ON THE LEAST SIGNIFICANT p ADIC DIGITS OF CERTAIN LUCAS NUMBERS

#A13 INTEGERS 14 (014) ON THE LEAST SIGNIFICANT ADIC DIGITS OF CERTAIN LUCAS NUMBERS Tamás Lengyel Deartment of Mathematics, Occidental College, Los Angeles, California lengyel@oxy.edu Received: 6/13/13, Acceted: /4/14, Published: 3/7/14 Abstract We calculate the least significant -ary digits of certain Lucas numbers V n V n (P, Q) with V 0, V 1 P and V n P V n 1 QV n for n. We base our study on an observation regarding these numbers: as m increases, more and more -adic digits match in V k m with integer k 1. We use multisection identities for generating functions and derive congruences for the underlying sequences. 1. Introduction Let n and k be ositive integers, and be a rime, and let (k) denote the highest ower of dividing k, i.e., the -adic order of k. The sequences U n U n (P, Q) with U 0 0, U 1 1 and U n P U n 1 QU n and V n V n (P, Q) with V 0, V 1 P and V n P V n 1 QV n for n are called Lucas (or Lucas and comanion Lucas, resectively) sequences associated with the air (P, Q) (cf. [7]), with U n and V n considered the generalization of the original Fibonacci and Lucas sequences, i.e., F n U(1, 1) and L n V (1, 1), resectively. Usually, we use the short notations U n and V n excet if exlicitly secifying the arameters might be helful. The corresonding generating functions are and U(x) V (x) U n x n x 1 P x + Qx n0 V n x n P x 1 P x + Qx. n0 We define the characteristic olynomial x P x + Q 0 associated with the Lucas sequences U n and V n, n 0. Its discriminant is D D(P, Q) P 4Q. If 3 and - P QD then we define the function () (, P, Q) where D

INTEGERS 14 (014) D is the Legendre symbol (cf. [7]), i.e., it is 1 if D is a quadratic residue modulo and 1 otherwise. We define the rank of aarition (or Fibonacci entry oint or restricted or fundamental eriod modulo n) (n) (n, P, Q) of integer n for the Lucas sequence U k (P, Q), k 0: if there exists m 1 such that n divides U m then (n) is the smallest such m (cf. [7]). Since Q 1, by observation (IV.19) in [7] we know that () exists for every odd rime. The same alies to by fact (IV.18) in [7] and all rime owers by fact (IV.0), i.e., ( e ) exists for any integer e 1. The determination of the exact ower of (F (e )) is discussed in [3] and that of (U (e )(P, Q)) in terms of (U () (P, Q)) follows by (IV.0). By (IV.9) we also know that n U m if and only if (n) m. (1.1) We define the modulo eriod (cf. []) or Pisano eriod modulo, () (, P, Q), as the smallest m so that U n+m (P, Q) U n (P, Q) (mod ) for every n 0. It is known that () (). We will use a basic relation among the members of a Lucas sequence V n V n Q n. (1.) We also rely on fact (IV.10) in [7], which rovides the identity n n n 1 V n P n + P n D + P n 4 D +... (1.3) 4 and its comanion identity n n 1 U n P n 1 + 1 n P n 3 D + 3 n P n 5 D +.... (1.4) 5 Our main goal is to rove that as m increases, more and more -adic digits match in V k m for di erent integers k 1, and establish the rate at which these digits match. Similar investigations have been done for the Stirling numbers of the second kind in [4] and [5], and Motzkin numbers in [6]. In Section we state the main Theorems 4, 5, 7, and 10. Their roofs are included in Section 3.. Main Results In [8], an alication of -Honda sequences is mentioned in Section 3.4. It can be summarized as follows. Proosition 1 (Beukers). Let M be a d d matrix with integer coe Define a n Tr(M n ) and cients. Z () {a/b : a Z, 0 6 b N, and b rime to }.

INTEGERS 14 (014) 3 Then for any rime, the sequence a n, n 1, is a -Honda sequence with a m a m/ (mod mz () ) if m. (.1) Corollary. The Lucas sequence L 0, L 1 1, L n+1 L n + L n 1 (n 1), is a -Honda sequence for any rime. We can easily extend this result and its roof (cf. sequence V n (P, Q), n 0. [8]) for the general Lucas Corollary 3. The Lucas sequence V n (P, Q) with V 0, V 1 P, V n+1 P V n QV n 1 (n 1), is a -Honda sequence for any rime. The following theorem follows by Corollary 3 and congruence (.1) after setting m k n+1. Theorem 4. For integers n 0 and k 1, we have that (V k n+1(p, Q) V k n(p, Q)) n + 1. The main goal of this aer is to find cases with Q 1 when Theorem 4 can be strengthened and the exact order (V k n+1(p, 1) V k n(p, 1)) and its congruential form can be determined. The results in this direction are summarized in Theorems 5 and 10, which deal with the -adic orders and high ower modulus congruences of the di erences, in articular for. Theorem 10 extends Theorem 5 from k 1 to other values and resents the cases in their congruential forms. We combine their roofs in Section 3. Theorem 5. For the Lucas sequence V k V k (P, 1), k 0, we set r m () (V m+1 V m), e m () (U (V m+1, 1)) and e m () (U (V m+1, 1). Let D P + 4 be the discriminant of the sequence. For 3 odd and m 0 we get V k m+1 V m+1 V k m V m ku k (V m+1(p, 1), 1) mod rm(), (.) while for and m V k m+1 V m+1 1 we get V k m V m ku k (V m+1(p, 1), 1) mod rm(). (.3)

INTEGERS 14 (014) 4 For any rime, if - D then e m () 0 for all integers m 0. In this case, if 3 then and if P holds too then while if then (V m+1 V n) m 1 + (V V ), for m 1, (.4) (V m+1 V m) m + (P ); (V m+1 V m) m 1 + (V 4 V ) m + 1, for m 1. (.5) If 3 is an odd rime and D then e m () 1 for all m 0. If m 1 then thus r m () > r m () + 1 + e m () r m () +, and r m () > 1 + e m (), (.6) (V m+1 V m) (m 1) + (V V ). (.7) If and D, i.e., P, then e m () 1 for all m 0. If m then r m () >, and (V m+1 V m) (m ) + (V 8 V 4 ) (m 1 + (P )) + 4-P. (.8) Remark 6. In Theorem 5 we need the smallest m that guarantees r m () > 1 + e m () (.9) and thus, the alicability of (.) and (.3) with k. By Theorem 4, we immediately get that r () 3, hence, m will be su ciently large in Theorem 5, however, if - D then r 1 () > 1 + e 1 () 1 and m 1 su ces. If D then we have two cases. If then it is easy to find examles when the smallest m is, e.g., if P. However, we could not find any such case when 3, and in fact, m 1 is su ciently large. It turns out that if D P + 4 then r 0 (), which makes r 1 () 4 > 1 + e 1 (). It also aears that if we relace D with D and even if r 0 () (V V 1 ) 1, these assumtions already imly that r 1 () (V V ) 3 > 1 + e 1 (). Thus, clearly m 1 is su cient in (.7) of Theorem 5 as stated in Theorem 7. Theorem 7. With the notation of Theorem 5, if 3 and D then r 0 () and r 1 () 4 > 1 + e 1 (). If D then r 1 () 3 > 1 + e 1 (). On the other hand, if - D then r 1 () > 1 + e 1 () 1. In each case, m 1 is su ciently large in Theorem 5. If then m is required exactly if (P ) 1.

INTEGERS 14 (014) 5 Remark 8. According to Theorem 5, if or 3 with P then the -adic orders can be determined without calculating any value of the Lucas sequence. Remark 9. There are two frequently occurring cases. If r 0 () (V V 1 ) 1 and e 0 () (U (V, 1)) 0 then (.4) imlies that (V n+1 V n) n + 1, for n 0 (.10) with r 0 () 1 taking care of the case n 0. If r 1 () (V V ) 3 (cf. Theorem 7), and e 1 () (U (V, 1)) 1 then the inequality (.6) is satisfied, and (.7) imlies that (V n+1 V n) n + 1, for n 1. (.11) Examle 1. With P 1 (and Q 1) we work with the original Lucas and Fibonacci sequences. If 5 then (5) 5, and we get that r 0 (5) 1, r 1 (5) 3, e 0 (5) e 1 (5) 1, and 5 (V 5 n+1 V 5 n) n + 1 for n 0, in fact, it follows by (.11) for n 1 and by r 0 (5) 1 for n 0. Theorem 10. With the notations of Theorem 5, for k 1 we get the following congruences and consequently, the aroriate -adic orders. If 3 and D then with m 1, we have that V k m+1 8 k 1 k (mod 1) P k 1 m (V V ) mod m 1+ (V V), >< if gcd(k, ) 1, V k m >: m (V V ) mod m+1+ (V V), if k, with the convention that 0 ale a (mod b) < b. If 3, - D, and - P then for m 1, we get that 8 m 1 m 1 D m+ (V V )ku k (V m+1, 1) mod (V V ), >< if gcd(k, ) 1 and V k m+1 V k m gcd(k, (, V m+1, 1)) 1, m m D (V V ) mod m+1+ (V V ), >: if k. If 3, - D, and P then for m 1, we have that V k m+1 8 m 1 (V V )ku k (V m+1, 1) mod m+ (V V), >< if gcd(k, ) 1 and k odd, V k m >: m (V V ) mod m+1+ (V V), if k.

INTEGERS 14 (014) 6 If and D, i.e., P, then with k 1 odd and m s, we have that V k m+1 V k m (m s) (V s+1 V s)ku k (V m+1, 1) mod (m s 1+rs()). If, - D, and gcd(k, 6) 1 then for m s 1, we get V k m+1 V k m ( ) m s (V s+1 V s)ku k (V m+1, 1) mod m+s+1. Remark 11. In the congruences of Theorem 10, the di erence between the - adic order of the right-hand side quantity and that of the modulus is 1 for 3, r s () s 4 + (P ) + 4-P s 1 3 if P (by (3.1)), and s 1 if - P. 3. Proofs P Q Proof of Corollary 3. We aly Proosition 1 to the matrix M. 1 0 The characteristic olynomial of M is x P x + Q, hence M P M + Q 0 by the Hamilton Cayley theorem. We deduce M n+ P M n+1 QM n (n 0). Since Tr(M 0 ) Tr(I ) and Tr(M) P, this roves that V n Tr(M n ) is the mentioned Lucas sequence. In the roofs of the main theorems we need some auxiliary results. The first result is due to Lehmer, cf. Theorem 1.6 in [1]. We need only its version stated as art of the facts (IV.0) and (IV.7) in [7]. Theorem 1. Let e 1, and let e be the exact ower of dividing U k. If - r and f 0 then e+f divides U kr f. Moreover, if - Q and e 6, then e+f is the exact ower of dividing U kr f. From now on we focus on the cases with Q will require the use of U k with Q 1). 1 only (although a subcase with Theorem 13. If P is odd and k is an odd multile of 3 then we have that (U k (V m+1(p, 1), 1)) m + 3, for all m 0. If P is odd then U (V m+1(p, 1), 1) V m+1(p, 1) 1 (mod m+ ) for all m 0, while if P is even then (U (V m+1(p, 1), 1) ) (V m+1(p, 1) ) (m 1 + (P )) + (V (P, 1) + ) (m + (P )) for all m 1, and thus, U (V m+1(p, 1), 1) (mod (m+ (P )) ) for all m 0. For an odd rime, if P, k even, and gcd(k, ) 1 then we have that U k (V m+1(p, 1), 1) m + 1 + (P ), for all m 0. Proof of Theorem 13. For any odd P and with V k V k (P, 1), we get that V 1 (P + ) 1 8 mod 16, and thus, (V 1) 3. By induction we can rove

INTEGERS 14 (014) 7 that (V m+1 1) m + 3 for any m 0. In fact, we just demonstrated the case with m 0. For the inductive ste with m 0, we use identity (1.): V m+ 1 (V m+1 ) 1 (V m+1 1)(V m+1 3), hence, (V 1) m + 4 by the inductive hyothesis. m+ We note that U 3 (P, 1) P 1. Since U 3 (V m+1, 1) V 1, it now follows m+1 that (U 3 (V m+1, 1)) (V 1) m + 3 > 1, and by Theorem 1 we derive m+1 that (U 3r (V m+1, 1)) m + 3 for - r. Similarly to the above stes, we can also rove that (V m+1 + 1) m + with V k V k (P, 1) for any m 0 and odd P by induction. We have (V + 1) for m 0 since with V 1 P 4k ± 1 we get that V + 1 P + 3 4 (mod 8). For the inductive ste with m 0, we use identity (1.) again: V m+ + 1 (V ) + 1 V m+1 1, and since m+1 (V 1) m + 3 as we have just m+1 seen it, we obtain that (V m+ + 1) m + 3; hence, V m+ 1 (mod m+3 ). On the other hand, if P is even then we rove that (U (V m+1(p, 1), 1) ) (V m+1(p, 1) ) (m 1 + (P )) + (V (P, 1) + ) (m + (P )) for m 1, and thus, U (V m+1(p, 1), 1) (mod (m+ (P )) ) by induction on m 0. Indeed, (P ) 1 and V P + (mod (P ) ) when m 0, and by identity (1.) we have for m 0 that V m+ (V ) (V m+1 m+1 )(V m+1 + ), hence, by the case m 0 and the inductive hyothesis, (V m+ ) (m + (P )) + (V (P, 1) + ) (m + 1 + (P )). For an odd rime with P, gcd(k, ) 1, and k even, we first observe that (1.3) imlies that (V m+1) m + 1 + (P ) for all m 0. By (IV.19) in [7] we know that U k (V m+1, 1) exactly if k is even. Moreover, by Theorem 1, we also know that (U k (V m+1, 1)) (U (V m+1, 1)) m + 1 + (P ) as long as gcd(k, ) 1, since U (V m+1, 1) V m+1. Now we resent the combined roof of Theorems 5 and 10. Proof of Theorems 5 and 10. We will rove the congruences (.) and (.3) that in turn, will imly the identities (.4), (.5), (.7), (.8), and the congruences in Theorem 10. We need some rearation. First we observe that with U k U k (P, 1) n1 nu n x n xu 0 (x) x (1 P x x ) x( P x) (1 P x x ) x(1 + x ) (1 P x x ). (3.1) By a multisection identity [], we get that with V k V k (P, 1) V kn x k V n x 1 V n x + ( 1) n x. (3.) k0

INTEGERS 14 (014) 8 Now we assume that 3 and n is odd. By (3.), we get that k0 V kn x k V nx 1 V n x x. We aly this with n m+1 and n m and, after normalizing by the factor V m+1 V m, we derive for the normalized di erence that V k m+1 V k m W m(x) x k 1 V m+1x V mx V m+1 V m V m+1 V m 1 V m+1x x 1 V mx x k0 It imlies that x + x 3 (1 V m+1x x )(1 V mx x ). x(1 + x ) W m (x) (1 V m+1x x ) (3.3) x(1 + x ) x(1 + x ) (1 V m+1x x ) (1 V m+1x x )(1 V mx x. (3.4) ) The term on the right-hand side of (3.3) corresonds to the generating function P 1 k0 ku k(v m+1, 1)x k by identity (3.1), while the other terms, i.e., those in (3.4), contribute x(1 + x ) (1 V mx x ) (1 V m+1x x ) (1 V m+1x x ) (1 V mx x ) x (1 + x 1 )(V m+1 V m) (1 V m+1x x ) (1 V mx x ) with terms that are multiles of V m+1 V m with -adic order of at least (V m+1 V m). Thus, by (3.3) and (3.4) we obtain (.). The case with is slightly di erent. We need the corresonding generating function for U k U k (P, 1) (rather than for U k (P, 1)) and thus, n1 U(x) U n x n x 1 P x + x n0 nu n x n xu 0 (x) x (1 P x + x ) x( P + x) (1 P x + x ) Also, the multisection identity (3.) yields that k0 V kn x k V nx 1 V n x + x x(1 x ) (1 P x + x ). (3.5)

INTEGERS 14 (014) 9 with V k V k (P, 1) and n even. We aly this with n m+1 and n m, m 1, and, after normalizing by the factor V m+1 V m, we derive for the normalized di erence that V W m(x) k m+1 V k m x k 1 V m+1x V mx V m+1 V m V m+1 V m 1 V m+1x + x 1 V mx + x k0 x x 3 (1 V m+1x + x )(1 V mx + x ). (3.6) Note that this is the reason for requiring m 1 in Theorems 5 and 10 if. Identity (3.6) imlies that x(1 x ) W m (x) (1 V m+1x + x ) (3.7) x(1 x ) x(1 x ) (1 V m+1x + x ) (1 V m+1x + x )(1 V mx + x. (3.8) ) The term on the right-hand side of (3.7) corresonds to the generating function P 1 k0 ku k(v m+1, 1)x k by identity (3.5), while the other terms, i.e., those in (3.8), contribute x(1 x ) (1 V mx + x ) (1 V m+1x + x ) (1 V m+1x + x ) (1 V mx + x ) x (1 x 1 )(V m+1 V m) (1 V m+1x + x ) (1 V mx + x ) with terms that are multiles of V m+1 V m with -adic order of at least (V m+1 V m). Thus, by (3.7) and (3.8), we obtain (.3). We are ready for the actual roof and start with the case when is an odd rime. Recall that we use the notation V k V k (P, 1). The basic idea is to rove that V m P mod for m 0. (3.9) To see this, we aly identity (1.3) modulo with the setting n m and observe that m i 0 (mod ) for m 1 and 1 ale i ale m 1. It follows that D 0 D(V m+1, 1) V m+1 + 4 P + 4 D(P, 1) mod, (3.10) and therefore, D if and only if D 0. (3.11) Note that the comanion identity (1.4) taken modulo imlies (cf. fact (IV.13) in [7]) that for an odd rime D U mod. (3.1)

INTEGERS 14 (014) 10 After these rearatory stes, we roceed with the roofs in three cases. Case 1: ( D, odd rime). First we assume that D D(P, 1). Note that in this case 5 since clearly, D P + 4 6 0 (mod 3). The roof is by induction on m 1. We set k in (.). If m 1 then the statement follows by inequality (.6), which holds by Theorem 7. Here we also use identity (1.4), now for U n (P 0, 1) n n n n 1 U n (P 0 ) n 1 + (P 0 ) n 3 D 0 + (P 0 ) n 5 (D 0 ) +... (3.13) 1 3 5 taken modulo, with n, P 0 V m+1 P (mod ) (by (3.9)), and D 0. Indeed, we obtain that 1 U (P 0 ) 1 + (P 0 ) 3 D 0 + (P 0 ) 5 (D 0 ) + + (D 0 ) 1 1 3 5 and U (V m+1, 1) P 1 mod since 3 D0, 5 (D0 ),..., (D0 ) 1 is divisible by for every rime 5 and m 0; thus, e m () (U (V m+1, 1)) 1 for D imlies - P. Theorem 7 yields (.6) with m 1. Since r m+1 () > r m (), the inequality (.6) remains true for all m. If (.7) holds with m 1 then the above argument also guarantees that it holds for m + 1, too. Note that we get that U (V m+1, 1) mod 3 (3.14) and similarly, for gcd(k, ) 1 and with n k in (3.13), we also obtain on the right-hand side of (.) that for m 0 Hence, for m 1, V k m+1 ku k (V m+1, 1) k 1 k (mod 1) P k 1 mod. 8 k 1 k (mod 1) P k 1 m (V V ) mod m 1+ (V V), >< if gcd(k, ) 1, V k m >: m (V V ) mod m+1+ (V V), if k, and V k m+1 V k m m + (V V ) if gcd(k, ) 1. Case : ( - D, odd rime). On the other hand, - D D(P, 1) imlies that - U (V m+1, 1). By (3.11) it follows that - D 0 D(V m+1, 1).

INTEGERS 14 (014) 11 Case.1: ( - P ). If - P and thus, - P 0 V m+1 by (3.9), then by - P 0 QD 0 and (, V m+1, 1) (, V m+1, 1), we get that gcd(, (, V m+1, 1)) 1, which yields that - U (V m+1, 1) by (1.1). In fact, D 0 U (V m+1, 1) mod (3.15) by (3.1), hence e m () 0 for m 0 and r m () > 1 + e m (), i.e., inequality (.9) holds for all m 1. We note that for m 0 we get that by (3.10), too. In general, the congruences become more comlicated if - D. For instance, we get that for m 1 8 m 1 m 1 D m+ (V V )ku k (V m+1, 1) mod (V V ), >< if gcd(k, ) 1 and V k m+1 V k m gcd(k, (, V m+1, 1)) 1, m D mod m+1+ (V V ), m (V V ) >: if k. In fact, a little more can be said. If besides gcd(k, ) 1 we have (, V m+1, 1) - k then the first congruence alies. On the other hand, if (, V m+1, 1) k then (U k (V m+1, 1)) seems to be increasing as m grows (and the fact that it is nondecreasing for any even k follows by (1.4) and the first congruence with k 1). Now we derive only the lower bound (V k m+1 V k m) m 1 + (V V ), although it can be imroved to (V k m+1 V k m) m + a with some constant a if (U k (V m+1, 1)) increases as m increases. Note, however, that checking a condition that involves the rank of aarition is more di cult than establishing gcd(k, (, V m+1, 1)) 1. D 0 D 0 D Case.: ( P ). If P and therefore, V m+1 P 0 by (3.9) and D 0 4 (mod ), which imlies that 1 for m 0. Then, by fact (IV.19) in [7], we D 0 get that U k (V m+1, 1) exactly if k is even, hence - U (V m+1, 1). Note that U (V m+1, 1) 1 (mod ) by (3.1). In both cases, we have e m () 0 D 0 for all m 0, and the result follows by the congruence (.) with k. In this case, for m 1 we obtain V k m+1 8 m 1 (V V )ku k (V m+1, 1) mod m+ (V V), >< if gcd(k, ) 1 and k odd, V k m >: (V V ) mod m+1+ k(v V ), if k. If gcd(k, ) 1 and k even, we get the lower bound (V k m+1 V k m) (m 1 + (V V )) (m + (P )) by Remark 16.

INTEGERS 14 (014) 1 Case 3: ( ). If then, with a little extra work and Theorem 13, we can imrove the congruences by increasing the exonent of the owers in the modulus. We have two cases. Case 3.1: ( P ). If P then V m+1 V m V P Q mod 4 by reeated alications of the recurrence (1.). It follows that U (V m+1, 1) V m+1 mod 4 which yields that e m () (U (V m+1, 1)) 1 for all m 0. Note that (.9) is satisfied for m since r m () 3. Let k 1 be an odd integer, then for m we also get by (.3) and U k k (mod ) (cf. fact (IV.18) in [7]) that (V k m+1 V k m) m 4 + (V 8 V 4 ) (m 1 + (P )) + 4-P. In the last equation we use r () (V 8 V 4 ) as it is given in (3.1). In order to obtain useful -adic congruences, we increase the exonent in the ower of in the modulus of (.3). With m s and by reeated alications of (.3) with k and Theorem 13, we obtain that V k m+1 V k m (m s) (V s+1 V s)ku k (V m+1, 1) mod (m s 1+rs()). This guarantees a di erence of r s () s 4 + (P ) + 4-P s 1 3 by (3.1) in the exonents of the owers of in the modulus and the right-hand side quantity. Case 3.: ( - P ). If - P then - D and V m+1 V m V 1 P 1 (mod ) by (1.) as above. We get that U (V m+1, 1) V m+1 1 (mod ) and e m () (U (V m+1, 1)) 0 for all m 0. By setting k, the reeated alication of the congruence (.3) yields the result (.5). As in the case with 3, the inequality (.9) is true for m 1. Note that r 1 () by (3.0). In addition, by the congruence (.3) it also follows that for any odd k which is not a multile of 3 and m 1 that (V k m+1 V k m) m 1+ (V 4 V ) and therefore, (V k m+1 V k m) m + 1 by (3.0) (cf. roof of Theorem 7) since if P 4k ± 1 then (3 + P ). Yet again, we increase the exonent in the ower of in the modulus and get useful -adic congruences by setting m s 1 and alying congruence (.3) with k and Theorem 13. For any k such that gcd(k, 6) 1, we get that V k m+1 V k m ( ) m s (V s+1 V s)ku k (V m+1, 1) mod m+s+1, which guarantees a di erence of s 1 in the exonents of the owers of in the modulus and the right-hand side quantity. In fact, U k is even exactly if k is a multile of 3 by fact (IV.18) in [7]. However, in this case (U k (V m+1, 1)) m + 3 by Theorem 13, while r m () m + 1, and thus, we get only the lower bound (V k m+1 V k m) (m + 1) for m 1 by (.3) and similarly to the above derivations. Although, numerical evidence seems to suort the conjecture that equality holds.

INTEGERS 14 (014) 13 Remark 14. In some cases, the above calculations contain a factor U k modulo a high rime ower with a large index k. The eriodicity of the sequence might make it easier to determine these factors. We note that an alternative derivation of some art of the Case 3 of Theorem 5 follows by the following lemma which is of indeendent interest. Lemma 15. If a > b 1 are integers and V k V k (P, 1), then V 4a V 4b is divisible by V a V b. In fact, V 4a V 4b V a V b V a + V b, and in the secial case using a k m and b k m 1 with m 1 and k 1 integers, we have that V k m+ V k m+1 V k m+1 V k m (V k m 1)(V k m + ). (3.16) Proof of Lemma 15. We use identity (1.) and derive that V 4a V 4b (Va ) (Vb ) (V a V b )(V a + V b ). Furthermore, in the mentioned secial case, we have that V k m+1 + V k m (Vk ) + V m k m (V k m 1)(V k m + ). We use (3.16) in its equivalent form V k m+ V k m+1 (V k m+1 V k m)(v k m 1)(V k m + ), (3.17) which suggests a recurrence for the -adic order of V k m+ V k m+1 after the -adic orders of the last two factors on the right-hand side are determined. For instance, the Case 3.1 can be treated as follows. If, P even, and k 1 odd, then we set g 4k (V 4k ) and get that g 4k (Vk 4) ((V k )(V k + )) 3. Identity (1.) imlies that V k m+ Vk 4 (V m+1 k m+1 )(V km+1 + ) (V k m )(V k m + )(V k m+1 + )... m+1 Y (V 4k ) (V k i + ), i and thus, (V k m+ ) g 4k 3 for m 0. It follows that (V k m+ + ) and (V k m+ 1) 0. By reeated alications of (3.17), we get that (V k m+ V k m+1) (m 1) + (V 8k V 4k ) which comletes the roof of (.8) with k 1. (We note that for any odd k 1, (V 8k V 4k ) (V 8 V 4 ) is guaranteed by Theorem 5.) We conclude with the roof of Theorem 7.

INTEGERS 14 (014) 14 Proof of Theorem 7. We start with the case of odd rimes. By the alication of (1.3), it follows that V V 1 P ( 1 P ) 1 1 mod. Since D P + 4 imlies that P 4 mod, we derive ( ( 1) P ) 1 1 ( 1) 1 ( 1) 1 0 mod. Note that P 4 mod (or for that matter, already P 4 mod ) yields that 4 is a quadratic residue modulo, however, 1 only if 1 mod 4. Hence, r 0 () (V V 1 ) and by the congruences (.) and (3.14), we have that r 1 () min{1 + e 0 () + r 0 (), r 0 ()} 4 > 1 + e 1 (). Now we are ready to do the harder roblem when we have only D. We use (1.3) modulo 3 with n and n, and get that 1 V P + P D mod 3 and 1 V P mod 3. (3.18) After some maniulations (in fact, after moving the owers of to the right-hand sides, then subtracting the former congruence from the latter one, and factoring out 3 +1 P ), we get that in order to rove that r 1 () (V V ) 3 we need that P P ( 1) ( 1) ( 1) 1 D 0 mod 3. (3.19) Now we write P 1 1 + P 1 1, and thus, P ( 1) ( 1) 1 + P 1 1 4 ( 1) ( 1) P 1 1 mod 3. We will use the congruences ( 1) 1 mod and ( 1) 1 mod, and derive that, since D and 1 mod 4, P 1 P 1 ( 4 + D) 1 ( 4) 1 + ( 4) 3 1 D mod, hence, since D, ( 1) P 1 1 3 1 D 1 8 D mod. On the left-hand side of congruence (3.19) it follows that P 1 8 D 1 D 1 8 D(P + 4) 1 8 D 0 mod 3. If - D then by Theorem 5, for all m 1, we have r m () > 1 + e m () since e m () 0 by (3.15). 1

INTEGERS 14 (014) 15 If then easy calculation shows that V 4 V P (3 + P ). (3.0) If - P then since e m () 0 for m 0 by Theorem 5, we derive that r 1 () (V 4 V ) > 1 + e 1 () and m 1 su ces. If P then e m () 1 for m 0, and m is necessary exactly if r 1 (), which haens exactly if (P ) 1 by (3.0). In fact, we can determine that since V 8 V 4 P (1 + P )(3 + P )(4 + P ). r () (P ) + + 4-P 5 (3.1) Remark 16. We note that if for some odd rime we have - D and P, then 1 by the arguments in the roof of Theorem 5, and in a similar fashion to the D 0 derivation of (3.18) from (1.3), we obtain that V V 3 +1 P (mod 3 ). Therefore, r 1 () (V V ) if (P ) 1 and r 1 () (V V ) 3, otherwise, and it gives the minimum m 1 or 0 so that r m () > 1+e m (). Similarly, if (P ) then r 1 () 3 also follows. In fact, in general, Theorem 13 yields that r 1 () (P ) + 1 by setting k with m 0 and 1. Note that in order to have a unified aroach in Theorems 5 and 10, we didn t searate the cases with m 0 and 1. Acknowledgment. The author wishes to thank Gregory P. Tollisen for his helful comments. References [1] D. H. Lehmer, An extended theory of Lucas functions, Ann. of Math. 31 (1930), 419 448. [] T. Lengyel, Divisibility roerties by multisection, Fibonacci Quart. 41 (003), 7 79. [3] T. Lengyel, The order of the Fibonacci and Lucas numbers, Fibonacci Quart. 33 (1995), 34 39. [4] T. Lengyel, On the -adic order of Stirling numbers of the second kind and their di erences, 1st International Conference on Power Series and Algebraic Combinatorics (FPSAC 009), Hagenberg, Austria, Discrete Math. Theor. Comut. Sci. Proceedings AK, 561 57, 009. Downloadable from: htt://www.dmtcs.org/dmtcs-ojs/index.h/roceedings/article/ view/dmak0147/. [5] T. Lengyel, On the least significant -adic and ternary digits of certain Stirling numbers, INTEGERS 13 (013), A13:47, 1 10. [6] T. Lengyel, Exact -adic orders for di erences of Motzkin numbers, Int. J. Number Theory, acceted. [7] P. Ribenboim, The new book of rime number records, Sringer, 3rd edition, 1996. [8] A. M. Robert, A course in -adic analysis, Graduate texts in mathematics 198, Sringer, 010.