GU J Sci 31(1): 213-220 (2018) Gazi University Journal of Science http://dergipark.gov.tr/gujs Pisano Periods For The K-Fibonacci And K-Lucas Sequences Mod 2ⁿ Dursun TAŞCI 1,*, Gül ÖZKAN KIZILIRMAK 2 1 Gazi University, Faculty of Sciences, Departments Mathematics, 06500, Ankara, Turkey 2Gazi University, Faculty of Sciences, Departments Mathematics, 06500, Ankara, Turkey Article Info Abstract The goal of this paper is to investigate period of 푘-Lucas sequence with related divisibility 푛 Received: 10/04/2017 properties and periods of 푘-Fibonacci and 푘-Lucas sequences mod 2 . Accepted: 23/12/2017 Keywords 푘-Fibonacci 푘-Lucas Pisano period 1. INTRODUCTION Some sequences of numbers have been studied over several years. In the literature, in mathematics and physics, there are a lot of integer sequences, which are used in almost every field modern sciences. The Fibonacci sequence is the famous integer sequence, which is defined by the following recurrence relation 퐹푛+1 = 퐹푛+퐹푛−1 With the initial conditions 퐹₀ = 0 and 퐹₁ = 1. Another well-known sequence is the Lucas sequence, which satisfies the following recurrence relation 퐿푛+1 = 퐿푛+퐿푛−1 with 퐿₀ = 2 and 퐿₁ = 1. There are many generalizations of the Fibonacci and Lucas sequences [1,2,4]. Two of them was given by Falcon and Plaza in [2,4] as follows: For any integer number 푘 ≥ 1, the 푘th Fibonacci sequences {퐹 } is defined as for n≥ 1 푘,푛 푛∈ℕ 퐹푘,푛+1 = 푘퐹푘,푛 + 퐹푘,푛−1 (1) with initial conditions 퐹푘,0 = 0, 퐹푘,1 = 1. If we take 푘 = 1 in (1), we get the Fibonacci sequence: {0, 1 ,1, 2, 3, 5, 8, . }. *Corresponding author, e-mail: [email protected] 214 Dursun TAŞCI, Gül ÖZKAN KIZILIRMAK/ GU J Sci, 31(1): 213-220 (2018) By setting 푘 = 2 in (1), we obtain the Pell sequence: {0, 1, 2, 5, 12, 29, 70, . }. The 푘-Lucas sequence {퐿 } is defined by the following recurrence relation for 푛, 푘 ≥ 1 푘,푛 푛∈ℕ 퐿푘,푛+1 = 푘퐿푘,푛 + 퐿푘,푛−1 (2) with 퐿푘,0 = 2, 퐿푘,1 = 푘. For 푘 = 1 in (2), the classical Lucas sequence is obtained: {2, 1, 3, 4, 7, 11, 18, … }. For 푘 = 2 in (2), the Pell-Lucas sequence is obtained: {2, 2, 6, 14, 34, 82, 198, . }. There are some properties for these numbers. Some of them are [2,4]: 2 2 For 푛 ∈ ℕ, 퐹푘,2푛+1 = (퐹푘,푛) +(퐹푘,푛+1) , (3) 2 푛 For 푛 ∈ ℕ, 퐹푘,푛−1 퐹푘,푛+1 − (퐹푘,푛) = (−1) , (4) 2 푛+푟 푛+1 For 푟 > 푛, 퐿푘,푛−푟퐿푘,푛+푟 − (퐿푘,푛) = (−1) 퐿푘,2푟 + 2(−1) , (5) For 푛 ∈ ℕ, 퐹푘,2푛 = 퐹푘,푛퐿푘,푛, (6) 푛 For 푛, 푚 ∈ ℕ, 퐿푘,푛퐿푘,푛+푚 = 퐿푘,2푛+푚 + (−1) 퐿푘,푚, (7) 2 For 푚 ≥ 1, 퐿푘,푛+1퐿푘,푚 + 퐿푘,푛퐿푘,푚−1 = (푘 + 4) 퐹푘,푛+푚. (8) The period of the Fibonacci sequence mod 푚 was first studied by Wall [12]. The recurrence part in the sequence creates a new sequence and gives the length of the periods of these sequences. Furthermore Kramer and Hoggatt [8] studied the periods of Fibonacci and Lucas sequences mod 2ⁿ. Falcon and Plaza [3] studied the period length of the 푘-Fibonacci sequence mod 푚. The period of such cyclic sequences is known as Pisano period and the period-length is denoted by 휋푘(푚). Motivated by the above papers, we study the Pisano period for the 푘-Lucas sequence and we obtain Pisano periods for the 푘-Fibonacci and 푘-Lucas sequences mod 2ⁿ. 2. PISANO PERIODS FOR THE K-FIBONACCI AND K-LUCAS SEQUENCES Theorem 2.1. {퐿 mod 푚 } is a simple periodic sequence . 푘,푛 푛∈ℕ Proof. From the defining relation we write, 퐿푘,푛−1 = 퐿푘,푛+1−푘퐿푘,푛 . If 퐿푘,푡+1 ≡ 퐿푘,푠+1 (mod 푚) and 퐿푘,푡 ≡ 퐿푘,푠 (mod 푚) , then 퐿푘,푡−1 ≡ 푘퐿푘,푠−1 (mod 푚). By continiuing this way, we get 퐿푘,푡−푠+1 ≡ 퐿푘,1 (mod 푚) and 퐿푘,푡−푠 ≡ 퐿푘,0 (mod 푚). So that {퐿 mod 푚 } is a simple periodic sequence with 푡 − 푠 period. 푘,푛 푛∈ℕ Corollary 2.2. For 푚 > 3 every Pisano period begins with 2, 3. 푒푖 Theorem 2.3. If the prime factorization of 푚 is 푚 = ∏ 푝푖 , then 푒푖 푒푖 휋푘(푙푐푚(푝푖 )) = 푙푐푚(휋푘(푝푖 )). 푒푖 Proof. The statement 휋푘(푝푖 ) is the length of the period of 퐿푘,푛 (mod p) implies that the sequence 푒푖 푒푖 퐿푘,푛 (mod 푝푖 ), repeats only after blocks of length 푐휋푘(푝푖 ) and the statement 휋푘(푚) is the period- 215 Dursun TAŞCI, Gül ÖZKAN KIZILIRMAK/ GU J Sci, 31(1): 213-220 (2018) 푒푖 length of the sequence 퐿푘,푛 (mod m), which is, 퐿푘,푛 (mod 푝푖 ) repeats after 휋푘(푚) terms for all values 푒푖 of 푖. Since any such number gives a period of 퐿푘,푛 (mod m), we conclude that 휋푘(푚) = 푙푐푚(휋푘(푝푖 )). Corollary 2.4. If 푟|푚 then 휋푘(푟)|휋푘(푚). 푒1 푒2 푒푘 Proof. If 푟 ∣ 푚, then 푚 = 푟. 푝1 푝2 … 푝푘 . From Theorem 2.3, we get 푒1 푒푘 휋푘(푚) = 푙푐푚(휋푘(푟), 휋푘(푝1 ), … , 휋푘(푝푘 )) and from lcm definition 휋푘(푟)|휋푘(푚). Lemma 2.5. If 푘 is an odd integer, then for 푛 ∈ ℕ i. 퐿푘,3푛 ≡ 0 (푚표푑 2) (9) ii. 퐹푘,3푛 ≡ 0 (푚표푑 2). (10) Proof. i. We can give the proof by induction. For 푛 = 1, 3 퐿푘,3 = 푘 + 3푘. Since 푘 is an odd number, 푘3 + 푘 is an even integer. Thus, 퐿푘,3 ≡ 0 (mod 2). Suppose 퐿푘,3푛 ≡ 0 (mod 2). So, 퐿푘,3(푛+1) = 푘퐿푘,3푛+2 + 퐿푘,3푛+1 = 푘(푘퐿푘,3푛+1 + 퐿푘,3푛) + 퐿푘,3푛+1 2 = (푘 + 1)퐿푘,3푛+1 + 푘퐿푘,3푛. Since (푘2 + 1) is an even integer and from induction hypothesis, 2 (푘 + 1)퐿푘,3푛+1 + 푘퐿푘,3푛 ≡ 0 (mod 2). Thus we get 퐿푘,3(푛+1) ≡ 0 (mod 2). 2 ii. We can give the proof by induction. For 푛 = 1, 퐹푘,3 = 푘 + 1 and thus 퐹푘,3 ≡ 0 (mod 2). Suppose 퐹푘,3푛 ≡ 0 (mod 2). So, 퐹푘,3(푛+1) = 푘퐹푘,3푛+2 + 퐹푘,3푛+1 = 푘(푘퐹푘,3푛+1 + 퐹푘,3푛) + 퐹푘,3푛+1 2 = (푘 + 1)퐹푘,3푛+1 + 푘퐹푘,3푛 and thus we have 퐹푘,3(푛+1) ≡ 0 (mod 2). Lemma 2.6. If 푘 is an even integer, then for 푛 ∈ ℕ i. 퐿푘,2푛 ≡ 0 (mod 2) (11) ii. 퐹푘,2푛 ≡ 0 (mod 2). (12) 216 Dursun TAŞCI, Gül ÖZKAN KIZILIRMAK/ GU J Sci, 31(1): 213-220 (2018) 2 Proof. i. We can give the proof by induction. For 푛 = 1, 퐿푘,2 = 푘 + 2 and thus 퐿푘,2 ≡ 0 (mod 2). Suppose 퐿푘,2푛 ≡ 0 (mod 2). For 푚 = 0 and 푛 is replaced by 2ⁿ, we have the Eq. (7) 2 2푛+1 퐿푘,2푛+1 = (퐿푘,2푛) + 2(−1) and thus 퐿푘,2푛+1 ≡ 0 (mod 2). ii. We can give the proof by induction. For 푛 = 1, 퐹푘,2 = 푘 and thus 퐹푘,2 ≡ 0 (mod 2). Suppose 퐹푘,2푛 ≡ 0 (mod 2). For 푛 is replaced by 2ⁿ, we get the Eq. (6) 퐹푘,2푛+1 = 퐹푘,2푛퐿푘,2푛. From the Eq. (11) and induction hypothesis can be formulated as 퐹푘,2푛+1 ≡ 0 (mod 2). Lemma 2.7. If 푘 is odd integer, 푛 i. 퐹푘,3.2푛−1 ≡ 0 (mod 2 ) (13) 푛 ii. 퐹푘,3.2푛−1+1 ≡ 1 (mod 2 ). (14) 2 Proof. i. We can give the proof by induction. For 푛 = 1, 퐹푘,3 = 푘 + 1 and 퐹푘,3 ≡ 0 (mod 2). 푛 Suppose 퐹푘,3.2푛−1 ≡ 0 (mod 2 ). For 푛 is replaced by 3. 2푛−1, we have the Eq. (6) 퐹푘,3.2푛= 퐹푘,3.2푛−1 퐿푘,3.2푛−1. 푛+1 From the Eq. (9) and induction hypothesis, 퐹푘,3.2푛 ≡ 0 (mod 2 ) is satisfies. 3 ii. We can give the proof by induction. For 푛 = 1, 퐹푘,4 = 푘 + 2푘 and thus 퐹푘,4 ≡ 1 (mod 2). 푛 Suppose 퐹푘,3.2푛−1+1 ≡ 1 (mod 2 ). For 푛 is replaced by 3. 2푛−1, we get the Eq. (3) 2 2 퐹푘,3.2푛+1 = (퐹푘,3.2푛−1 ) + ( 퐹푘,3.2푛−1+1) (15) 217 Dursun TAŞCI, Gül ÖZKAN KIZILIRMAK/ GU J Sci, 31(1): 213-220 (2018) From the Eq. (10) and Eq. (13), 2 푛+1 (퐹푘,3.2푛−1) ≡ 0 (mod 2 ) is satisfies. For 푛 is replaced by 3. 2푛−1, we have the Eq. (4) 2 3.2푛−1 (퐹푘,3.2푛−1+1 )(퐹푘,3.2푛−1−1) − (퐹푘,3.2푛−1) = (−1) = 1. 푛 Since 퐹푘,3.2푛−1−1 = 퐹푘,3.2푛−1+1 − 푘 퐹푘,3.2푛−1 and 퐹푘,3.2푛−1+1 ≡ 1 (mod 2 ), then 푛+1 퐹푘,3.2푛−1+1퐹푘,3.2푛−1 ≡ 0 (mod 2 ) is satisfies. Since, 2 2 (퐹푘,3.2푛−1+1 )(퐹푘,3.2푛−1+1 − 푘 퐹푘,3.2푛−1)– (퐹푘,3.2푛−1) = ( 퐹푘,3.2푛−1+1) − 푘 퐹푘,3.2푛−1+1 퐹푘,3.2푛−1 2 −(퐹푘,3.2푛−1) 2 푛+1 and (퐹푘,3.2푛−1) ≡ 0 (mod 2 ) , then we get 2 2 푛+1 (퐹푘,3.2푛−1+1 )(퐹푘,3.2푛−1+1 − 푘 퐹푘,3.2푛−1)– (퐹푘,3.2푛−1) ≡ ( 퐹푘,3.2푛−1+1) (mod 2 ) 푛+1 ≡ 1 (mod 2 ) .