Unknown Journal Series Volume 000, Number 0, 0000 Arrays, Numeration Systems and Games Aviezri S. Fraenkel Department of Applied Mathematics and Compu

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1 Unknown Journal Series Volume, Number, Arrays, Numeration Systems and Games Aviezri S. Fraenkel Department of Applied Mathematics and Computer Science Weizmann Institute of Science Rehovot 761, Israel Abstract. We dene an innite array A of nonnegative integers based on a linear recurrence, which produces basis elements of an exotic ternary numeration system. Using the numeration system we explore many properties of A. Further, we propose and analyze a family of 2-player games on a variable number of heaps of tokens, and present a winning strategy based on certain subarrays of A. Though the strategy looks easy, it is actually computationally hard. The numeration system is then used to decide whether the family has an ecient strategy or not. 1. Introduction Consider a doubly innite array (matrix) A = fa n j : j; n 1g of nonnegative integers whose rst few entries are displayed in Table 1. To dene its formation rule, we introduce a little notation. Denote by Z and Z + the set of nonnegative integers and positive integers respectively. If S is any nite subset of Z, denote by mex S the least nonnegative integer in the complement of S with respect to Z, i.e., the least nonnegative integer not occurring in S. Note that the mex of the empty set is. The term mex, used in [BCG1982], stands for Minimum Ecluded value. For n 2 Z, the entries of the array are dened as follows. (1) A n = mexfa i j : i < n; j g; (2) A n 1 = 2A n + n (n ); A n j = 3A n j?1? A n j?2 (j 2; n ): c American Mathematical Society???-????/ $1. + $.25 per page 1

2 2 Table 1: A doubly infinite array of nonnegative integers. n A n A n 1 A n 2 A n 3 A n 4 A n 5 A n : : : We further introduce a special ternary numeration system U. Its basis elements are dened by u = 1, u 1 = 3, u i = 3u i?1? u i?2 (i 2). Theorem I. Every positive integer n has a unique representation over U, in the form n = P i d iu i, where the digits d i assume values in f; 1; 2g, subject to the following special condition: if for some j < l; d j = d l = 2, then there exists k satisfying j < k < l (so actually l? j 2), such that d k =. Theorem I is a special case of Theorem 4, x4, stated and proved in [Fr1985]. The representation of the rst few positive integers over U is given in Table P 2. We write m the representation of n both in terms of its basis elements, n = d i= iu i, and in its \ternary" form n = d m : : : d, the same as is customary for more conventional numeration systems, such as decimal or binary (528 = ). Table 2 shows, for example, that 41 = 1211; and 42 = 2 rather than 1212, because of the the special condition. Similarly, 55 = 1, not Some of my best friends are nonsemitic, among them referees and readers of my articles. A number of them have commented to me that in a table such as Table 2, the basis elements 1; 3; 8; 21; 55 should be written from left to right rather than from right to left. I disagree. The \ternary" number n = dm : : : d, now easily readable from the table, would be reversed! There is a discrepancy in nonsemitic languages, often ignored, between text, including mathematical formulas, and \digital" numbers. Though all of these are both written and read from left to right, the basis elements of the latter, which are usually implicit but here explicit, nevertheless increase from right to left. (There is an even greater discrepancy when embedding formulas and digital numbers in semitic language texts, but it is well-known and acknowledged. Moreover, word processors have long since learned to overcome it; human beings still have diculties with it.)

3 3 Table 2: A special ternary representation of integers n n n

4 4 Lastly, we dene the following two-person game G, played on k heaps of tokens of sizes x ; : : : ; x k?1. The positions of G have the following form: (3) = (x ; : : : ; x k?1); < x x k?1; in addition to (; : : : ; ). There are two types of moves. I Remove a positive number of tokens from k? 1 heaps, at least one of them not an entire heap, such that each reduced heap is of size 6= x i (i ). One of the reduced heaps may be split into a number of new heaps, each of size 6= x i (i ). II Remove a positive number of tokens from every heap, say < n n k?1. Then n k?1 should not be too large; namely, (4) n k?1 2n k?2 + n k?3 + + n : The player rst unable to move loses, and the opponent wins. Notice that if k = 1, there is no move of type I. There is also no move of type II, since there are no n k?2; n k?; : : : for which (4) could hold. So single heaps, and of course (; : : : ; ), are end positions. The number of heaps can decrease or increase during play; but the total number of tokens is decreased at each move. Therefore play ends, and no position is repeated. In a move of type II, no heap-splitting is permitted. Example. Let = (1; 3). By (4), a move to (; ) is not possible. Also a move to 1 or 3 is not permitted. Thus the only possible moves are to (1; 2) or (1; 1). In either case, player II moves to (; ) and wins. We shall show that certain subarrays of the array A are the so-called \losing positions" of our heap game G. For proving this it is helpful to use some of the properties of A. Essentially, A is a splitting of Z +, but to state the result precisely, some further notions need to be introduced. Dene the P operators L (Left shift) and R (Right shift) on representations over m U: if n = d i= iu i for P some n 2 Z +, P P P m then L(n) = d m i= il(u i ) = d i= iu i+1 m (i ); and R(n) = d i=1 ir(u i ) = d i1 iu i?1 is dened if d =. In other words, if n = d m : : : d, then L(n) = d m : : : d, and, if i 1 (i.e., d = ), then R(n) = d m : : : d 1. In particular: L(u i ) = u i+1 (i ); and R(u i ) = u i?1 (i 1). S 1 n=1 An j, j ; and the n-th rown-th row is A n = S 1 The j-th column of A, excluding the in the rst row, is denoted by A j = j= An j, n 1. If n = P d i iu i with d 6=, we say that n is reduced. A reduced number n has no right shift. The golden section is the positive root of the polynomial equation x 2? x? 1 =, so = (1 + p 5)=2 and 2 = + 1. In x2 we prove, Theorem 1. The array A is a splitting of Z + : every positive integer appears precisely once in A. Moreover, for every j, the column A j consists precisely of all positive integers whose representation ends in j s. In particular, A n is reduced for all n 2 Z +. The proof leans heavily on properties of the special ternary numeration system U, which are also explored in x2. The system U is even more useful: the winning

5 5 strategy of G, based on subarrays of A, is inecient (exponential). The system U enables one to decide whether there is or there isn't a dierent, ecient (polynomial) strategy. This is taken up in x4. Some further remarkable properties of A are listed in Theorem 2, also proved in x2. Theorem 2. (i) For j; n 2 Z + ; A n j = ba n j?1 2 c + 1 = L(A n j?1 ). (ii) For all n 1; A n = b(n? 1)c + 1 is reduced. (iii) Let j 1. There are no real numbers ;, such that for all n 1; A n j = bn + c. Properties of A are also presented in Lemmas 1, 2, 3, 5, 6 in x2. The formulation of a winning strategy for G needs a few technical concepts, so is best postponed to x3, where the precise result is stated and proved. We begin with a simple result. 2. Some Properties of the Array Lemma 1. For all j; n 2 Z + ; A n j = 2An j?1 + An j?2 + + An + n. Proof. Induction on j, for arbitrary but xed n. By the rst part of (2), the assertion holds for j = 1. Suppose it holds for some j 1. By the second part of (2), A n j+1 = 2A n j + (A n j? A n j?1) = 2A n j + A n j?1 + + A n + n: The following is the main lemma used for proving both Theorem 1 and Theorem 2. Lemma 2. Let n 1, S n = S m<n S 1 j= Am j. In every row An of A, the element A n is the smallest reduced element not in S n, and A n j+1 is the left shift of An j for all j. Proof. Since u 1 = 2u + 1 and u i = 3u i?1? u i?2 (i 2), the same as the recurrence (2), and A 1 = 1 = u, the row A 1 consists of the basis elements of U, for which the statement clearly holds. Suppose it holds for all m < n (n 2). If A n would not be reduced, then R(A n ) would be a smaller element than A n. Moreover, R(A n ) =2 S n, otherwise also A n = LR(A n ) would be in S n, by the induction hypothesis, contradicting (1). Thus A n is the smallest reduced element not in S n. P Let A n?1 = d i iu i be the P representation of A n?1 over U. By the induction hypothesis, A n?1 1 = L(A n?1 ) = d i iu i+1 and A n?1 is reduced. In particular, d 6=. We consider two cases. (i) There exists P j 1 such that d i = 1 for all i < j, but d j =. Then A n?1 + 1 = d i1 iu i + (d + 1)u is reduced (by Theorem I, with least signicant digit 2), so the rst part of the proof implies A n?1 +1 = A n. Now, A n 1 = 2A n + n = 2(A n?1 + 1) + n = 2A n?1 + (n? 1) + 3 = A n? = d i u i+1 + u 1 = d i u i+1 + (d + 1)u 1 = L(A n ): i i1

6 6 (ii) There exists j such that d i = 1 for all i < j, but d j = 2. By Theorem I, d j+1 1. By Lemma 1 with n = P P 1, A n?1 + 1 = d ij+2 iu i + (d j+1 + 1)u j+1 is not reduced, but A n?1 + 2 = d ij+2 iu i + (d j+1 + 1)u j+1 + u = A n is reduced. Then by Lemma 1 (n = 1), A n 1 = 2A n + n = 2(A n?1 + 2) + n = 2A n?1 + (n? 1) + 5 = A n? = d i u i = i ij+2 = ij+2 = ij+2 j+1 d i u i+1 + i= d i u i d i u i+1 + d j+1 u j+2 + (2u j+1 + u j + + u 1 ) + (u ) d i u i+1 + (d j+1 + 1)u j+2 + u 1 = L(A n ): It remains only to show that A n = j+1 L(An j ) for all j 1. Suppose we already showed this P for all j < m. For m = 1 this was just done. So consider A n m+1. Let A n m?1 = d i iu i be the representation of A n m?1. By the induction hypothesis and (2), A n m+1 = 3A n m? A n m?1 = i d i (3u i+1? u i ) = i d i u i+2 = L(A n m): Proof of Theorem 1. By Lemma 2, A n = j+1 L(An j ). Therefore the representation of A n j+1 has one additional at its tail end than that of A n j. Since the representations of positive integers over U are unique (Theorem I), all entries in A are indeed distinct. Finally, every positive integer appears in A in view of (1). For proving (i) of Theorem 2, we prove, more generally, Lemma 3. Let m 2 Z +, n = bm 2 c + 1. Then n = L(m). Proof. Let m P P r = d i= iu i be the representation of m, for suitable r 2 Z. We have r to show: n = d i= iu i+1. It suces to show that P P < m 2 r + 1 = d i= iu i+1 +, r for some < < 1. So it suces to show that < d i= i(u i+1? u i 2 ) < 1. The characteristic equation of the second recurrence of (2) is x 2? 3x + 1 =, with solutions 2 = (3 + p 5)=2 and conjugate?2 = (3? p 5)=2. From this it follows that for n, (5) u n = 2n+2??(2n+2) p 5 : Then u i+1? u i P 2 =?2i (1??4 )= p 5 >. Note that, due to the special condition r of Theorem I, d i= i(u i+1? u i 2 ) is largest when d = 2 and d i = 1 for all i 1. Thus, < r i= d i (u i+1? u i 2 ) < (1??4 ) p 5 (1 + 1 i=?2i ) = (1??4 ) p 5 2 = 1:

7 7 Proof of Theorem 2. From Lemma 3 we have, in particular, ba n j?1 2 c + 1 = L(A n ). By Lemma 2, this is also the same as j?1 An j for all j 1, proving (i). S Since?1 +?2 = 1, it follows from Theorem II of [Fra1969] that if S = 1 S (bnc + 1), T = 1 n=1 n=1 (bn2 c + 1), then S, T are 2-upper complementary, i.e., S [ T = Z + n f1g and S \ T = ;. By Lemma 3, T contains only non reduced numbers. Hence S contains all the reduced numbers. Replacing n by n? 1, (ii) follows from (1). Let f = 1, f 1 = 2, f n = f n?1 + f n?2 (n 2) be the sequence of Fibonacci numbers. (It is easily seen that the u i dened in x1 are precisely the \even" Fibonacci numbers, i.e., u i = f 2i for all i, but this fact isn't needed here.) For proving (iii), we use an additional property of the array A, enunciated in Lemma 5. Before that a technical result. Lemma 4. Let, be real numbers with irrational and positive. Then there exist m; n 2 Z + such that N m = b(m+1)+c?bm+c = bc; N n = b(n+1)+c?bn+c = bc+1: Proof. For r 2 Z +, we clearly have, bc N r bc + 1. Both extremal values are assumed, since irrational implies that the fractional values (n) are dense in (; 1) (Kronecker's Theorem; see e.g., [HaWr1989] Ch. 23). Lemma 5. There exist innitely many m; n 2 Z such that for all j 1; A m+1 A m j = f 2j, A n+1 j? A n j = f 2j+1. j? Proof. Induction on j. For j =, this follows directly from Lemma 4 and (ii) of Theorem 2 with =, = 1. For j = 1 we have by (2), for r 2 Z, A r+1 1? A r 1 = 2(A r+1? A r ) f2f + 1; 2f 1 + 1g = ff 2 ; f 3 g. Suppose it holds for all i < j (j 2). By (2), A m+1 j? A m j = 3(A m+1 j?1? Am j?1)? (A m+1 j?2? Am j?2) = 3f 2j?2? f 2j?4 = f 2j : The same argument shows that also A n+1 j? A n j = f 2j+1. By Lemma 4, a necessary condition for the existence of real, with positive and irrational such that A j = bn + c, is that A n+1 j? A n j 2 fbc; bc + 1g. In particular, A n+1 j? A n j has to assume two consecutive integer values. But by Lemma 5, the two assumed values are f 2j and f 2j+1 ; which are consecutive if and only if j =. This proves (iii). Remark. Theorem 2 can be used to give an independent proof of Theorem 1, because the former implies, using the uniqueness of representation (Theorem I), that all entries of A are distinct, A j+1 = L(A j ), and also every positive integer is assumed. We conclude this section with another property of A.

8 8 Lemma 6. For n > m we have, (6) A n j+1? A m j+1 = 2(A n j? A m j ) + j?1 i= (A n i? A m i ) + n? m > A n j? A m j + n? m > A n j? A m j > ; for all j. Proof. By (1), A n = mex(s 1 [ S 2 ), where S 1 = fa i j : i < m; j g, S 2 = fa i j : m i < n; j g. Since Am = mex S 1, it follows that the set S 1 [ S 2 contains all nonnegative integers A m. Hence A n > Am. Therefore (2) implies A n 1? A m 1 = 2(A n? A m ) + n? m > A n? A m >. Assume that A n h? Am h > An? h?1 Am + n? m > h?1 An? h?1 Am h?1 > for all h j. Since all summands are positive, Lemma 1 now implies (6). 3. A Winning Strategy for the Game G We denote the -vector by = (; : : : ; ). Its dimension depends on the context, and it seems best to leave it unspecied. Informally, a position u in a game such as G is called a P -position, if the Previous player can win, i.e., the player who moved to u. It is an N-position, if the Next player can win, i.e., the player moving from u. The position or any single-heap position, are P -positions, since player I cannot even make a move, so player II wins by default. By F (u) we denote the set of all immediate followers of u, i.e., the set of all positions reachable from u by a single move. Note that F (u) = ; if u is a leaf, i.e., an end position. Denote by P the set of all P -positions, and by N the set of all N-positions. The informal denition of P - and N-positions implies, (7) u 2 P () F (u) N ; u 2 N () F (u) \ P 6= ;: All of these things can be done formally. See [Fra2] 2. The main result of this section is Theorem 3. The P -positions of the game G are given by P = 1[ n= 1[ k=2 (A n ; : : : ; An k?1) [ Z + : Proof. Let W = S 1 n= S 1 k=2 (An ; : : : ; An k?1 )S Z +. As was pointed out in x1, any position u 2 Z + [ fg is a leaf, i.e., F (u) = ;, and so is a P -position by (7). It turns out that in view of (7), it suces to demonstrate the following two properties for all other positions. (A) Every move from a position in W produces a position not in W. (B) From every position not in W there exists a move producing a position in W. 2 Successfully debugged the Y2K bug.

9 9 (A) Let (A n ; : : : ; A n ) 2 W. For a move of type I, there is a heap k?1 An j which remains xed, and a heap H which is either reduced, or formed by a split. In either case, the resulting position contains A n j and a heap H 6= An i for all i, so it is not in W. Now consider a move of type II. Suppose there is a move (A n ; : : : ; A n )! k?1 (A m ; : : : ; A m j?1) 2 W. Then m < n, j k. Suppose that j < k. If m = (so (A m; : : : ; Am j?1 ) = ), we have, using Lemma 1, k?3 k?3 (8) A n k?1 = 2A n k?2 + A n i + n > 2A n k?2 + A n i ; i= contradicting condition (4). This contradiction holds a fortiori if m >, because then the terms to the right of A n k?1 in (8) are even smaller, but the left side is still A n k?1 if j < k. We conclude that j = k. The presumed move is thus (A n ; : : : ; A n )! k?1 (Am ; : : : ; A m k?1 ). By Lemma 1, i= k?3 A n? k?1 Am k?1 = 2(A n? k?2 Am k?2) + (A n i? Am i ) + n? m i= k?3 > 2(A n? k?2 Am k?2) + (A n i? Am i ): By Lemma 6, this inequality contradicts (4), so this move is not possible. (B) Given a position of the form (3), =2 W. We show that there is a single move to a position in W. By complementarity (Theorem 1), x = A n j?1 for some j; n 2 Z +. We may also assume k > 1, since for k = 1, 2 W. Assume P rst j > 1. Since k > 1, there is x 1 x. P By Lemma 1, x 1 x = 2A n + j?3 j?2 i= An i + n. Thus we can reduce x j?2 1 to size i= An i, and then split it into piles of sizes A n ; : : : ; A n j?2. Any x i with i > 1 is put to. We have made a type I move to (A n; : : : ; An ) 2 W. j?1 We may thus assume x = A n. Then there exists j 2 such that x i = A n i for i < j? 1, but x j?1 6= A n. If x j?1 j?1 > A n, move x j?1 j?1! A n and put x j?1 i! for all i > j? 1. So we may assume x j?1 < A n j?1. We consider the following cases. (i) j = k, so x j?1 = x k?1. We have x k?1 = A n k?1? t for some t 1. We claim that = (A n; : : : ; An ; x k?2 k?1)! (A n?t ; : : : ; A n?t ; k?2 An?t k?1 ) 2 W is a legal move for t < n; and!, for t n. For t < n we have by (6) (with m = n? t and j replaced by j? 2 = k? 2), x k?1? A n?t = k?1 An? k?1 An?t? t > k?1 An? k?2 An?t k?2. Then by Lemma 1, k?3 x k?1? A n?t = k?1 An? k?1 An?t? t = k?1 2(An? k?2 An?t) + k?2 (A n i? An?t i ); i= i=

10 1 which satises (4). If t n, then by Lemma 1, x k?1 A k?1? n = 2A n k?2 + P k?3 i= An i, so! satises (4). (ii) j < k. (Recall that x j?1 < A n j?1.) We rst dispose of two subcases. a. If there is r > j? 1 with x r > A n and x j?1 i 6= A n j?1 for all i, then make the type I move x r! A n j?1 and x i! for all i > j? 1, i 6= r, resulting in (A n; : : : ; An ) 2 W. j?1 b. If x i A n j?1 for all i j? 1, then = (A n ; : : : ; A n ; x j?2 j?1; : : : ; x k?1)! is a legal move. Indeed, x j?1 A n ; and Lemma 1 implies j?2 An j?2 n. Hence, j?3 x k?1 A n j?1 = 2A n j?2 + is a legal type II move by (4). i= So we may assume that has the form j?2 A n i + n 2x j?1 + i= k?3 A n i 2x k?2 + = (A n ; : : : ; A n j?2; x j?1; : : : ; A n j?1; : : : ; A n j?1; : : : ; A n j+s; : : : ; A n j+s; x t ; : : : ; x k?1); where each A n j+i appears at least once for all i s, s?1, but A n j+s+1 does not appear. Here are the two nal subcases. c. x k?1 > A n j+s+1. Then move, x k?1! A n j+s+1, x t ; : : : ; x k?2!, and multiple copies A n j+i! (type I move), resulting in the position (An; : : : ; An ) 2 W. j+s+1 d. x k?1 < A n. Then! is a legal type II move. Indeed, x j+s+1 j?1 n, so A n j?2 j+s?1 k?3 x k?1 < A n j+s+1 = 2A n j+s + A n i 2x k?2 + x i : We have shown that W = P. i= i= 4. Does the game G have a Polynomial Strategy? The statement of Theorem 3 enables one to decide whether any given position of the form (3) of the game G is a P -position or an N-position, and the proof clearly indicates a winning move from any N-position. These two things together constitute a winning strategy for the game. Given any position of the form (3) of G. To decide whether 2 P or 2 N, we have to compute the entries of A only up to the rst encounter of x. Thus it is readily seen that Theorem 2(ii) implies that A n j has to be computed only for n x (? 1); and (5) implies that j < 1 log 2 ( p 5(x + 1))? 1. So the array has to be computed only up to (x ), which implies a strategy computation linear in x, which looks good. P k?1 However, the input size for G is ( log x i= i). So unless either k or x k?1 are exponentially larger than x, the indicated strategy is actually exponential. But only the construction of the table needs exponential time and, in fact, exponential space. The rest of the algorithm embodied in the proof of Theorem 3 is polynomial. A winning strategy is polynomial only if both of its parts are polynomial. i= x i ;

11 11 It follows from [Fra1985] that the computation of the representation of a positive integer N over the numeration system U can be done by a greedy Euclidean algorithm, namely always dividing the remainder r (initially: r = N), by the largest basis element u n r. This is a polynomial process. In particular, expressing a game position of the form (3) over U can be done in polynomial time. It can then be observed in linear time whether or not x is reduced, and all the other steps of the winning algorithm indicated in the proof of Theorem 3 can also be done in polynomial time. Thus the numeration system U actually enables us to formulate a polynomial strategy for the game G. The game G proposed here belongs to the family of succinct games, i.e., their input size is logarithmic. Normally an extra eort is required for showing that such games have a polynomial strategy. Dierent families of succinct games seem to require dierent methods of strategy computations. For example, in octal games, invented by Guy and Smith [GuSm1956], a linearly ordered string of beads may be split and or reduced according to rules encoded in octal. See also [BCG1982, Ch. 4], [Con1976, Ch. 11]. The standard method for showing that an octal game is polynomial, is to demonstrate that its Sprague- Grundy function (the s of which constitute the set of P -positions) is periodic. Periodicity has been established for a number of octal games. Some of the periods and or preperiods may be very large; see [GaPl1989]. Another way to establish polynomiality is to show that the Sprague-Grundy function values obey some other simple rule, such as forming an arithmetic sequence, as for Nim. For the present class of heap games, polynomiality was established by a nonstandard method. An arithmetic procedure, based on a class of special numeration systems, was the key to polynomiality. In [Fra1998] a game was proposed and analysed, and another numeration system was used there to establish polynomiality. For Wytho's game [Wyt197], [Cox1953], [YaYa1967], the Zeckendorf numeration system [Zec1972] can be used to establish polynomiality. But for Wytho's game, this can be done also using the integer value function. From Theorem 2(iii) it follows that this cannot be done for G. In [Fra1998] it was also proved that the integer value function cannot be used to establish polynomiality for the game dened there. But the question remains whether here or there there is some polynomial algorithm not based on numeration systems. 5. Epilogue We recap the main properties of the array A. (a) A n = mexfa i j : i < n; j g (n ); A n 1 = 2A n + n (n ); A n j = 3A n j?1? An j?2 (j 2; n ): (The denition.) (b) For all j; n 2 Z + ; A n j = 2An + j?1 An + + j?2 An + n. (Lemma 1.) (c) A is a splitting of Z + : every positive integer appears precisely once in A. Moreover, for every j, the column A j consists precisely of all positive integers whose representation ends in j s. In particular, A n is reduced for all n 2 Z +. (Theorem 1.)

12 12 (d) (i) For j; n 2 Z + ; A n j = ba n j?1 2 c + 1 = L(A n j?1). (ii) For all n 1; A n = b(n? 1)c + 1 is reduced. (iii) Let j 1. There are no real numbers ;, such that for all n 1; A n j = bn + c. (Theorem 2.) (e) There exist innitely many m; n 2 Z such that for all j 1; A m+1 j? A m j = f 2j, A n+1 j? A n j = f 2j+1. (Lemma 5.) The numeration system U was used both for proving the most important of these properties, and for deciding the polynomiality question of the strategy of the game G. The reason our title contains the term \arrays", whereas we have presented only a single array, is that we allude to an innite family of arrays, based on some linear recurrence of the form (9) u = 1; u n = b 1 u n?1 + + b m u m ; where the b i are constants, except that b 1 = b 1 (n) may depend on n, with given initial integer values u?m+1; : : : ; u?1. If (1) 1 b m b 1 ; then there is also an associated numeration system [Fra1985]. Replacing in (9) the elements u j by columns A j the recurrence is used to construct A (with possibly a special construction for the rst initial values of j). 3 A rst (to my knowledge) \Fibonacci array" has been dened in [Sto1977]. Other \Stolarsky arrays" were dened in papers such as [Kim1995] and [FrKi1994], and there are innitely many such arrays. But we have not seen any applications of these arrays. Perhaps the present use for a winning strategy to a new class of games is the rst application? Is there a natural innite family of combinatorial games, matching the innite family of arrays? And what's the nature of these arrays and their uses if (1) is violated? It seems that the array dened here was not given before. Its antidiagonal hasn't appeared in [Slo1998] until we sent it in there recently; and its columns A j and its rows A n do not seem to appear in it for j > 1 and n > 3. Several comments can be made about recurrences such as (2). We shall briey relate to two items. I. The second recurrence of (2) can be considered to be the recurrence of the convergents of the quasiregular (or semiregular halbregelmassig) continued fraction? ? 1 3 +? 1... In [Per195, Ch. 5] it is shown that every quasiregular continued fraction converges. In the present case it converges to 2. Many of the above properties of A 3 If the referees object to such futurist allusions, I'll change the title to \An Array, : : : ".

13 13 can be deduced from this observation; also other properties not mentioned above, such as u 2 n? u n?1u n+1 = 1 for all n, and somewhat more complicated identities for elements in the other rows of A. II. In [BBDD1998], the authors quote [BSS1993]: \: : : the recurrence f n+1 = 6f n? f n?1 cries out for a combinatorial interpretation. Finding this interpretation is an open problem." [BBDD1998] gives such an interpretation. We remark that in [Fra1985, x4] a class of regular (simple) continued fractions is dened whose convergents satisfy recurrences including the above. In particular, the numerators of the even-indexed convergents of the simple continued fraction p 1 2 = [1; 2; 2; : : : ] = constitute the sequence 1; 7; 41; 239; : : : with initial values f 1 = 1, f 2 = 7 considered in [BBDD1998]. Needless to say that each such recurrence also denes an exotic numeration system. Perhaps these facts constitute a \combinatorial interpretation". The game of Welter consists of a nite collection of heaps of nitely many tokens of distinct sizes. Two players alternate in selecting a heap and reducing it so as to preserve the property that all heap sizes are distinct. The player rst unable to move loses, and the opponent wins. The winning strategy is intricate. See [Con1976, Ch. 13]. Moreover, it seems very dicult to generalize it. The game proposed here is not a generalization of Welter, but the type I move is reminiscent of several moves of Welter taken simultaneously. References 1. [BBDD1998] E. Barcucci, S. Brunetti, A. Del Lungo and F. Del Ristoro [1998], A combinatorial interpretation of the recurrence f n+1 = 6f n?f n?1, Discrete Math. 19, 235{ [BCG1982] E.R. Berlekamp, J.H. Conway and R.K. Guy [1982], Winning Ways (two volumes), Academic Press, London. 3. [BSS1993] J. Bonin, L. Shapiro and R. Simion [1993], Some q-analogues of the Schroder numbers arising from combinatorial statistics on lattice paths, J. Statist. Plann. Inference 34, 35{ [Con1976] J.H. Conway [1976], On Numbers and Games, Academic Press, London. 5. [Cox1953] H.S.M. Coxeter [1953], The golden section, phyllotaxis and Wytho's game, Scripta Math. 19, 135{ [Fra1969] A.S. Fraenkel [1969], The bracket function and complementary sets of integers, Canadian J. Math. 21, 6{ [Fra1985] A.S. Fraenkel [1985], Systems of numeration, Amer. Math. Monthly 92, 15{ [Fra1998] A.S. Fraenkel [1988], Heap games, numeration systems and sequences, Ann. of Combinatorics 2, 197{21.

14 14 9. [Fra2] A.S. Fraenkel [2], Adventures in Games and Computational Complexity, to appear in Graduate Studies in Mathematics, Amer. Math. Soc., Providence, RI. 1. [FrKi1994] A.S. Fraenkel and C. Kimberling [1994], Generalized Wytho arrays, shues and interspersions, Discrete Mathematics 126, 137{ [GaPl1989] A. Gangolli and T. Plambeck [1989], A note on periodicity in some octal games, Internat. J. Game Theory 18, 311{ [GuSm1956] R.K. Guy and C.A.B. Smith [1956], The G-values of various games, Proc. Camb. Phil. Soc. 52, 514{ [HaWr1989] G.H. Hardy [1989], An Introduction to the Theory of Numbers, 5th ed., Clarendon Press, Oxford. 14. [Kim1995] C. Kimberling [1995], Stolarsky interspersions, Ars Combinatoria 39, 129{ [Per195] O. Perron [195], Die Lehre von den Kettenbruchen, Chelsea, New York. 16. [Slo1998] N.J.A. Sloane [1998], Sloane's On-Line Encyclopedia of Integer Sequences, njas/sequences/. 17. [Sto1977] K.B. Stolarsky [1977], A set of generalized Fibonacci sequences such that each natural number belongs to exactly one, Fibonacci Quart. 15, [Wyt197] W.A. Wytho [197], A modication of the game of Nim, Nieuw Arch. Wisk. 7, 199{ [YaYa1967] A.M. Yaglom and I.M. Yaglom [1967], Challenging Mathematical Problems with Elementary Solutions, translated by J. McCawley, Jr., revised and edited by B. Gordon, Vol. II, Holden-Day, San Francisco. 2. [Zec1972] E. Zeckendorf [1972], Representation des nombres naturels par une somme de nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liege 41, 179{182.