Sign-reductions, p-adic valuations, binomial coefficients modulo p k and triangular symmetries

Transcription

1 Sign-redctions, p-adic valations, binomial coefficients modlo p k and trianglar symmetries Mihai Prnesc Abstract According to a classical reslt of E. Kmmer, the p-adic valation v p applied to a binomial coefficient ( ) a+b a yields the nmber of carries occrring while adding a and b in basis p. We show that for all m N the nmbers v ( a p( b) ) bild a pattern with trianglar symmetry for 0 b a p m 1. This fact will be compared with the trianglar symmetry of the patterns ( a p( b) mod p) for 0 b a p m 1, where p is the sign-redction: p(x) = x if 0 x p/2 and (x) = p x if p/2 < x < p. It is shown that n = 4 is the only one composite nmber sch that ( a n( b) mod n) has trianglar symmetry. In this special case the two patterns coincide. It is also shown that the last non-zero digits of the binomial coefficients written in basis p bild a pattern with trianglar symmetry. A combined pattern nifies all proven featres. Key Words: binomial coefficient, p-adic valation, trianglar symmetry, Kmmer s theorem abot carries, Pascal s Triangle modlo p k, atomatic 2-dimensional seqence, Zaphod Beeblebrox. A.M.S.-Classification: 11A07, 05E11, 28A80, 68Q45. 1 Introdction Let p be a prime. According to a classical reslt of Ernst Kmmer, if p c ( ) a+b a bt p c+1 ( ) a+b a, then the nmber of carries that occr dring the digital addition of a with b in basis p is c. If k Z and p c k bt p c+1 k, one says that v p (k) = c. The fnction v p is called p-adic valation. One takes by convention v p (0) =. Three general properties of the general notion of valation will be sed here. A valation is a homomorphism, i. e. v p (ab) = v p (a) + v p (b), satisfying triangle s ineqality for ltra-metrics, i.e. v p (a + b) min(v p (a), v p (b)). Moreover, if v p (a) v p (b), then v p (a + b) = min(v p (a), v p (b)). A direct conseqence of Kmmer s Theorem abot carries is the following: Corollary 1.1 If 0 v p m 1, then 0 v p ( ( ) m 1. Both bonds are taken as vales over the given set. Kmmer s Theorem can be seen as a reslt concerning the complexity of addition. It corresponds to the sal hman experience that an instance of addition, that needs more carries prodces the sensation of being more difficlt. If the vales v p ( ( s a) ) are written down in a trianglar lattice, in the same way as ( s a), this lattice can be seen as a map of the difficlty to add nmbers written in basis p. The nmbers on row s express the difficlty to get the sm s from smmands a and s a. The first goal of this note is to show that for all m N, the first p m rows of Pascal s Triangle, which are nmbered from 0 to p m 1, bild a pattern with trianglar symmetry. This fact implies Brain Prodcts, Freibrg, Germany and Simion Stoilow Institte of Mathematics of the Romanian Academy, Research nit 5, P. O. Box 1-764, RO Bcharest, Romania. mihai.prnesc@math.ni-freibrg.de. 1

2 that every Pascal s Triangle modlo p k is the nion of an ascendant chains of symmetric trianglar blocks of edge p m, where m N. This is done in Section 3. In Section 4 this symmetry is compared with another symmetry occrring when the binomial coefficients are projected onto a finite set in connection with a prime nmber. To reach the second symmetry, the binomial coefficients modlo p are projected onto the set {0,..., (p 1)/2} by a fnction p, defined sch that p (x) = x if 0 x (p 1)/2 and (x) = p x if (p 1)/2 < x < p. We call the fnction p sign-redction and we observe that it is a kind of absolte vale. From a philosophical point of view absolte vales and valations are related. Unhappily, for the moment we cannot give more deep evidence of a relation between the two symmetries. In [9] the athor described the sqare dihedral symmetry of the seqences a(i, j) mod p for a(i, j) = (a(i, j 1) + ma(i 1, j 1) + a(i 1, j)) mod p with constant initial conditions and m 0. This symmetry was also bazed on sign-redction. The reader will observe that Lemmas 5.3 and 5.4 in [9] are related with Lemma 4.2 in the present article. The trianglar dihedral symmetry for the case m = 0 (i. e. ( ) i+j i mod p) was not noticed by the athor becase it did not match in the sqare grid sed in [9]. In Section 4 this gap is filled. The trianglar symmetry was empirically observed by other athors, at least for the fndamental block given by the first p rows - see [3], where some images and comments are displayed. Section 5 contains a more or less coincidential interference between both symmetries in a limit case. It is shown that the nmber n = 4 is the only one composite nmber sch that the triangle n ( ( mod n) consists of an ascendant chain of symmetric triangles of edge n m, where m N. This pattern has been also stdied by A. Granville in [5] and [6]. A constrction in Section 6 is possible only becase one symmetric pattern is active where the other one is not. They are complementary. Also, some natral connections with atomatic seqences arise dring the paper. 2 Prereqisites Definition 2.1 A trianglar lattice Θ is a set of doble indexed points P (, of the plane,, v N, 0 v, sch that all triangles in the set {P (, P ( + 1, P ( + 1, v + 1) 0 v } {P (, P (+1, v +1)P (, v +1) 0 v } are disjoint and congrent eqilateral triangles. Up to similarity there is only one trianglar lattice, that will be called Θ. Definition 2.2 Let S be a set. A triangle over S is an application T : Θ S. For some n N, let Θ(n) be the sbset consisting of the first n rows of Θ, indexed from 0 to n 1. Let T (n) be the restriction of T to the set Θ(n). The fnction T (n) will be also called a triangle. The vale T (P (, ) will be shortly written down as T (,. Pascal s Triangle and Pascal s Triangle modlo n are examples of triangles, with T (, = ( ) v and T (, = ( mod n respectively. Both triangles are niqely determined by the initial conditions T (, 0) = T (, ) = 1 and by the recrrence T ( + 1, v + 1) = T (, + T (, v + 1) for all, v N with 0 v. The same recrrence mst be considered once over Z and once over the finite cyclic grop Z/nZ. There is also another way to define Pascal s Triangles, considering a sqare lattice N N and a recrrent doble seqence given by the initial conditions a(i, 0) = a(0, j) = 1 and a(i, j) = a(i, j 1)+a(i 1, j). In this case a(i, j) = ( ) ( i+j i or respectively i+j ) i mod n, see athor s papers [9] and [11]. To change the coordinates fron the sqare lattice coordinates to the trianglar lattice coordinates, observe that: a(i, j) = T (i + j, i), 2

3 for all 0 i, j and 0 v. T (, = a(v,, The sqare lattice representation of the binomial coefficients has some advantages. As proven in [9], if m F p, if the seqence a(i, j) satisfies the conditions a(i, 0) = a(0, j) = 1 and a(i, j) = a(i, j 1) + ma(i 1, j 1) + a(i 1, j), and if for m N we define the matrix A m = {a(i, j) 0 i, j < p m }, then: A m = A 1 (A 1 A 1 ) = A n 1. Here means the (Kronnecker-) tensor prodct of matrice. If S is some mltiplicative monoid, A M n,m (S) and B M s,t (S), then the matrix A B belongs to M ns,mt (S) and is the matrix with block-wise representation (a(i, j)b) 0 i<n,0 j<m. The -monomial A n is defined as A A (n 1). A( 1 was called fndamental block. If m = 0 the recrrent doble seqence a(i, j) is exactly the i+j ) i as remarked above. The tensor prodct representation of this doble seqence follows also directly from the classical theorem of Lcas concerning the vale of ( a b) mod p as a fnction of their digits in basis p. At this point shold be mentioned that Pascal s Triangle modlo p k is not a limit of tensor powers of matrices if k 2. However, Pascal s Triangles modlo p k are p-atomatic, and conseqently can be prodced by matrix sbstittion and are projections of doble seqences prodced by two-dimensional morphisms. See [1] and [2]. Definition 2.3 A triangle T (n) : Θ(n) S is called symmetric if for all 0 v n 1, T (, = T (, and T (, = T (n 1 + v, n 1). To nderstand this definition, consider the applications S, R : Θ(n) Θ(n), given by S(, = (, and R(, = (n 1 v, for all, v N with 0 v n 1. It is only pre comptation to prove that R 3 = S 2 = id and that S 1 RS = R 1. In fact, S is a reflection of Θ(n) across a median, R is a rotation with 120 of Θ(n) arond its center, and the grop generated by S and R is the whole dihedral grop D 6, the symmetry grop of the eqilateral triangle. This grop has six elements. Under the action of D 6, Θ(n) splits in orbits of length 6, 3 or 1. If n 3k + 1 there is no central element, so no orbit of length 1 does occr. Instead of T (, = T (n 1 + v, n 1), one can check that T (, = T (n v 1,. This is jst the other rotation. Using the correspondence between trianglar and sqare lattice coordinates, one can adapt this definition for triangles presented in sqare lattice coordinates. Definition 2.4 Let A M n,n (S) be a sqare matrix. The set T 1 (A) = {a(i, j) 0 i, j < n 0 i + j n 1} is called the first triangle of A. The complemetary set T 2 (A) = {a(i, j) 0 i, j < n i + j > n 1} is the second triangle of A. T 1 (A) is called symmetric if it satisfies the identities a(i, j) = a(j, i) and a(i, j) = a(j, n 1 i j) for all i, j 0 with i + j n 1. Instead of a(i, j) = a(j, n 1 i j) one can check that a(i, j) = a(j, n 1 i j). This is again the other rotation. 3 p-adic valation Let p be a prime and v p : Z N { } the p-adic valation. 3

4 Lemma 3.1 For 1 i p m and 0 k p m i the following holds: [ ( p m ) i ] [ ( ) i 1 + k ]. v p = vp k i 1 Proof: By definition, ( p m ) i = (pm i (k 1)) (p m i) k k! ( ) i 1 + k = i 1 i (i + (k 1)). k! By the grop homomorphism property of valations, it mst be shown that v p ((p m i) (p m i (k 1))) = v p (i (i + (k 1))). For, it wold be enogh that for all i x i + (k 1), v p (p m x) = v p (x). Indeed, x i + (p m i) 1 = p m 1 < p m, so v p (x) < v p (p m ) = m. Hence, v p (p m x) = min(v p (p m ), v p (x)) = v p (x). Theorem 3.2 The patterns {v p ( ( ) 0 v < p m } have trianglar symmetry for all m 0. A possible name for the pattern bilt by the p-valation applied to binomial coefficients, i. e. {v p ( ( ) v ) 0 v }, cold be the Pascal - Kmmer Triangle. The set with 0 v p m contains the vales {0,..., m 1} according to Corollary 1.1. An example is displayed in Figre 1. Proof: By Lemma 3.1, the pattern is preserved by a rotation with 120 arond its center. By the identity ( ) ( v =, it is preserved by a reflection across its median. According to the definition 2.3 and its conseqences, the pattern has trianglar symmetry. The next Lemma has been pblished by I. Tomesc as a problem proposed to the readers of Gazeta Matematică in [12]. Lemma 3.3 Let p be a prime and n = n k p k + + n 0, with n k,..., n 0 {0,..., p 1}. The nmber of binomial coefficients ( n 0), ( n 1),..., ( n n) that are mltiples of p is: n + 1 (n 0 + 1) (n k + 1). Proof: Let a = a k p k + + a 0, with a k,..., a 0 {0,..., p 1}. By Kmmer s Theorem, p ( ) n a if and only if for all i, n i a i. So for all i, a i can be chosen in n i + 1 ways. The following Lemma provides spplementary information abot this pattern and will be also applied in a later section. It has been given as a problem to be solved dring a mathematical contest in Lxembrg, For both Lemmas 3.3 and 3.4 and other nice pzzles, see [8]. Lemma 3.4 v p ( ( ) = 0 for all v {0,..., } iff = zp m 1, m 0 and z {1,..., p 1}. Proof: By Lemma 3.3, if = zp m 1, with m 0 and z {1,..., p 1}, then the nmber of binomial coefficients in row that are not divisible with p is zp m (z 1+1)(p 1+1) (p 1+1) = 0. For the converse, if a nmber contains a digit n i < p 1 in its inner or at the end, one can prodce a carry over in addition by choosing a nmber v with a bigger digit v i > n i for this position. So only the first digit might be different from p 1. By Lemma 3.4 we know exactly which are the constant lines in the Pascal-Kmmer Triangle. The Pascal-Kmmer Triangle is not atomatic, becase the vales of v p ( ( ) are not bonded. To srpass this inconvenient, one has to adapt the notion of valation for rings of classes of remainders, like Z/p k Z. The reslting notion is not standard, becase valation theory has been developped for fields, and the rings Z/p k Z are not domains. However, this non-standard notion is very natral in the present context. We recall that all ideals in Z/p k Z have the form p i Z/p k Z and that they bild a descending finite chain of ideals: Z/p k Z = p 0 Z/p k Z > pz/p k Z > > p k 1 Z/p k Z > p k Z/p k Z = 0. 4

5 Figre 1: The first 64 rows of the Pascal-Kmmer Triangle v 2 ( ( ). Definition 3.5 For a prime p and for k 1 we define v p : Z/p k Z {0, 1,..., k} as: { s x p s Z/p k Z x / p s+1 Z/p k Z s < k, v p (x) = k x = 0. Corollary 3.6 The patterns {v p ( ( mod p k ) 0 v < p m } have trianglar symmetry for all m 0. Moreover, the two-dimensional seqence {v p ( ( mod p k ) 0 v } is p-atomatic. Proof: The trianglar symmetry of the patterns follows directly from Theorem 3.2. The 2- dimensional seqence is p-atomatic becase the 2-dimensional seqence ( ( mod p k ) is p- atomatic, and that the p-atomatic seqences are closed nder projections. See the monograph [1] for both properties. However, sing Kmmer s Theorem, one can very easily constrct an atomaton generating the same seqence in sqare coordinates - i.e. a(i, j) = ( ) i+j j mod p k. The inpt alphabet is Σ = {0,..., p 1} {0,..., p 1}. The set of states is Z = {z 0, z 1,..., z k 1 } {w 1,..., w k 1 } {f}. For t < k, z t means that t many carries have been conted so far, bt in the moment there is no carry to add. Similarly, w t means that t many carries have been conted so far and the digit addition done in the last step prodced a carry. In 5

6 state f a nmber of k carries have been already conted. In this case it is no more important whether in the last step a carry has been prodced or not. The digits of the inpt (i, j) come in starting with the less significant pair (i 0, j 0 ). The otpt fnction ω assigns to each state the nmber of carries: ω(z t ) = ω(w t ) = t, ω(f) = k. Corollary 3.7 The patterns { ( mod 2 0 v < 2 m } have trianglar symmetry for all m 0. Proof: Indeed, for v 2 : Z/2Z {0, 1} hold v 2 (1) = 0 and v 2 (0) = 1. So p to a permtation of vales, v 2 ( ( mod 2) prodces the same pattern as ( mod 2. 4 Sign-redction In the next definition the elements of the ring Z/nZ are intentionally identified with their canonical representatives from the set {0, 1,..., n 1}. The order sed in the definition is the order of natral nmbers. Definition 4.1 Let n be a natral nmber. The sign-redction n : Z/nZ Z/nZ is defined as: { x 0 x n/2, n (x) = n x n/2 < x n 1. Lemma 4.2 Let p be a prime. For 1 i p and 0 k p i the following congrence holds: ( ) ( ) p i i 1 + k ( 1) k mod p. k i 1 Proof: For i = 1 this reslt is folklore - bt maybe not the following proof. The (p + 1)-th row of Pascal s Triangle consists of ( p k) and they are mltiples of p for k = 1,..., p 1. The p-th and (p + 1)-th rows start in the field F p as follows: 1 x y z We apply the recrrence T (+1, v +1) = T (, +T (, v +1) and get sccessively x = 1, y = 1, z = 1 and so on. So ( ) p 1 k ( 1) k ( 1) k( ) 1 1+k 1 1 mod p, as it was to prove. Now we contine by indction. Sppose that we have already shown that the row p i consists of elements respectively congrent with ( 1) k( ) i 1+k i 1 mod p for 0 k p i, and sppose that in the row p i 1 we have already shown that ( ) p i 1 ( ) k ( 1) k i+k i mod p. The next binomial coefficient is ( ) p i 1 k+1 and has the following position in Pascal s Triangle: Conseqently: ( p i 1 k ) ( p i 1 ) ) k+1 ( p i k+1 ( ) ( ) ( ) ( ) ( ) p i 1 p i p i 1 i 1 + k + 1 i + k = ( 1) k+1 ( 1) k = k + 1 k + 1 k i 1 i = ( 1) k+1[( ) ( ) ( ) ( ) i + 1 i + k ] + = ( 1) k+1 i + k + 1 (i + 1) 1 + (k + 1) = ( 1) k+1 mod p. i 1 i i (i + 1) 1 6

7 Figre 2: 11 ( ( ) v mod 11) with 0 v 10 and 13 ( ( mod 13) with 0 v 12. Lemma 4.3 The pattern T 1 (A 1 ) = { p ( ( mod p) 0 v < p} has trianglar symmetry. ( Proof: Sign-redction over the identity in Lemma 4.2 yields: p i ) ( k mod p = i 1+k ) i 1 mod p. ( This means that pattern is preserved by a rotation with 120 arond its center. By the identity ( =, the pattern is preserved by a reflection across its median. According to the definition 2.3 and its conseqences, the pattern has trianglar symmetry. See Figre 2 for two examples. Lemma 4.4 In p ( ( mod p) the configration: a a a is possible only if p 3 or a = 0. Conseqently, the central configrations in T for p 5 are described below. Here always 0 a b 0: a a a b a a a if p = 4k + 1, b a b a a b if p = 4k + 3. Proof: Verify the eight possible relations (ɛ 1 p+( 1) 1+ɛ1 a)+(ɛ 2 p+( 1) 1+ɛ2 a) = (ɛ 3 p+( 1) 1+ɛ3 a) for ɛ 1, ɛ 2, ɛ 3 {0, 1}. The cases 100 and 011 lead to 3a = 0, so a = 0 or p = 3. The cases 000 and 101 lead to a = 0. The other cases lead to ±p = a or 2p = a possible only if p {2, 3} and a = 0. The central configrations depend of the trianglar symmetry, of the existence of a central element and of this condition. For the next steps we need the tensor prodct strctre of Pascal s Triangle mod p as it has been recalled in the Introdction. Let F p = F p \ {0} be the mltiplicative grop pf the field F p. 7

8 We observe that the application: p : F p {1, 2,..., (p 1)/2} := H p, has #p 1 (a) = 2, for all a H p, that 1 p (1) = {1, 1} and that for all a, b H p and for all x 1 p (a), y 1 p (b), p (xy) does not depend of the choice of the representatives x and y. Conseqently one can define a new mltiplication over H p by a b = p ( 1 p (a) 1 p (b)). This operation indces a strctre of grop (H p,, 1) sch that p : F p H p is a homomorphism of grops with kernel {1, 1} =< 1 >. This yields: H p = F p / < 1 >. If we complete now this mltiplication in a natral way with a 0 = 0 a = 0, we get: Lemma 4.5 If A m = { p ( ( ) i+j i mod p) 0 i, j < p m } M pm,p m(h p {0}), then: A m = A 1 (A 1 A 1 ) = A n 1, where the tensor prodct is defined according to the mltiplication on H p {0} and the tensor prodct monomial is indctively defined by A n = A A (n 1). Lemma 4.6 Let (J, ) be some associative monoid containing an element 0 with the property that for all x J, x 0 = 0 x = 0. Let A M m,m (J) and B M n,n (J) two matrices, sch that T 1 (A), T 1 (B) have both trianglar symmetry and T 2 (A), T 2 (B) consist both only of zeros. (Compare with Definition 2.4). Then for the matrix A B M mn,mn (J) holds: T 1 (A B) has trianglar symmetry and T 2 (A B) consists only of zeros. Proof: Pre comptation sing Definition 2.4. More interesting than the proof is maybe the fact that if one tries to prove Lemma 4.6 in the form: if both T i (A) have trianglar symmetry and both T i (B) have trianglar symmetry, then both T i (A B) have trianglar symmetry, it jst does not work. Indeed, very easy conterexamples with m = n = 2 for T 2 and with m = 3 and n = 2 for both T i can be constrcted. Theorem 4.7 The patterns { p ( ( mod p) 0 v < p m } have trianglar symmetry for all m 0. Proof: By indction in m 0. The case m = 0 is trivial. For the case m = 1 we apply Lemma 4.3. Now we trn to sqare coordinates and we observe that T 1 (A 1 ) has trianglar symmetry and that T 2 (A 1 ) consists only of zeros. Indeed, for 0 < i, j < p with 2p > i + j p, p ( ) i+j i. This means by Lemma 4.5 and by Lemma 4.7 that all A m = A n 1 are sch that T 1 (A m ) has trianglar symmetry and T 2 (A m ) consists only of zeros. Bt the patterns in qestion are exactly T 1 (A m ). For an example, see Figre 3. We observe that the tensor prodct strctre confirms Lemma 3.4. Another conseqence of the tensor prodct strctre is that the doble seqence p ( ( mod p) is p-atomatic. This follows again by the fact that the p-atomatic seqence ( mod p is projected onto the finite set Hp {0}. In fact we know more: the seqnce is a 2-dimensional morphic seqence, with start-letter 1 and with sbstittions a a A 1 for all a H p {0}, where is the appropriate mltiplication. All elements of H p {0} occr in the pattern, and can be fond already in the second diagonal, near the edge. 8

9 Figre 3: The first 121 rows of 11 ( ( mod 11), bilding T1 (A 1 A 1 ). T 1 (A 1 ) mltiplied with different grop elements from H 11 yields new blocks with permted colors. 5 A property of the nmber 4 In this section we show that the nmber n = 4 is the only one composite nmber with the property that the triangles { n ( ( mod n) 0 v n m } have trianglar symmetry for all m 0. Lemma 5.1 is folklore. It can be fond e.g. in the preprint [7], whose athor has got the statement dring a night-dream. For statements like Lemma 5.1 and mch stronger, see Granville s remarkable article [4]. This Lemma is strong enogh for or needs. Lemma 5.1 For all m, n N and prime p, ( np mp) ( m n) mod p 2. Proof: (from [7]) In (1 + X) np = [(1 + X) p ] n the coefficient of X mp is: ( ) np ( ) p =. mp k i i 0 k i p k k n = mp 9

10 Modlo p 2 contribte only those terms with at least n 1 many k i eqal 0 or p. The sm of k i being mltiple of p, all of them mst be 0 or p. So m of n many k i mst be p, and the nmber of possible choices is ( n m). Theorem 5.2 The niqe composite n N sch that the patterns { n ( ( mod n) 0 v n m } have trianglar symmetry for all m N is n = 4. In its case all patterns { 4 ( ( ) v mod 4) 0 v 2 m } have trianglar symmetry, and the patterns are the same as those given by {v 2 ( ( mod 4) 0 v 2 m }. Proof: The proof is strctred in a seqence of Claims. Claim: 1. If all patterns { n ( ( mod n) 0 v < n m } have trianglar symmetry, then n mst be a prime-power. Let n = p n1 1 pn2 2 pns s be the prime factor decomposition of n. By the Chinese Remainder Theorem the following rings are isomorphic: Z/nZ = Z/p n1 1 Z Z/pn2 2 Z Z/pns s Z, by x mod n (x mod p k1 1 pks s ) and by this isomorphism 1 Z/nZ corresponds to (1, 1,..., 1). Sppose that the given sets have trianglar symmetry. This implies that all ( ) n m 1 k = ±1 for m N and 0 k n m. In particlar, v p ( ( ) n m k ) = 0 for all 0 k n m. If we focs on p 1 and apply Lemma 3.4, it follows that there is a seqence (x m ) taking vales in {1,..., p 1 1} and an increasing seqence (k m ) of natral nmbers sch that for all m N, x m p km 1 = p mn1 1 p mn2 2 p mns s. The seqence (x m ) has a constant sbseqence; let x be its constant vale. It trns ot that x has not a niqe prime factor decomposition, nless p 2 = = p s = 1. Claim: 2. If n = p k sch that all patterns { n ( ( mod n) 0 v < n m } have trianglar symmetry and k 2, then p cannot be an odd prime. Sppose that n = p k, k 2 and p is an odd prime. By Lemma 5.1, ( ) ( ) p k p p k 1 = p mod p 2, 1 so ( p k p k 1 ) ap 2 + p mod p k. If the row p k 1 consists only of ±1 mod p k, then ap 2 + p mod p k mst belong to the set {±2, 0}, which is the set of possible sms of two elements of row p k 1. This is impossible, becase p mod p 2 mst be then ±2, which implies p = 2. Claim: 3. If n = 2 k sch that all patterns { n ( ( mod n) 0 v < n m } have trianglar symmetry, then k 2. Sppose n = 2 k and k 3. It follows: ( 2 k ) 1 = (2k 1)(2 k 2) = 2 2k 1 2 k 2 k k 1 mod 2 k. 2 2 For k 3, 2 k 1 mod 2 k cannot be ±1 mod 2 k. Claim: 4. All patterns { 4 ( ( mod 4) 0 v 2 m } have trianglar symmetry, and are the same as those given by {v 2 ( ( mod 4) 0 v 2 m }. If we compare the fnctions v 2 : Z/4Z {0, 1, 2} with 4 : Z/4Z {0, 1, 2} we see that: 2 x = 0, v 2 (x) = 0 x = 1 x = 3, 1 x = 2, 0 x = 0, 4 (x) = 1 x = 1 x = 3, 2 x = 2. So p to a permtation of vales, v 2 and 4 are eqal and prodce the same pattern. This pattern is symmetric by the Theorem

11 Figre 4: The first 16 rows of 4 ( ( ) v mod 4) or of v2 ( ( mod 4). First 16 lines of this pattern can be seen in Figre 4. The fnction 4 has been also considered by Zaphod Beeblebrox in the nice papers [4] and [5] by A. Granville. A little bit in this spirit, we show now a complete description of the pattern. Corollary 5.3 The doble seqence 4 ( ( ) i+j i mod 4) consists of the minors: ) A 1 = ( ) 1 1, A = ( , A 3 = ( ) 2 2, A = ( ) Moreover, the whole doble seqence can be generated starting with A 1 and sccessively applying the following sbstittion rles: ( ) ( ) ( ) ( ) A1 A A 1 2 A1 A, A A 2 A 2 2 A3 A, A 3 A 2 A 3 3 A4 A, A 3 A 3 A A 4 A 4 Proof: The athor displayed a sbstittion with eight minors generating the pattern ( ( mod 4) in [10]. If we apply the fnction 4 on these eight minors element-wise, two of them yield the minors called here A 3 and A 4 (which starting with A 3 wold generate alone a pattern isomorphic with ( mod 2), other two of them redce to A1 and for of them redce to A 2. The big srprise comes when one applies 4 also on the rles of sbstittion. Withot any contradiction, they fall together onto the rles given here, exactly like the minors: one, one, two and for at a time. 6 The last non-zero digit symmetry The fnctions v p and p are complementary in the sense that one of them is active exactly over the places where the other one is constant. We can gle them together by considering their vales as natral nmbers and bilding the sm p (x) + v p (x). This fnction generate symmetric patterns if applied over ( with 0 v < p m for all m N, bt has the disadvantage, that vales 0, 1,..., min(m 1, (p 1)/2) have not a niqe interpretation anymore. Another idea is to fix the vale of m and to consider the fnction f(, = m 1 v p ( ( ) v ) + p ( ( mod p). Now the two complementary patterns gle well and vales have a niqe interpretation. Unhappily, p is not too creative for p 5 and v p becomes interesting when m 4. Bt one can get mch more if one applies p on the last non-zero digit of ( written in basis p. 11

12 Definition 6.1 Let w p : Z \ {0} Z, given by w p (x) = x/p vp(x), the p-free part of x. Lemma 6.2 (Anton - Stickelberger - Hensel) Let p be prime, and m, n N with n m. Let r = n m. Let n = n 0 + n 1 p + + n d p d with 0 n i < p, and similarly for m and n with digits m i and r i respectively. Finally, let v p ( ( n m) ) = k. Then: ( ) n w p ( m ) ( 1) k( n 0! m 0!r 0! )( n 1! ) ( n d! ) m 1!r 1! m d!r d! mod p. Proof: See [4] for the proof of a stronger identity, modlo p k. Figre 5: 11 (w 11 ( ( ) v ) mod 11) + 5v11 ( ( ) with 0 v 121. Theorem 6.3 Let p be a prime. trianglar symmetry for all m N. The patterns { p (w p ( ( ) mod p) 0 v p m } have Proof: Fix some m N. Like before, it is enogh to prove that one rotation conserve the pattern. We se this time the rotation (, (n 1 v,. It is to show that: ( ( ( ) n ) p wp mod p ) ( ( ( p m ) 1 s ) = p wp mod p ). s n s 12

13 In order to se Lemma 6.2, let r = n s, and n i, r i, s i their digits in basis p, with 0 i m 1. We observe that p m 1 in basis p consists of the repeated digit p 1 only, and that p m 1 s consists of the digits p 1 s i. Moreover (p m 1 s) (n s) = p m 1 n, that consists of the digits p 1 n i. Also recall that p and the projection mod p are mltiplicative homomorphisms. One has to show that: p ( n 0! r 0!s 0! = p ( (p 1 s 0 )! r 0!(p 1 n 0 )! mod p ) p ( n m 1! r m 1!s m 1! mod p ) = mod p ) p ( (p 1 s m 1 )! r m 1!(p 1 n m 1 )! mod p ). Now we focs on some factor p ( (p 1 si)! r i!(p 1 n i)! mod p). p ( (p 1 s i )! (p 1 n i )! mod p ) = p ( 1 2 (p s i 2)(p s i 1) 1 2 (p n i 2)(p n i 1) mod p ). Recall that by definition p (x) = p (p x). We apply this identity on every factor. One gets: p ( (p 1) (p 2) (s i + 2)(s i + 1) (p 1) (p 2) (n i + 2)(n i + 1) mod p ). Bt according to Wilson s Theorem, (p 1)! 1 mod p, so the last term displayed is eqal with: ( ( 1)/s i! p mod p ) (n i! = p mod p ). ( 1)/n i! s i! Now the eqality to show follows by eqality factor-wise. Corollary 6.4 The patterns { p (w p ( ( ) v ) mod p) + vp ( ( )(p 1)/2 0 v p m } have trianglar symmetry for all m N. Proof: This follows directly from Theorem 3.2 and Theorem 6.3. An application of the Corollary 6.4 can be seen in Figre 5. The advantage of this fnction is that it does not represent only the last non-zero digit, bt also the p-adic valation. Those patterns sggest that all sets v p ( ( ) = k have a strctre of tensor prodct of matrices, exactly like the set v p ( ( ) = 0. This is a qestion to stdy. References [1] Jean-Pal Alloche, Jeffrey Shallit: Atomatic seqences - theory, applications, generalizations. Cambridge University Press, [2] J.-P. Alloche, F. von Haeseler, H.-O. Peitgen, A. Petersen, G. Skordev: Atomaticity of doble seqences generated by one-dimensional linear celllar atomata. Theoretical Compter Science, 188, , [3] Mike Bardzell, Jennifer Bergner, Kathleen Shannon, Don Spickler, Tyler Evans: PascGalois Abstract Algebra Classroom Resorces. Math DL - The MAA Mathematical Sciences Digital Library, [4] Andrew Granville: Arithmetic Properties of Binomial Coefficients I: Binomial coefficients modlo prime powers. Canadian Mathematical Society Conference Proceedings 20, ,

14 [5] Andrew Granville: Zaphod Beeblebrox brain and the fifty-ninth row of Pascal s triangle. American Mathematical Monthly, 99, 4, , [6] Andrew Granville: Correction to: Zaphod Beeblebrox brain and the fifty-ninth row of Pascal s triangle. American Mathematical Monthly, 104, 9, [7] Mike Swarbrick Jones: A binomial congrence. Preprint [8] Dorel Miheţ: Legendre s and Kmmer s theorems again. Resonance, 15, 12, , [9] Mihai Prnesc: Self-similar carpets over finite fields. Eropean Jornal of Combinatorics, 30, 4, , [10] Mihai Prnesc: Recrrent two-dimensional seqences generated by homomorphisms of finite abelian p-grops with periodic initial conditions. Fractals, 19, 4, , [11] Mihai Prnesc: F p -affine recrrent n-dimensional seqences over F q are p-atomatic. To appear in Eropean Jornal of Combinatorics, 34, , See also [12] Ioan Tomesc: Problem proposed to the readers. Gazeta Matematică, XCIV,