Characterization of the Line Complexity of Cellular Automata Generated by Polynomial Transition Rules. Bertrand Stone

Transcription

1 Characterization of the Line Coplexity of Cellular Autoata Generated by Polynoial Transition Rules Bertrand Stone

2 Abstract Cellular autoata are discrete dynaical systes which consist of changing patterns of sybols on a grid. The changes are specified in such a way that the sybol in a given position is deterined by the sybols surrounding that position in the previous state. Despite the siplicity of their definition, cellular autoata have been applied in the siulation of coplex phenoena as disparate as biological systes and universal coputers. In this paper, we investigate the line coplexity a T (k), or nuber of accessible coefficient blocks of length k, for cellular autoata arising fro a polynoial transition rule T. We first derive recursion forulas for the sequence a T (k) associated to polynoials of the for 1+x+x n where n 3 and the coefficients are taken odulo 2. We then derive functional relations for the generating functions associated to these polynoials. Extending to a ore general setting, we investigate the asyptotics of a T (k) by considering a generating function φ(z) = k=1 α(k)zk which satisfies a certain general functional equation relating φ(z) and φ(z p ) for soe prie p. We show that for positive integral sequences s k which are dependent upon ( a real nuber x [1/p,1] and for which li k logp s k s k ) 1 = log p x, the ratio α(s k )/s 2 k tends to a piecewise quadratic function of x. Suary Cellular autoata are discrete systes which consist of changing patterns of sybols on a grid. Despite the siplicity of their definition, cellular autoata have been eployed in applications as disparate as the siulation of biological systes and the odelling of coputers. The purpose of this project is to investigate the line coplexity, or the nuber of accessible coefficient strings of a given length, for cellular autoata generated by iteratively ultiplying polynoials of a specific class and reducing the coefficients odulo 2. We consider the recursive properties of the line coplexity sequence, and consider the asyptotic properties of a related sequence in a ore general setting.

3 1 Introduction A cellular autoaton is a syste which consists of patterns of sybols on a grid. These patterns change at discrete tie intervals. A given rule specifies the next state in such a anner that each sybol in the next state is deterined by the surrounding sybols in the current state. Von Neuann, who initiated the study of cellular autoata, investigated their connections to the odelling of biological systes [1]. Wolfra observes in [2] that although cellular autoata are often constructed fro identical siple coponents, they can nonetheless exhibit coplex behavior. As Willson notes in [3], cellular autoata can be used to odel chaotic phenoena because their discrete structure facilitates exact coputation. An exaple of a siple cellular autoaton, Pascal s triangle odulo 2, is illustrated in Figure 1. Figure 1: Pascal s Triangle odulo 2 A particular state for a cellular autoaton is called a configuration. A d-diensional cellular autoaton is one whose configurations are graphed on a d-diensional grid. A configuration ω for a one-diensional cellular autoaton can be expressed by its Laurent series a i x i, where the superscripts correspond to the locations of the values a i. For exaple, the expression 1

4 1 + 3x + 2x 3 + x 7 represents the string A configuration is finite if a i is nonzero for only finitely any integers i. Given a configuration ω, the transition rule R for a cellular autoaton deterines a new configuration Rω in such a way that the value at a given index i in Rω is deterined by values near i in ω. An additive transition rule is specified by a Laurent polynoial and it acts upon a configuration by ultiplication, where the coefficients are taken odulo soe prie p. For exaple, (1 + x) 2 = 1 + x 2 (od 2). Applying the transition rule R = 1 + x iteratively to the initial state ω 0 = 1 and taking the coefficients odulo 2 produces Pascal s Triangle (see Figure 2). Figure 2: Revisiting Pascal s Triangle odulo 2 Sequences of length k which appear in soe configuration are called k-accessible blocks. For exaple, the block appears in line 5 of the autoaton shown in Figure 1, and is thus accessible. Denote by a R (k) the nuber of k-accessible blocks for a given transition rule R. This sequence is called the line coplexity of the autoaton. In this paper, we investigate the recursive and asyptotic properties of the sequence a T (k) for polynoials of the for T = 1 + x + x n with n 3 and the coefficients taken odulo 2. Garbe [4] considered the sequence a R (k) for R = 1 + x with the coefficients taken odulo pries p, as well as R = 1 + x + x 2 with the coefficients taken odulo sall pries p, and investigated the asyptotic behavior of a R (k)/k 2. We consider the case p = 2 and explore the ore general polynoials described above. For certain sequences s k (x), where x [1/p,1], we will show that li k a T (s k (x))/s k (x) 2 is a concatenation of quadratic functions of x. 2

5 In Section 2 we derive recursion forulas for a T (k), where T is as above. We consider the even and odd cases separately. In Section 3 we consider the generating functions associated to the sequences a T (k) for even and odd n, and derive functional relations which will be used later. In Section 4 we develop the asyptotic properties of a T (k)/k 2 in significant generality. 2 Recursion Forulas for a T (k) Let a T (k) denote the line coplexity of the cellular autoaton whose initial state is ω 0 = 1 and whose transition rule is given by T = 1+x+x n, where n 3 and the coefficients are taken odulo 2. In this section, we will derive recursion forulas for the sequence a T (k). We will denote by A the set of accessible blocks. We will use exponents to denote repetitions in strings of coefficients; for exaple, = We first consider the case where n 4 is even. We will assue that k n 2 /2 + n. For the purposes of this discussion, we will write a(k) for a T (k). We observe that if line r of the autoaton (i. e., the configuration which results fro applying the transition rule r ties to the initial state) is given by ω r = x 0 x 1 x k, where k = nr, then line r + 1 is given by T ω r = x 0 (x 0 + x 1 )(x 1 + x 2 ) (x n 2 + x n 1 )(x 0 + x n 1 + x n )(x 1 + x n + x n+1 ) (x k n + x k 1 + x k )(x k n+1 + x k )x k+2 n x k+3 n x k 1 x k. In view of the identity T (s 2 ) = T (s) 2 for additive transition rules T odulo 2, line 2r has the for x 0 0x 1 0 0x k. It follows that line 2r + 1 has the for x 0 x 0 x 1 x 1 x n/2 1 x n/2 1 (x 0 + x n/2 )x n/2 (x 1 + x n/2+1 ) (x k n/2 + x k )x k x k n/2+1 0x k n/ x k. Let A ={x 0 0x 1 0 0x k 1 0 x 0 x 1 x k 1 A } B ={0x 0 0x 1 0 0x k 1 x 0 x 1 x k 1 A } C ={(x 0 + x n/2 )x n/2 (x 1 + x n/2+1 ) x n/2+k 1 x 0 x n/2+k 1 A } D ={x n/2 (x 1 + x n/2+1 ) x n/2+k 1 (x k + x n/2+k ) x 1 x n/2+k A }. 3

6 Then the accessible 2k-blocks, since they ust appear either on odd or even rows, belong to the sets A, B, C, or D. In order to deterine the nuber of accessible 2k-blocks, it is necessary to account for the cardinalities of the sets A, B, C, D, and their intersections. We have a(2k) = A B C D = A + B + C + D A B A C A D B C B D C D + A B C + A B D + A C D + B C D A B C D. We observe that, by the definitions given above, A = a(k), B = a(k), C = a(n/2 + k), and D = a(n/2 + k). We observe that all the intersections are nonepty (they contain 0 2k ). For convenience of notation, we define A i = {x 0 x 1 x i 1 } A, and define B i, C i, and D i siilarly. The following leas will be used later. Lea 1. We have A B = A B C = A B D = A B C D = 1. Proof. We observe that A B = {0 2k } = 1. The lea follows. We will now prove two leas which will be used in the proof of Lea 4. Lea 2. The block a = 10 d 10 n is not accessible for even d n/2 2. Proof. Since d is even, a / A n+d+2,b n+d+2. Suppose a C n+d+2 (the case a D n+d+2 is siilar). If w 0 w 1 w n+d/2 A, we have 1 = w 0 + w n/2,..., 1 = w n/2+d/2 and w n/2+d/2+i = 0, w d/2+i + w n/2+d/2+i = 0 for 1 i n/2. Take i = n/2. Then w n+d/2 = 0 and w n/2+d/2 + w n+d/2 = 0, so that w n/2+d/2 = 0. This contradicts the assuption that w n/2+d/2 = 1. Lea 3. Let d, 0 be integers. If b = 10 2d A, then 10 d 10 A. Proof. If b appears on an even row r, then 10 d 10 appears on row r/2. If, on the other hand, b appears on an odd row, we can write 1 = z 0 + z n/2, 0 = z n/2,..., 0 = z d + z n/2+d, 0 = z n/2+d, 1 = z d+1 + z n/2+d+1, 0 = z n/2+d+1,..., 0 = z s++1 + z n/2+s++1, 0 = z n/2+s++1. This iplies that z 0 = z d+1 = 1 and z i = 0 for all other i. Thus, z 0 z 1 z n/2+s++1 = 10 d Since z 0 z 1 z n/2+s++1 A by hypothesis, 10 d 10 A. 4

7 Lea 4. Suppose k n 2 /2 + n. Then A C = n/ Proof. Strings in A C consist of nubers which satisfy the equations x 0 = y 0 + y n/2, 0 = y n/2, x 1 = y 1 + y n/2+1, 0 = y n/2+1,..., x k 1 = y k 1 + y n/2+k 1, 0 = y n/2+k 1. It follows that x i = y i for n/2 i k 1. We will show that it is ipossible to have x i = x j = 1 for 1 i, j n/2 1 and i j. We now clai that the block described above can be expressed in the for c = x 0 x 1 x i 1 x i 0 2p (d+1) 1 x j 0 2p (n+1) 1, where x i and x j are the last two occurrences of the digit 1, p 0 is an integer, and d is even. For this we ust have a sufficient nuber of terinal zeros, and hence k ust be sufficiently large. In particular, we ust have c n/2 + k. We have 2 p 2 p (d + 1) n/2 and we ust have k 2 p (n + 1) 1. Cobining these inequalities gives k n 2 (n + 1) 1. It is true by hypothesis that k n 2 /2 + n, which is sufficient. Given the for of the block c, we now ay iteratively construct the inaccessible block a fro Lea 2, leading to a contradiction in view of Lea 3. Thus, the digit 1 ust appear at ost once in x 0 x 1 x n/2 1. There are n/2 + 1 possible ways this could occur. Thus, A C = n/ The lea follows. Lea 5. Suppose k n 2 /2 + n. Then B D = n/ Proof. We have the equations 0 = y n/2, x 0 = y 1 +y n/2, 0 = y n/2+1, x 1 = y 2 +y n/2+2, 0 = y n/2+2,..., 0 = y n/2+k 1, x k 1 = y k +y n/2+k, which iply that y i = 0 for n/2 i n/2+k 1. As in the case of A C, we know that the nuber 1 can appear at ost once in y 1 y n/2 1. We ust consider the case of y n/2+k. We therefore consider the accessibility of the blocks y 1 y 2 y n/ y n/2+k and x 0 x 1 x n/ x k 1. We note that there are k and k n/2 zeros, respectively. If x k 1 = y n/2+k = 1 and two of the other digits (e.g., x 0 = y 1 ) are also 1, then there are k +L and k +L n/2 zeros, where 0 L n/2 2. If n/2 is odd, then either k + L or k + L n/2 is even, leading to a contradiction as in the case of A C. If n/2 is even, then either k +L and k +L n/2 are both even, leading to a contradiction, or both odd. In this case, we note that these blocks arise fro blocks 5

8 with k+l 1 2 and k+l 1 2 n 4 zeros, respectively. We iterate this process until the nuber of zeros is even. We thus have n/2 1 cases as in the case of A C, as well as the trivial case and the case It follows that B D = n/ Lea 6. Suppose k n 2 /2 + n. Then A D = B C = A C D = B C D = 1. Proof. Consider the set A D. We have x 0 = y n/2, 0 = y 1 + y n/2+1, x 1 = y n/2+1,..., x k 1 = y n/2+k 1, 0 = y k + y n/2+k, so that y i = y n/2+i = x i for 1 i k 1. By the periodicity relations, the eleents of A D arise fro strings of the for (x 0 x n/2 1 ) d for soe d 0. Suppose (x 0 x n/2 1 ) d A, where the lengths of eleents in A are dn/2 (the case of B is siilar). If n/2 is odd, then x 1 = x 3 = = x n/2 2 = x 0 = 0, so that x 0 = x n/2 = 0, x 2 = x n/2+2 = 0, etc. Thus, x i = 0 for all i. If n/2 is even, we apply the sae arguent to the block (x 0 x 2 x n/2 2 ) d. If (x 0 x n/2 1 ) d C (defined siilarly), then we have x 0 = z 0 + z n/2, x 1 = z n/2, x 2 = z 1 + z n/2+1, etc. for soe z 0,z 1,... We have z i = x 2i + x 2i+1 for all i. Thus, z n/2+i = x n+2i + x n+2i+1 = x 2i = x 2i+1 = z i, so that x 2i = z i + z n/2 + i = 0. This essentially reduces to the previous case. The case of D is siilar. We require d n + 2. Since dn/2 = k, k n 2 /2 + n is sufficient. Thus, A D = 1. It follows by siilar reasoning that B C = 1, and hence that A C D = B C D = 1. Lea 7. We have C D = a(n). Proof. We have the systes of equations { x i + x n/2+i = y n/2+i x n/2+i = y i+1 + y n/2+i+1, where 0 i k 1, as well as x k +x n/2+k = y n/2+k. It follows by adding the pairs of equations that x i = y i+1 + y n/2+i + y n/2+i+1. In particular, if 1 i k 1, we also have x i + x n/2+i + x n/2+i 1 = y n/2+i +y i +y n/2+i = y i, which iplies that y n/2+i = x n/2+i +x n 1+i +x n+i. It follows that x n/2+i + x n 1+i + x n+i = x i + x n/2+i. Hence x n+i = x n+i 1 + x i. The nubers x i are thus all deterined by x 0,...,x n 1. It follows that C D a(n). 6

9 We clai that C D a(n). We will show this in the case of n even; the proof should be siilar if n is odd. We ust show that for any n 1, there are at least a(n) blocks of the for x 0 x 1 x, where x 0 x 1 x n 1 A and x k = x k 1 + x k n (1) for n k. This is clearly true for = n 1. Now suppose this is the case for soe n 1. We will proceed by induction. Let x 0 x 1 x be an accessible block in row r, say, that satisfies equation (1). Then there is a block of the for 0x 0 0x x 0 in row 2r. It follows that line 2r + 1 contains a block of the for x n/2 1 (x 0 + x n/2 )x n/2 (x 1 + x n/2+1 ) (x n/2 + x )x. Write y 0 y 1 y 2 n+2 for this block, that is, y 2i = x i 1+n/2 for i = 0,1,..., n/2 + 1, y 2i+1 = x i + x i+n/2 for i = 0,1,..., n/2. We clai that y 0 y 1 y 2 n+2 satisfies (1). For, suppose that k n/2. Then y 2k = x k 1+n/2, y 2k 1 = x k 1 + x k 1+n/2, and y 2k n = x k 1, so that y 2k = y 2k 1 + y 2k n. Moreover, we have y 2k+1 = x k +x k+n/2, y 2k = x k 1+n/2, y 2k n+1 = x k n/2 +x k, so that y 2k+1 y 2k y 2k n+1 = x k+n/2 x k+n/2 1 x k n/2, and the last expression is 0, by (1). It follows that y 0 y 1 y 2 n+2 is deterined by y 0 y 1 y n 1. Since the ap that takes x 0 x n 1 to y 0 y n 1 is injective, the x 0 x are apped to different y 0 y 2 n+2, so that the nuber of accessible blocks of the for y 0 y 2 n+2 is at least a(n) by the induction hypothesis. Since 2 n , the clai holds for + 1. It follows that C D = a(n). 7

10 Theore 1. Suppose n 4 is even. Then, given k n 2 /2 + n and the base cases a T (1), a T (2),..., a T (n 2 /2 + n 1), we have a T (2k) = 2a T (k) + 2a T (n/2 + k) a T (n) n 2 and a T (2k + 1) = a T (k) + a T (k + 1) + a T (n/2 + k) + a T (n/2 + k + 1) a T (n) n 2. Proof. We recall that a(2k) = a(k)+a(k)+a(n/2+k)+a(n/2+k) A B A C A D B C B D C D + A B C + A B D + A C D + B C D A B C D fro the arguent described above, and apply the leas. The forula for a(2k) follows. The forula for a(2k + 1) follows by siilar reasoning. The following theore for odd n is obtained by siilar reasoning. Theore 2. Suppose n 3 is odd. Then, given k n(n+3) 2 = n 2 /2 + 3n/2 and the base cases a T (1), a T (2),..., a T (n 2 /2 + 3n/2 1), we have a T (2k) = 2a T (k) + a T ( n 1 2 ) ( ) n k + a T + k a T (n) n 2 2 and a T (2k + 1) = a T (k) + a T (k + 1) + 2a T ( n ) + k a T (n) n 2. 3 Generating Functions In this section we will investigate generating functions for the sequences a T (k). We will derive functional equations which will be useful in the next section. The odd and even cases will be considered separately, as in the previous section. 8

11 For coplex z with z < 1 2 and for n 4 even, write N = n2 /2 + n and define f n (z) = a T (k)z k. k=2n Note that a T (k) 2 k, so that the right-hand expression is defined. Theore 3. Let 0 < z < 1 2 and let n 4 be even. Then f n (z) = P n (z) z2n (a T (n) + n + 2) 1 z + 1 z n+1 (1 + zn )(1 + z) 2 f n (z 2 ), where P n (z) is a polynoial. Proof. We will again write a(k) for a T (k) in the proof. We have f n (z) = Applying Theore 1 gives Therefore, f n (z) = f n (z) =2 k=n + z k=n a(k)z k = k=2n k=n (2a(k) + 2a(n/2 + k))z 2k (1 + z) k=n + z a(2k)z 2k + z k=n a(2k + 1)z 2k. (a(k + 1) + a(k) + a(n/2 + k) + a(n/2 + k + 1))z 2k k=n a(k)z 2k + 2 z n k=n a(k)z 2k + 1 z n 1 (a(n) + n + 2)z 2k. a(k)z 2k + 1 z a(k)z 2k k=n+n/2 k=n+1 a(k)z 2k + 1 z n+1 k=n+n/2 a(k)z 2k k=n+n/2+1 9

12 Collecting ters, we have 2N 1 f n (z) =2 ( N 1 1 (1 + z)(a(n) + n + 2) 1 z 2 z ). 2k k=n 2N 1 + z + k=n a(k)z 2k + 2 z n a(k)z 2k + 1 z n 1 k=0 2N 1 a(k)z 2k + 1 2N 1 z a(k)z 2k k=n+n/2 k=n+1 ( z n + 1 z + z + 1 z n z n+1 2N 1 + (a(n) + n + 2) k=0 z k =P n (z) z2n (a(n) + n + 2) 1 z 2N 1 a(k)z 2k + 1 z n+1 k=n+n/2 ) f n (z 2 ) 2N 1 a(k)z 2k k=n+n/2+1 (1 + z)(a(n) + n + 2) 1 z z n+1 (1 + zn )(1 + z) 2 f n (z 2 ), where N+n/2 1 P n (z) =2 + k=n ( z + 1 z a T (k)z 2k + (2 + 2z ) 2N 1 n a T (k)z 2k k=n+n/2 ) 2N 1 ( 1 a T (k)z 2k + z k=n+1 n z n+1 + (a T (N) + a T (N + n/2))z 2N+1. ) 2N 1 a T (k)z 2k k=n+n/2+1 For coplex z with z < 1 2 and for n 3 odd, write M = n(n+3) 2 and define f n (z) = a T (k)z k. k=2m By reasoning siilar to that above, we obtain the following theore. 10

13 Theore 4. Let 0 < z < 1 2 and let n 3 be odd. Then f n (z) = Q n (z) z2m (a T (n) + n + 2) 1 z + 1 z n+1 (1 + zn )(1 + z) 2 f n (z 2 ), where Q n (z) is a polynoial. 4 Asyptotic Behavior of a T (k)/k 2 For the discussion of asyptotic behavior, we will work in a uch ore general fraework. We will consider a generating function φ which is assued to satisfy a general functional equation. The ain result we derive in this section will include the case of a T (k). We shall require the following facts about power series. (a) If R is the radius of convergence of the power series, then in z < R the su of the series is analytic and its derivative has the sae radius of convergence (see [5], Ch. 2, 2.4, Theore 2(iii)). (b) If f has a power series developent in a disk, then the coefficients are uniquely deterined (see [5], p. 40). Let p be prie and let D = {z C z < 1/p}. For z < 1, let 1 λ(z) = k=0 γ(k)zk, where γ(0) = 1 and γ(k) = Ck 2 + f (k), where C > 0 is constant and li k f (k)log p k k 2 be a function given by the power series expression φ(z) = k=1 α k z k, = 0. Let φ : D C where α k p k, and assue that φ satisfies λ(z)φ(z) = R(z) + λ(z p )φ(z p ), where R : C C is a polynoial with R(1) = 0. Note that R(0) = 0. 11

14 Proposition 1. We have φ(z) = 1 λ(z) k=0 ( R z pk). Proof. We have that λ(z)φ(z) = R(z) + λ(z p )φ(z p ). Iterating this equation gives λ(z)φ(z) = ( R(z) + R(z p ) + + R z pk) ( + λ z pk+1) ( φ z pk+1). We now note that since λ and φ are analytic and hence continuous in D, we have ( li λ z pk+1) ( φ z pk+1) = r(0)φ(0) = 0. k Thus, λ(z)φ(z) = k=0 R (z pk). The result follows. We now use the above proposition to develop an explicit forula for the coefficients α k for sufficiently large k. Theore 5. Set = degr, and write R(z) = c jz j. Then for k, α k = k j t=0 c j γ(k jp t ). Proof. We have φ(z) = ( q=0γ(q)z q R z pk) = k=0 q=0γ(q)z q z j pk. k=0 We now note that f (q) < q 2, for otherwise we would not have f (q)(log p q)/q 2 0. Thus q=0 γ(q) z q q=0 ( Cq 2 p q + f (q) ) p q (C + 1) q=1 q 2 p q. It follows that the series q=0 γ(q)zq is absolutely convergent. We thus ay for the Cauchy product of the series q=0 γ(q)zq and k=0 z jpk as follows (see [6], Theore 3.50): φ(z) = k j c k=1 i=0 γ(k i)b i, j z k, 12

15 where b i, j = { 1 if i = jp t for soe integer t 0, 0 otherwise. Write S = {t N {0} jp t k}. It follows that k=1 α k z k = 1 k ( k=1 i=0 c j γ(k i)b i, j )z k + Thus for k, α k = t S c j γ(k jp t ) = /0). This copletes the proof. k=( t S k j c j γ(k i)b i, j )z k. t=0 c j γ(k jp t ) (if k, then S Reark 1. Define r n (z) = (1 z n )(1 z) 2 for all z C, n N. We note that in the case of the cellular autoaton with transition rule 1 + x + x n for even n 4, we ay take p = 2, φ n (z) = ( ) f n (z)/z n+1, and R n (z) = r n(z) P z n+1 n (z) z2n (a T (n)+n+2) 1 z. Proposition 2. Write 1 r(z) = 1 r n (z) = k=0 η(k)zk. Then η(k) = ( 1 + )( k k + 1 n n 2 ) k. n Proof. We observe that 1 r(z) = z n (1 z) 2 = β k z k + 1)z k=0 q=0(q q, where β k = { 1 if k = nv for soe integer v 0, 0 otherwise. We note that the first series is doinated by the geoetric series and is hence absolutely conver- 13

16 gent. We again for the Cauchy product, obtaining 1 r(z) = k=0 k q=0 (k q + 1)β q z k = k/n k=0 v=0 (k nv + 1)z k. Thus, η(k) = k/n v=0 (k nv + 1) = (1 + k/n)(k + 1 n 2 k/n). This copletes the proof. Reark 2. If φ is the generating function for a cellular autoaton with line coplexity a R (k), then a R (k + M) = α(k) α k for soe M. We ay thus consider the asyptotic behavior of α(k)/k 2 to deterine that of a R (k)/k 2. For exaple, if T is as in Theore 1, then a T (k + n + 1) = α(k). Reark 3. Write δ(k) = k k n n kn 2 2 k 2 2n. Observe that δ(k + n) = δ(k), and if 0 k n, then n 2 δ(k) 0. Thus δ(k) = O(1). Since η(k) = k2 2n +( k + 1 n 2 kn + kn ) +δ(k), we have η(k) = k2 2n + O(k). We now turn to the ain result regarding the asyptotic behavior of α(k)/k 2. For y R, we will denote the fractional part of y by y = y y. Theore 6. For 1 p x 1, let {s k(x)} be a sequence such that s k (x) s k N, s k, and li k logp s k = logp 1 x = logp 1 x. Then where degr α(s k ) li k s 2 = C k c j ( σ j (p) 2 p 2 1 x2 + 2σ j(p) 1 p x log p j ε j ), and 1 {1 if log ε j = p x < log p j 0 otherwise, σ j (p) = p 1+ log p j ε j. 14

17 Proof. By Theore 5, we have Thus, α(s k ) s 2 k = = c j t=0 s log kj p c j t=0 ( C(sk jp t ) 2 s 2 k + f (s k jp t ) ) s 2 k C(s k jp t ) 2 ( f (sk )log p s k + O s 2 k s 2 k ). α(s k ) li k s 2 = li k k c j t=0 C(s k jp t ) 2 s 2 k C k c j = li t=0 (1 2 jpt s k + j2 p 2t ) s 2 k ( ) = li k C s k c j log p + 1 j C c j t=0 2 jp t s k +C c j t=0 j 2 p 2t. s 2 k We note that C ( ) s k c j log p + 1 j = C ) s k c j (1 log p + log j p s k log p j. Since R(1) = 0, we have c j = 0. The above expression reduces to We have C s k c j log p j C c j log p j. { s k logp s k logp j if log log p = p s k logp j j logp s k logp j + 1 otherwise, s so that for all sufficiently large k, log kj p = log p s k logp j + ε j (for exaple, to show that 15

18 1 log p x < log p j iplies log p s k < logp j for sufficiently large k, take K to be so large that logp s 1 k logp x < log p j 1 log p x for all k > K). Therefore, ( ) li C s k k c j log p + 1 = li j We have that = C = C k C c j logp s k +C c j ( logp j ε j log p j ) 1 c j log p x +C ( c j logp j ) log p j ε j c j ( logp j + ε j ). t=0 2 jp t s k = 2 t=0 1 p log p p t = 2 p log p p 1 p +1 1 = 2 ) (p s log kj s p +1 p log kj p = 2 ( p log p s k p log p j p 1 ε j j ). 1 p 1 p s k Thus, C k c j li t=0 2 jp t s k = C 2c j 1 p p log p 1 x p 1+ log p j ε j = C 2c j σ j (p) 1 p x. 16

19 Siilarly, t=0 ( ) j 2 ( p 2 ) t 1 = s k p 2log p = 1 1 p 2 = 1 1 p 2 1 p 2 s log kj p +2 1 p 2 ( ) p 2log p p 2 2 s log kj p ( j 2 p 2 2 log p s k +2 log p j 2ε j ). s 2 k Thus, li C k c j The theore follows. t=0 ( ) j 2 p 2t = C s k = C c j 1 p 2 p2log p x p 2+2 log p j 2ε j c j σ j (p) 2 p 2 1 x2. Corollary 1. If T is as in Theore 1, 2 1 x 1, and R n(z) = degr n v j z j, then a T (s k ) li k s 2 = 1 degr n 2n k ( σ j (2) 2 v j x 2 2σ j (2)x log 3 2 j ε j ). Exaple 1. Let T = 1 + x + x 3. Then if we denote the liit function above by f 3 (x), we obtain the following representation (using a coputer) 17

20 Figure 3: Plot of the liit function for the case n = x x if 1 2 x < x x + 2 f 3 (x) = if 2 3 x < x2 3 4 x if 4 5 x < x2 3 8 x We note that the axiu and iniu of f 3 are li sup k a T (k) k 2 = if 8 9 x and 246, respectively. Thus, and li inf k a T (k) k 2 = to f 3. See Figure 3 for a graph of f 3. See Figure 4 for an illustration of the convergence of a T (s k )/s 2 k 18

21 Figure 4: We plot f 3 (2 y ) (above) and a T ( y)/ y 2 (below) versus log 2 y (horizontal) for various points y [2 5,2 14 ). Exaple 2. Let T = 1 + x + x 4. Then, denoting the liit function by f 4 (x), we obtain the following representation (again using a coputer) x x if 1 2 x < x x + 2 if 8 15 x < 4 7 f 4 (x) = x x + 17 if x < x x if 8 13 x < x x We note that the axiu and iniu of f 4 are given by li sup k a T (k) k 2 = if 4 5 x and 980, respectively. Thus, 19

22 Figure 5: f 4 (2 y ) (above) and a T ( y)/ y 2 (below) versus log 2 y (horizontal) and li inf k a T (k) k 2 = See Figure 5 for an illustration of the convergence of a T (s k )/s 2 k to f 4. 5 Conclusion We have found recursion relations for a T (k), where T = 1 + x + x n for n 3 and the coefficients are taken odulo 2. We have proven functional relations for the generating functions associated to a T (k) for even and odd n. Finally, we have proven that if φ(z) = k=1 α(k)zk satisfies a certain functional equation relating φ(z) and φ(z p ), and if for 1 p x 1, s k(x) s k is a sequence divergent to infinity such that log p s k converges to 1 logp x, then the sequence α(s k )/s 2 k tends to a piecewise quadratic function of x. Possible directions of future research include investigating the following conjectures: 20

23 Conjecture 1. Let T be given by a polynoial of degree n which cannot be expressed as a power of another polynoial, and take the coefficients odulo soe prie p. Then a T (k) = p 1 j=0 where C is a constant depending upon T. p 1 r=0 ( ) k + jn + r a T C, p Conjecture 2. If a T (k) arises fro an irreducible polynoial transition rule of degree n with coefficients taken odulo a prie p, then if φ is a generating function for a T (k), we have r n (z)φ(z) = R(z) + r n (z p )φ(z p ) for soe polynoial R(z). 6 Acknowledgents I would like to thank Mr. Chiheon Ki for entoring this project and Prof. Pavel Etingof for supervising this project. I would also like to thank the Center for Excellence in Education, the Massachusetts Institute of Technology, and the MIT Math Departent for aking RSI possible. I would like to thank Mr. Antoni Rangachev and Dr. Tanya Khovanova for their advice regarding this paper. Finally, I would like to thank y sponsors, Mr. Steven Ferrucci of the Aerican Matheatical Society, Dr. Donald McClure of the Aerican Matheatical Society, Mr. and Mrs. Rayond C. Kubacki, Mr. Piotr Mitros, Mr. and Mrs. Steven Scott, Dr. To Leighton of Akaai Technologies, and Dr. Bonnie Berger of MIT. 21

24 References [1] J. von Neuann and A. W. Brooks. Theory of Self-Reproducing Autoata. University of Illinois Press, Urbana, [2] S. Wolfra. Cellular autoata as odels of coplexity. Nature, 311: , [3] S. J. Willson. Coputing fractal diensions for additive cellular autoata. Physica 24D, pages , [4] K. Garbe. Patterns in the coefficients of powers of polynoials over a finite field. Preprint, [5] L. V. Ahlfors. Coplex Analysis: An Introduction to the Theory of Analytic Functions of One Coplex Variable, Third Edition. McGraw-Hill, New York, [6] W. Rudin. Principles of Matheatical Analysis, Third Edition. McGraw-Hill, New York,