Ratio Mathematica                                               Volume 41, 2021, pp. 71-78 

 

 
 

71 

 

 

 

Blocks within the period of                 

Lucas sequence 

 
 

Rima P. Patel* 

Devbhadra V. Shah† 
 

 

 

Abstract 

In this paper, we consider the periodic nature of the sequence of 

Lucas numbers 𝐿𝑛 defined by the recurrence relation 𝐿𝑛 = 𝐿𝑛−1 +
𝐿𝑛−2; for all 𝑛 ≥ 2; with initial condition 𝐿0 = 2 and 𝐿1 = 1. For 
any modulo 𝑚 > 1, we introduce the ‘blocks’ within this sequence 
by observing the distribution of residues within a single period of 

Lucas sequence. We show that length of any one period of the 

Lucas sequence contains either 1,2 or 4 blocks.  
Keywords: Fibonacci sequence, Lucas sequence, Periodicity of 

Lucas sequence 

2010 AMS subject classification‡: 11B37, 11B39, 11B50 
 

 
 

 

 

 

 

 
* Rima P. Patel (Mahavir Swami College of Polytechnic, Bhagwan Mahavir University, Surat, 

India); rimapatel25@gmail.com 
†Devbhadra V. Shah (Department of Mathematics, Veer Narmad South Gujarat University, 

Surat, India); drdvshah@yahoo.com 
‡ Received on September 21, 2021. Accepted on December 5, 2021. Published on December 

31, 2021. doi: 10.23755/rm.v41i0.664. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. 

This paper is published under the CC-BY licence agreement. 



R. Patel and D. Shah 

72 

 

 

1. Introduction 
 

The Fibonacci sequence and the Lucas sequence are well-known 

sequences among all the integer sequences. The Fibonacci sequence {𝐹𝑛} 
satisfies the recurrence relation 𝐹𝑛 = 𝐹𝑛−1 + 𝐹𝑛−2, with the initial conditions 
𝐹0 = 1 and 𝐹1 = 1. Lucas sequence {𝐿𝑛} is considered as the ‘twin sequence’ 
of Fibonacci sequence which satisfies the similar recursive relation             

𝐿𝑛 = 𝐿𝑛−1 + 𝐿𝑛−2, with the initial conditions 𝐿0 = 2 and 𝐿1 = 1. 
On the other hand, some researchers have conducted important research on 

the period of these two recursive sequences [1, 3, 4, 5, 6]. Wall [1] defines the 

length of period of the Fibonacci sequence by reducing it through the modulo 

any positive integer 𝑚 > 1. Kramer and Hoggatt Jr. [3] also defined the length 
of the period of the Lucas sequence obtained by reducing the sequence through 

modulo any positive integer 𝑚 > 1. 
In this paper, we take deep insight in to the periodic nature of Lucas 

sequence and introduce the concept of ‘blocks’ by observing the distribution of 

residues within a single period of Lucas sequence when considered modulo any 

positive integer 𝑚 > 1. 
We denote the sequence of least non-negative residues of the terms of {𝐿𝑛} 

taken modulo 𝑚 (𝑚 ≥ 2) by 𝐿(𝑚𝑜𝑑 𝑚). If we examine the sequence of final 
digits of {𝐿𝑛}, then we notice an interesting pattern that the sequence 
𝐿(𝑚𝑜𝑑 10) = {2,1,3,4,7,1,8,9,7,6,3,9,2,1,3,…} repeats after the 12 
terms again and again all the way. For any modulo 𝑚, it is easy to observe that 
the sequence {𝐿𝑛} is always periodic and it repeats from starting values 0 and 
1. By 𝑘𝐿 = 𝑘𝐿(𝑚), we mean the lengths of the period of {𝐿𝑛} modulo any 
positive integer 𝑚. These leads us to the following easy consequences. 
 

Lemma 𝟏.𝟏. (a) 𝐿𝑘𝐿(𝑚)−1 ≡ −1(𝑚𝑜𝑑𝑚)    (b) 𝐿𝑘𝐿(𝑚) ≡ 2(𝑚𝑜𝑑𝑚) 

           (c) 𝐿𝑘𝐿(𝑚)+1 ≡ 1(𝑚𝑜𝑑𝑚)       (d)  𝐿𝑘𝐿(𝑚)+2 ≡ 3(𝑚𝑜𝑑𝑚) 

     (e) 𝐿𝑘𝐿(𝑚)+𝑛𝑟 ≡ 𝐿𝑛(𝑚𝑜𝑑 𝑚), for all 𝑟 ∈ ℤ. 

 The following is an important result which speaks about the divisibility 

property of  𝑘𝐿(𝑚). 
 

Fact 𝟏.𝟐. For any 𝑚 > 1, since 𝐿(𝑚𝑜𝑑 𝑚) is always periodic, we conclude 
that if 𝐿𝑛 ≡ 2 (𝑚𝑜𝑑 𝑚) and 𝐿𝑛 + 1 ≡ 1(𝑚𝑜𝑑 𝑚), then 𝑘𝐿(𝑚) | 𝑛. 
 



Blocks of Lucas sequence 
 

73 

 

2.  Blocks within 𝑳(𝒎𝒐𝒅 𝒎) 
 

In this article, we restrict our attention to the behavior of the blocks within 

the residues for a given modulus and consequently some interesting 

relationships will be derived. 

 

Definition 2.1. By 𝛼(𝑚) we mean the smallest positive value of the index 𝑛 of 
Lucas numbers such that 𝐿𝑛 ≡ 2𝐿𝑛+1(𝑚𝑜𝑑 𝑚), when 𝑛 > 1. We call 𝛼(𝑚) the 
restricted period of 𝐿(𝑚𝑜𝑑 𝑚).  
 

 Equivalently, 𝛼(𝑚) is the position of the first repeated term in the 
sequence 𝐿(𝑚𝑜𝑑 𝑚). Thus, 𝐿𝛼(𝑚) ≡ 2𝐿𝛼(𝑚)+1 when considered (𝑚𝑜𝑑 𝑚).  

As an illustration, if we consider 𝐿(𝑚𝑜𝑑 3) =
{2,1,0,1,1,2,0,2,2,1,0,1,1…}, then it is appearent that 𝐿4 ≡ 1(𝑚𝑜𝑑 3) and 
𝐿5 ≡ 2(𝑚𝑜𝑑 3). Thus 𝐿4 ≡ 2𝐿5(𝑚𝑜𝑑 3), which gives 𝛼(3) = 4. 
  We call the finite sequence 𝐿0, 𝐿1,… , 𝐿𝛼(𝑚)−1 to be the first block 

occurring in 𝐿(𝑚𝑜𝑑 𝑚). It may happen that 𝛼(𝑚) = 𝑘𝐿(𝑚). In such case, we 
call 𝐿(𝑚𝑜𝑑 𝑚) to be without restricted period.  

For 𝐿(𝑚𝑜𝑑 4) = {2,1,3,0,3,3,2,1,3…}, clearly 𝛼(4) = 𝑘𝐿(4) = 6, and 
thus 𝐿(𝑚𝑜𝑑 4) has no restricted period. 

 

Definition 2.2. By 𝑠(𝑚) we mean the second positive residue ′𝑡′, which 
appears after the first block in 𝐿(𝑚𝑜𝑑 𝑚).  
 

 This clearly means that 2𝑠(𝑚) ≡ 𝐿𝛼(𝑚) (𝑚𝑜𝑑 𝑚);  𝑠(𝑚) = 𝐿𝛼(𝑚)+1. 

Using the definition of 𝐿𝑛, we now conclude that 𝐿𝛼(𝑚)+2 = 3𝑠(𝑚),

𝐿𝛼(𝑚)+3 = 4𝑠(𝑚), 𝐿𝛼(𝑚)+4 = 7𝑠(𝑚),…  . Also the first block ends with     

𝑚 − 𝑠(𝑚). Thus,  
 

(𝐿𝛼(𝑚), 𝐿𝛼(𝑚)+1, 𝐿𝛼(𝑚)+2, 𝐿𝛼(𝑚)+3,…) = 𝑠(𝑚)(2, 1, 3, 4, 7,…)(𝑚𝑜𝑑 𝑚). 

 

This implies that the successive terms in 𝐿(𝑚𝑜𝑑 𝑚) after the first block are the 
multiples of 𝑠(𝑚). We therefore call 𝑠(𝑚) to be a multiplier. 

Again, in the sequence 𝐿(𝑚𝑜𝑑 𝑚), the blocks are of the form 2,1,…,𝑚 −
𝑠(𝑚),2𝑠(𝑚),𝑠(𝑚),… , 𝑚 − 𝑥,𝑥,𝑥,…; where 2,1,…,𝑚 − 𝑠(𝑚) is the first 
block, 2𝑠(𝑚),𝑠(𝑚),…,𝑚 − 𝑥 is the second block, and so on. The occurrence 
of 3 − 2𝑚,𝑚 − 1 in 𝐿(𝑚𝑜𝑑 𝑚) will indicate that the end of the period has 
been reached and there after repetition begins, since the next two terms will be 

2,1. Here we note that each block contains the same (that is 𝛼(𝑚)) number of 
terms and the subscripts are in arithmetic progression. Thus, 𝐿𝛼(𝑚)𝑢 ≡



R. Patel and D. Shah 

74 

 

2𝐿𝛼(𝑚)𝑢+1(𝑚𝑜𝑑 𝑚), for each positive integer 𝑢. Since 𝐿𝑘𝐿(𝑚) ≡

2𝐿𝑘𝐿(𝑚)+1(𝑚𝑜𝑑 𝑚), we conclude that 𝛼(𝑚)𝑢 = 𝑘𝐿(𝑚), where 𝑢 is a positive 

integer, which implies that 𝛼(𝑚) | 𝑘𝐿(𝑚). Later in the paper we show that the 
value of 𝑢 is either 1 or 2 or 4. 
  

Definition 2.3. By 𝛽(𝑚) we mean the order of 𝑠(𝑚), when considered modulo 

𝑚. That is, 𝑠(𝑚)𝛽(𝑚) ≡ 1 (𝑚𝑜𝑑 𝑚) and if 𝑛 < 𝛽(𝑚) then 𝑠(𝑚)𝑛 ≢
1 (𝑚𝑜𝑑 𝑚). 
  

 To illustrate above definitions, we consider the following three examples: 

(1) For 𝐿(𝑚𝑜𝑑 4) = {2,1,3,0,3,3,2,1, . . . }, clearly 𝑘𝐿(4) = 6. Also, 
the restricted period 𝛼(4) is 6 and multiplier 𝑠(4) is 1. Thus, the 
order of  𝑠(4) = 1 is 1 and hence 𝛽(4) = 1. 

(2) If we consider 𝐿(𝑚𝑜𝑑 6) = {2,1,3,4,1,5,0,5,5,4,3,1,4,5,3,2, 
5,1,0,1,1,2,3,5,2,1,3,4,1,…} , then clearly 𝑘𝐿(6) is 24, 𝛼(6) is 
12 and 𝑠(𝑚) is 5. Since 52 ≡ 1 (𝑚𝑜𝑑 6), we get 𝛽(11) = 2.    

(3) If we consider 𝐿(𝑚𝑜𝑑 13) = 2,1,3,4,7,11,18,3,8,11,6,4,10,1, 
11,12,10,9,6,2,8,10, 5,2,7,9,3,12,2,1,3,4,… , then 𝑘𝐿(13) =
28, 𝛼(13) = 7 and 𝑠(𝑚) = 8. Since 84 ≡ 1 (𝑚𝑜𝑑 13), we have 
𝛽(13) = 4.   

 The following theorem ties together the three functions 𝑘𝐿(𝑚), 𝛼(𝑚) and 
𝛽(𝑚).  

Theorem 2.4. 𝑘𝐿(𝑚) = 𝛼(𝑚)× 𝛽(𝑚). 
Proof: We first divide the single period of 𝐿(𝑚𝑜𝑑 𝑚) into smaller finite 
subsequences, say 𝑅0,𝑅1,𝑅2,…,𝑅𝑛,… as shown below: 

2,1,… 3𝑠1 −𝑚,𝑚 − 𝑠1⏞              ,

𝑅0

2𝑠1,𝑠1, . . . ,3𝑠2 − 𝑚,𝑚 − 𝑠2⏞                  
𝑅1

,2𝑠2,𝑠2, . . . ,3𝑠3 − 𝑚,𝑚 −𝑠3⏞                  
𝑅2

…

..2𝑠𝑛,𝑠𝑛, . . . ,3− 𝑚,𝑚 −1⏞                
𝑅𝑛

,2,1,…3𝑠1 − 𝑚,𝑚 − 𝑠1⏞              ,

𝑅𝑛+1

… ,                                  (2.1) 

where 𝑠1 = 𝑠(𝑚). 

  Obviously each finite subsequence ‘𝑅𝑖’ has 𝛼(𝑚) terms and it contains 
exactly one block. Hence every subsequence 𝑅𝑖(𝑖 ≥ 1) is a multiple of ‘𝑅0’. 
Therefore, we have the following congruences modulo 𝑚: 

𝑅1 = 𝑠1𝑅0 ; 𝑅2 = 𝑠2𝑅0 ; 𝑅3 = 𝑠3𝑅0 ; ⋯ ; 𝑅𝑛−1 = 𝑠𝑛−1𝑅0 ; 𝑅𝑛 = 𝑠𝑛𝑅0. 
Since the first term of 𝑅1 is 𝑚 − 𝑠2 and that of 𝑅0 is 𝑚 − 𝑠1 and we also have 
𝑅1 = 𝑠1𝑅0, we get 𝑚 − 𝑠2 = 𝑠1(𝑚 − 𝑠1). If we consider the modulo 𝑚, we get 
𝑠2 = 𝑠1 × 𝑠1. According to similar arguments, when considering the modulo 



Blocks of Lucas sequence 
 

75 

 

𝑚, we have 𝑠3 = 𝑠2 × 𝑠1,𝑠4 = 𝑠3 × 𝑠1,𝑠5 = 𝑠4 ×𝑠1,⋯ ,𝑠𝑛 = 𝑠𝑛−1 × 𝑠1. 
Therefore, we have  

 

𝑠𝑛 = 𝑠𝑛−1 × 𝑠1 
     = (𝑠𝑛−2 ×𝑠1) × 𝑠1 
     = (𝑠𝑛−3 ×𝑠1) × 𝑠1 × 𝑠1 
                     ⋮ 

     = (𝑠𝑛−(𝑛−1) ×𝑠1) × 𝑠1 ×𝑠1 × …×𝑠1⏞          
(𝑛−2) 𝑡𝑖𝑚𝑒𝑠

 

Thus, 𝑠𝑛 = 𝑠1
𝑛. 

 

Now since the order of 𝑠1 is 𝛽(𝑚), we can write single period of 𝐿(𝑚𝑜𝑑 𝑚) as 
follows: 

 

2,1,3,4,7,…,3𝑠1 − 𝑚,𝑚 − 𝑠1,𝑠1,…,3𝑠1
2 − 𝑚,𝑚 −  𝑠1

2,  𝑠1
2,…,3𝑠1

3 − 𝑚,

𝑚 − 𝑠1
3,… ,3 − 𝑚,𝑚 − 1,  𝑠1

𝛽(𝑚)−1
,…,2,1. 

 

Therefore, 𝛽(𝑚) can be interpreted differently as the number of blocks in a 
single period of 𝐿(𝑚𝑜𝑑 𝑚). It now follows easily that 𝑘𝐿(𝑚) = 𝛼(𝑚)×𝛽(𝑚). 
 

  Following are some interesting consequences which follows from these 

results. 

 

Corollary 2.5. 𝐿𝑛×𝛼(𝑚)+𝑟 ≡ (𝑠(𝑚))
𝑛
× 𝐿𝑟 (𝑚𝑜𝑑 𝑚). 

Proof: From the previous theorem, we have 𝑅𝑛 ≡ 𝑠𝑛𝑅0(𝑚𝑜𝑑 𝑚) and 𝑠𝑛 ≡
𝑠1
𝑛(𝑚𝑜𝑑 𝑚) . Thus, we have 

 

𝑅𝑛 ≡ 𝑠1
𝑛𝑅0 (𝑚𝑜𝑑 𝑚).                                 (2.2) 
 

This shows that the 𝑟𝑡ℎ term of 𝑅𝑛 is equal to 𝑠1
𝑛 times the 𝑟𝑡ℎ term of 𝑅0, 

when considering the modulo 𝑚. Also, from the definition of 𝑠(𝑚), an 
immediate conclusion that would be drawn is 𝑠1 = 2𝐿𝛼(𝑚) when considering 

modulo 𝑚. Therefore, from lemma 1.1 and above arguments, we can say that 

𝐿𝑛×𝛼(𝑚)+𝑟 ≡ (𝐿𝛼(𝑚))
𝑛
× 𝐿𝑟 (𝑚𝑜𝑑 𝑚). This finally gives 𝐿𝑛×𝛼(𝑚)+𝑟 ≡

(𝑠(𝑚))
𝑛
× 𝐿𝑟 (𝑚𝑜𝑑 𝑚).  

 

Corollary 2.6. 𝑔𝑐𝑑(𝑚,𝑠𝑖) = 1; for all 𝑖 ≥ 1. 
Proof: From the definition, when considered modulo 𝑚 we have 𝑠𝑛 = 𝑠1

𝑛. 

Therefore, we write 𝑠𝑖
𝛽(𝑚) ≡ (𝑠1

𝑖)
𝛽(𝑚)

≡ (𝑠1
𝛽(𝑚)

)
𝑖

(𝑚𝑜𝑑 𝑚). Thus, since 



R. Patel and D. Shah 

76 

 

𝑠1
𝛽(𝑚)

≡ 1(𝑚𝑜𝑑 𝑚), we have (𝑠1
𝛽(𝑚)

)
𝑖

≡ 1(𝑚𝑜𝑑 𝑚). This gives 𝑠𝑖
𝛽(𝑚)

≡

1(𝑚𝑜𝑑 𝑚). Now suppose 𝑔𝑐𝑑(𝑚,𝑠𝑖) = 𝑑. Then, 𝑑 | 𝑚 and 𝑑 | 𝑠𝑖, which gives 

𝑑 | 𝑠𝑖
𝛽(𝑚)

. Also, 𝑚 | (𝑠𝑖
𝛽(𝑚)

− 1). Using both together, we have 𝑑 | (𝑠𝑖
𝛽(𝑚)

−

(𝑠𝑖
𝛽(𝑚)

− 1)). This gives, 𝑑 = 1. Thus, 𝑔𝑐𝑑(𝑚,𝑠𝑖) = 1.  

 

Corollary 2.7. 𝑠𝑛
𝑟 ≡ 𝑠𝑛×𝑟(𝑚𝑜𝑑 𝑚). 

Proof: From the definition of 𝑠(𝑚), we have 𝑠𝑛 ≡ 𝑠1
𝑛(𝑚𝑜𝑑 𝑚). Then we can 

write 𝑠𝑛
𝑟 ≡ (𝑠1

𝑛)𝑟 ≡ 𝑠1
𝑛×𝑟 ≡ 𝑠𝑛×𝑟(𝑚𝑜𝑑 𝑚). It now follows that 𝑠𝑛

𝑟 ≡
𝑠𝑛×𝑟(𝑚𝑜𝑑 𝑚).  
 

 The following theorem doesn’t seem to give us an immediate idea about 

𝐿(𝑚𝑜𝑑 𝑚), but some good results follow. The evidence comes from Robinson 
[2] but admits that Morgan Wood knew the result in the early 1930’s. 
 

Theorem 2.8. 𝑘𝐿(𝑚) = gcd(2,𝛽(𝑚))× 𝑙𝑐𝑚[2,𝛼(𝑚)], for 𝑚 > 2. 

Proof: Koshy [5] proved that 𝐿𝑛
2 − 𝐿𝑛−1𝐿𝑛+1 = 5(−1)

𝑛. Taking 𝑛 = 𝛼(𝑚), 
we get 

 

𝐿𝛼(𝑚)
2 − 𝐿𝛼(𝑚)−1𝐿𝛼(𝑚)+1 = 5(−1)

𝛼(𝑚).                      (2.3) 

 

Now, 𝐿𝛼(𝑚) ≡ 2𝑠(𝑚)(𝑚𝑜𝑑 𝑚), 𝐿𝛼(𝑚)−1 ≡ −𝑠(𝑚)(𝑚𝑜𝑑 𝑚) and 𝐿𝛼(𝑚)+1 ≡

𝑠(𝑚)(𝑚𝑜𝑑 𝑚). Therefore, by (2.2) we have 
 

(2𝑠(𝑚))2 −(−𝑠(𝑚))(𝑠(𝑚)) ≡ 5(−1)𝛼(𝑚)(𝑚𝑜𝑑 𝑚). 

 

This gives 

 

5(𝑠(𝑚))2 ≡ 5(−1)𝛼(𝑚)(𝑚𝑜𝑑 𝑚).                        (2.4) 
 

 Thus (𝑠(𝑚))
2
 and (−1)𝛼(𝑚) has same order modulo 𝑚. But the order of 

−1 is 2 and the order of 𝑠(𝑚) is 𝛽(𝑚) modulo 𝑚. Thus, 

𝛽(𝑚)

gcd(2,𝛽(𝑚))
=

2

gcd(2,𝛼(𝑚))
 . 

Thus, 

 

𝑘𝐿(𝑚) = 𝛼(𝑚)𝛽(𝑚) = 𝛼(𝑚)
2gcd(2,𝛽(𝑚))

gcd(2,𝛼(𝑚))
= gcd(2,𝛽(𝑚))× 𝑙𝑐𝑚[2,𝛼(𝑚)], 

 

as required. 



Blocks of Lucas sequence 
 

77 

 

 Finally, we calculate the possible values of 𝛽(𝑚). 
 

Theorem 2.9. 𝛽(𝑚) = 1 or 2 or 4; for any 𝑚 ≥ 2. 
Proof: By above theorem we have 

 

𝜇(𝑚) = gcd(2,𝛽(𝑚))× 𝑙𝑐𝑚[2,𝛼(𝑚)],= (1 𝑜𝑟 2) ×(𝛼(𝑚) 𝑜𝑟 2𝛼(𝑚)). 
 

Therefore, 𝜇(𝑚) = 𝛼(𝑚) or 2𝛼(𝑚) or 4𝛼(𝑚). Thus, we have 𝛽(𝑚) = 1 or 2 
or 4; for any 𝑚 ≥ 2. 
 

We conclude the paper by noting the following obvious result which is a 

direct consequence of theorem 2.4 and theorem 2.9. 
 

Corollary 2.10. 𝑘(𝑚) = 𝛼(𝑚) or 2𝛼(𝑚) or 4𝛼(𝑚). 
 

 

 

3. Conclusion 

 In this article we had introduced the ‘blocks’ within the period of the 

Lucas sequence and shown that length of any one period of the Lucas sequence 

contains either 1,2 or 4 blocks.  
 

 

 

 

References 

 

[1] D. D. Wall. Fibonacci series modulo 𝑚. The American Mathematical 

Monthly, 67, 525 – 532, 1960. 

[2] D. W. Robinson. The Fibonacci matrix modulo 𝑚. The Fibonacci 

Quarterly, 1, 29 – 36, 1963. 

[3] J. Kramer, V. E. Jr. Hoggatt. Special cases of Fibonacci Periodicity. The 

Fibonacci Quarterly, 1:5, 519 – 522, 1972. 

[4] K. Thomas. Fibonacci and Lucas Numbers with Applications. John Wiley 

and Sons,Inc., New York, 2001. 



R. Patel and D. Shah 

78 

 

[5] R. Marc. Properties of the Fibonacci sequence under various moduli. 

Master’s Thesis, Wake Forest University, 1996. 

[6] R. P. Patel, D. V. Shah. Periodicity of generalized Lucas numbers and the 

length of its period under modulo 2𝑒. The Mathematics Today, 33, 67 – 

74, 2017.