Approach of the value of a rent when non-central moments of the capitalization factor are known: an R application with interest rates following normal and beta distributions


Ratio Mathematica                                            Volume 41, 2021, pp. 181-196 

 

              
 

181 

 

 

 

The sequence of trifurcating Fibonacci 

numbers 
 

Parimalkumar A. Patel * 

Devbhadra V. Shah † 
  

 

 

Abstract  

One of the remarkable generalizations of Fibonacci sequence is a 

𝑘-Fibonacci sequence and subsequently generalized into the 
‘Bifurcating Fibonacci sequence’. In this paper, further 

generalization as the sequence of ‘trifurcating Fibonacci numbers’ 

is studied and Binet-like formula for these numbers is obtained. 

The analogous of Cassini’s identity, Catalan’s identity, d’Ocagne’s 

identity and some fundamental identities for the terms of this 

sequence has also been investigated. 

Keywords: Fibonacci sequence, bifurcating Fibonacci sequence, 

generalization of Fibonacci sequence, Binet formula, identities 

related with the Fibonacci sequence. 

2010 AMS subject classification‡: 11B39, 11B83, 11B99. 
 

 
 

 
* Veer Narmad South Gujarat University, Surat, India; parimalpatel4149@gmail.com. 
† Veer Narmad South Gujarat University, Surat, India; drdvshah@yahoo.com. 
‡ 1Received on September 30, 2021. Accepted on December 10, 2021. Published on December 

31, 2021.doi: 10.23755/rm.v41i0.668. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. 

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



Parimalkumar A. Patel, Devbhadra V. Shah 

182 

 

1. Introduction  
 

The Fibonacci sequence {𝐹𝑛}𝑛≥0 is a sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 
34, 55, 89, … ,   where each term is the sum of two preceding terms. The 

corresponding recurrence relation is 𝐹𝑛 = 𝐹𝑛−1 + 𝐹𝑛−2 ;  𝑛 ≥ 2. Various 
generalizations of this sequence have appeared in recent years [5, 7, 8]. 

Related to this work (i) some work alters the first two terms of the sequence 

from 0,1 to any arbitrary integers 𝑎, 𝑏 while maintaining the recurrence 
relation (ii) some more work preserves the first two terms of the sequence but 

alters the recurrence relation (iii) even the combined approach of altering the 

initial terms as well as recurrence relation was considered by several authors. 

For further details about this sequence, one can refer Koshy [6], Patel, Shah 

[7], Singh, Sikhwal, Bhatnagar [8] and related papers available in the 

literature. 

 

One interesting generalization depending on exactly one real parameter 𝑘 
is the sequence of 𝑘-Fibonacci numbers {𝐹𝑘,𝑛} which is defined using a linear 

recurrence relation 𝐹𝑘,𝑛 = 𝑘𝐹𝑘,𝑛−1 + 𝐹𝑘,𝑛−2 ; 𝑛 ≥ 2 where 𝐹𝑘,0 = 0 and 𝐹𝑘,1 =
1. For 𝑘 = 1, we get the standard Fibonacci sequence and for 𝑘 = 2, we get 
the sequence of Pell numbers. This sequence was studied by Arvadia and Shah 

[1]. Edson and Yayenie [4] generalized this sequence to a sequence which 

depends on two real parameters 𝑎, 𝑏. They defined the bifurcating sequence  

{𝐹𝑛
(𝑎,𝑏) }

𝑛≥0
 by the recurrence relation 

𝐹𝑛
(𝑎,𝑏)

= {
𝑎𝐹𝑛−1

(𝑎,𝑏)
+ 𝐹𝑛−2

(𝑎,𝑏)
; if 𝑛 is even

𝑏𝐹𝑛−1
(𝑎,𝑏)

+ 𝐹𝑛−2
(𝑎,𝑏)

; if 𝑛 is odd
 ; 𝑛 ≥ 2  

where 𝐹0
(𝑎,𝑏)

= 0, 𝐹1
(𝑎,𝑏)

= 1.  
 

 Diwan and Shah [2, 3], Yayenie [10], Verma and Bala [9] studied this 

sequence extensively and obtained significant results. It is easy to observe that 

(i) by considering 𝑎 = 𝑏 = 1, we get standard Fibonacci sequence (ii) by 
considering 𝑎 = 𝑏 = 2, we get the sequence of Pell numbers and (iii) by 
considering 𝑎 = 𝑏 = 𝑘, we get the sequence of 𝑘-Fibonacci numbers. In this 
paper, we further generalize this sequence to a sequence of trifurcating 

Fibonacci numbers, which depends on the three real parameters 𝑎, 𝑏, 𝑐. 
 

Definition: For any three nonzero positive integers 𝑎, 𝑏 and 𝑐, the trifurcating 

Fibonacci sequence {𝐹𝑛
(𝑎,𝑏,𝑐)

}
𝑛≥0

 is defined recursively by 𝐹0
(𝑎,𝑏,𝑐)

=

0, 𝐹1
(𝑎,𝑏,𝑐)

= 1 and the recurrence relation  



The sequence of trifurcating Fibonacci numbers 

183 

 

𝐹𝑛
(𝑎,𝑏,𝑐)

= {

𝑎𝐹𝑛−1
(𝑎,𝑏,𝑐)

+ 𝐹𝑛−2
(𝑎,𝑏,𝑐)

; if 𝑛 ≡ 0(𝑚𝑜𝑑 3)

𝑏𝐹𝑛−1
(𝑎,𝑏,𝑐)

+ 𝐹𝑛−2
(𝑎,𝑏,𝑐)

; if 𝑛 ≡ 1(𝑚𝑜𝑑 3)

𝑐𝐹𝑛−1
(𝑎,𝑏,𝑐)

+ 𝐹𝑛−2
(𝑎,𝑏,𝑐)

; if 𝑛 ≡ 2(𝑚𝑜𝑑 3)

. 

 

 To avoid cumbersome notation, we denote 𝐹𝑛
(𝑎,𝑏,𝑐)

 by 𝑃𝑛. Few terms of 
this trifurcating generalized Fibonacci sequence are shown in the Table 1. In 

this paper we obtain various interesting results for this sequence. 

 

𝒏 𝑷𝒏 

0 0 

1 1 

2 𝑐 

3 𝑎𝑐 + 1 

4 𝑎𝑏𝑐 + 𝑏 + 𝑐 

5 𝑎𝑏𝑐2 + 𝑏𝑐 + 𝑐2 + 𝑎𝑐 + 1 

6 𝑎2𝑏𝑐2 + 2𝑎𝑏𝑐 + 𝑎𝑐2 + 𝑎2𝑐 + 𝑎 + 𝑏 + 𝑐 

7 𝑎2𝑏2𝑐2 + 2𝑎𝑏2𝑐 + 2𝑎𝑏𝑐2 + 𝑎2𝑏𝑐 + 𝑎𝑏 + 2𝑏𝑐 + 𝑏2 + 𝑐2 + 𝑎𝑐 + 1 

Table 1 

2. Fundamental identities for the trifurcating 

sequence {𝑷𝒏}𝒏≥𝟎 
  

 In this section, we derive some interesting identities for the terms of the 

sequence {𝑃𝑛}𝑛≥0. We first show that any two consecutive terms of {𝑃𝑛}𝑛≥0 are 
always relatively prime. 

 

Theorem 2.1. gcd(𝑃𝑛, 𝑃𝑛−1) = 1; for all 𝑛 = 1, 2, … . 
 

Proof. We prove the result by considering three cases when 𝑛 = 3𝑘, 3𝑘 + 1 or 
3𝑘 + 2. We present the proof only for the case 𝑛 = 3𝑘 and other cases follows 
accordingly. Now Euclidean algorithm leads to the following system of 

equations: 

 

𝑃3𝑘 = 𝑎𝑃3𝑘−1 + 𝑃3𝑘−2          
𝑃3𝑘−1 = 𝑐𝑃3𝑘−2 + 𝑃3𝑘−3        
𝑃3𝑘−2 = 𝑏𝑃3𝑘−3 + 𝑃3𝑘−4        
           ⋮  



Parimalkumar A. Patel, Devbhadra V. Shah 

184 

 

𝑃4 = 𝑏𝑃3 + 𝑃2                        
𝑃3 = 𝑎𝑃2 + 𝑃1                       
𝑃2 = 𝑐𝑃1 + 0               
         

It now easily follows from Euclidean algorithm that gcd(𝑃𝑛, 𝑃𝑛−1) = 𝑃1 = 1. 
  

 We now prove certain summation formulae for the terms of {𝑃𝑛}𝑛≥0. 
 

Lemma 2.2. 

a) 𝑃3𝑛+2 = (𝑏𝑐 + 1)(𝑃3 + 𝑃6 + ⋯ + 𝑃3𝑛) + (𝑃2 + 𝑃5 + ⋯ + 𝑃3𝑛−2) + 𝑐 
b) 𝑃3𝑛+1 = (𝑎𝑏 + 1)(𝑃2 + 𝑃5 + 𝑃8 + ⋯ + 𝑃3𝑛−1) 
                +(𝑏 − 1)(𝑃1 + 𝑃4+. . . +𝑃3𝑛−2) + 1 
c) 𝑃3𝑛 = (𝑎𝑐 + 1)(𝑃1 + 𝑃4+. . . +𝑃3𝑛−2) + (𝑎 − 1)(𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛−3). 
 

Proof. Since, 𝑃3𝑛+2 = 𝑐𝑃3𝑛+1 + 𝑃3𝑛 we get 
𝑃2 = 𝑐𝑃1 + 𝑃0 
𝑃5 = 𝑐𝑃4 + 𝑃3 
𝑃8 = 𝑐𝑃7 + 𝑃6 

 ⋮ 
𝑃3𝑛−1 = 𝑐𝑃3𝑛−2 + 𝑃3𝑛−3 
𝑃3𝑛+2 = 𝑐𝑃3𝑛+1 + 𝑃3𝑛. 
 

Adding all these equations we get 
𝑃2 + 𝑃5 + 𝑃8+. . . +𝑃3𝑛+2 = 𝑐(𝑃1 + 𝑃4+. . . +𝑃3𝑛+1) + (𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛)       (2.1) 
 

 Again, 𝑃3𝑛 = 𝑎𝑃3𝑛−1 + 𝑃3𝑛−2 gives 
𝑃3 = 𝑎𝑃2 + 𝑃1 
𝑃6 = 𝑎𝑃5 + 𝑃4 
𝑃9 = 𝑎𝑃8 + 𝑃7 

           ⋮ 
𝑃3𝑛−3 = 𝑎𝑃3𝑛−4 + 𝑃3𝑛−5 
𝑃3𝑛 = 𝑎𝑃3𝑛−1 + 𝑃3𝑛−2 
 

 Adding these equations, we get 
𝑃3 + 𝑃6 + 𝑃9 + ⋯ + 𝑃3𝑛 = 𝑎(𝑃2 + 𝑃5 + 𝑃8 + ⋯ + 𝑃3𝑛−1) 
                                         +(𝑃1 + 𝑃4+. . . +𝑃3𝑛−2)                                                  (2.2) 
 

 Also, since 𝑃3𝑛+1 = 𝑏𝑃3𝑛 + 𝑃3𝑛−1, we have 
𝑃4 = 𝑏𝑃3 + 𝑃2 
𝑃7 = 𝑏𝑃6 + 𝑃5 
𝑃10 = 𝑏𝑃9 + 𝑃8 

           ⋮ 



The sequence of trifurcating Fibonacci numbers 

185 

 

𝑃3𝑛−2 = 𝑏𝑃3𝑛−3 + 𝑃3𝑛−4 
𝑃3𝑛+1 = 𝑏𝑃3𝑛 + 𝑃3𝑛−1 
 

 Adding all these equations we get 
𝑃4 + 𝑃7 + ⋯ + 𝑃3𝑛−2 + 𝑃3𝑛+1 
                                    = 𝑏(𝑃3 + 𝑃6 + ⋯ + 𝑃3𝑛) + (𝑃2 + 𝑃5+. . . +𝑃3𝑛−1)            (2.3) 

 

 Using (2.3) in (2.1) we get 

𝑃2 + 𝑃5 + 𝑃8 + ⋯ + 𝑃3𝑛+2 

     = 𝑐(1 + 𝑏(𝑃3 + 𝑃6 + ⋯ + 𝑃3𝑛) + (𝑃2 + 𝑃5 + ⋯ + 𝑃3𝑛−1)) 

     +(𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛)                              
     = 𝑐 + (𝑏𝑐 + 1)(𝑃0 + 𝑃3 + 𝑃6 + ⋯ + 𝑃3𝑛) + 𝑐(𝑃2 + 𝑃5+. . . +𝑃3𝑛−1) 
 

This finally gives 

𝑃3𝑛+2 = (𝑏𝑐 + 1)(𝑃0 + 𝑃3 + 𝑃6 + ⋯ + 𝑃3𝑛) 
           +(𝑐 − 1)(𝑃2 + 𝑃5+. . . +𝑃3𝑛−1) + 𝑐                                                       (2.4) 
 

 Other results can be proved accordingly by considering the pair of 

equations (2.1), (2.2) and further (2.1), (2.3) together.  

 

Lemma 2.3. 

a)  ∑ 𝑃3𝑖
𝑛
𝑖=0 =

𝑎𝑃3𝑛(𝑏−1)(𝑐−1)−(𝑃3𝑛+1−1)(𝑎𝑐+1)(𝑐−1)+(𝑃3𝑛+2−𝑐)(𝑎𝑏+1)(𝑎𝑐+1)

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
. 

b)  ∑ 𝑃3𝑖+1 =
𝑛
𝑖=0

(𝑃3𝑛+1−1)(𝑎−1)(𝑐−1)−(𝑃3𝑛+2−𝑐)(𝑎𝑏+1)(𝑎−1)+𝑎𝑃3𝑛(𝑎𝑏+1)(𝑏𝑐+1)

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
. 

c) ∑ 𝑃3𝑖+2 =
(𝑃3𝑛+2−𝑐)(𝑎−1)(𝑏−1)−𝑎𝑃3𝑛(𝑏𝑐+1)(𝑏−1)+(𝑃3𝑛+1−1)(𝑎𝑐+1)(𝑏𝑐+1)

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
𝑛
𝑖=0 . 

 

Proof. We only prove result (a) here and other two results can be proved in a 

similar way. Using (b) and (c) of lemma 2.2, we get 

𝑎𝑃3𝑛 =
(𝑎𝑐 + 1)

(𝑏 − 1)
{(𝑃3𝑛+1 − 1) − (𝑎𝑏 + 1)(𝑃2 + 𝑃5 + 𝑃8 + ⋯ + 𝑃3𝑛−1)} 

         +(𝑎 − 1)(𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛) 

Then 𝑎𝑃3𝑛 −
(𝑎𝑐+1)

(𝑏−1)
(𝑃3𝑛+1 − 1) − (𝑎 − 1)(𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛) 

          = −
(𝑎𝑐+1)(𝑎𝑏+1)

(𝑏−1)
(𝑃2 + 𝑃5 + 𝑃8 + ⋯ + 𝑃3𝑛−1)  

Using (2.4) we get 

 𝑎𝑃3𝑛 −
(𝑎𝑐+1)

(𝑏−1)
(𝑃3𝑛+1 − 1) − (𝑎 − 1)(𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛) 

          = −
(𝑎𝑏+1)(𝑎𝑐+1)

(𝑏−1)(𝑐−1)
((𝑃3𝑛+2 − 𝑐) − (𝑏𝑐 + 1)(𝑃0 + 𝑃3 + 𝑃6 + ⋯ + 𝑃3𝑛))                               

Then, ((𝑎𝑏 + 1)(𝑎𝑐 + 1)(𝑏𝑐 + 1) + (𝑎 − 1)(𝑏 − 1)(𝑐 − 1))(𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛) 

 = 𝑎(𝑏 − 1)(𝑐 − 1)𝑃3𝑛 − (𝑎𝑐 + 1)(𝑐 − 1)(𝑃3𝑛+1 − 1) 



Parimalkumar A. Patel, Devbhadra V. Shah 

186 

 

   +(𝑎𝑏 + 1)(𝑎𝑐 + 1)(𝑃3𝑛+2 − 𝑐). 
Hence, 𝑃0 + 𝑃3 + ⋯ + 𝑃3𝑛 

                                =
𝑎𝑃3𝑛(𝑏−1)(𝑐−1)−(𝑃3𝑛+1−1)(𝑎𝑐+1)(𝑐−1)+(𝑃3𝑛+2−𝑐)(𝑎𝑏+1)(𝑎𝑐+1)

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
.  

 

We now obtain the sum of first 𝑘 trifurcating Fibonacci numbers. 
 

Theorem 2.4.  

∑ 𝑃𝑛
𝑘
𝑛=1 =

{(𝑏−1)(𝑐−1)−⌊1−
𝜒(𝑘)

3
⌋(𝑏𝑐+1)(𝑏−1)}⌊

4−𝜒(𝑘)

3
⌋𝑎𝑃

3⌊
𝑘
3
⌋−3

+{⌊
1+𝜒(𝑘)

3
⌋(𝑏−1)(𝑐−1)+(𝑎𝑏+1)(𝑏𝑐+1)−⌊

2+𝜒(𝑘)

3
⌋(𝑏𝑐+1)(𝑏−1)}𝑎𝑃

3⌊
𝑘
3
⌋

+{⌊1−
𝜒(𝑘)

3
⌋[(𝑎−1)(𝑐−1)+(𝑎𝑐+1)(𝑏𝑐+1)]−(𝑎𝑐+1)(𝑐−1)}⌊

4−𝜒(𝑘)

3
⌋(𝑃

3⌊
𝑘
3
⌋−2

−1)

+⌊
2+𝜒(𝑘)

3
⌋{(𝑎−1)(𝑐−1)+(𝑎𝑐+1)(𝑏𝑐+1)−⌊

1+𝜒(𝑘)

3
⌋(𝑎𝑐+1)(𝑐−1)}

+⌊
4−𝜒(𝑘)

3
⌋{⌊1−

𝜒(𝑘)

3
⌋(𝑎−1)(𝑏−1)+(𝑎𝑏+1)(𝑎𝑐+1)}(𝑃

3⌊
𝑘
3
⌋−1

−𝑐)

+{⌊
2+𝜒(𝑘)

3
⌋(𝑎−1)(𝑏−1)+⌊

1+𝜒(𝑘)

3
⌋(𝑎𝑏+1)(𝑎𝑐+1)−(𝑎𝑏+1)(𝑎−1)}(𝑃

3⌊
𝑘
3
⌋+2

−𝑐)

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
 . 

 

Proof.  We first obtain the value of ∑ 𝑃𝑛
𝑘
𝑛=1  in three cases when 𝑘 is of the 

form 3𝑚 − 2, 3𝑚 − 1 and 3𝑚 and then combine the results to obtain a single 
result.  

 

 For 𝑘 = 3𝑚 − 2, using the above lemma we get 
 ∑ 𝑃𝑛

𝑘
𝑛=1 = (𝑃3 + ⋯ + 𝑃3𝑚−3) + (𝑃1 + 𝑃4 + ⋯ + 𝑃3𝑚−2) 

               +(𝑃2 + 𝑃5+. . . +𝑃3𝑚−4)  

              =

{

𝑎𝑃3𝑚−3(𝑏−1)(𝑐−1)−(𝑃3𝑚−2−1)(𝑎𝑐+1)(𝑐−1)+(𝑃3𝑚−1−𝑐)(𝑎𝑏+1)(𝑎𝑐+1)+
(𝑃3𝑚−2−1)(𝑎−1)(𝑐−1)−(𝑃3𝑚+2−𝑐)(𝑎𝑏+1)(𝑎−1)+𝑎𝑃3𝑚(𝑎𝑏+1)(𝑏𝑐+1)+

(𝑃3𝑚−1−𝑐)(𝑎−1)(𝑏−1)−𝑎𝑃3𝑚−3(𝑏𝑐+1)(𝑏−1)+(𝑃3𝑚−2−1)(𝑎𝑐+1)(𝑏𝑐+1)

}

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
 

On simplification, we get, 

∑ 𝑃𝑛
𝑘
𝑛=1 =

{

𝑎𝑃3𝑚−3{(𝑏−1)(𝑐−1)−(𝑏𝑐+1)(𝑏−1)}+(𝑃3𝑚−2−1){(𝑎−1)(𝑐−1)+
(𝑎𝑐+1)(𝑏𝑐+1)−(𝑎𝑐+1)(𝑐−1)}+(𝑃3𝑚−1−𝑐){(𝑎−1)(𝑏−1)+(𝑎𝑏+1)(𝑎𝑐+1)}

−(𝑃3𝑚+2−𝑐)(𝑎𝑏+1)(𝑎−1)+𝑎𝑃3𝑚(𝑎𝑏+1)(𝑏𝑐+1)
}

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
  

 

Next, for the case 𝑘 = 3𝑚 − 1, we get 
∑ 𝑃𝑛

𝑘
𝑛=1 = (𝑃3 + ⋯ + 𝑃3𝑚−3) + (𝑃1 + 𝑃4 + ⋯ + 𝑃3𝑚−2)  
         +(𝑃2 + 𝑃5+. . . +𝑃3𝑚−1)  
 



The sequence of trifurcating Fibonacci numbers 

187 

 

                     =

{

𝑎𝑃3𝑚−3(𝑏−1)(𝑐−1)−(𝑃3𝑚−2−1)(𝑎𝑐+1)(𝑐−1)+(𝑃3𝑚−1−𝑐)(𝑎𝑏+1)(𝑎𝑐+1)+
(𝑃3𝑚+1−1)(𝑎−1)(𝑐−1)−(𝑃3𝑚+2−𝑐)(𝑎𝑏+1)(𝑎−1)+𝑎𝑃3𝑚(𝑎𝑏+1)(𝑏𝑐+1)+

(𝑃3𝑚+2−𝑐)(𝑎−1)(𝑏−1)−𝑎𝑃3𝑚(𝑏𝑐+1)(𝑏−1)+(𝑃3𝑚+1−1)(𝑎𝑐+1)(𝑏𝑐+1)

}

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
   

Simplifying this we get 

∑ 𝑃𝑛
𝑘
𝑛=1 =

𝑎𝑃3𝑚−3(𝑏−1)(𝑐−1)+(𝑃3𝑚−1−𝑐)(𝑎𝑏+1)(𝑎𝑐+1)−(𝑃3𝑚−2−1)(𝑎𝑐+1)(𝑐−1)+

{(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑐−1)}(𝑃3𝑚+1−1)+(𝑃3𝑚+2−𝑐){(𝑎−1)(𝑏−1)

−(𝑎𝑏+1)(𝑎−1)}+𝑎𝑃3𝑚{(𝑎𝑏+1)(𝑏𝑐+1)−(𝑏𝑐+1)(𝑏−1)}

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
  

 

   Finally, for 𝑘 = 3𝑚, we get 
∑ 𝑃𝑛

𝑘
𝑛=1 = (𝑃3 + ⋯ + 𝑃3𝑚) + (𝑃1 + 𝑃4 + ⋯ + 𝑃3𝑚−2)  
              +(𝑃2 + 𝑃5+. . . +𝑃3𝑚−1) 

             =

{

𝑎𝑃3𝑚(𝑏−1)(𝑐−1)−(𝑃3𝑚+1−1)(𝑎𝑐+1)(𝑐−1)+(𝑃3𝑚+2−𝑐)(𝑎𝑏+1)(𝑎𝑐+1)+
(𝑃3𝑚+1−1)(𝑎−1)(𝑐−1)−(𝑃𝑚𝑛+2−𝑐)(𝑎𝑏+1)(𝑎−1)+𝑎𝑃3𝑚(𝑎𝑏+1)(𝑏𝑐+1)+

(𝑃3𝑚+2−𝑐)(𝑎−1)(𝑏−1)−𝑎𝑃3𝑚(𝑏𝑐+1)(𝑏−1)+(𝑃3𝑚+1−1)(𝑎𝑐+1)(𝑏𝑐+1)

}

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
  

This on simplification gives 

∑ 𝑃𝑛
𝑘
𝑛=1 =  

                

𝑎𝑃3𝑚{(𝑏−1)(𝑐−1)+(𝑎𝑏+1)(𝑏𝑐+1)−(𝑏𝑐+1)(𝑏−1)}+(𝑃3𝑚+1−1){(𝑎−1)(𝑐−1)+(𝑎𝑐+1)

(𝑏𝑐+1)−(𝑎𝑐+1)(𝑐−1)}+(𝑃3𝑚+2−𝑐){(𝑎−1)(𝑏−1)+(𝑎𝑏+1)(𝑎𝑐+1)−(𝑎𝑏+1)(𝑎−1)}

(𝑎𝑏+1)(𝑎𝑐+1)(𝑏𝑐+1)+(𝑎−1)(𝑏−1)(𝑐−1)
 

Combining all these three results, we finally get the required result.   

 

 The following are the interesting identities related with the summation of  

trifurcating Fibonacci numbers as well as its squares. 

 

Theorem 2.5. ∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋𝑛

𝑘=0 𝑃𝑘 = 𝑃𝑛 + 𝑃𝑛+1 − 1. 
 

Proof. Using the definition of trifurcating Fibonacci numbers, we have 

𝑃3𝑛 = 𝑎𝑃3𝑛−1 + 𝑃3𝑛−2 ; 𝑃3𝑛+1 = 𝑏𝑃3𝑛 + 𝑃3𝑛−1; 𝑃3𝑛+2 = 𝑐𝑃3𝑛+1 + 𝑃3𝑛 
This can be written as 

𝑎𝑃3𝑛−1 = 𝑃3𝑛 − 𝑃3𝑛−2 ;  𝑏𝑃3𝑛 = 𝑃3𝑛+1 − 𝑃3𝑛−1 ;  𝑐𝑃3𝑛+1 = 𝑃3𝑛+2 − 𝑃3𝑛 
Thus, we have the following system of equations: 

𝑐𝑃1 = 𝑃2 − 𝑃0  
𝑎𝑃2 = 𝑃3 − 𝑃1  
𝑏𝑃3 = 𝑃4 − 𝑃2  
            ⋮  
𝑎𝑃𝑛 = 𝑃𝑛+1 − 𝑃𝑛−1; if 𝑛 ≡ 0(𝑚𝑜𝑑 3)  
𝑏𝑃𝑛 = 𝑃𝑛+1 − 𝑃𝑛−1; if 𝑛 ≡ 1(𝑚𝑜𝑑 3)  
𝑃𝑛 = 𝑃𝑛+1 − 𝑃𝑛−1; if 𝑛 ≡ 2(𝑚𝑜𝑑 3)  
 

Adding all the above equations and using the fact that 𝑃0 = 0 and 𝑃1 =

1, we get ∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋𝑛

𝑘=0 𝑃𝑘 = 𝑃𝑛 + 𝑃𝑛+1 − 1. 



Parimalkumar A. Patel, Devbhadra V. Shah 

188 

 

 

Theorem 2.6. ∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋𝑛

𝑘=0 𝑃𝑘
2 = 𝑃𝑛𝑃𝑛+1. 

 

Proof. We prove this result only for the case 𝑛 ≡ 0(𝑚𝑜𝑑 3) and remaining 
cases can be proved accordingly. We let 𝑛 = 3𝑚 and apply induction on 𝑚. 
 

 For 𝑚 = 1, we have 

∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋3

𝑘=0 𝑃𝑘
2 = 𝑏𝑃0

2 + 𝑐𝑃1
2 + 𝑎𝑃2

2 + 𝑏𝑃3
2  

Since 𝑃0 = 0, 𝑃1 = 1, 𝑃2 = 𝑐 and 𝑃3 = (1 + 𝑎𝑐), we get 

∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋3

𝑘=0 = 𝑐(1 + 𝑎𝑐) + 𝑏(1 + 𝑎𝑐)
2 = 𝑃3𝑃4  

 

   We next assume that the result holds for some positive integer 𝑚 = 𝑙 > 1. 

That is let  ∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋3𝑙

𝑘=0 𝑃𝑘
2 = 𝑃3𝑙𝑃3𝑙+1 holds. 

 Now, ∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋3(𝑙+1)

𝑘=0 𝑃𝑘
2  

              = ∑ 𝑎
⌊1−

𝜒(𝑘+1)

3
⌋
𝑏

⌊1−
𝜒(𝑘)

3
⌋
𝑐

⌊1−
𝜒(𝑘+2)

3
⌋3𝑙

𝑘=0 𝑃𝑘
2 + 𝑐𝑃3𝑙+1

2 + 𝑎𝑃3𝑙+2
2 + 𝑏𝑃3𝑙+3

2 . 

              = 𝑃3𝑙𝑃3𝑙+1 + 𝑐𝑃3𝑙+1
2 + 𝑎𝑃3𝑙+2

2 + 𝑏𝑃3𝑙+3
2  

              = 𝑃3𝑙+1𝑃3𝑙+2 + 𝑎𝑃3𝑙+2
2 + 𝑏𝑃3𝑙+3

2  

  = 𝑃3𝑙+2𝑃3𝑙+3 + 𝑏𝑃3𝑙+3
2 = 𝑃3(𝑙+1)𝑃3(𝑙+1)+1 

Thus, by induction, the result to be proved holds for every positive integer 𝑛. 
 

3. Binet-like formula for the trifurcating 

Fibonacci sequence: 
 

Generating function is used to solve the linear homogeneous recurrence 

relations. In this section, the generating function for the trifurcating Fibonacci 

sequence is derived and it is used to obtain Binet-like formula for these 

numbers. We first prove a result which will be needed to obtain the generating 

function of 𝑃𝑛. 
 

Lemma 3.1.  𝑃𝑛+3 − (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑃𝑛 + 𝑃𝑛−3 = 0. 
 

Proof. We prove the result by considering the three cases when 𝑛 = 3𝑘, 3𝑘 +
1 or 3𝑘 + 2. We present the proof only for the case 𝑛 = 3𝑘 and other cases 
follows accordingly.  

 

 Using the definition of 𝑃𝑛, we get 
𝑃3𝑘+3 = 𝑎𝑃3𝑘+2 + 𝑃3𝑘+1 = 𝑎(𝑐𝑃3𝑘+1 + 𝑃3𝑘) + 𝑏𝑃3𝑘 + 𝑃3𝑘−1 



The sequence of trifurcating Fibonacci numbers 

189 

 

                                          = 𝑎𝑐𝑃3𝑘+1 + (𝑎 + 𝑏)𝑃3𝑘 + 𝑐𝑃3𝑘−2 + 𝑃3𝑘−3. 
  

 Now, by definition we have 𝑃3𝑘+1 = 𝑏𝑃3𝑘 + 𝑃3𝑘−1. Multiplying this by 
𝑎𝑐 we get 𝑎𝑐𝑃3𝑘+1 = 𝑎𝑏𝑐𝑃3𝑘 + 𝑎𝑐𝑃3𝑘−1 = 𝑎𝑏𝑐𝑃3𝑘 + 𝑐(𝑃3𝑘 − 𝑃3𝑘−2) 
Substituting this value in above equation we get 

𝑃3𝑘+3 = (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑃3𝑘 − 𝑃3𝑘−3 
Hence, 𝑃3𝑘+3 − (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑃3𝑘 + 𝑃3𝑘−3 = 0. 
 

Lemma 3.2. The generating function of the subsequence {𝑃𝑚}𝑚≥0 of {𝑃𝑛}𝑛=0
∞  

is 

(i) 𝑓(𝑥) =
𝑐𝑥2+𝑥5

(1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6)
 ; when 𝑚 ≡ 2(𝑚𝑜𝑑 3) 

(ii) 𝑔(𝑥) =
𝑥−𝑎𝑥4

(1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6)
 ; when 𝑚 ≡ 1(𝑚𝑜𝑑 3) 

(iii) ℎ(𝑥) =
(𝑎𝑐+1)𝑥3

(1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6)
 ; when 𝑚 ≡ 0(𝑚𝑜𝑑 3). 

 

Proof. We present the proof only for the case 𝑚 ≡ 2(𝑚𝑜𝑑 3). The other cases 

can be proved accordingly. We first let 𝑓(𝑥) = ∑ 𝑃3𝑛−1𝑥
3𝑛−1∞

𝑛=1 = 𝑃2𝑥
2 +

𝑃5𝑥
5 + 𝑃8𝑥

8 + ⋯. Then 
(𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑥3 𝑓(𝑥) 
                                        = (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑃2𝑥

5 + (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑃5𝑥
8 

 +(𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑃8𝑥
11 … 

Using lemma 3.1, we get 

(1 − (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑥3 − 𝑥6)𝑓(𝑥)  
= 𝑐𝑥2 + 𝑥5 − ∑ (𝑝3𝑘+2 − (𝑎𝑏𝑐 + 𝑎 + 𝑏 + 𝑐)𝑝3𝑘−1 − 𝑝3𝑘−4)

∞
𝑚=2 = 𝑐𝑥

2 + 𝑥5. 

Thus, 𝑓(𝑥) =
𝑐𝑥2+𝑥5

(1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6)
, as required. 

 

The following result gives the generating function for 𝑃𝑛. 
 

Theorem 3.3. The generating function for the trifurcating Fibonacci sequence 

{𝑃𝑛} is 𝐹(𝑥) =
𝑥(1+𝑐𝑥+𝑥2+𝑎𝑐𝑥2−𝑎𝑥3+𝑥4)

1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6
. 

 

Proof. We begin with the formal power series representation of the generating 

function for {𝑃𝑛}. Let  
𝐹(𝑥) = 𝑃0 + 𝑃1𝑥 + 𝑃2𝑥

2 + ⋯ + 𝑃𝑘𝑥
𝑘 + ⋯ = ∑ 𝑃𝑚𝑥

𝑚∞
𝑚=0   

Then, 𝑐𝑥𝐹(𝑥) = 𝑐𝑃0𝑥 + 𝑐𝑃1𝑥
2 + 𝑐𝑃2𝑥

3 + ⋯ + 𝑐𝑃𝑘𝑥
𝑘+1 + ⋯ 

                        = ∑ 𝑐𝑃𝑚𝑥
𝑚+1∞

𝑚=0 = ∑ 𝑐𝑃𝑚−1𝑥
𝑚∞

𝑚=1 . 

Also, 𝑥2𝐹(𝑥) = 𝑃0𝑥
2 + 𝑃1𝑥

3 + 𝑃2𝑥
2 + ⋯ + 𝑃𝑘𝑥

𝑘 + ⋯ = ∑ 𝑃𝑚𝑥
𝑚∞

𝑚=0  

Since 𝑃3𝑘+2 = 𝑐𝑃3𝑘+1 + 𝑃3𝑘, we get 
(1 − 𝑐𝑥 − 𝑥2)𝐹(𝑥) = 𝑥 + ∑ (𝑃3𝑛 − 𝑐𝑃3𝑛−1 − 𝑃3𝑛−2)𝑥

3𝑛∞
𝑛=1   



Parimalkumar A. Patel, Devbhadra V. Shah 

190 

 

                          + ∑ (𝑃3𝑛+1 − 𝑐𝑃3𝑛 − 𝑃3𝑛−1)𝑥
3𝑛+1∞

𝑛=1  

Since 𝑃3𝑘 = 𝑎𝑃3𝑘−1 + 𝑃3𝑘−2 and 𝑃3𝑘+1 = 𝑏𝑃3𝑘 + 𝑃3𝑘−1 we get 
(1 − 𝑐𝑥 − 𝑥2)𝐹(𝑥) = 𝑥 + (𝑎 − 𝑐) ∑ 𝑃3𝑛−1𝑥

3𝑛∞
𝑛=1 + (𝑏 − 𝑐) ∑ 𝑃3𝑛𝑥

3𝑛+1∞
𝑛=1 . 

For convenience, we let 𝑓(𝑥) = ∑ 𝑃3𝑛−1𝑥
3𝑛∞

𝑛=1  and 𝑔(𝑥) = ∑ 𝑃3𝑛𝑥
3𝑛∞

𝑛=1  

Using lemma 3.2 (a) and 3.2 (b) we get 

(1 − 𝑐𝑥 − 𝑥2)𝐹(𝑥) = 𝑥 + (𝑎 − 𝑐)
(𝑎𝑐+1)𝑥3

(1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6)
  

                                 +(𝑏 − 𝑐)
𝑐𝑥2+𝑥5

(1−(𝑎𝑏𝑐+𝑎+𝑏+𝑐)𝑥3−𝑥6)
 

On simplification, we get the required result. 

 

We now obtain the Binet-like formula for the sequence of trifurcating 

Fibonacci numbers. 

 

Theorem 3.4. The terms of the trifurcating Fibonacci sequence {𝑃𝑛} are given 

by 𝑃𝑛 =
𝛾(𝑛)𝛼

⌊
𝑛
3
⌋
−𝛿(𝑛)𝛽

⌊
𝑛
3
⌋

𝛼−𝛽
  

where 𝛾(𝑛) = (⌊
𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
𝛼 + (−1)𝜒(𝑛)𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋
+ ⌊1 −

𝜒(𝑛)

3
⌋) 

and 𝛿(𝑛) = (⌊
𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
𝛽 + (−1)𝜒(𝑛)𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋
+ ⌊1 −

𝜒(𝑛)

3
⌋) 

with 𝛼 =
𝑢+√𝑢2+4

2
, 𝛽 =

𝑢−√𝑢2+4

2
, 𝑢 = 𝑎 + 𝑏 + 𝑐 + 𝑎𝑏𝑐 and 

𝜒(𝑛) = {

0  𝑖𝑓 𝑛 ≡ 0(𝑚𝑜𝑑 3)
1  𝑖𝑓 𝑛 ≡ 1(𝑚𝑜𝑑 3)
2  𝑖𝑓 𝑛 ≡ 2(𝑚𝑜𝑑 3)

 . 

 

Proof. From the generating function of {𝑃𝑛}, we have  

𝐹(𝑥) = −
𝑥(1+𝑐𝑥+(1+𝑎𝑐)𝑥2−𝑎𝑥3+𝑥4)

(𝑥3+𝛼)(𝑥3+𝛽)
  

This can be rewritten as 

𝐹(𝑥) = −
1

𝛼−𝛽
[
(1+𝑎𝑐)𝛼−(𝑎𝛼+1)𝑥+(𝛼−𝑐)𝑥2

(𝑥3+𝛼)
−

(1+𝑎𝑐)𝛽−(𝑎𝛽+1)𝑥+(𝛽−𝑐)𝑥2

(𝑥3+𝛽)
]  

Using McLaurin series expansion, we get 

𝐹(𝑥) = −
1

𝛼−𝛽

[
 
 
 
 ∑

(−1)𝑛(1+𝑎𝑐)𝛼

𝛼𝑛+1
∞
𝑛=0 𝑥

3𝑛 − ∑
(−1)𝑛(𝑎𝛼+1)

𝛼𝑛+1
∞
𝑛=0 𝑥

3𝑛+1 

+ ∑
(−1)𝑛(𝛼−𝑐)

𝛼𝑛+1
∞
𝑛=0 𝑥

3𝑛+2 − ∑
(−1)𝑛(1+𝑎𝑐)𝛽

𝛽𝑛+1
∞
𝑛=0 𝑥

3𝑛

+ ∑
(−1)𝑛(𝑎𝛽+1)

𝛽𝑛+1
∞
𝑛=0 𝑥

3𝑛+1 − ∑
(−1)𝑛(𝛽−𝑐)

𝛽𝑛+1
∞
𝑛=0 𝑥

3𝑛+2
]
 
 
 
 

        



The sequence of trifurcating Fibonacci numbers 

191 

 

         = −
1

𝛼−𝛽

[
 
 
 
 
 ∑ (−1)

𝑛(1 + 𝑎𝑐) (
𝛽𝑛−𝛼𝑛

(𝛼𝛽)𝑛
)∞𝑛=0 𝑥

3𝑛

− ∑ (−1)𝑛 (
(𝑎𝛼+1)𝛽𝑛+1−(𝑎𝛽+1)𝛼𝑛+1

(𝛼𝛽)𝑛+1
)∞𝑛=0 𝑥

3𝑛+1

+ ∑ (−1)𝑛 (
(𝛼−𝑐)𝛽𝑛+1−(𝛽−𝑐)𝛼𝑛+1

(𝛼𝛽)𝑛+1
)∞𝑛=0 𝑥

3𝑛+2
]
 
 
 
 
 

  

Now, if 𝛼, 𝛽 are the roots of 1 − (𝑎 + 𝑏 + 𝑐 + 𝑎𝑏𝑐)𝑥 − 𝑥2 = 0 then 

𝛼 =
𝑢+√𝑢2+4

2
, 𝛽 =

𝑢−√𝑢2+4

2
. 

If we let 𝑢 = 𝑎 + 𝑏 + 𝑐 + 𝑎𝑏𝑐, then it is easy to observe that 𝛼𝛽 = −1, 𝛼 +

𝛽 = 𝑢 and 𝛼 − 𝛽 = √𝑢2 + 4 .  
 

Then 

𝐹(𝑥) = [
∑ (1 + 𝑎𝑐) (

𝛼𝑛−𝛽𝑛

𝛼−𝛽
)∞𝑛=0 𝑥

3𝑛 − ∑ (
(𝑎𝛽+1)𝛼𝑛+1−(𝑎𝛼+1)𝛽𝑛+1

𝛼−𝛽
)∞𝑛=0 𝑥

3𝑛+1

+ ∑ (
(𝛽−𝑐)𝛼𝑛+1−(𝛼−𝑐)𝛽𝑛+1

𝛼−𝛽
)∞𝑛=0 𝑥

3𝑛+2
]  

 

Thus,  

𝐹(𝑥) = ∑
1

𝛼−𝛽

(

 
 
 
 
 (

⌊1 −
𝜒(𝑛)

3
⌋ + (−1)𝜒(𝑛)𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋

+ ⌊
𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
𝛼

) 𝛼⌊
𝑛
3
⌋

− (
⌊1 −

𝜒(𝑛)

3
⌋ + (−1)𝜒(𝑛)𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋

+ ⌊
𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
𝛽

) 𝛽⌊
𝑛
3
⌋

)

 
 
 
 
 

∞
𝑛=0 𝑥

𝑛  

 

For convenience if we write 𝜒(𝑛) = {

0   𝑖𝑓 𝑛 ≡ 0(𝑚𝑜𝑑 3)
1   𝑖𝑓 𝑛 ≡ 1(𝑚𝑜𝑑 3)
2   𝑖𝑓 𝑛 ≡ 2(𝑚𝑜𝑑 3)

 and 

𝛾(𝑛) = (⌊
𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
𝛼 + (−1)𝜒(𝑛)𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋
+ ⌊1 −

𝜒(𝑛)

3
⌋), 

𝛿(𝑛) = (⌊
𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
𝛽 + (−1)𝜒(𝑛)𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋
+ ⌊1 −

𝜒(𝑛)

3
⌋) then 

𝐹(𝑥) can be written as 𝐹(𝑥) = ∑
𝛾(𝑛)𝛼

⌊
𝑛
3
⌋
−𝛿(𝑛)𝛽

⌊
𝑛
3
⌋

𝛼−𝛽
∞
𝑛=0 𝑥

𝑛 

This gives 𝑃𝑛 =
𝛾(𝑛)𝛼

⌊
𝑛
3
⌋
−𝛿(𝑛)𝛽

⌊
𝑛
3
⌋

𝛼−𝛽
 , as desired. 

 

 The following results are the easy consequence from this theorem. 

 

Corollary 3.5. (i) 𝑃3𝑛 = (1 + 𝑎𝑐) (
𝛼𝑛−𝛽𝑛

𝛼−𝛽
) 



Parimalkumar A. Patel, Devbhadra V. Shah 

192 

 

                       (ii) 𝑃3𝑛+1 = (
(𝑎𝛽+1)𝛼𝑛+1−(𝑎𝛼+1)𝛽𝑛+1

𝛼−𝛽
) 

                      (iii) 𝑃3𝑛+2 = (
(𝛽−𝑐)𝛼𝑛+1−(𝛼−𝑐)𝛽𝑛+1

𝛼−𝛽
). 

 

4. Some more identities relating trifurcating 

Fibonacci numbers: 
 

 In this section, we use the above Binet-like formula to derive some 

interesting properties for the terms of trifurcating Fibonacci sequence. If we let 

𝑤 = 𝛾(𝑛)𝛿(𝑛), then we observe that 

𝑤 = ⌊1 −
𝜒(𝑛)

3
⌋
2

+ 2(−1)𝜒(𝑛) ⌊1 −
𝜒(𝑛)

3
⌋ 𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊1−
𝜒(𝑛)

3
⌋
  

    + ⌊1 −
𝜒(𝑛)

3
⌋ ⌊

𝜒(𝑛)+2

3
⌋ 𝑐

⌊
𝜒(𝑛)+1

3
⌋
(𝛼 + 𝛽) + 𝑎

2⌊
4−𝜒(𝑛)

3
⌋
𝑐

2⌊1−
𝜒(𝑛)

3
⌋
 

+(−1)𝜒(𝑛) ⌊
𝜒(𝑛)+2

3
⌋ 𝑎

⌊
4−𝜒(𝑛)

3
⌋
𝑐

⌊
𝜒(𝑛)+1

3
⌋+⌊1−

𝜒(𝑛)

3
⌋
(𝛼 + 𝛽) + ⌊

𝜒(𝑛)+2

3
⌋
2

𝑐
2⌊

𝜒(𝑛)+1

3
⌋
𝛼𝛽  

 

This on simplification gives the value of 𝑤 = 𝛾(𝑛)𝛿(𝑛) as  

𝑤 = {

(1 + 𝑎𝑐)2              ; if 𝑛 ≡ 0(𝑚𝑜𝑑 3) 
(1 + 𝑎𝑐)(1 + 𝑏) ; if 𝑛 ≡ 1(𝑚𝑜𝑑 3)
(1 + 𝑎𝑐)(1 + 𝑏𝑐); if 𝑛 ≡ 2(𝑚𝑜𝑑 3)

 

This can be further written as 𝑤 = {

𝑃3
2               ; if 𝑛 ≡ 0(𝑚𝑜𝑑 3) 

(1 + 𝑏)𝑃3 ; if 𝑛 ≡ 1(𝑚𝑜𝑑 3)
(1 + 𝑏𝑐)𝑃3; if 𝑛 ≡ 2(𝑚𝑜𝑑 3)

 

 

We first obtain an identity for the terms of {𝑃𝑛}, which is analogous to that 
of Catalan’s identity for Fibonacci numbers. 

 

Theorem 4.1. For any two nonnegative integers 𝑘 and 𝑟 (≤
𝑘

3
), we have  

𝑃𝑘−3𝑟𝑃𝑘+3𝑟 − 𝑃𝑘
2 = (−1)𝑙−𝑟+1𝑤 (

𝑃3𝑟

𝑃3
)
2

. 

 

Proof. If 𝑘 ≡ 0(𝑚𝑜𝑑 3) then taking 𝑘 = 3𝑙 and using corollary 3.5, we get 
𝑃3𝑙−3𝑟𝑃3𝑙+3𝑟 − 𝑃3𝑙

2  

     = (1 + 𝑎𝑐) (
𝛼𝑙−𝑟−𝛽𝑙−𝑟

𝛼−𝛽
) (1 + 𝑎𝑐) (

𝛼𝑙+𝑟−𝛽𝑙+𝑟

𝛼−𝛽
) − (1 + 𝑎𝑐)2 (

𝛼𝑙−𝛽𝑙

𝛼−𝛽
)
2

   

     =
(1+𝑎𝑐)2

(𝛼−𝛽)2
{(𝛼𝑙−𝑟 − 𝛽𝑙−𝑟)(𝛼𝑙+𝑟 − 𝛽𝑙+𝑟) − (𝛼𝑙 − 𝛽𝑙)2}                              

     =
(1+𝑎𝑐)2

(𝛼−𝛽)2
{𝛼2𝑙 − (−1)𝑙 (

𝛽

𝛼
)
𝑟

− (−1)𝑙 (
𝛼

𝛽
)
𝑟

+ 𝛽2𝑙 − (𝛼2𝑙 − 2(−1)𝑙 + 𝛽2𝑙)}             



The sequence of trifurcating Fibonacci numbers 

193 

 

     =
(1+𝑎𝑐)2

(𝛼−𝛽)2
(−1)𝑙−𝑟+1{𝛼2𝑟 − 2(−1)𝑟 + 𝛽2𝑟} 

     =
(1+𝑎𝑐)2

(𝛼−𝛽)2
(−1)𝑙−𝑟+1(𝛼𝑟 − 𝛽𝑟)2 = (−1)𝑙−𝑟+1 (

(1+𝑎𝑐)(𝛼𝑟−𝛽𝑟)

(𝛼−𝛽)
)
2

  

Thus, 𝑃3𝑙−3𝑟𝑃3𝑙+3𝑟 − 𝑃3𝑙
2 = (−1)𝑙−𝑟+1𝑃3𝑟

2  

 

 Next, if we let 𝑘 ≡ 1(𝑚𝑜𝑑 3) then by considering 𝑘 = 3𝑙 + 1, we get 
𝑃3𝑙−3𝑟+1𝑃3𝑙+3𝑟+1 − 𝑃3𝑙+1

2  

= (
(𝑎𝛽+1)𝛼𝑙−𝑟+1−(𝑎𝛼+1)𝛽𝑙−𝑟+1

𝛼−𝛽
) (

(𝑎𝛽+1)𝛼𝑙+𝑟+1−(𝑎𝛼+1)𝛽𝑙+𝑟+1

𝛼−𝛽
)                           

 − (
(𝑎𝛽+1)𝛼𝑙+1−(𝑎𝛼+1)𝛽𝑙+1

𝛼−𝛽
)
2

 

=
1

(𝛼−𝛽)2

[
 
 
 
 
 

𝑎2(𝛼𝑙−𝑟 − 𝛽𝑙−𝑟)(𝛼𝑙+𝑟 − 𝛽𝑙+𝑟)

−𝑎(𝛼𝑙−𝑟 − 𝛽𝑙−𝑟)(𝛼𝑙+𝑟+1 − 𝛽𝑙+𝑟+1)

− 𝑎(𝛼𝑙−𝑟+1 − 𝛽𝑙−𝑟+1)(𝛼𝑙+𝑟 − 𝛽𝑙+𝑟)

+(𝛼𝑙−𝑟+1 − 𝛽𝑙−𝑟+1)(𝛼𝑙+𝑟+1 − 𝛽𝑙+𝑟+1)

−{𝑎2(𝛼𝑙 − 𝛽𝑙)2 − 2𝑎(𝛼𝑙 − 𝛽𝑙)(𝛼𝑙+1 − 𝛽𝑙+1) + (𝛼𝑙+1 − 𝛽𝑙+1)2}]
 
 
 
 
 

  

=
1

(𝛼−𝛽)2
[

𝑎2(−1)𝑙−𝑟+1[𝛽2𝑟 + 𝛼2𝑟 + 2(−1)𝑟]

−(−1)𝑙−𝑟+1[𝛽2𝑟 + 𝛼2𝑟 + 2(−1)𝑟]

−𝑎(−1)𝑙+1 [(
𝛽

𝛼
)
𝑟
(𝛼 + 𝛽) + (

𝛼

𝛽
)
𝑟
(𝛼 + 𝛽) + 2(𝛼 + 𝛽)]

]  

=
(−1)𝑙−𝑟+1

(𝛼−𝛽)2
{𝑎2(𝛼𝑟 − 𝛽𝑟)2 − 𝑎𝑢[𝛽2𝑟 + 𝛼2𝑟 + 2(−1)𝑟] − (𝛼𝑟 − 𝛽𝑟)2}  

=
(−1)𝑙−𝑟+1

(𝛼−𝛽)2
(𝛼𝑟 − 𝛽𝑟)2{𝑎2 − 𝑎𝑢 − 1} =

(−1)𝑙−𝑟+1

(1+𝑎𝑐)2
{𝑎2 − 𝑎𝑢 − 1}𝑃3𝑟

2   

Thus, 𝑃3𝑙−3𝑟+1𝑃3𝑙+3𝑟+1 − 𝑃3𝑙+1
2 = (−1)𝑙−𝑟+1𝑤𝑃3𝑟

2 𝑃3
−2. 

 

 Finally, if 𝑘 ≡ 2(𝑚𝑜𝑑 3) then by considering 𝑘 = 3𝑙 + 2, we get 
𝑃3𝑙−3𝑟+2𝑃3𝑙+3𝑟+2 − 𝑃3𝑙+2

2  

= (
(𝛽−𝑐)𝛼𝑙−𝑟+1−(𝛼−𝑐)𝛽𝑙−𝑟+1

𝛼−𝛽
) (

(𝛽−𝑐)𝛼𝑙+𝑟+1−(𝛼−𝑐)𝛽𝑙+𝑟+1

𝛼−𝛽
) − (

(𝛽−𝑐)𝛼𝑙+1−(𝛼−𝑐)𝛽𝑙+1

𝛼−𝛽
)
2

.  

=
1

(𝛼−𝛽)2

[
 
 
 
 
(𝛼𝑙−𝑟 − 𝛽𝑙−𝑟)(𝛼𝑙+𝑟 − 𝛽𝑙+𝑟) + 𝑐(𝛼𝑙−𝑟 − 𝛽𝑙−𝑟)(𝛼𝑙+𝑟+1 − 𝛽𝑙+𝑟+1)

+𝑐(𝛼𝑙−𝑟+1 − 𝛽𝑙−𝑟+1)(𝛼𝑙+𝑟 − 𝛽𝑙+𝑟)

+𝑐2(𝛼𝑙−𝑟+1 − 𝛽𝑙−𝑟+1)(𝛼𝑙+𝑟+1 − 𝛽𝑙+𝑟+1)

−(𝛼𝑙−𝑟 − 𝛽𝑙−𝑟)2 − (𝛼𝑙 − 𝛽𝑙)(𝛼𝑙+1 − 𝛽𝑙+1) − 𝑐2(𝛼𝑙+1 − 𝛽𝑙+1)2 ]
 
 
 
 

    

=
1

(𝛼−𝛽)2
[

(−1)𝑙−𝑟+1[𝛽2𝑟 + 𝛼2𝑟 + 2(−1)𝑟]

−𝑐2(−1)𝑙−𝑟+1[𝛽2𝑟 + 𝛼2𝑟 + 2(−1)𝑟]

−𝑐(−1)𝑙+1 [(
𝛽

𝛼
)
𝑟
(𝛼 + 𝛽) + (

𝛼

𝛽
)
𝑟
(𝛼 + 𝛽) − 2(𝛼 + 𝛽)]

]   

=
(−1)𝑙−𝑟+1

(𝛼−𝛽)2
{(𝛼𝑟 − 𝛽𝑟)2 − 𝑐𝑢[𝛽2𝑟 + 𝛼2𝑟 + 2(−1)𝑟] − 𝑐2(𝛼𝑟 − 𝛽𝑟)2}  



Parimalkumar A. Patel, Devbhadra V. Shah 

194 

 

=
(−1)𝑙−𝑟+1

(𝛼−𝛽)2
(𝛼𝑟 − 𝛽𝑟)2{1 + 𝑐𝑢 − 𝑐2} =

(−1)𝑙−𝑟+1

(1+𝑎𝑐)2
{1 + 𝑐𝑢 − 𝑐2}𝑃3𝑟

2   

Thus, 𝑃3𝑙−3𝑟+2𝑃3𝑙+3𝑟+2 − 𝑃3𝑙+2
2 = (−1)𝑙−𝑟+1𝑤𝑃3𝑟

2 𝑃3
−2 

 

 Hence, in general we write 𝑃𝑘−3𝑟𝑃𝑘+3𝑟 − 𝑃𝑘
2 = (−1)𝑙−𝑟+1𝑤𝑃3𝑟

2 𝑃3
−2. 

 

The following identity is analogous to the Cassini’s identity for Fibonacci 

numbers which follows easily from the above theorem. 

 

Corollary 4.2. 𝑃𝑘−3𝑃𝑘+3 − 𝑃𝑘
2 = (−1)𝑛𝑤 for any integer 𝑘 ≥ 3.  

 

 The following identity is similar to d’Ocagne’s identity of Fibonacci 

numbers. 

 

Theorem 4.3. 𝑃𝑚𝑃𝑛+3 − 𝑃𝑚+3𝑃𝑛 = (−1)
𝑛𝑃𝑚−𝑛 (

𝑤

𝑃3
) where 𝑚, 𝑛 are 

nonnegative integers such that 𝑚 ≥ 𝑛 and 𝑚 ≡ 𝑛(𝑚𝑜𝑑 3). 
 

Proof. Since 𝑚 ≡ 𝑛(𝑚𝑜𝑑 3), we first let both 𝑚, 𝑛 to be of the form 3𝑗, 3𝑘 
respectively, for positive integers 𝑗 and 𝑘 ≤ 𝑗. Then 

𝑃3𝑗𝑃3𝑘+3 − 𝑃3𝑗+3𝑃3𝑘 = (1 + 𝑎𝑐) (
𝛼𝑗−𝛽𝑗

𝛼−𝛽
) (1 + 𝑎𝑐) (

𝛼𝑘+1−𝛽𝑘+1

𝛼−𝛽
)  

               −(1 + 𝑎𝑐) (
𝛼𝑗+1−𝛽𝑗+1

𝛼−𝛽
) (1 + 𝑎𝑐) (

𝛼𝑘−𝛽𝑘

𝛼−𝛽
)  

                       =
(1+𝑎𝑐)2

(𝛼−𝛽)2
{𝛼𝑗𝛽𝑘(𝛼 − 𝛽) − 𝛼𝑘𝛽𝑗(𝛼 − 𝛽)}  

                        = (1 + 𝑎𝑐)2(−1)𝑘 (
𝛼𝑗−𝑘−𝛽𝑗−𝑘

𝛼−𝛽
) = (−1)

⌊
𝑛−1

3
⌋
𝑤𝑃𝑚−𝑛𝑃3

−1 

 

       If 𝑚, 𝑛 are of the form 3𝑗 + 1 and 3𝑘 + 1 respectively, then for positive 
integers 𝑗 and 𝑘 ≤ 𝑗, we have 
𝑃3𝑗+1𝑃3𝑘+3+1 − 𝑃3𝑗+3+1𝑃3𝑘+1 

= [
(
(𝑎𝛽+1)𝛼𝑗+1−(𝑎𝛼+1)𝛽𝑗+1

𝛼−𝛽
) (

(𝑎𝛽+1)𝛼𝑘+2−(𝑎𝛼+1)𝛽𝑘+2

𝛼−𝛽
)

− (
(𝑎𝛽+1)𝛼𝑗+2−(𝑎𝛼+1)𝛽𝑗+2

𝛼−𝛽
) (

(𝑎𝛽+1)𝛼𝑘+1−(𝑎𝛼+1)𝛽𝑘+1

𝛼−𝛽
)
]   

=
1

(𝛼−𝛽)2

[
 
 
 
 𝑎

2 (𝛼𝑗𝛽𝑘(𝛼 − 𝛽) − 𝛼𝑘𝛽𝑗(𝛼 − 𝛽))

−𝑎 (𝛼𝑗𝛽𝑘(𝛼2 − 𝛽2) − 𝛼𝑘𝛽𝑗(𝛼2 − 𝛽2))

−(𝛼𝑗𝛽𝑘(𝛼 − 𝛽) − 𝛼𝑘𝛽𝑗(𝛼 − 𝛽) ]
 
 
 
 

   

=
1

(𝛼−𝛽)
[(𝑎2 − 𝑎𝑢 − 1)𝛼𝑘𝛽𝑘(𝛼𝑗−𝑘 − 𝛽𝑗−𝑘)]  

= (−1)𝑘(𝑎2 − 𝑎𝑢 − 1) (
𝛼𝑗−𝑘−𝛽𝑗−𝑘

𝛼−𝛽
) = (−1)

⌊
𝑛−1

3
⌋
𝑤𝑃𝑚−𝑛𝑃3

−1  



The sequence of trifurcating Fibonacci numbers 

195 

 

 Finally, if 𝑚, 𝑛 are of the form 3𝑗 + 2 and 3𝑘 + 2 respectively, then for 
positive integers 𝑗 and 𝑘 ≤ 𝑗, we have 

 𝑃3𝑗+2𝑃3𝑘+3+2 − 𝑃3𝑗+3+2𝑃3𝑘+2 = (
(𝛽−𝑐)𝛼𝑗+1−(𝛼−𝑐)𝛽𝑗+1

𝛼−𝛽
) (

(𝛽−𝑐)𝛼𝑘+2−(𝛼−𝑐)𝛽𝑘+2

𝛼−𝛽
) 

       − (
(𝛽−𝑐)𝛼𝑗+2−(𝛼−𝑐)𝛽𝑗+2

𝛼−𝛽
) (

(𝛽−𝑐)𝛼𝑘+1−(𝛼−𝑐)𝛽𝑘+1

𝛼−𝛽
)                 

=
1

(𝛼−𝛽)2
[
𝛼𝑗𝛽𝑘(𝛼 − 𝛽) − 𝛼𝑘𝛽𝑗(𝛼 − 𝛽) − 𝑐2(𝛼𝑗𝛽𝑘(𝛼 − 𝛽) − 𝛼𝑘𝛽𝑗(𝛼 − 𝛽)

+𝑐 (𝛼𝑗𝛽𝑘(𝛼2 − 𝛽2) − 𝛼𝑘𝛽𝑗(𝛼2 − 𝛽2))
]  

=
1

(𝛼−𝛽)
[(1 + 𝑐𝑢 − 𝑐2)𝛼𝑘𝛽𝑘(𝛼𝑗−𝑘 − 𝛽𝑗−𝑘)]  

= (−1)𝑘(𝑎2 − 𝑎𝑢 − 1) (
𝛼𝑗−𝑘−𝛽𝑗−𝑘

𝛼−𝛽
) = (−1)

⌊
𝑛−1

3
⌋
𝑤𝑃𝑚−𝑛𝑃3

−1  

Combining all the above cases, we finally get  

𝑃𝑚𝑃𝑛+3 − 𝑃𝑚+3𝑃𝑛 = (−1)
⌊
𝑛−1

3
⌋
𝑤𝑃𝑚−𝑛𝑃3

−1.  

 

 We use above Binet-like formula to prove the following identity which 

combines four consecutive 𝑃𝑛’s. 
 

Theorem 4.4.  

𝑎
⌊1−

𝜒(𝑘+2)
3

⌋
𝑏

⌊1−
𝜒(𝑘+1)

3
⌋
𝑐

⌊1−
𝜒(𝑘)

3
⌋
𝑃𝑘+1

2 + 𝑎
⌊1−

𝜒(𝑘)
3

⌋
𝑏

⌊1−
𝜒(𝑘+2)

3
⌋
𝑐

⌊1−
𝜒(𝑘+1)

3
⌋
𝑃𝑘+2

2  

= 𝑃𝑘+2𝑃𝑘+3 − 𝑃𝑘𝑃𝑘+1. 
 

Proof. We prove the result only for the case 𝑘 = 3𝑛 and the remaining cases  
𝑘 = 3𝑛 + 1 and 𝑘 = 3𝑛 + 2 can be handled accordingly.  
 

Now, 𝑐𝑃3𝑛+1
2 + 𝑎𝑃3𝑛+2

2  

= 𝑐 {(
(𝑎𝛽+1)𝛼𝑛+1−(𝑎𝛼+1)𝛽𝑛+1

𝛼−𝛽
)}

2

+ 𝑎 {(
(𝛽−𝑐)𝛼𝑛+1−(𝛼−𝑐)𝛽𝑛+1

𝛼−𝛽
)}

2

.  

=
1

(𝛼−𝛽)2

[
 
 
 
 
 

𝑎2𝑐(𝛼2𝑛 − 2(−1)𝑛 + 𝛽2𝑛)

−2𝑎𝑐(𝛼2𝑛+1 − (−1)𝑛𝛼 − (−1)𝑛𝛽 + 𝛽2𝑛+1)

+𝑐(𝛼2𝑛+2 − 2(−1)𝑛+1 + 𝛽2𝑛+2)

+2𝑎𝑐(𝛼2𝑛+1 − (−1)𝑛𝛽 + (−1)𝑛𝛼 + 𝛽2𝑛+1)

+𝑐2(𝛼2𝑛+2 − 2(−1)𝑛+1 + 𝛽2𝑛+2) + 𝑎(𝛼2𝑛 − 2(−1)𝑛 + 𝛽2𝑛)]
 
 
 
 
 

  

=
(1+𝑎𝑐)

(𝛼−𝛽)2
{𝑎(𝛼𝑛 − 𝛽𝑛)2 + 𝑐(𝛼𝑛+1 − 𝛽𝑛+1)2}  

 

 Also, 𝑃3𝑛+2𝑃3𝑛+3 − 𝑃3𝑛𝑃3𝑛+1 

= (
(𝛽−𝑐)𝛼𝑛+1−(𝛼−𝑐)𝛽𝑛+1

𝛼−𝛽
) (1 + 𝑎𝑐) (

𝛼𝑛+1−𝛽𝑛+1

𝛼−𝛽
)  

−(1 + 𝑎𝑐) (
𝛼𝑛−𝛽𝑛

𝛼−𝛽
) (

(𝑎𝛽+1)𝛼𝑛+1−(𝑎𝛼+1)𝛽𝑛+1

𝛼−𝛽
)      



Parimalkumar A. Patel, Devbhadra V. Shah 

196 

 

=
(1+𝑎𝑐)

(𝛼−𝛽)2
[
(𝛼𝑛 − 𝛽𝑛)(𝛼𝑛+1 − 𝛽𝑛+1) + 𝑐(𝛼𝑛+1 − 𝛽𝑛+1)2

+𝑎(𝛼𝑛 − 𝛽𝑛)2 − (𝛼𝑛 − 𝛽𝑛)(𝛼𝑛+1 − 𝛽𝑛+1)
]  

=
(1+𝑎𝑐)

(𝛼−𝛽)2
{𝑎(𝛼𝑛 − 𝛽𝑛)2 + 𝑐(𝛼𝑛+1 − 𝛽𝑛+1)2}  

This proves the required result.  

 

4.  Conclusions 

 

 In this paper, we considered the sequence of ‘trifurcating Fibonacci 

numbers’ and obtained its Binet-like formula. We also obtained the analogous 

of Cassini’s identity, Catalan’s identity, d’Ocagne’s identity and some 

fundamental identities for the terms of this sequence. 

 

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