p-adic valuation of the Morgan Voyce sequence and p-regularity

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1 Proc Indian Acad Sci (Math Sci Vol 127, No 2, April 2017, pp DOI /s p-adic valuation of the Morgan Voyce sequence and p-regularity LYES AIT-AMRANE 1,2, 1 USTHB/Faculty of Mathematics, LATN Laboratory, DG-RSDT, BP 32, El Alia, 16111, Bab Ezzouar, Algiers, Algeria 2 Ecole nationale Supérieure d Informatique (ESI, BP 68M Oued Smar, 16270, El Harrach, Algiers, Algeria * Corresponding author l_ait_amrane@esidz; lyesait@gmailcom MS received 17 August 2015; revised 18 May 2016 Abstract We characterize the p-adic valuation of the Morgan Voyce sequence and its companion sequence Further, we show that the p-adic valuation of the Morgan Voyce sequence is a p-regular sequence and we determine its rank explicitly Keywords p-adic valuation; k-regular sequences; Morgan Voyce sequence; linear recurrence sequences 2010 Mathematics Subject Classification Primary: 12J20; Secondary: 11A41, 11B39, 11B50 1 Introduction In dealing with electrical ladder networks, Morgan Voyce [22] defined the two sequences (V n n and (M n n by V0 = 1, V 1 = 1, V n = (2 + tv n 1 V n 2, (n 2 and M0 = 0, M 1 = 1, M n = (2 + tm n 1 M n 2, (n 2, (11 where t is a parameter that we assume to be in Z As for the Fibonacci sequence, each element of the Morgan Voyce sequences (V n n 1 and (M n n 1 can be expressed in Pascal s triangle as follows [1, 3, 4, 27]: n 1 ( n + k 1 V n = t k, (n 1 2k k=0 c Indian Academy of Sciences 235

2 236 Lyes Ait-Amrane and n 1 ( n + k M n = t k, (n 1 (12 2k + 1 k=0 The coefficients ( n+k 1 ( 2k and n+k 2k+1 are well known in the study of electrical networks The coefficients ( n+k 1 2k are exactly the lines of the DFF triangle given in Table 1 [10], while the coefficients ( n+k 2k+1 form the lines of the DFFz triangle given in Table 2 [11] Several works on linear recurrence sequences were done, especially on the Fibonacci sequence In particular, Lengyel [19] studied the p-adic valuation of Fibonacci numbers Medina and Rowland [21] used the results of Lengyel [19] to study the p-regularity of the p-adic valuation of Fibonacci numbers In this paper we extend these works to the Morgan Voyce sequence (M n n 0 defined by (11 Its companion sequence (C n n 0 is defined by C0 = 2, C 1 = 2 + t, (13 C n = (2 + tc n 1 C n 2, (n 2 The sequences (M n n 0 and (C n n 0 satisfy a number of interesting properties, see [1, 3, 4, 15, 27] The characterization of the divisibility properties of combinatorial quantities like Fibonacci numbers [19], factorials and binomial coefficients [26], Stirling numbers Table 1 DFF triangle n k Table 2 DFFz triangle n k

3 p-adic valuation of the Morgan Voyce sequence and p-regularity 237 [6, 9, 18, 20] has been studied by a number of researchers The analysis of the periodicity of these numbers modulo any integer [12, 17, 23 26, 29, 30] helps in exploring their divisibility properties [18, 19] The periodic properties of Morgan Voyce sequences have been studied in [1] Here we use some of these properties to find ν p (M n and ν p (C n Let k 2 be an integer The k-kernel of a sequence (S n n 0, denoted by ker k (S n n 0, is the set of all subsequences (S k e n+i n 0, where e and i are integers such that e 0 and 0 i k e 1, ie, ker k (S n n 0 := (S k e n+i n 0 e 0, 0 i k e 1} An integer sequence (S n n 0 is said to be k-regular if the Z-module ker k (S n n 0 generated by its k-kernel is finitely generated The rank of a k-regular sequence is the rank of this Z-module The k-regular sequences were introduced by Allouche and Shallit [2] in order to generalize automatic sequences The automatic sequence is the central concept at the intersection of formal language theory and number theory It was introduced by Cobham [7, 8] and has been extensively studied by Christol et al [5] and other writers The companion matrix of the Morgan Voyce sequence is ( t This matrix is of determinant 1 and is therefore invertible modulo any integer m 2 We deduce that the Morgan Voyce sequence modulo m is simply periodic [28] We denote the period of the sequence (M n mod m n 0 by k(m, ie, the smallest positive integer k such that M k 0modm and M k+1 1modm We denote by d(m the smallest positive integer k such that M k 0modm Note that d(m > 1 since M 1 = 1 It is known that the terms for which M n 0modmhave subscripts that form a simple arithmetic progression (Theorem 24 of [1] Thus we have M n 0modm d(m n (14 In 2, we give some identities satisfied by the sequence (M n n and prove some results, some of which will be used in 3 to characterize the p-adic valuation of the Morgan Voyce sequence (M n n 1 and its companion sequence (C n n 1 Some of these results are inspired by the work of Halton [14] Section 4 is devoted to the p-regularity of the p-adic valuation of the Morgan Voyce sequence (M n n 1 2 Some properties Let F(x = x 2 (2 + tx + 1 be the characteristic polynomial of the sequence (M n ( n and let = t 2 + 4t be its discriminant The complex roots of F(x are α = 1/2 (2 + t + ( and β = 1/2 (2 + t If = 0 (ie, t = 0, 4, we have Binet s formula [15] M n = αn β n, (n 0 (21 α β Identity (21 gives a natural extension of the sequence (M n n 0 to negative values of n Using the identity α n β n = 1, we find that M n = M n (22 If t = 0, then M n = n for any n 0 and if t = 4, then M n = ( 1 n+1 n for any n 0 It is easy to see that identity (22 also holds for t 0, 4} We deduce that the

4 238 Lyes Ait-Amrane recurrence relation M n = (2 + tm n 1 M n 2 holds for the extended sequence for any t Z The other identities that will be needed can be listed as follows: M 2 n M n 1M n+1 = 1, (n Z, (23 M n+h = M n+1 M h M n M h 1, (n,h Z, (24 M 2n = M n (M n+1 M n 1, (n Z, (25 Mn 1 2 (2 + tm n 1M n + Mn 2 = 1, (n Z, (26 C n = M n+1 M n 1 = α n + β n, (n Z, (27 M 2n = M n C n, (n Z, (28 M nk+r = M n = ( 1 2 n ( n ( 1 n j M j k j Mn j k 1 M r+j, (n,k 0,r Z, (29 n 1 n 1 2 ( n (2 + t n (2j+1 j, (n 1, (210 2j + 1 ( M an = 2 1 a M n Mn 2 K + aca 1, for a 1 and some integer K (211 n Identity (26 follows from identities (11 and (23, identity (25 follows from (24 and identity (28 comes from identities (25 and (27 From identity (27, we get that C n = C n for any integer n From identity (23, we deduce that gcd(m n 1,M n = 1 Identities (11 and (27 givec n = (2 + tm n 2M n 1, from which gcd(c n,m n = 1 or 2 is deduced Identity (210 is easy to verify if t 0, 4} For = 0 and n 1, we have from (21, M n = = = 1 2 n ([(2 + t + ] n t[(2 + t ] n 1 n ( n 2 n (2 + t n k ( k ( 1 k k k ( 1 2 k=0 n 1 n 1 2 ( n (2 + t n (2j+1 j 2j + 1 To prove identity (211, assume that t 0, 4} From identities (21 and (27, we have Mn = α n β n C n = α n + β n Mn + C n = 2α n, M n + C n = 2β n, α n = 2 1 ( M n + C n, β n = 2 1 ( M n + C n

5 p-adic valuation of the Morgan Voyce sequence and p-regularity 239 Thus we deduce from (21, that for an integer a 1, M an = αan β an = 2 a ( M n + C n a 2 a ( M n + C n a a ( a a = 2 a = 2 a j odd = 2 1 a j odd j ( a j ( a j = 2 1 a M n [ac a 1 n j j M n Cn a j 2 j M j n C a j n (j 1/2 M j n C a j n + M 2 n K], ( a j ( 1 j j M j n C a j n for some integer K Ift = 0or 4, it is easy to see that identity (211 is verified for K = 0, since for t = 0wehaveM n = n and C n = 2forn 0 For t = 4, we have M n = ( 1 n+1 n and C n = ( 1 n 2forn 0 For the other identities, see [1, 15, 27] We will sometimes need to divide by M n, so let us analyse when this is possible, ie, when M n = 0 First, note the following three special cases where the sequence (M n n 0 is periodic: (1 If t = 1, the sequence (M n n 0 is 0, 1, 1, 0, 1, 1, 0, 1, Thus,wehave M 3k = 0, M 6k+1 = M 6k+2 = 1 and M 6k+4 = M 6k+5 = 1 (2 If t = 2, the sequence (M n n 0 is 0, 1, 0, 1, 0, 1, Thus,wehaveM 2k = 0, M 4k+1 = 1 and M 4k+3 = 1 (3 If t = 3, the sequence (M n n 0 is 0, 1, 1, 0, 1, Thus,wehaveM 3k = 0, M 3k+1 = 1 and M 3k+2 = 1 When t Z \ 1, 2, 3}, wehavem n = 0 if and only if n = 0 Indeed, for n 1, it is easy to see from (12 that M n+1 M n = n ( n+k k=0 n k t k We deduce that, for t 0, we have M n+1 >M n for n 0 In particular, we have M n 2forn 2 Using identity (22 we deduce that M n = 0forn = 0 and M m = M n for m = n We obtain from (11 that M n 2 + t M n 1 M n 2 for n 2 From this, we can prove by induction, for t 4, that M n+1 > M n for n 0 In particular, we obtain M n 2forn 2 Using identity (22 we deduce that M n = 0forn = 0 and M m = M n for m = n Still for t 4, we have from identity (26 that sign(m n+1 = sign(m n for n 1 and we deduce for k>0that M 2k 1 > 0 and M 2k < 0 For the extended sequence we have M 2k+1 = M 2k 1 < 0 and M 2k = M 2k > 0 If we put r = 0 in identity (29, we get n ( n M nk = M k ( 1 n j M j 1 k M n j j k 1 M j, (212 j=1 from which we deduce, using identity (22, that M k M kn, for k, n Z with M k = 0 (213

6 240 Lyes Ait-Amrane PROPOSITION 21 Let m and n be integers such that (M m,m n = (0, 0 Then gcd(m m,m n = M gcd(m,n Proof Obvious if t 1, 2, 3} Thus, assume that t 1, 2, 3}, in this case the condition (M m,m n = (0, 0 is equivalent to (m, n = (0, 0 First, assume that m and n are two positive integers and let g = gcd(m, n and G = gcd(m m,m n There exists integers α and β (not both negative such that g = αm + βn Assume that α 0, then from (29 wehave α ( α M g = M αm+βn = ( 1 α j MmM j α j j m 1 M βn+j 0modG, since from (213 wehavem n divides M βn and G divides both M m and M n Thus M g is divisible by G Since g divides m and n, wehavefrom(213 that M g divides M m and M n Thus G is divisible by M g We deduce that G = M g Identity (22 allows us to extend this result to m and n in Z not both zero Remark 1 If t 0, then G = M g, while if t 4, then G = M g if g is odd and G = M g if g is even For the following corollary, assume that t Z \ 1, 2, 3} COROLLARY 211 Let m and n be integers such that m = 0 IfM m M n then m n Proof If M m M n, then from Proposition 21 we have M gcd(m,n =gcd(m m,m n = M m Thus, since for t 1, 2, 3} we have M n+1 > M n for n 0 and M n = M n, we deduce that gcd(m, n = m, hence m n Remark 2 From identity (213 and Corollary 211, we deduce that in the case where t Z \ 1, 2, 3}, wehaveforn = 0, m n M m M n When t 1, 2, 3}, we have the next theorem which is an analogy of Matijasevich s lemma [13] for the Morgan Voyce case Theorem 22 Let n 2 be an integer, then M kn 0modM 2 n k 0modM n Proof Since for t Z \ 1, 2, 3} we have M n 2forn 2, we get for n 2 that M n mod M 2 n = M n

7 p-adic valuation of the Morgan Voyce sequence and p-regularity 241 Then, we have from identity (24 that M 2n = M n M n+1 M n 1 M n = M n M n+1 [(2 + tm n M n+1 ]M n 2M n M n+1 mod M 2 n and M 2n+1 = M 2 n+1 M2 n M2 n+1 mod M2 n These last two congruences with identity (24 allow us to calculate M 3n = M n M 2n+1 M n 1 M 2n M n M 2 n+1 M n 1(2M n M n+1 mod M 2 n M n M 2 n+1 [(2 + tm n M n+1 ](2M n M n+1 mod M 2 n 3M n M 2 n+1 mod M2 n and M 3n+1 = M 2n+1 M n+1 M 2n M n M 3 n+1 (2M nm n+1 M n mod M 2 n M 3 n+1 mod M2 n Then we can show by induction on k 1 that we have and M kn km n M k 1 n+1 mod M2 n (214 M kn+1 M k n+1 mod M2 n (215 Now, since gcd(m n,m n+1 = 1, we have M kn 0modM 2 n km n 0modM 2 n k 0modM n Lemma 21 Let p be an odd prime If p (2 + t, then p divides only one of M p 1, M p and M p+1 ; namely, M p ( /p, where ( /p is the Legendre symbol If p (2 + t, then p M p+1, p M p 1 and p M p Proof Let p be an odd prime From identity (210 wehave (p 1/2 ( p (2 + t p (2j+1 j = 2 p 1 M p M p mod p 2j + 1 Since p divides ( p 2j+1 for 0 j<(p 1/2, we deduce that M p (p 1/2 mod p (216

8 242 Lyes Ait-Amrane But we know that (p 1/2 ( /p mod p (Proposition 512 of [16], we deduce that p divides M p if and only if ( /p = 0, ie, when p divides (note that p implies that p (2 + t Again according to identity (210, we have (p 1/2 Since ( p+1 ( 2j+1 = p ( p + 1 (2 + t p 2j j = 2 p M p+1 2M p+1 mod p 2j + 1 2j+1 + ( p 2j and p divides ( p l for 0 <l<p, we get 2M p+1 (2 + t ( 1 + (p 1/2 mod p (217 We deduce from (11, (216 and (217 that ( 2M p 1 (2 + t (p 1/2 1 mod p (218 If p (2 + t, then we get from identities (216, (217 and (218 that p M p 1, p M p+1 and p M p This case can also be deduced from the following identity, which can be found in Proposition 21(a of [1], M n = n 1 2 k=0 ( 1 k ( n k 1 k (2 + t n 2k 1, (n 1 Now assume that p (2 + t and p, then we have from (23 and (216 that M p 1 M p+1 = M 2 p 1 0modp We deduce from (11 that p divides only one of M p 1 and M p+1 From(218 we get M p 1 0modp (p 1/2 1modp ( /p = 1 From (217 we get M p+1 0modp (p 1/2 1modp ( /p = 1 Theorem 23 Let p be an odd prime, then d(p (p ( /p Proof If p (2 + t, the result follows from (14 and Lemma 21 If p (2 + t, then the sequence (M n mod p n 0 is 0, 1, 0, We deduce that d(p = 2 Since by hypothesis p>2and p (2 + t,wehavep (t 2 + 4t =, hence ( /p =±1 We deduce that d(p = 2 (p ( /p = p ± 1 3 p-adic valuation For a prime p and an integer s, we denote the p-adic order of s by ν p (s, ie, the exponent of the highest power of p which divides s

9 p-adic valuation of the Morgan Voyce sequence and p-regularity 243 Theorem 31 Let l = d(2, that is, l = 2 if 2 t and l = 3 if 2 t, then and ν 2 (M n = ν 2 (C n = ν2 (n + ν 2 (M l ν 2 (l if n 0modl 0 if n 0modl, 1 if n 0modl ν 2 (2 + t if n 0modl (31 (32 Proof From (14 we know that 2 M n if and only if l n, from this we deduce part two of (31 To prove the first part of (31, assume that n 0modl Letj = ν 2 (n/l = ν 2 (n ν 2 (l and let us write n = 2 j+1 kl + 2 j l for a nonnegative integer k Wehaveto prove that ν 2 (M 2 j+1 kl+2 j l = ν 2(M l + j, for k, j 0 For j = 0, let us show by induction on k that ν 2 (M 2kl+l = ν 2 (M l, for k 0 This is obvious for k = 0 Assume that we have ν 2 (M 2kl+l = ν 2 (M l for an integer k 0 and let us prove that ν 2 (M 2(k+1l+l = ν 2 (M l Using identity (24, we get ν 2 (M 2(k+1l+l = ν 2 (M (2kl+l 1+(2l+1 = ν 2 (M 2kl+l M 2l+1 M 2kl+l 1 M 2l, since ν 2 (M 2kl+l 1 = ν 2 (M 2l+1 = 0,ν 2 (M 2l = ν 2 (M l +1 and by induction hypothesis ν 2 (M 2kl+l = ν 2 (M l, we obtain that ν 2 (M 2(k+1l+l = ν 2 (M l Next, let k 0beany integer and assume that ν 2 (M 2 j+1 kl+2 j l = ν 2(M l + j for an integer j 0 and let us prove that ν 2 (M 2 j+2 kl+2 j+1 l = ν 2 (M l + j + 1 Using identities (11 and (25, we get ν 2 (M 2 j+2 kl+2 j+1 l = ν 2(M 2(2 j+1 kl+2 j l = ν 2 (M 2 j+1 kl+2 j l + ν 2(M 2 j+1 kl+2 j l+1 M 2 j+1 kl+2 j l 1 = ν 2 (M l + j + ν 2 [(2 + tm 2 j+1 kl+2 j l 2M 2 j+1 kl+2 j l 1 ] = ν 2 (M l + j + 1 To prove (32, we use identity (28 from which we deduce that ν 2 (C n = ν 2 (M 2n ν 2 (M n Assume that l n and let n = lh, then ν 2 (C n = ν 2 (M 2lh ν 2 (M lh = ν 2 [M lh (M lh+1 M lh 1 ] ν 2 (M lh = ν 2 (M lh + ν 2 [(2 + tm lh 2M lh 1 ] ν 2 (M lh = ν 2 [(2 + tm lh 2M lh 1 ] If 2 t, then l = 2, ν 2 [(2+tM 2h ] 2 and ν 2 (2M 2h 1 = 1 We deduce that ν 2 (C n = 1 If 2 t, then l = 3 In this case, it is easy to see that ν 2 (M 3 2 and we distinguish two cases If t 1, 3}, then from (213 we get ν 2 (M 3h 2 and if t 1, 3}, then M 3h = 0 Since ν 2 (2M 3h 1 = 1, we deduce that ν 2 (C n = 1

10 244 Lyes Ait-Amrane Now assume that l n, then ν 2 (C n = ν 2 (M 2n Ifl = 2 then n is odd, say n = 2h + 1 Thus ν 2 (C n = ν 2 (M 4h+2 = ν 2 (M 2 = ν 2 (2 + t If l = 3, then 3 2n Thus, since in this case 2 t,wehave ν 2 (C n = ν 2 (M 2n = 0 = ν 2 (2 + t Theorem 32 Let p be a prime such that p Ifp 5, then ν p (M n = ν p (n for all n 0 Ifp 3, then ν p (C n = 0 for all n 0 Proof This is obvious for n = 0 Thus, assume that n 1 The hypothesis p implies that p (2 + t, we deduce that for k 1, we have (( (( n n 1 ν p (2 + t n (2k+1 k = ν p (n ν p (2k ν p k 2k + 1 2k ν p (n ν p (2k k >ν p (n and for k = 0, we have (( n ν p (2 + t n (2k+1 k = ν p (n 2k + 1 Now, identity (210 impliesthatν p (M n = ν p (n The period of the sequence (C n mod p n 0 is 2 with the cycle 2, 2 + t}; therefore, p can never be a divisor of C n Remark 3 When p = 3, it is not always true that k ν p (2k + 1 >0fork 1 Indeed, for k = 1, we have k ν p (2k + 1 = 0 and therefore the above proof does not work for p = 3 To be more precise, it does not work when one has ν 3 ( = 1 as one can see, for example, when t = 6 In this case we have ν 3 ( = ν 3 (60 = 1 and ν 3 (M 3 = ν 3 ( + 3 = ν 3 (63 = 2 = 1 = ν 3 (3, also,ν 3 (M 6 = ν 3 (30744 = 2 = 1 = ν 3 (6 To avoid complications, we use another method in the case where p = 3 through which we get the following theorem that complements the above theorem Theorem 33 Assume that 3, then ν 3 (M n = ν3 (n + ν 3 (M 3 1 if n 0mod3, 0 if n 0mod3 (33 Proof Since 3, we get that d(3 = 3 and from identity (14 we deduce that 3 M n if and only if 3 n Thus, to prove the first part of (33, assume that n 0 mod 3 Let j = ν 3 (n 1 and let us write n = 3 j k with 3 k We have to prove, by induction on j, that ν 3 (M 3 j k = ν 3(M 3 + j 1, j 1

11 p-adic valuation of the Morgan Voyce sequence and p-regularity 245 For j = 1, let us show that ν 3 (M 3k = ν 3 (M 3 From Theorem 32, we know that ν 3 (C n = 0 for all n 0 Thus, from identity (211 we deduce that ν 3 (M 3k = ν 3 [2 1 k M 3 (M3 2 K + kck 1 3 ] =ν 3 (M 3 Now assume that ν 3 (M 3 j k = ν 3(M 3 + j 1 for an integer j 1, then from identity (211, we get ν 3 (M 3 j+1 k = ν 3[2 2 M 3 j k (M2 3 j k K + 3Ck 1 3 ] = ν 3 (M 3 j k + 1 = ν 3 (M 3 + j Remark 4 If ν 3 ( > 1, then ν 3 (M 3 = ν 3 ( +3 = 1 Thus, we obtain ν 3 (M n = ν 3 (n as in Theorem 32 for p 5 Let e = ν p (M d(p, hence M d(p 0modp e and M d(p 0modp e+1 Note that d ( p i = d(p for all 1 i e We have gcd(p, d(p = 1 Indeed, if p d(p then from Theorem 23, we get for p that p p ± 1, which is a contradiction Theorem 34 Let p be an odd prime such that p and n be a positive integer, then and ν p (M n = ν p (C n = νp (n + ν p (M d(p if n 0modd(p, 0 if n 0modd(p, νp (n + ν p (M d(p if d(p n and d(p 2n, 0 otherwise (34 (35 Proof If d(p n then from identity (14 we deduce that ν p (M n = 0 Assume that d(p n and let n = cd(pp α, where c 1 and α 0 are integers with gcd(c, p = 1 If α = 0, identity (214 gives M n = M cd(p cm d(p M c 1 d(p+1 mod p2e (36 From identity (23 we have M 2 d(p M d(p 1M d(p+1 = 1 We deduce that gcd(p, M d(p+1 = 1 Since ν p (c = ν p (M d(p+1 = 0, we deduce from identity (36 that ν p (M n = ν p (M cd(p = ν p (M d(p = ν p (n + ν p (M d(p If α 1, identity (211 impliesthat M cd(pp α = 2 1 p M cd(pp α 1(M 2 cd(pp α 1 K + pc p 1 cd(pp α 1 = 2 1 p M cd(pp α 1(p 2 K + pc p 1, cd(pp α 1 for some integer K Since gcd(c n,m n = 1 or 2, we get ν p (M cd(pp α = ν p (M cd(pp α 1 + 1,

12 246 Lyes Ait-Amrane and inductively ν p (M cd(pp α = ν p (M cd(p + α (37 Since ν p (M cd(p = ν p (M d(p and α = ν p (n, we get from identity (37 that ν p (M n = ν p (n + ν p (M d(p Now we use identity (28 to prove (35 The following three cases can be distinguished: either d(p n, ord(p n and d(p 2n, ord(p n and d(p 2n In the first case we have ν p (M 2n = ν p (2n + ν p (M d(p = ν p (n + ν p (M d(p = ν p (M n, which gives ν p (C n = 0 In the second case we have ν p (M 2n = ν p (M n = 0, thus ν p (C n = 0 In the third case d(p is necessarily even and we can write n = sd(p/2 where s is an odd integer Thus, since ν p (M n = 0 and gcd(p, d(p = 1, we have ν p (C n = ν p (M 2n = ν p (M sd(p = ν p (s+ν p (M d(p = ν p (n+ν p (M d(p 4 p-regularity Recall that for an integer k 2 and a ring R containing a commutative Noetherian ring R, Allouche and Shallit [2] defined k-regular sequences A sequence (S n n 0 with values in R is said to be (R,k-regular, or just k-regular, if the R -module generated by its k-kernel, which is defined by ker k (S n n 0 := (S k e n+i n 0 e 0 and 0 i k e 1}, is a finitely generated R -submodule of S (R, where S (R is the set of sequences with values in R The rank of a k-regular sequence is the rank of this R -module Allouche and Shallit (Theorem 22 of [2] gave several alternative characterizations of k-regular sequences The one we will use is the following: A sequence (S n n 0 is k-regular if and only if there exist an integer r and r sequences (S n n 0 = (S 1,n n 0,(S 2,n n 0,,(S r,n n 0 such that for 1 i r the k-sequences (S i,kn+a n 0, 0 a k 1areR -linear combinations of (S i,n n 0 We will need the following result in the remainder of this section Theorem 41 Let k 2 be an integer, then (ν k (n + 1 n 0 is a k-regular sequence of rank 2 Proof See Theorem 15 of [21] We now begin our study of p-regularity of the sequence (ν p (M n+1 n 0 We will show, by differentiating several cases and using results of the previous sections, that the sequence (ν p (M n+1 n 0 is (Z,p-regular and we determine its rank in each case Theorem 42 Let p be a prime such that p and assume that ν 3 (M 3 = 1 for p = 3 and ν 2 (M 2 = 1 for p = 2 Then (ν p (M n+1 n 0 is a p-regular sequence of rank 2

13 p-adic valuation of the Morgan Voyce sequence and p-regularity 247 Proof Follows from Theorems 31, 32, 33 and 41 Note that from Theorem 31, if ν 2 (M 2 = 1 then ν 2 (M n = ν 2 (n for all n 0 and from Theorem 33, if ν 3 (M 3 = 1 then, ν 3 (M n = ν 3 (n for all n 0 Theorem 43 The sequence (ν 2 (M n+1 n 0 is 2-regular of rank 4 if 2 t and of rank 3 if 2 t and ν 2 (M 2 >1 Proof First, assume that 2 t and ν 2 (M 2 >1 Then from Theorem 31, we have for all n 0, ν2 (n ν ν 2 (M n+1 = 2 (M 2 1 if n + 1 0mod2, 0 if n + 1 0mod2 Consider the set B =(ν 2 (M n+1 n 0, (ν 2 (M 2n+2 n 0, 1 = (1 n 0 } Then we have ker 2 (ν 2 (M n+1 n 0 = B Indeed, we have for all n 0, ν 2 (M (2n+0+1 = 0 ν 2 (M (2n+1+1 = ν 2 (M 2n+2, ν 2 (M 2(2n+0+2 = ν 2 (M 2 ν 2 (M 2(2n+1+2 = ν 2 (M 2n We deduce that for each sequence (S n n 0 B,wehave(S 2n+a n 0 B for 0 a 1 Finally, one checks easily that the elements of B are linearly independent We now assume that 2 t, then from Theorem 31, we have for all n 0, ν2 (n ν 2 (M 3 if n 2mod3, ν 2 (M n+1 = 0 ifn 2mod3 Consider the set B =(ν 2 (M n+1 n 0 } (u j,n n 0, 0 j 2}, where 1 ifn j mod 3, u j,n = 0 ifn j mod 3 We claim that ker 2 (ν 2 (M n+1 n 0 = B We have for all n 0, ν 2 (M (2n+0+1 = ν 2 (M 3 u 1,n ν 2 (M (2n+1+1 = ν 2 (M n+1 + u 2,n, u 0,2n = u 0,n, u 0,2n+1 = u 1,n, u 1,2n = u 2,n, u 1,2n+1 = u 0,n, u 2,2n = u 1,n, u 2,2n+1 = u 2,n This shows that for each sequence (S n n 0 B,(S 2n+a n 0 B for each 0 a 1 The fact that the sequences in B are linearly independent is left to the reader Theorem 44 If ν 3 (M 3 >1, then the sequence (ν 3 (M n+1 n 0 is 3-regular of rank 3 Proof From Theorem 33, we have for all n 0, ν3 (n ν 3 (M 3 1 ifn + 1 0mod3, ν 3 (M n+1 = 0 ifn + 1 0mod3

14 248 Lyes Ait-Amrane Consider the set B =(ν 3 (M n+1 n 0,(ν 3 (M 3n+3 n 0, 1 = (1 n 0 } Then we have ker 3 (ν 3 (M n+1 n 0 = B Indeed, we have for all n 0, ν 3 (M (3n+0+1 = 0 ν 3 (M (3n+1+1 = 0 ν 3 (M (3n+2+1 = ν 3 (M 3n+3 ν 3 (M 3(3n+0+3 = ν 3 (M 3 ν 3 (M 3(3n+1+3 = ν 3 (M 3 ν 3 (M 3(3n+2+3 = ν 3 (M 3n We deduce that for each sequence (S n n 0 B,wehave(S 3n+a n 0 B for 0 a 2 The fact that the sequences in B are linearly independent is left to the reader Theorem 45 Let p be an odd prime such that p, then (ν p (M n+1 n 0 is a p-regular sequence of rank d(p + 1 Proof Since we have Theorem 23 and Theorem 34, the same proof as in the Fibonacci case holds (see Theorems 23, 33 and 34 of [21] References [1] Ait-Amrane L, Belbachir H and Betina K, Periods of Morgan Voyce sequences and elliptic curves, to appear in Math Slovaca [2] Allouche J P and Shallit J, The ring of k-regular sequences, Theoret Comput Sci 98(2 ( [3] Belbachir H, Komatsu T and Szalay L, Characterization of linear recurrences associated to rays in Pascal s triangle in: Diophantine analysis and related fields 2010, AIP Conf Proc (2010 (Melville, NY: Amer Inst Phys vol 1264, pp [4] Belbachir H, Komatsu T and Szalay L, Linear recurrences associated to rays in Pascal s triangle and combinatorial identities, Math Slovaca 64(2 ( [5] Christol G, Kamae T, Mends France M and Rauzy G, Suites algbriques, automates et substitutions, Bull Soc Math France 108(4 ( [6] Clarke F, Hensel s lemma and the divisibility by primes of Stirling-like numbers, J Number Theory 52(1 ( [7] Cobham A, On the base-dependence of sets of numbers recognizable by finite automata, Math Systems Theory 3 ( [8] Cobham A, Uniform tag sequences, Math Systems Theory 6 ( [9] Davis D M, Divisibility by 2 of Stirling-like numbers, Proc Amer Math Soc 110(3 ( [10] Ferri G, Faccio M and Amico D A, A new numerical triangle showing links with Fibonacci numbers, Fibonacci Quart 29(4 ( [11] Ferri G, Faccio M and Amico D A, Fibonacci numbers and ladder networks impedance, Fibonacci Quart 30(1 ( [12] Gessel I, Congruences for Bell and tangent numbers, Fibonacci Quart 19(2 ( [13] Graham R L, Knuth D E and Patashnik O, Concrete mathematics, A foundation for computer science, second edition (1994 (Reading, MA: Addison-Wesley Publishing Company [14] Halton J H, On the divisibility properties of Fibonacci numbers, Fibonacci Quart 4 ( [15] Horadam A F, New aspects of Morgan-Voyce polynomials in: Applications of Fibonacci numbers (1998 (Dordrecht, Graz, 1996: Kluwer Acad Publ vol 7, pp [16] Ireland K and Rosen M, A classical introduction to modern number theory, Second edition, Graduate Texts in Mathematics, No 84 (1990 (New York: Springer-Verlag

15 p-adic valuation of the Morgan Voyce sequence and p-regularity 249 [17] Kwong Y H H, Periodicities of a class of infinite integer sequences modulo M, J Number Theory 31(1 ( [18] Lengyel T, On the divisibility by 2 of the Stirling numbers of the second kind, Fibonacci Quart 32(3 ( [19] Lengyel T, The order of the Fibonacci and Lucas numbers, Fibonacci Quart 33(3 ( [20] Lundell A T, A divisibility property for Stirling numbers, J Number Theory 10(1 ( [21] Medina L A and Rowland E, p-regularity of the p-adic valuation of the Fibonacci sequence, Fibonacci Quart 53(3 ( [22] Morgan-Voyce A M, Ladder network analysis using Fibonacci numbers, IRE Trans Circuit Theory 6 ( [23] Nijenhuis A and Wilf H S, Periodicities of partition functions and Stirling numbers modulo p, J Number Theory 25(3 ( [24] Ribenboim P, The little book of big primes (1991 (New York: Springer-Verlag [25] Robinson D W, The Fibonacci matrix modulo m, Fibonacci Quart 1(2 ( [26] Sved M, Divisibility-with visibility, Math Intelligencer 10(2 ( [27] Swamy M N S, Properties of the polynomials defined by Morgan Voyce, Fibonacci Quart 4(1 ( [28] Vince A, Period of a linear recurrence, Acta Arith 39(4 ( [29] Wall D D, Fibonacci series modulo m, Amer Math Monthly 67 ( [30] Wilcox H J, Fibonacci sequences of period n in groups, Fibonacci Quart 24(4 ( COMMUNICATING EDITOR: S D Adhikari