m u k+1 m = u k+1, an impossibility. At this point we have shown that the greedy algorithm always provides a representation of the desired kind. To show that there is only one such representation we first give a lemma which is interesting in its own right. Lemma 2.3. To start with let t be any nonnegative integer and write t+1 = qm + r with 0 r < m where m is the m of Theorem 2.2. (i) Consider the following representation of a nonnegative integer f(t) in 10

11 the system S of Eq. 5: f(t) = b (t+1) 1 u t + b 2 u t b m 1 u t+2 m + (b m 1)u t+1 m + b (t+1 m) 1 u t m + b 2 u t 1 m + b m 1 u t+2 2m + (b m 1)u t+1 2m + + b (t+1 qm) 1 u t (q 1)m + b 2 u t 1 (q 1)m + + b m 1 u t+2 qm + (b m 1)u t+1 qm + b (t+1 qm) 1 u t qm + b 2 u t 1 qm + + b r 1 u t+2 r qm + d 0 u 0, d 0 = b m 1 (if r = 0); d 0 = m r j=0 b r+ju j 1 (if r > 0). Then f(t) = u t+1 1. (ii) Let N 1 = n i=0 d iu i and N 2 = l i=0 c iu i + n i=l+1 d iu i, where c l < d l, for l any integer in [0, n], and the c i, d i are arbitrary digits satisfying the condition of Theorem 2.2. Then N 2 < N 1. Part (i) of the lemma states that f(t) is the analog of (t s) in the decimal system. Part (ii) states that if two representations differ in the lth position, being identical to the left of it, then the representation with the larger digit in position l corresponds to the larger number. This is, of course, trivial for the decimal system. In the present more general case, however, this result is not a priori clear. Suppose, for example, that (for m > 3) three consecutive digits of some representation of some number satisfy (d k, d k 1, d k 2 ) = (b (k+1) 1, b 2, b 3 ). By the condition of Theorem 2.2 these digits are maximal, in the sense that none of them can be replaced by a larger one without violating the condition. However, replacing, say, d k 1 = b 2 by b 2 1 enables replacing b k 2 = b 3 by b (k 1) 1, which may be considerably larger than b 3. Part (ii) of the Lemma states that such a replacement yields a smaller, not a larger number as might be expected. Proof. It is straightforward to verify the following two facts: (a) The representation of f(t) as given in (i) satisfies the condition of Theorem 2.2. (b) Increasing any digit of this representation while leaving the digits to its left unchanged results in a representation violating the condition of Theorem

12 For the proof of (i) note that Eq. 5 implies directly that f(t) = (u t+1 u t+1 m ) + (u t+1 m u t+1 2m ) + + (u t+1 (q 1)m u t+1 qm ) + (u t+1 qm 1) = u t+1 1. Next, we prove (ii) by induction on the digital position l. The result is clear for l = 0. Suppose that the result holds for any two representations satisfying the condition of theorem 2.2 whose leftmost differing digits are in some position l. We have to show that, given any representation N 1 = n i=0 d iu i, we have N 3 < N 1, where N 3 = n i=l+2 l+1 d i u i + g i u i, g l+1 < d i+1, i=0 and the representations of N 1 and N 3 satisfy the condition of Theorem 2.2. Subtracting, we get N 1 N 3 = (d l+1 g l+1 )u l+1 + l (d i g i )u i u i+1 i=0 l g i u i. i=0 Let g(l) = l i=0 g iu i. If the representatons of f(l) = l i=0 f iu i as given in (i) of the lemma (with t replaced by l) and g(l) disagree, then the leftmost disagreement occurs in position k for some k [0, l], that is, f k g k, but f i = g i for k < i l. Since the representation g(l) = l i=0 g iu i satisfies the condition of Theorem 2.2, fact (b) implies g k < f k. by our induction hypothesis we thus have g(l) f(l). Since by (i) f(l) = u l+1 1, we must have N 1 N 3 1; that is N 3 < N 1. We now resume the proof of Theorem 2.2. Let N = n k i u i be a representation satisfying the condition of Theorem 2.2. If the representations of f(n) and N are different, then fact (b) implies d k < f k for some k [0, n]. By (ii) and (i) of Lemma 2.3, we thus have N f(n) < u n+1, so inequality Eq. 3 is satisfied. Hence the representation n i=0 d iu i of N is unique by Theorem

13 3 Combinatorial Numeration Theorem 3.1. Fix a positive integer k. Then each nonnegative integer n has a unique representation of k terms of the form ( ) ( ) ( ) b1 b2 bk n = + + +, 1 2 k in which the combinatorial digits b i are subject to 0 b 1 < b 2 < b 3 < < b k. The combinatorial digits b i of n are found recursively as follows: b k is the greatest integer such that ( b k k ) n, b k 1 is the greatest integer such that ( b k 1) n ( bk k ),. b r is the greatest integer such that ( b r r ) n k j=r+1 ( bj j ),. b 1 is the greatest integer such that b 1 n k ( bj ) j=2 j. Proof. It should be clear that the greedy algorithm given above to define the b j does give n = ( b 1 1 ) + ( b2 2 ) + + ( bk k ), for some nonnegative values of the b j. It will take some work to show that 0 b 1 < b 2 < b 3 < < b k and that such a representation is unique. Recall the well-known identity n ( ) ( ) r + j r + n + 1 =. (8) j n j=0 Suppose for now that this has been shown and let ( ) ( ) ( ) b1 b2 bk n = k be the representation obtained by the greedy algorithm, and let ( ) ( ) ( ) a1 a2 ak n = k 13

14 be any other representation satisfying 0 a 1 < a 2 < < a k < b k. Clearly b k has the largest possible value in any such representation of n. Then ( ) a1 + 1 ( ) a ( ) ak k k ( ) ( ) ak (k j) ak + 1 = 1 < j k j=1 ( ) bk. k Here we used the equality in Eq. 8. This shows that if there is a representation satisfying the conditions of the theorem, it must be unique. We now just have to show that the greedy algorithm gives such a representation. So suppose that the greedy algorithm gives n = ( ) b1 1 + ( ) b ( ) bk. k This means that b k is the largest integer such that ( b k ) k n. Suppose that b k 1 b k. Then ( ) ( ) ( ) ( ) ( ) bk 1 bk bk bk bk + 1 n + + =, k 1 k k 1 k k which is a contradiction. The same argument works at each stage for b k 2 down to b 1, so that indeed 0 b 1 < b 2 < < b k as desired. The article [1] goes on to study several types of numeration systems and to hint at uses of them for solving a variety of problems. However, we now turn to a study of the results in [3]. 4 Numeration with forbidden substrings Let U = {u n } n 0 be an increasing sequence of natural numbers starting with u 0 = 1. Also, from now on we assume that U is defined by a linear recurrence of the form u n = m a l u n l + u n m, 0 a i Z, 1 i m. (9) l=1 So all the information about a sequence U is encoded in the word a = (a 1, a 2,..., a m ) = a 1 a 2... a m : the coefficients of the order m recurrence are 14

15 a 1, a 2,..., a m 1, a m + 1. Later we will give the appropriate initial terms for counting words with specific forbidden substrings. Without loss of generality we ignore the case in which a is periodic, i.e., there exists a non-empty word b and an exponent s > 1 for which a = b s. If a = b s, the recurrence defined by a is equivalent to that defined by b. Given a sequence as above, we may associate a natural number with each finite word in the alphabet N {0} as follows. Denote the word by w = (d k, d k 1,..., d 0 ) = d k d k 1... d 0. The number k i=0 d iu i corresponding, in this base, to the word w will be denoted by N(w). If d k 0, the word w is a called a U-representation, or simply a representation, for the number N = N(w). As a consequence, the number 0 is represented by the empty word ε. We shall use the words (sub)word and (sub)string interchangeably, and all our subwords will be contiguous, i.e., u is a subword of w only if there are possibly empty words a and b such that w = aub. We saw in Section 2 that the greedy algorithm gives a uniquely defined representation of N in the system U characterized by N = n d i u i with i=0 i d j u j < u i+1 (i 0). The greedy U-representation of N will be denoted by N U. j=0 A word a = (a 1, a 2,..., a m ) = a 1 a 2 a m is said to be anti-lyndon provided it is greater than each of its other cyclic shifts. So 312 and 3231 are anti-lyndon, but 2131 and 3131 are not. In addition to the usual lexicographic order on words (denoted by < ) we will use an order called LenLex denoted. Specifically, let v, w be words in an alphabet over which a total order < is defined which induces a lexicographic order on the words. Then w v if l(w) > l(v) or if l(w) = l(v) and w > v in the lexicographic order. Given a sequence U, we are interested in the set of (words that are) greedy U-representations for the natural numbers. This set is denoted L(U). Let a = a 1 a 2 a m be any given anti-lyndon word. Let F a denote the set of words of length at most m that are lexicographically greater than a. A word is called F a -free provided it contains no substrings belonging to F a, i.e., provided each of its substrings is lexicographically less than or equal to a. In practice we use as our alphabet the set {0, 1,..., a 1 }. For example, if the alphabet is A = {0, 1, 2, 3}, if a = 312, the F a -free words are the words not containing as substrings the strings 32, 33 and

16 The word w = a k a 1... a r (where a k is the word consisting of a = a 1... a m repeated k times) is the lexicographically largest F a -free word of its length. Indeed, w is F a -free and any lexicographically larger word of the same length has a prefix a h a 1... a s in common with w, followed by a digit b greater than a s+1 : so a 1... a s b is in F a. Recall that the greedy expressions of natural numbers in Fibonacci base are exactly the binary words beginning with 1 and with no 11 substring, that is, the F a -free words for a = 10 (and beginning with 1 ). Generalizing the fact that, conversely, the n-th Fibonacci number enumerates the binary words beginning with 1 without 11 of length up to n, or equivalently, the n-th Fibonacci number enumerates all the binary words without 11, regardless of their leftmost digit, of length exactly n, we shall see that for an important class of recurrences, the n-th term u n counts the F a -free words of length n. 5 The dominant root of the recurrence relation The sequences U = {u n } with which we are dealing arise from recurrences of the form u n = b 1 u n b m 1 u n m+1 + b m u n m with nonnegative coefficients. The characteristic polynomials of these recurrences are of the form p(x) = x m b 1 x m 1 b m 1 x b m with b i 0 and b m 0. Such polynomials are called nonnegative. (See Chapter 5 of [2] for these ideas and results.) A dominant root is a root of the characteristic polynomial having largest modulus. A useful result is given by the following. Theorem 5.1. Let u n = b 1 u n 1 + +b m 1 u n m+1 +b m u n m be a recurrence such that b 1 b i 0 for 2 i m, and b m 0. Let p(x) = x m b 1 x m 1 b m 1 x b m be its characteristic polynomial. Then p has a unique dominant root and this root lies in (b 1, b 1 + 1). Proof. We start by showing that p(b 1 ) < 0. p(b 1 ) = b m 1 b 1 b m 1 1 b m 1 b 1 b m, 16

17 where the first wo terms cancel out and the other ones are negative or zero. Next we show p(b This proof proceeds by induction on m. If m = 1, the p(x) = x b 1, so p(b 1 + 1) = (b 1 + 1) b 1 1. Suppose m > 1 the result is true for m 1. Keep in mind that b 1 b i for i > 1 both for the polynomial of degree m and the polynomial of degree m 1 in the inductive step. Then we may write p(b 1 + 1) = (b 1 + 1) m = (b 1 + 1) [ m 1 i=1 b i (b 1 + 1) m i b m (b 1 + 1) m 1 (b 1 + 1) b m 1. m 1 i=1 b i (b 1 + 1) m i 1 ] b m By the induction hypothesis we know that the expression in large brackets is greater than or equal to 1, so we have proved that for all p we have p(b 1 + 1) 1. Descartes rule of signs states in this case that p has at most one positive root, so p has a unique positive root β. In polynomials with coefficient signs as given here the unique positive root is dominant (see [2]), so β is a dominant root. (See the preprint [3] for further comments about the dominant root.) Call a polynomial primitive if the greatest common divisor of the indices of its non-zero coefficients is 1. In particular, this happens if b 1 > 0, so our p is primitive. If a nonnegative polynomial is primitive, then it has a unique root of largest modulus ([2], page 106), and it is real. This concludes the proof. 6 Initial conditions Let U be as described above and put L(U) equal to the set of greedy expressions N U. We have seen that L(U) is the set of words such that each suffix of length n is less than or equal to (in the lexicographic order) u n 1 U. For instance, for the Fibonacci sequence, the numbers in the form u n 1 (excluding 1-1) are 1, 2, 4, 7, 12, 20,... which have greedy representations 1, 10, 101, 1010, 10101, , respectively. Note that these words are of the form a k where a = 10, possibly truncated. So in this case, as noted earlier, we have the fact that when writing numbers in the Fibonacci base greedily, all binary words with no substring 11 appear. 17

18 If the recurrence is defined by the word a = a 1... a m, we choose initial conditions for the sequence U so that the word a k a 1... a r is a U-expression for u s 1, where s = km + r with 0 r < m, and r = 0 means that the word is just a k. It turns out that these are also the correct initial conditions for enumerative purposes. Lemma 6.1. Consider a sequence U = {u n } n 0 defined by a recurrence of degree m and used as a base for numeration. Take as initial conditions for the numbers u 0,..., u m 1 the following: u 0 = 1; u 1 = a 1 +1 = N(a 1 )+1; u 2 = N(a 1 a 2 )+1;... u m 1 = N(a 1 a 2... a m 1 )+1. Then for s 0, if s = km + r with 0 r < m, with the convention that when r = 0, the word a 1... a r denotes the empty word, we have N(a k a 1... a r ) = u s 1. (10) Proof. The proof is by induction on k. If k = 0, the result is given by the hypothesis. For k > 0, just note that m u s 1 = a i u s i + u s m 1, i=1 by the recurrence defining the sequence, and the induction hypothesis says So N(a k 1 a 1... a r ) = u s m 1. N(a k a 1... a r ) = N(a ) + N(a k 1 a 1... a r ) m = a i u s i + u s m 1 i=1 = u s 1, where in N(a00 0) exactly s m zeros appear. Hence we now have a U-representation for u s 1, namely a k a 1... a r, since N(a k a 1... a r ) = u s 1. But we need the greedy representation for u s 1. It turns out that the expression a k a 1... a r is indeed the greedy representation if (and only if) the word a is anti-lyndon. This choice of initial conditions is, in the present context, the only right one. When a is anti-lyndon, since u n will count the number of F a -free words, the sequence starts off as given in the following lemma. 18

19 Lemma 6.2. If a is anti-lyndon, then for 0 i m 1, the number of F a -free words of length i is N(a 1... a i ) + 1. Proof. The claim of the lemma is trivially true for i = 0. (Note that regardless of which word a is, the empty word ε is F a -free. Suppose we are constructing all F a -free words of a fixed length i > 0, going from the left to the right. If we choose as the leftmost letter any of the a 1 digits 0, 1,..., a 1 1, we may complete the word with any of the u i 1 F a -free words of length i 1 without the risk of obtaining an occurrence of an F a -subword. So for the moment we have a 1 u i 1 words. We have still to count those words beginning with a 1. If the second letter is any of the a 2 digits less than a 2, the word can be completed with any one of the u i 2 F a -free words of length of length i 2. Now we have a 2 u i 2 more words, getting a 1 u i 1 + a 2 u i 2 words so far. We go on in the same way, but successively fixing the first k digits of the words to be constructed in common with the first k digits of a. So we construct all the words of length i, obtaining a 1 u i 1 + a 2 u i a i 1 u 1 + (a i + 1)u 0 of them (since the rightmost digit of any digit from 0 to a i can appear). 7 Numeration system corresponding to an anti- Lyndon word Let a = a 1 a 2 a m be an anti-lyndon word. The number system U determined by a is described as follows. At this point we merely state what the initial conditions are, but the next couple results will justify this choice. u 0 = 1 u 1 = N(a 1 ) + 1 = a u 2 = N(a 1 a 2 ) + 1 = a 1 u 1 + a 2 u u m 2 = N(a 1 a 2... a m 2 ) + 1 = a 1 u m 3 + a 2 u m a m 2 u u m 1 = N(a 1... a m 1 ) + 1 = a 1 u m 2 + a 2 u m a m 1 u u n = m i=1 a iu n i + u n m for n m. Consider again the anti-lyndon word a = 312. The associated number system U has u 0 = 1; u 1 = a = 4; u 2 = N(a 1 a 2 ) + 1 = 14; u 3 = 19

20 N(a 1 a 2 a 3 ) + 1 = 49. Then u n = 3u n 1 + u n 2 + 2u n 3 + u n 3. In particular, u 4 = 173. Consider the map N that assigns to a string w the integer it corresponds to (i.e., represents) in a particular numeration system. We first note that it does not preserve order. For example, N(1100) = 63 and N(1033) = 64, but The first representation is greedy and F a -free, but the second is neither. (The greedy and F a -free representation of 64 is 1101.) On the other hand, the following lemmas shows that for F a -free words the order is preserved. Lemma 7.1. Let w and v be distinct F a -free words not beginning with zero. If w v, then N(w) > N(v). Proof. Let u, v begiven as described in the lemma. The proof is by induction on l = l(w). If l = 1, the result is obvious. So suppose that l > 1. First suppose that l = l(w) is greater than l(v). Consider the following chain of identities: N(w) N( ) (l 1 zeros) > N((a 1 a 2... a m ) k a 1... a r ) N(v), where l 1 = km + r and 0 r < m. The first inequality holds because we are actually removing some (positive) summands, while the second one follows from Lemma 6.1. The last inequality holds by induction because the words (1 1 a 2... a m ) k a 1... a l 1 mk and v have length less than l, and the hypotheses of the lemma are satisfied: both words are F a -free, and the fact that v is F a -free implies that v is smaller than or equal to (in the LenLex order) the word (a 1... a m ) k a 1... a l 1 mk ) (indeed either the former is shorter than the latter or is lexicographically smaller than or equal to it by the penultimate paragraph of Section 4). This concludes the proof in case l(w) > l(v). Suppose that both words have length l, say w = w l 1... w 0 and v = v l 1... v 0 with w l 1 v l1. Consider then the word w obtained by substituting w l 1 v l 1 for w l 1 in w and then by stripping it of the leading zeros. Define analogously v starting from v. Now N(w ) = N(w) v l 1 u l 1 and N(v ) = N(v) v l 1 u l 1, the new words are still F a -free and w v. If w l 1 > v l 1, the new words have different lengths, and we may apply the argument given for different-length words. If w l 1 = v l 1, then these new words both have length less than l, and we may apply induction. 20

21 This lemma implies that there is at most one F a -free representation of a given natural number, since different F a -free words represent different numbers. We can also use the last lemma to prove that the greedy expression of a number n is F a -free: the next lemma shows that an expression containing a subword in F a does not satisfy the greediness condition in Eq. 3. Lemma 7.2. Let N = k l=0 d lu l, and assume that in this expression for N, a word in F a appears starting in the s-th position. That is, for some index s k and for some number j 1 of digits we have a i = d s+1 i for i = 1, 2,..., j 1, while a j < d s j+1. Then s l=0 d lu l u s+1, so that the given expression for N is not greedy. Proof. The proof is by induction on s. The basis of the induction is given by the cases s = 0, 1,..., m 1, that is, when the subword in F a starts in one of the m rightmost digits. The base cases will be covered later. For the inductive step we shall consider the subword starting with d s and show that it represents a number greater than or equal to u s+1. Now, d s u s + d s 1 u s d 0 u 0 d s u s + d s 1 u s d s j+1 u s j+1 a 1 u s + a 2 u s a j u s j+1 + u s j+1 = a 1 u s + a 2 u s a j u s j+1 + a 1 u s j + a 2 u s j+1 + a m u s j a m u s j (m 1) + u s j (m 1), (11) where in the last step we have applied the recurrence to u s j+1. Consider now the word w = a 1 a 2... a m with s j (m 1) zeros after a m. This word corresponds to the terms a 1 u s j + + a m u s j (m 1). The word w is greater lexicographically than a j+1 a j+2... a m a 1... a j (same number of zeros as above; the sums in the indices are taken modulo m) by the anti-lyndon condition, so we may apply Lemma 7.1. Hence the last sum in Eq. 11 is greater than or equal to a 1 u s + a 2 u s a j u s j+1 + a j+1 u s j + + a m u s m+1 +a 1 u s m + + a j u s j (m 1) + u s j (m 1). We now carry out the actual inductive step, by applying the result with s m in place of s and taking, as new d i s : d s m = a 1, s s m 1 = a 2, 21

22 ..., d s m (j 2) = a j 1, d s j (m 1) = a j + 1, d s j m = 0,..., d 0 = 0. In this way we obtain the inequality a 1 u s m + + a j u s j (m 1) + u s j (m 1) u s m+1. Thus the last previously displayed sum is greater than or equal to a 1 u s + a 2 u s a j u s j+1 + a j+1 u s j + + a m u s m+1 + u s m+1, which is equal to u s+1, as claimed, completing the general inductive step. The basis of induction works similarly, but when we rewrite the first few terms of the sequence or compare numbers represented using them we must take into account the fact that the recurrence has not yet kicked in. More precisely, as u s j+1 is one of the first m terms, it is given by the definition of the starting terms rather than by the recurrence: a 1 u s j + a 2 u s j a s j+1 u And when we use the inequality derived by the anti-lyndon condition, it involves now a word shorter than the full a. Howevr, the anti- Lyndon condition automatically implies that a prefex of a is lexicographically greater than or equal to the same length prefix of each cyclic shift. We have now proved that each natural integer has an F a -free expression, the greedy one. The uniqueness was proved in Lemma 7.1. This completes a proof of the following theorem. Theorem 7.3. Let U be a sequence defined by the recurrence m u n = a l u n l + u n m l=1 with a = a 1... a m an anti-lyndon word and with the initial conditions u 0 = 1, u 1 = a 1 u = N(a 1 ) + 1, u 2 = a 1 u 1 + a 2 u = N(a 1 a 2 ) + 1,..., u m 1 = a 1 u m 2 + a 2 u m 3 + a m 1 u = N(a 1 a 2... a m 1 ) + 1. Then each positive integer n has a unique F a -free representation, which turns out to be the greedy one. Moreover, considering the set of F a -free strings not beginning by zero with the LenLex order, the nth string represents the number n. This theorem has an enumerative counterpart. Corollary 7.4. If a = a 1... a m is an anti-lyndon word and U = {u n } n 0 is the sequence as defined in Theorem 7.3, then the number of F a -free words of length k is u k. 22

23 Proof. The proof is straightforward from the theorem, keeping in mind that we are now counting all words (including those beginning by zero). Remark 7.5. Note that, while in the context of numeration systems we are generally only interested in words without leading zeros (which in this context would be non-significant digits), when we enumerate words with prescribed conditions, those are just ordinary words. But this is not a problem, since counting all F a -free words of a certain length k is equivalent to counting the F a words not beginning with zero of length at most k (just pad the latter ones with zeros). 8 The lazy expression When representing natural numbers in terms of a given sequence we may act lazily rather than greedily, taking the smallest expression for a given integer, and if the sequence is given by an anti-lyndon word, then it is easy to obtain a characterization for the strings appearing as lazy expressions for the natural numbers, much in the same way as shown above for greedy expressions. 8.1 Example The lazy representation of an integer using Fibonacci numbers turns out to be its (unique) expression as a sum of distinct Fibonacci numbers without ever skipping more than one in a row. In terms of binary strings, this corresponds to strings with no consecutive zeros. For instance, the lazy expression of 22 is while the greedy one is (still another expression is ). It is easily seen that there is a simple procedure to get the lazy expression of N, which basically dualizes the greedy one. We describe the procedure for the case studied earlier, i.e., for a recurrent sequence arising from an anti-lyndon word a = a 1... a m, thus using the alphabet {0,..., a 1 }. Perform the following steps to obtain the lazy expression for a natural number N: (i) Determine the smallest k such that the number N(a 1 a 1... a 1 ) (k digits) is greater than or equal to N; 23

24 (ii) Find the greedy expression b k 1... b 0 of the number N(a 1 a 1... a 1 ) N (padded on the left with zeros if its length is less than k); (iii) Subtract termwise the string b k 1... b 0 from a 1... a 1. The result is the lazy expression for N. 8.2 Example Consider once more the sequence arising from the word a = a 1 a 2 a 3 = 312. So u 0 = 1, u 1 = 4, u 2 = 14, u 3 = 49, u 4 = 173. We want to apply the above procedure to N = 100. Then N(3) = 3, N(33) = 15, N(333) = 57, N(3333) = 204. So the k we need is k = 4. The greedy expression of = 104 is Hence the lazy expression for 100 is = Check: 100 = When the word is anti-lyndon, we have seen that the strings obtained with the greedy procedure have a simple characterization in terms of forbidden substrings. From the procedure described above, it follows easily that the same is true in the lazy case. Here the forbidden strings are the complements to a 1 a 1... a 1. For instance, in the example just given, the forbidden subwords in the greedy case are 32, 33, 313. Therefore in the lazy case the strings are characterized by forbidding 01, 00, 020. Recalling that, for an anti-lyndon word a = a 1... a m, the set F a was defined to be the set of words of length at most m lexicographically greater than a, we define F a to be the set of words of length at most m which are lexicographically less than a = (a 1 a 2 )(a 1 a 2 )... (a 1 a m ). With this notation, the above can be summarized in the following theorem. Theorem 8.3. If a = a 1... a m is an anti-lyndon word and {u n } is the sequence determined by the word a, the set of words that are lazy expressions of natural numbers is the set of F a -free words in the alphabet {0, 1,..., a 1 }. Note that, while this theorem mirrors clearly the situation for greedy strings, there are other slightly less obvious parallels between the greedy and the lazy cases. When we write numbers greedily, it is immediate to find the least natural number expressed with a given number k of digits: it is simply u k, the k-th term of the sequence, whose greedy representation is given by 1 followed by k zeros. The largest k-digit expression, on the other hand, is less immediate, and is described in Lemma

25 For lazy expressions, on the other hand, it is immediate to write the lexicographically largest one with k digits, which is simply a 1 a 1... a 1, while the smallest one can be once more described in terms of repetitions of the word a. For instance, as the threes complement of 312 is 021, the least number with a given number of digits in the base defined by this word has the form The Non-anti-Lyndon case The gist of the previous results culminating with Theorem 7.3 is that, when the recurrence is defined by an anti-lyndon word a, the structure of the set L(U) of greedy expressions of natural numbers is encoded in the greedy expression a k a 1... a r of the terms of the form u n 1, where n = mk + r. Indeed, (recall the first paragraph of Section 6), the greedy expression for u n 1 characterizes completely L(U): if a is anti-lyndon, the greedy expression has this neat periodic form. This is no longer true when a is not anti-lyndon. If the word a defining a recurrence is not anti-lyndon, then a k a 1... a r is by Lemma 6.1 a representation of the number u n 1 but is not its greedy expression. For instance, if a = 3132, then the first terms of its sequence are 1, 4, 14, 50, 179,.... The word 3132 represents u 4 1 = 178, but the greedy representation of 178 is Lemma 9.1. Let a = a 1... a m be a non-anti-lyndon (non-periodic) word. Let i + 1 (with i + 1 > 1) be the starting index of the anti-lyndon shift of a, so a i+1 a i+2... a m a 1... a i is an anti-lyndon word). Let a i+1... a i+k be the prefix of the anti-lyndon shift such that a i+1 = a 1, a i+2 = a 2,..., a i+k 1 = a k 1, a i+k > a k. Then the word a 1... a i a i+1... a i+k number N(a 1... a i a i+1... a i+k ). is not the greedy representation of the Proof. It suffices to show that a i+1... a i+k is not greedy. We have a i+k a k + 1. Then N(a i+1... a i+k ) N(a 1... a k ) + 1, which is denoted greedily by (k zeros). So we obtain the following characterization. 25

26 Theorem 9.2. Let U be the sequence obtained with the recurrence induced by the word a and the usual initial conditions. Then one has u n 1 U = a k a 1... a r if and only if a is an anti-lyndon word. 10 Fibonacci Nim Just for fun this time! RULES FOR FIBONACCI NIM 1. Play starts with one pile of counters having at least 2 counters. The first player may remove any positive number of counters, but not the whole pile. Thereafter, each player must remove at least one counter and may remove at most twice the number of counters his opponent took on the previous move. 2. The player who removes the last counter wins. NOTE: If both players play perfectly, the first player can win if the number of counters in the initial pile is not a Fibonacci number. If it is a Fibonacci number, the second player can win. STRATEGY Let s say you are playing first and your opponent is named Nim. Here is how to always win using Fibonacci numbers: On the first move, leave Nim a Fibonacci number of coins, or Nim will leave you with a Fibonacci number of coins and force you to lose. If possible without losing, always leave Nim a Fibonacci number of coins. But don t take too many, or Nim will take the rest. Otherwise, count the number of coins remaining. (If the count is a Fibonacci number, Nim will win no matter what you do.) Write down the number of coins on the board as the SUM of nonconsecutive Fibonacci numbers. Find the smallest Fibonacci number in the sum. Remove that number of coins from the board. Repeat this process each time you play. For example, if it is your turn and 15 coins remain on the board, write 15 as the sum of Then remove 2 coins leaving 13. Nim can then only remove 1 to twice as many coins as you removed: that would be 1 to 4 coins. Suppose Nim removes 2 coins leaving 11. Write 11 as 8+3. Then remove 3 coins leaving 8. Nim can now remove 1 to 6 coins (2x3=6). Suppose Nim 26

27 removes 2 coins leaving 6. Write 6 as 5+1. Then remove 1 coin leaving 5. Nim can only remove 1 or 2 coins now. Suppose Nim removes 1 coin leaving 4. Write 4 as 3+1. Then take 1 coin leaving 3. If Nim takes 1 coin, you will take 2 coins and win! If Nim takes 2 coins, you will take the last coin and win! Using this strategy, you will always win if you leave Nim a number of coins equal to the sum of Fibonacci numbers. Remember also that every Fibonacci number is the sum of Fibonacci numbers. So, if possible leave Nim a Fibonacci number of coins. Just make sure Nim cannot take all the remaining coins. For a proof that the strategy given above really works, see the article by Holshouser, Reiter and Rudzinski, [4]. References [1] Aviezri S. Fraenkel, Systems of numeration, Amer. Math. Monthly 92 (1985), [2] P. Cull, M. Flahive, R. Robson, Difference Equations From Rabbits to Chaos, Springer, [3] Daniele A. Gewurz and Francesca Merola, Numeration and enumeration, Proc. 4th Pythagorean [4] Arthur Holshouser, Harold Reiter and James Rudzinski, Dynamic one-pile Nim, Fibonacci Quarterly, 41 #3 (June-July) (2003), [5] D. H. Lehmer, The machine tools of combinatorics, in Applied Combinatorial Mathematics (E. F. Beckenbach, ed), Wiley, 1964, [6] E. Zeckendorf, Représentation des nombres naturels par une somme de nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sc i. Liège, 41 (1972),