Ratio Mathematica                                                              Volume 42, 2022 

 

              
 

 

 

 

Insertion of terms satisfying the 

recurrence relations of Horadam 

sequence and Bifurcating Fibonacci 

sequences 

 
 

Khushbu Das * 

Devbhadra Shah † 
 

 

 

Abstract  

In this article, we consider the problem of finding general formula 

for the terms introduced between two given positive integers ‘𝑎’ 
and ‘𝑏’ in such a way that the terms of newly formed finite sequence 
satisfy the recurrence relations of Horadam sequence and some 

bifurcating Fibonacci sequences. 

Keywords: Generalized Fibonacci sequences, Horadam sequence,   

Bifurcating sequence, Inserted terms, Missing terms. 

2010 AMS Subject Classification: 11B37, 11B39, 11B99. ‡ 
 

 
 

 

 

 

 

 
* Khushbu Das (Department of Mathematics, Veer Narmad South Gujarat University, Surat, 

India); khushbudas14@gmail.com. 
† Devbhadra Shah (Department of Mathematics, Veer Narmad South Gujarat University, Surat, 

India); drdvshah@yahoo.com. 
‡ Received on September 15th, 2021. Accepted on April 17th, 2022. Published on June 30th, 

2022. doi: 10.23755/rm.v39i0.639. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This 

paper is published under the CC-BY licence agreement. 

7



Khushbu J. Das and Devbhadra V. Shah 

 

 

1  Introduction 
 

Consider any two fixed given positive integers 𝑎 and 𝑏 such that                             
𝑎 < 𝑏. We insert 𝑘 number of terms between 𝑎 and 𝑏 so that every term inserted 
between follows the given recurrence relation of any generalized Fibonacci 

sequence. Then we say 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) to be the inserted term (or missing term, 
as defined by some of the authors) in the finite (generalized Fibonacci-like) 

sequence 𝑎,𝑥1,𝑥2,𝑥3,…. ,𝑥𝑘,𝑏. 

The problem of insertion of terms in arithmetic, harmonic, geometric 

sequence is considered to be elementary which occurs in the high school 

mathematics. However, not much work is done regarding the general formula 

for the inserted terms in a Fibonacci-like sequences. Howell [8] presented a 

proof for finding the 𝑛th term of Fibonacci sequence using vectors and 
eigenvalues. But he said nothing for the Fibonacci-like sequences. Agnes, et al. 

[1] provided a formula for inclusion of three consecutive missing terms in 

Fibonacci-like sequence.  

Horadam [7] defined a linear recurrence sequence of second order 𝐹𝑛(𝑝,𝑞), 
acknowledged as Horadam sequence, by the recurrence relation                         

𝐹𝑛
(𝑝,𝑞)

= 𝑝𝐹𝑛−1
(𝑝,𝑞)

+ 𝑞𝐹𝑛−2
(𝑝,𝑞)

 with the initial conditions 𝐹0
(𝑝,𝑞)

= 𝑎,𝐹1
(𝑝,𝑞)

= 𝑐, 
where 𝑎,𝑐 and 𝑝,𝑞 are arbitrary positive integers. First few terms of this 
sequence are shown in the Table 1.  

 

n 𝑭𝒏
(𝒑,𝒒)

 

0 𝑎 

1 𝑐 

2 𝑝𝑐 + 𝑞𝑎 

3 𝑝2𝑐 + 𝑝𝑞𝑎 + 𝑞𝑐 

4 𝑝3𝑐 + 𝑝2𝑞𝑎 + 2𝑝𝑞𝑐 + 𝑞2𝑎 

5 𝑝4𝑐 + 𝑝3𝑞𝑎 + 3𝑝2𝑞𝑐 + 2𝑝𝑞2𝑎 + 𝑞2𝑐 

6 𝑝5𝑐 + 𝑝4𝑞𝑎 + 4𝑝3𝑞𝑐 + 3𝑝2𝑞2𝑎 + 3𝑝𝑞2𝑐 + 𝑞3𝑎 
 

Table 1 

8



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

The Binet-type explicit formula for the terms of this sequence is given by               

𝐹𝑘
(𝑝,𝑞)

=
(𝑐−𝑎𝛽)𝛼𝑘−(𝑐−𝑎𝛼)𝛽𝑘

𝛼−𝛽
, where 𝛼 =

𝑝+√𝑝2+4𝑞

2
,𝛽 =

𝑝−√𝑝2+4𝑞

2
. Here we note 

that 𝛼 − 𝛽 = √𝑝2 + 4𝑞,  𝛼 + 𝛽 = 𝑝 and 𝛼𝛽 = −𝑞.   
 

 Diwan, Shah [4,5] considered the sequence {𝐹𝑛
(𝑝,𝑞,𝑟,𝑠)

} defined by the 

recurrence relation 

        𝐹𝑛
(𝑝,𝑞,𝑟,𝑠)

= 𝑝𝜒(𝑛)𝑞1−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,𝑟,𝑠)

+ 𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)𝐹𝑛−2
(𝑝,𝑞,𝑟,𝑠)

,        (1.1) 

 

where 𝑝,𝑞,𝑟,𝑠 are any fixed positive integers and 𝜒(𝑛) = {
1;if 𝑛 is odd

 0;if 𝑛 is even
.  

They studied this sequence extensively for some explicit values of 𝑝,𝑞,𝑟,𝑠. 
Verma and Bala, Bilgici, Yayenie [2,9,10] also studied this sequence for some 

specific values of 𝑝,𝑞,𝑟,𝑠. It is easy to observe that this sequence is bifurcating 
sequence depending on the parity of 𝑛. 
 

 In this paper we derive the general formula which gives the value of any 

inserted term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) for the sequence {𝐹𝑛(𝑝,𝑞)} and various 

bifurcating subsequence of {𝐹𝑛
(𝑝,𝑞,𝑟,𝑠)

} by considering some fixed values of 

𝑝,𝑞,𝑟,𝑠. Throughout we assume that the positive integers 𝑎,𝑏 are the given first 
and last term respectively in the sequence to be considered.   

 

2 Insertion of terms satisfying the recurrence 

relation of Horadam sequence 
 

In this section, we find the formula for the inserted terms between the given 

fixed positive integers 𝑎,𝑏 so that terms of the sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 
satisfies the recurrence relation of 𝐹𝑛(𝑝,𝑞). Before finding the general formula, 
we find the formula for the first term 𝑥1 in the finite sequence 
𝑎,𝑥1,𝑥2,𝑥3,…. ,𝑥𝑘,𝑏. This value of 𝑥1 will be further used to find the general 
formula for any 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘). 

 

In [7], Horadam obtained only the formula for the first missing term 

(inserted term) 𝑥1 when 𝑘 terms are inserted between given two positive 

9



Khushbu J. Das and Devbhadra V. Shah 

 

 

integers 𝑎 and 𝑏, so that all 𝑥𝑖’s satisfy the recurrence relation of 𝐹𝑛
(𝑝,𝑞)

. In fact, 

he proved that 

 𝑥1 =
𝑏−𝑎𝑞𝐹𝑘

(𝑝,𝑞)

𝐹
𝑘+1
(𝑝,𝑞) .                                               (2.1) 

 

In the following theorem, we obtain the general formula for any arbitrary 

term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) when 𝑘 terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between 

positive integers 𝑎 and 𝑏, so that all 𝑥𝑖’s satisfy the recurrence relation of 𝐹𝑛
(𝑝,𝑞)

. 

 

Theorem 2.1. When the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given 
integers 𝑎 and 𝑏 in such a way that the finite sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 

satisfy the recurrence relation of 𝐹𝑛
(𝑝,𝑞)

, then the general formula for any 

arbitrary term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) is given by  
 

𝑥𝑛 =
𝐹𝑛
(𝑝,𝑞)

𝑏(𝛼−𝛽)−𝑎𝑞𝐹𝑛
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼𝑘−(𝑐−𝑎𝛼)𝛽
𝑘
}+𝑎𝑞𝐹𝑛−1

(𝑝,𝑞)
{(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽

𝑘+1
}

(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽
𝑘+1 .  (2.2) 

 

Proof. We prove the result by the principle of mathematical induction. For we 
𝑘 = 1, we get 𝑛 = 1 and   

𝑥1 =
𝐹1
(𝑝,𝑞)

𝑏(𝛼−𝛽)−𝑎𝑞𝐹1
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼−(𝑐−𝑎𝛼)𝛽}+𝑎𝑞𝐹0
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼2−(𝑐−𝑎𝛼)𝛽
2
}

(𝑐−𝑎𝛽)𝛼2−(𝑐−𝑎𝛼)𝛽
2 =

𝑏−𝑎𝑞𝑐

𝑐𝑝+𝑎𝑞
 , 

which is same as (2.1).   

Next, we assume that (2.2) holds for some positive integer not exceeding 

𝑛. Then both the following results hold:  

𝑥𝑛−1 =
𝐹𝑛−1
(𝑝,𝑞)

𝑏(𝛼−𝛽)−𝑎𝑞𝐹𝑛−1
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼𝑘−(𝑐−𝑎𝛼)𝛽
𝑘
}+𝑎𝑞𝐹𝑛−2

(𝑝,𝑞)
{(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽

𝑘+1
}

(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽
𝑘+1   

and 

𝑥𝑛 =
𝐹𝑛
(𝑝,𝑞)

𝑏(𝛼−𝛽)−𝑎𝑞𝐹𝑛
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼𝑘−(𝑐−𝑎𝛼)𝛽
𝑘
}+𝑎𝑞𝐹𝑛−1

(𝑝,𝑞)
{(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽

𝑘+1
}

(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽
𝑘+1   

Now consider  

𝑝𝑥𝑛 + 𝑞𝑥𝑛−1 = 

(𝑝𝐹𝑛
(𝑝,𝑞)

+𝑞𝐹𝑛−1
(𝑝,𝑞)

)𝑏(𝛼−𝛽)−𝑎𝑞(𝑝𝐹𝑛
(𝑝,𝑞)

+𝑞𝐹𝑛−1
(𝑝,𝑞)

){(𝑐−𝑎𝛽)𝛼𝑘−(𝑐−𝑎𝛼)𝛽𝑘}

+𝑎𝑞(𝑝𝐹𝑛−1
(𝑝,𝑞)

+𝑞𝐹𝑛−2
(𝑝,𝑞)

){(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽𝑘+1}

(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽𝑘+1
     

                     =

𝐹𝑛+1
(𝑝,𝑞)

𝑏(𝛼−𝛽)−𝑎𝑞𝐹𝑛+1
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼𝑘−(𝑐−𝑎𝛼)𝛽𝑘}

+𝑎𝑞𝐹𝑛
(𝑝,𝑞)

{(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽𝑘+1}

(𝑐−𝑎𝛽)𝛼𝑘+1−(𝑐−𝑎𝛼)𝛽𝑘+1
 .  

It can be observed that the right side of this result simplifies to 𝑥𝑛+1. Thus, 
by (2.2) is true for every positive integer 𝑛. 

 

10



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

3 Insertion of terms satisfying the recurrence 

relation of bifurcating sequence 𝑭𝒏
(𝒑,𝒒,𝟏,𝟏)

  
 

In this section, we find the formula for the inserted terms between the 

numbers 𝑎 and 𝑏 so that terms of the sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 satisfies the 
recurrence relation (1.2).  

If we let 𝑝 = 𝑞 = 𝑟 = 𝑠 = 1, the sequence {𝐹𝑛
(𝑝,𝑞,𝑟,𝑠)

} is the sequence of 

usual Fibonacci numbers. If we define 𝐹0
(𝑝,𝑞,𝑟,𝑠)

= 0,𝐹1
(𝑝,𝑞,𝑟,𝑠)

= 1, then first 
few terms of this sequence are shown in Table 2.  

 

n 𝑭𝒏
(𝒑,𝒒,𝒓,𝒔)

 

0 0 

1 1 

2 𝑞 

3 𝑝𝑞 + 𝑟 

4 𝑝𝑞2 + 𝑞𝑟 + 𝑞𝑠 

5 𝑝2𝑞2 + 2𝑝𝑞𝑟 + 𝑝𝑞𝑠 + 𝑟2 

6 𝑝2𝑞3 + 2𝑝𝑞2𝑟 + 2𝑝𝑞2𝑠 + 𝑟2𝑞 + 𝑠𝑞𝑟 + 𝑞𝑠2 
 

Table 2 

If we consider 𝑟 = 𝑠 = 1 in (1.1), we get the sequence {𝐹𝑛
(𝑝,𝑞,1,1)

} whose 

terms are defined by the recurrence relation 

𝐹𝑛
(𝑝,𝑞,1,1)

= 𝑝𝜒(𝑛)𝑞1−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

+ 𝐹𝑛−2
(𝑝,𝑞,1,1)

,  

where 𝐹0
(𝑝,𝑞,1,1)

= 0,𝐹1
(𝑝,𝑞,1,1)

= 1. This can be written in the form 
 

     𝐹𝑛
(𝑝,𝑞,1,1)

= {
𝑝𝐹𝑛−1

(𝑝,𝑞,1,1)
+ 𝐹𝑛−2

(𝑝,𝑞,1,1)
 ;when 𝑛 is odd

 𝑞𝐹𝑛−1
(𝑝,𝑞,1,1)

+ 𝐹𝑛−2
(𝑝,𝑞,1,1)

 ;when 𝑛 is even
(𝑛 ≥ 2)               (3.1) 

 

with the initial conditions 𝐹0
(𝑝,𝑞,1,1)

= 0,𝐹1
(𝑝,𝑞,1,1)

= 1. First few terms of 
this sequence are shown in Table 3. 

 This sequence was studied in detail by Diwan, Shah [4] as well as Edson, 

Yayenie [6]. They derived the Binet-type explicit formula for the terms of this 

sequence as 

 𝐹𝑘
(𝑝,𝑞,1,1)

=
𝑞1−𝜒(𝑘)

(𝑝𝑞)
[
𝑘
2
]
(
𝛼𝑘−𝛽𝑘

𝛼−𝛽
), where 𝛼 =

𝑝𝑞+√𝑝2𝑞2+4𝑝𝑞

2
,𝛽 =

𝑝𝑞−√𝑝2𝑞2+4𝑝𝑞

2
. 

Here we note that 𝛼 − 𝛽 = √𝑝2𝑞2 + 4𝑝𝑞,  𝛼 + 𝛽 = 𝑝𝑞 and 𝛼𝛽 = −𝑝𝑞.  

11



Khushbu J. Das and Devbhadra V. Shah 

 

 

n 𝑭𝒏
(𝒑,𝒒,𝟏,𝟏)

 

0                                  0 

1                                  1 

2 𝑞 

3 𝑝𝑞 + 1 

4 𝑝𝑞2 + 2𝑞 

5 𝑝2𝑞2 + 3𝑝𝑞 + 1 

6 𝑝2𝑞3 + 4𝑝𝑞2 + 3𝑞 
 

Table 3 
 

When we consider the sequence 𝑎,𝑥1,𝑏, where 𝑥1 is the only inserted term 

between 𝑎 and 𝑏, then using 𝐹2
(𝑝,𝑞,1,1)

= 𝑞𝐹1
(𝑝,𝑞,1,1)

+ 𝐹0
(𝑝,𝑞,1,1)

, we get               

𝑏 = 𝑞𝑥1 + 𝑎. Thus 𝑥1 =
𝑏−𝑎

𝑞
 .  

When we consider the finite sequence 𝑎,𝑥1,𝑥2,𝑏, so that it satisfies (3.1), 

we observe that 𝑥1 =
𝑥2−𝑎

𝑞
 and 𝑥2 =

𝑏−𝑥1

𝑝
. This gives 𝑥2 =

𝑏𝑞+𝑎

𝑝𝑞+1
 and 𝑥1 =

𝑏−𝑎𝑝

𝑝𝑞+1
. 

Further, considering the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑏, it is now easy to observe 

that 𝑥1 =
𝑥2−𝑎

𝑞
,𝑥2 =

𝑥3−𝑥1

𝑝
 and 𝑥3 =

b−𝑥2

𝑞
. Solving these three equations in 

three variables 𝑥1,𝑥2,𝑥3 we get 𝑥1 =
𝑏−𝑎(𝑝𝑞+1)

𝑝𝑞2+2𝑞
,𝑥2 =

𝑏𝑞+𝑎𝑞

𝑝𝑞2+2𝑞
,𝑥3 =

𝑏−𝑎+𝑏𝑝𝑞

𝑝𝑞2+2𝑞
 . 

If we continue extending the above finite sequence one more time, then we 

get the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑥4,𝑏. Using the similar approach as above, we 
obtain  

𝑥1 =
𝑏−𝑎(𝑝2𝑞+2𝑝)

𝑝2𝑞2+3𝑝𝑞+1
,𝑥2 =

𝑏𝑞+𝑎𝑝𝑞+𝑎

𝑝2𝑞2+3𝑝𝑞+1
,𝑥3 =

𝑏−𝑎𝑝+𝑏𝑝𝑞

𝑝2𝑞2+3𝑝𝑞+1
 and 𝑥4 =

𝑎+2𝑏𝑞+𝑏𝑝𝑞2

𝑝2𝑞2+3𝑝𝑞+1
 . 

 

 We mention these results in the Table 4. 
 

Number of 

inserted 

terms 

Formula 

x1 x2 x3 x4 

1 
𝑏 − 𝑎

𝑞
 --- --- --- 

2 
𝑏 − 𝑎𝑝

𝑝𝑞 + 1
 

𝑏𝑞 + 𝑎

𝑝𝑞 + 1
 --- --- 

3 
𝑏 − 𝑎(𝑝𝑞 + 1)

𝑝𝑞2 + 2𝑞
 

𝑏𝑞 + 𝑎𝑞

𝑝𝑞2 + 2𝑞
 

𝑏 − 𝑎 + 𝑏𝑝𝑞

𝑝𝑞2 + 2𝑞
 --- 

4 
𝑏 − 𝑎(𝑝2𝑞 + 2𝑝)

𝑝2𝑞2 + 3𝑝𝑞 + 1
 

𝑏𝑞 + 𝑎𝑝𝑞 + 𝑎

𝑝2𝑞2 + 3𝑝𝑞 + 1
 

𝑏 − 𝑎𝑝 + 𝑏𝑝𝑞

𝑝2𝑞2 + 3𝑝𝑞 + 1
 
𝑎 + 2𝑏𝑞 + 𝑏𝑝𝑞2

𝑝2𝑞2 + 3𝑝𝑞 + 1
 

 

Table 4 

12



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

 

From this table, we observe that there is a similar pattern for the first 

inserted term in case of any number of inserted terms. We thus generalize it for 

the case of any number of inserted terms and when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 
are inserted between given two positive integers 𝑎 and 𝑏, we can now write the 
general formula for 𝑥1 as 

 

                           𝑥1 =
𝑏−𝑎𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹

𝑘
(𝑝,𝑞,1,1)

𝐹
𝑘+1
(𝑝,𝑞,1,1)  .                                (3.2) 

        

Next, when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given two 
positive integers 𝑎 and 𝑏, we obtain the general formula for 𝑛th inserted term 
(1 ≤ 𝑛 ≤ 𝑘) using the recurrence relation (3.1).  

 

Theorem 3.1. When the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given 

integers 𝑎 and 𝑏, so that all 𝑥𝑖’s satisfy the recurrence relation of 𝐹𝑛
(𝑝,𝑞,1,1)

, the 

general formula for any arbitrary term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) is given by  

𝑥𝑛 =
𝐹𝑛
(𝑝,𝑞,1,1)

𝑏−𝑎𝐹𝑛
(𝑝,𝑞,1,1)

𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

𝐹𝑘+1
(𝑝,𝑞,1,1)

𝐹
𝑘+1
(𝑝,𝑞,1,1) . 

 

Proof. We use the principle of mathematical induction to prove the result. Since 

by (3.2), we have 𝑥1 =
𝑏−𝑎𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹

𝑘
(𝑝,𝑞,1,1)

𝐹
𝑘+1
(𝑝,𝑞,1,1)  , which proves the result for  

𝑛 = 1. We next assume that it is true for all positive integers not exceeding 𝑛. 
Then the following holds: 

𝑥𝑛−1 =
𝐹𝑛−1
(𝑝,𝑞,1,1)

𝑏−𝑎𝐹𝑛−1
(𝑝,𝑞,1,1)

𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎𝑝𝜒(𝑛−1)𝑞−𝜒(𝑛−1)𝐹𝑛−2
(𝑝,𝑞,1,1)

𝐹𝑘+1
(𝑝,𝑞,1,1)

𝐹
𝑘+1
(𝑝,𝑞,1,1)   

𝑥𝑛−2 =
𝐹𝑛−2
(𝑝,𝑞,1,1)

𝑏−𝑎𝐹𝑛−2
(𝑝,𝑞,1,1)

𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎𝑝𝜒(𝑛−2)𝑞−𝜒(𝑛−2)𝐹𝑛−3
(𝑝,𝑞,1,1)

𝐹𝑘+1
(𝑝,𝑞,1,1)

𝐹
𝑘+1
(𝑝,𝑞,1,1) . 

Now, 𝑝𝜒(𝑛)𝑞1−𝜒(𝑛)𝑥𝑛−1 + 𝑥𝑛−2 
 

           =
{
 
 

 
 (𝑝

𝜒(𝑛)𝑞1−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

+𝐹𝑛−2
(𝑝,𝑞,1,1)

)𝑏−𝑎(𝑝𝜒(𝑛)𝑞1−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

+𝐹𝑛−2
(𝑝,𝑞,1,1)

)

×𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎(𝑝𝜒(𝑛)𝑞1−𝜒(𝑛)𝑝𝜒(𝑛−1)𝑞−𝜒(𝑛−1)𝐹𝑛−2
(𝑝,𝑞,1,1)

+𝑝𝜒(𝑛−2)𝑞−𝜒(𝑛−2)𝐹𝑛−3
(𝑝,𝑞,1,1)

)𝐹𝑘+1
(𝑝,𝑞,1,1)

}
 
 

 
 

𝐹
𝑘+1
(𝑝,𝑞,1,1)   

 

           =

{
𝐹𝑛
(𝑝,𝑞,1,1)

𝑏−𝑎𝐹𝑛
(𝑝,𝑞,1,1)

𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎(𝑝𝜒(𝑛)+𝜒(𝑛−1)𝑞1−(𝜒(𝑛)+𝜒(𝑛−1))𝐹𝑛−2
(𝑝,𝑞,1,1)

+𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−3
(𝑝,𝑞,1,1)

)𝐹𝑘+1
(𝑝,𝑞,1,1)

}

𝐹
𝑘+1
(𝑝,𝑞,1,1)   

13



Khushbu J. Das and Devbhadra V. Shah 

 

 

            =

{
𝐹𝑛
(𝑝,𝑞,1,1)

𝑏−𝑎𝐹𝑛
(𝑝,𝑞,1,1)

𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎(𝑝𝐹𝑛−2
(𝑝,𝑞,1,1)

+𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−3
(𝑝,𝑞,1,1)

)𝐹𝑘+1
(𝑝,𝑞,1,1)

}

𝐹
𝑘+1
(𝑝,𝑞,1,1)  . 

 

Now, when 𝑛 is odd, we get 

 𝑝𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−3
(𝑝,𝑞,1,1)

 

= 𝑝𝜒(𝑛)𝑞1−𝜒(𝑛)𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−3
(𝑝,𝑞,1,1)

 

= 𝑝𝜒(𝑛)𝑞−𝜒(𝑛) (𝑞𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝐹𝑛−3
(𝑝,𝑞,1,1)

) 

= 𝑝𝜒(𝑛)𝑞−𝜒(𝑛) (𝑝𝜒(𝑛−1)𝑞1−𝜒(𝑛−1)𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝐹𝑛−3
(𝑝,𝑞,1,1)

) 

            = 𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

. 
 

Also, when 𝑛 is even, we get 

𝑝𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−3
(𝑝,𝑞,1,1)

 

= 𝑝1−𝜒(𝑛)𝑞𝜒(𝑛)𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−3
(𝑝,𝑞,1,1)

 

= 𝑝𝜒(𝑛)𝑞−𝜒(𝑛) (𝑝1−2𝜒(𝑛)𝑞2𝜒(𝑛)𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝐹𝑛−3
(𝑝,𝑞,1,1)

) 

= 𝑝𝜒(𝑛)𝑞−𝜒(𝑛) (𝑝𝜒(𝑛−1)𝑞1−𝜒(𝑛−1)𝐹𝑛−2
(𝑝,𝑞,1,1)

+ 𝐹𝑛−3
(𝑝,𝑞,1,1)

) 

= 𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

 
 

Therefore, we have  

𝑥𝑛 = 𝑝
𝜒(𝑛)𝑞1−𝜒(𝑛)𝑥𝑛−1 + 𝑥𝑛−2 =

{
𝐹𝑛
(𝑝,𝑞,1,1)

𝑏−𝑎𝐹𝑛
(𝑝,𝑞,1,1)

𝑝1−𝜒(𝑘)𝑞𝜒(𝑘)−1𝐹𝑘
(𝑝,𝑞,1,1)

+𝑎(𝑝𝜒(𝑛)𝑞−𝜒(𝑛)𝐹𝑛−1
(𝑝,𝑞,1,1)

)𝐹𝑘+1
(𝑝,𝑞,1,1)

}

𝐹
𝑘+1
(𝑝,𝑞,1,1)  , 

which proves the result for every positive integer 𝑛. 
 

4 Insertion of terms satisfying the recurrence  

relation of bifurcating sequence 𝑭𝒏
(𝒑,𝟏,𝟏,𝒔)

 
 

If we consider 𝑞 = 𝑟 = 1 in (1.1), the sequence {𝐹𝑛
(𝑝,1,1,𝑠)

} whose terms are 

defined by the recurrence relation 

𝐹𝑛
(𝑝,1,1,𝑠)

= 𝑝𝜒(𝑛)𝐹𝑛−1
(𝑝,1,1,𝑠)

+ 𝑠1−𝜒(𝑛)𝐹𝑛−2
(𝑝,1,1,𝑠)

,  

where 𝐹0
(𝑝,1,1,𝑠)

= 0,𝐹1
(𝑝,1,1,𝑠)

= 1. This can be written in the form  
 

𝐹𝑛
(𝑝,1,1,𝑠)

= {
𝑝𝐹𝑛−1

(𝑝,1,1,𝑠)
+ 𝐹𝑛−2

(𝑝,1,1,𝑠)
 ;when 𝑛 is odd

 𝐹𝑛−1
(𝑝,1,1,𝑠)

+ 𝑠𝐹𝑛−2
(𝑝,1,1,𝑠)

 ;when 𝑛 is even
(𝑛 ≥ 2)         (4.1) 

 

14



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

with the initial conditions 𝐹0
(𝑝,1,1,𝑠)

= 0,𝐹1
(𝑝,1,1,𝑠)

= 1. First few terms of this 
sequence are shown in Table 5. 
 

 

n 𝑭𝒏
(𝒑,𝟏,𝟏,𝒔)

 

0 0 

1 1 

2 1 

3 1 + 𝑝 

4 1 + 𝑝 + 𝑠 

5 1 + 2𝑝 + 𝑝𝑠 + 𝑝2 

6 1 + 𝑠 + 2𝑝 + 2𝑝𝑠 + 𝑝2 + 𝑠2 
 

Table 5 
 

This sequence was studied by Diwan, Shah [4]. They obtained Binet-type 

explicit formula for the terms of this sequence as  

𝐹𝑛
(𝑝,1,1,𝑠)

=
(𝛼−𝑠)𝜒(𝑛)𝛼

[
𝑛
2
]
−(𝛽−𝑠)𝜒(𝑛)𝛽

[
𝑛
2
]

𝛼−𝛽
, 

where 𝛼 =
(𝑝+𝑠+1)+√(𝑝+𝑠+1)2−4𝑠

2
,𝛽 =

(𝑝+𝑠+1)−√(𝑝+𝑠+1)2−4𝑠

2
. This gives                       

𝛼 − 𝛽 = √(𝑝 + 𝑠 + 1)2 − 4𝑠,𝛼 + 𝛽 = 𝑝 + 𝑠 + 1 and 𝛼𝛽 = 𝑠. 
In this section, we find the formula for the inserted terms between the 

numbers 𝑎 and 𝑏 so that terms of the sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 satisfies the 
recurrence relation (4.1). 

When we consider the sequence 𝑎,𝑥1,𝑏, by using 𝐹2
(𝑝,1,1,𝑠)

= 𝐹1
(𝑝,1,1,𝑠)

+

𝑠𝐹0
(𝑝,1,1,𝑠)

, we get 𝑏 = 𝑥1 + 𝑠𝑎. Thus 𝑥1 = 𝑏 − 𝑠𝑎. This basic formula will be 
used to find the other terms.  

When we consider the finite sequence as 𝑎,𝑥1,𝑥2,𝑏, so that it satisfies (4.1), 

we observe that 𝑥1 = 𝑥2 − 𝑠 and 𝑥2 =
𝑏−𝑥1

𝑝
. This gives 𝑥2 =

𝑏+𝑠𝑎

1+𝑝
 and                                  

𝑥1 =
𝑏−𝑠𝑎𝑝

1+𝑝    
. 

Further, considering the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑏, it is now easy to observe 

that 𝑥1 = 𝑥2 − 𝑠𝑎,𝑥2 =
𝑥3−𝑥1

𝑝
 and 𝑥3 = 𝑏 − 𝑠𝑥2. Solving these three equations 

in three variables 𝑥1,𝑥2,𝑥3 we get 𝑥1 =
𝑏−𝑎𝑠(𝑝+𝑠)

1+𝑝+𝑠
,𝑥2 =

𝑏+𝑠𝑎

1+𝑝+𝑠
,𝑥3 =

𝑏+𝑏𝑝−𝑎𝑠2

1+𝑝+𝑠
. 

If we continue extending the above finite sequence one more time, then we 

get the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑥4,𝑏. Using the similar approach as above, we 

15



Khushbu J. Das and Devbhadra V. Shah 

 

 

obtain 𝑥1 =
𝑏−𝑠𝑎(𝑝2+𝑝+𝑝𝑠)

1+2𝑝+𝑝𝑠+𝑝2
,𝑥2 =

𝑏+𝑠𝑎𝑝+𝑠𝑎

1+2𝑝+𝑝𝑠+𝑝2
,𝑥3 =

𝑏+𝑝𝑏−𝑠2𝑝𝑎

1+2𝑝+𝑝𝑠+𝑝2
 and                            

𝑥4 =
𝑏+𝑝𝑏+𝑠𝑏+𝑠2𝑎

1+2𝑝+𝑝𝑠+𝑝2
. 

We mention these results in the Table 6. 
 

Number of 

inserted 

terms 

Formula 

𝑥1 𝑥2 𝑥3 𝑥4 

1 𝑏 − 𝑠𝑎 --- --- --- 

2 
𝑏 − 𝑠𝑎𝑝

1 + 𝑝    
 

𝑏 + 𝑠𝑎

1 + 𝑝
 --- --- 

3 
𝑏 − 𝑠𝑎(𝑝 + 𝑠)

1 + 𝑝 + 𝑠
 

𝑏 + 𝑠𝑎

1 + 𝑝 + 𝑠
 

𝑏 + 𝑏𝑝 − 𝑎𝑠2

1 + 𝑝 + 𝑠
 --- 

4 
𝑏 − 𝑠𝑎(𝑝2 + 𝑝 + 𝑝𝑠)

1 + 2𝑝 + 𝑝𝑠 + 𝑝2
 

𝑏 + 𝑠𝑎𝑝 + 𝑠𝑎

1 + 2𝑝 + 𝑝𝑠 + 𝑝2
 

𝑏 + 𝑝𝑏 − 𝑠2𝑝𝑎

1 + 2𝑝 + 𝑝𝑠 + 𝑝2
 
𝑏 + 𝑝𝑏 + 𝑠𝑏 + 𝑠2𝑎

1 + 2𝑝 + 𝑝𝑠 + 𝑝2
 

 

Table 6 
 

 From this Table, we observe that there is a similar pattern for the first 

inserted term in case of any number of inserted terms. We thus generalize it for 

the case of 𝑘 number of inserted terms and we can now write the general 
formula for 𝑥1 as  

                      

                                            𝑥1 =
𝑏−𝑠𝑎{𝐹

𝑘+1
(𝑝,1,1,𝑠)

−𝐹
𝑘−1
(𝑝,1,1,𝑠)

}

𝐹
𝑘+1
(𝑝,1,1,𝑠) .                               (4.2) 

 

Next, when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given two 
positive integers 𝑎 and 𝑏, we obtain the general formula for 𝑛th inserted term 
(1 ≤ 𝑛 ≤ 𝑘) using the recurrence relation (4.1).  

 

Theorem 4.1. When the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given 

integers 𝑎 and 𝑏, so that all 𝑥𝑖’s satisfy the recurrence relation of 𝐹𝑛
(𝑝,1,1,𝑠)

, the 

general formula for any arbitrary term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) is given by  

𝑥𝑛 =
𝐹𝑛
(𝑝,1,1,𝑠)

𝑏+𝑠𝑎𝐹𝑛
(𝑝,1,1,𝑠)

𝐹𝑘−1
(𝑝,1,1,𝑠)

−𝑠𝑎𝐹𝑛−2
(𝑝,1,1,𝑠)

𝐹𝑘+1
(𝑝,1,1,𝑠)

𝐹
𝑘+1
(𝑝,1,1,𝑠) . 

 

Proof. We use the principle of mathematical induction to prove the result. Since 

by (4.2), we have 𝑥1 =
𝑏−𝑠𝑎{𝐹

𝑘+1
(𝑝,1,1,𝑠)

−𝐹
𝑘−1
(𝑝,1,1,𝑠)

}

𝐹
𝑘+1
(𝑝,1,1,𝑠)  , which proves the result for          

𝑛 = 1. We next assume that it is true for all positive integers not exceeding 𝑛. 
Then the following holds: 

16



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

𝑥𝑛−1 =
𝐹𝑛−1
(𝑝,1,1,𝑠)

𝑏+𝑠𝑎𝐹𝑛−1
(𝑝,1,1,𝑠)

𝐹𝑘−1
(𝑝,1,1,𝑠)

−𝑠𝑎𝐹𝑛−3
(𝑝,1,1,𝑠)

𝐹𝑘+1
(𝑝,1,1,𝑠)

𝐹
𝑘+1
(𝑝,1,1,𝑠)  and 

𝑥𝑛−2 =
𝐹𝑛−2
(𝑝,1,1,𝑠)

𝑏+𝑠𝑎𝐹𝑛−2
(𝑝,1,1,𝑠)

𝐹𝑘−1
(𝑝,1,1,𝑠)

−𝑠𝑎𝐹𝑛−4
(𝑝,1,1,𝑠)

𝐹𝑘+1
(𝑝,1,1,𝑠)

𝐹
𝑘+1
(𝑝,1,1,𝑠) . 

 

Now, 𝑝𝜒(𝑛)𝑥𝑛−1 + 𝑠
1−𝜒(𝑛)𝑥𝑛−2 

 

=

{
(𝑝𝜒(𝑛)𝐹𝑛−1

(𝑝,1,1,𝑠)
+𝑠1−𝜒(𝑛)𝐹𝑛−2

(𝑝,1,1,𝑠)
)𝑏+𝑠𝑎(𝑝𝜒(𝑛)𝐹𝑛−1

(𝑝,1,1,𝑠)
+𝑠1−𝜒(𝑛)𝐹𝑛−2

(𝑝,1,1,𝑠)
)𝐹𝑘−1

(𝑝,1,1,𝑠)

−𝑠𝑎(𝑝𝜒(𝑛)𝐹𝑛−3
(𝑝,1,1,𝑠)

+𝑠1−𝜒(𝑛)𝐹𝑛−4
(𝑝,1,1,𝑠)

)𝐹𝑘+1
(𝑝,1,1,𝑠)

}

𝐹
𝑘+1
(𝑝,1,1,𝑠)   

=

{𝐹𝑛
(𝑝,1,1,𝑠)

𝑏+𝑠𝑎𝐹𝑛
(𝑝,1,1,𝑠)

𝐹𝑘−1
(𝑝,1,1,𝑠)

−𝑠𝑎(
𝑝𝜒(𝑛−2)𝐹𝑛−3

(𝑝,1,1,𝑠)

+𝑠1−𝜒(𝑛−2)𝐹𝑛−4
(𝑝,1,1,𝑠)

)𝐹𝑘+1
(𝑝,1,1,𝑠)

}

𝐹
𝑘+1
(𝑝,1,1,𝑠)   

=
𝐹𝑛
(𝑝,1,1,𝑠)

𝑏+𝑠𝑎𝐹𝑛
(𝑝,1,1,𝑠)

𝐹𝑘−1
(𝑝,1,1,𝑠)

−𝑠𝑎𝐹𝑛−2
(𝑝,1,1,𝑠)

𝐹𝑘+1
(𝑝,1,1,𝑠)

𝐹
𝑘+1
(𝑝,1,1,𝑠) = 𝑥𝑛, which proves the result for 

every positive integer 𝑛. 
 

5 Insertion of terms satisfying the recurrence  

relation of bifurcating sequence 𝑭𝒏
(𝟏,𝒒,𝒓,𝟏)

 
 If we consider 𝑝 = 𝑠 = 1 in (1.1), then we get the sequence {𝐹𝑛

(1,𝑞,𝑟,1)
} 

whose terms are defined by the recurrence relation 

𝐹𝑛
(1,𝑞,𝑟,1)

= 𝑞𝜒(𝑛)𝐹𝑛−1
(1,𝑞,𝑟,1)

+ 𝑟1−𝜒(𝑛)𝐹𝑛−2
(1,𝑞,𝑟,1)

,  

where 𝐹0
(1,𝑞,𝑟,1)

= 0,𝐹1
(1,𝑞,𝑟,1)

= 1.This can be written in the form 
 

𝐹𝑛
(1,𝑞,𝑟,1)

= {
𝐹𝑛−1
(1,𝑞,𝑟,1)

+ 𝑟𝐹𝑛−2
(1,𝑞,𝑟,1)

 ;when 𝑛 is odd

 𝑞𝐹𝑛−1
(1,𝑞,𝑟,1)

+ 𝐹𝑛−2
(1,𝑞,𝑟,1)

 ;when 𝑛 is even
(𝑛 ≥ 2)           (5.1) 

 

with the initial conditions 𝐹0
(1,𝑞,𝑟,1)

= 0,𝐹1
(1,𝑞,𝑟,1)

= 1. First few terms of this 
sequence are shown in Table 7. 

Diwan, Shah [4] studied this sequence and obtained the Binet-type explicit 

formula for the terms of this sequence as  

𝐹𝑛
(1,𝑞,𝑟,1)

= 𝑞1−𝜒(𝑛) {
(𝛼−1)𝜒(𝑛)𝛼

[
𝑛
2
]
−(𝛽−1)𝜒(𝑛)𝛽

[
𝑛
2
]

𝛼−𝛽
}, 

where 𝛼 =
(𝑟+𝑞+1)+√(𝑟+𝑞+1)2−4𝑞

2
,𝛽 =

(𝑟+𝑞+1)−√(𝑟+𝑞+1)2−4𝑞

2
 with                                   

𝛼 − 𝛽 = √(𝑟 + 𝑞 + 1)2 − 4𝑞,𝛼 + 𝛽 = 𝑟 + 𝑞 + 1,𝛼𝛽 = 𝑟. 
 

17



Khushbu J. Das and Devbhadra V. Shah 

 

 

 

n 𝑭𝒏
(𝟏,𝒒,𝒓,𝟏)

 

0                                  0 

1                                  1 

2 𝑞 

3 𝑟 + 𝑞 

4 𝑞 + 𝑟𝑞 + 𝑞2 

5 𝑞 + 2𝑞𝑟 + 𝑟2 + 𝑞2 

6 𝑞 + 𝑟𝑞 + 2𝑟𝑞2 + 2𝑞2 + 𝑟2𝑞 + 𝑞3 
 

Table 7 
 

In this section, we find the formula for the inserted terms between the 

numbers 𝑎 and 𝑏 so that terms of the sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 satisfies the 
recurrence relation (5.1). 

When we consider the sequence 𝑎,𝑥1,𝑏, using 𝐹2
(1,𝑞,𝑟,1)

= 𝑞𝐹1
(1,𝑞,𝑟,1)

+

𝐹0
(1,𝑞,𝑟,1)

, we get  𝑏 = 𝑞𝑥1 + 𝑎. Thus 𝑥1 =
𝑏−𝑎

𝑞
.            

When we consider the sequence as 𝑎,𝑥1,𝑥2,𝑏, so that it satisfies (5.1), we 

observe that 𝑥1 =
𝑥2−𝑎

𝑞
 and 𝑥2 = 𝑏 − 𝑟𝑥1. This gives 𝑥2 =

𝑞𝑏+𝑟𝑎

𝑟+𝑞
 and                          

𝑥1 =
𝑏−𝑎

𝑟+𝑞
 . 

Further, considering the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑏, it is now easy to observe 

that 𝑥1 =
𝑥2−𝑎

𝑞
, 𝑥2 = 𝑥3 − 𝑟𝑥1, 𝑥2 = 𝑥3 − 𝑟𝑥1 and 𝑥3 =

𝑏−𝑥2

𝑞
 . Solving these 

three equations in three variables 𝑥1,𝑥2,𝑥3 we get 𝑥1 =
𝑏−𝑎(𝑞+1)

𝑞2+𝑞+𝑟𝑞
, 𝑥2 =

𝑏𝑞+𝑟𝑞𝑎

𝑞2+𝑞+𝑟𝑞
,𝑥3 =

𝑏𝑞+𝑏𝑟−𝑟𝑎

𝑞2+𝑞+𝑟𝑞
. 

If we continue extending the above finite sequence one more time, then we 

get the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑥4,𝑏. Using the similar approach as above, we 
obtain 

𝑥1 =
𝑏−𝑎(1+𝑟+𝑞)

𝑞2+𝑟2+2𝑟𝑞+𝑞
,𝑥2 =

𝑏𝑞+𝑎𝑟2+𝑎𝑟𝑞

𝑞2+𝑟2+2𝑟𝑞+𝑞
,𝑥3 =

𝑏𝑞+𝑟𝑏−𝑎𝑟

𝑞2+𝑟2+2𝑟𝑞+𝑞
,𝑥4 =

𝑏𝑞2+𝑟𝑏𝑞+𝑞𝑏+𝑎𝑟2

𝑞2+𝑟2+2𝑟𝑞+𝑞
. 

We mention these results in the Table 8. From this table, we observe that 

there is a similar pattern for the first inserted term in case of any number of 

inserted terms. We thus generalize it for the case of any number of inserted 

terms and when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given two 
positive integers 𝑎 and 𝑏, we can now write the general formula for 𝑥1 as 
 

                  𝑥1 =
𝑏−𝑎{[(

1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

𝐹
𝑘+1
(1,𝑞,𝑟,1) .                          (5.2) 

 

18



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

Number 

of 

inserted 

term 

Formula 

𝑥1 𝑥2 𝑥3 𝑥4 

1 
𝑏 − 𝑎

𝑞
 --- --- --- 

2 
𝑏 − 𝑎

𝑟 + 𝑞
 

𝑞𝑏 + 𝑟𝑎

𝑟 + 𝑞
 --- --- 

3 
𝑏 − 𝑎(𝑞 + 1)

𝑞2 + 𝑞 + 𝑟𝑞
 

𝑏𝑞 + 𝑟𝑞𝑎

𝑞2 + 𝑞 + 𝑟𝑞
 

𝑏𝑞 + 𝑏𝑟 − 𝑟𝑎

𝑞2 + 𝑞 + 𝑟𝑞
 --- 

4 
𝑏 − 𝑎(1 + 𝑟 + 𝑞)

𝑞2 + 𝑟2 + 2𝑟𝑞 + 𝑞
 

𝑏𝑞 + 𝑎𝑟2 + 𝑎𝑟𝑞

𝑞2 + 𝑟2 + 2𝑟𝑞 + 𝑞
 

𝑏𝑞 + 𝑟𝑏 − 𝑎𝑟

𝑞2 + 𝑟2 + 2𝑟𝑞 + 𝑞
 
𝑏𝑞2 + 𝑟𝑏𝑞 + 𝑞𝑏 + 𝑎𝑟2

𝑞2 + 𝑟2 + 2𝑟𝑞 + 𝑞
 

 

 Table 8 
  

Next, when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given two 
positive integers 𝑎 and 𝑏, we obtain the general formula for 𝑛th inserted term 
(1 ≤ 𝑛 ≤ 𝑘) using the recurrence relation (5.1).  

 

Theorem 5.1 When the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given 

integers 𝑎 and 𝑏, so that all 𝑥𝑖’s satisfy the recurrence relation of 𝐹𝑛
(1,𝑞,𝑟,1)

, the 

general formula for any arbitrary term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) is given by  

𝑥𝑛 =
{
 
 

 
 𝐹𝑛

(1,𝑞,𝑟,1)
𝑏−𝐹𝑛

(1,𝑞,𝑟,1)
𝑎{[(

1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

+𝑎{[(
1−𝑟

𝑞
)𝐹𝑛−2

(1,𝑞,𝑟,1)
+𝐹𝑛−1

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛)

[
1

𝑞
𝐹𝑛−1
(1,𝑞,𝑟,1)

]
𝜒(𝑛)

}𝐹
𝑘+1
(1,𝑞,𝑟,1)

}
 
 

 
 

𝐹
𝑘+1
(1,𝑞,𝑟,1)  . 

 

Proof. We use the principle of mathematical induction to prove the result.                      

By (5.2), we have 𝑥1 =
𝑏−𝑎{[(

1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

𝐹
𝑘+1
(1,𝑞,𝑟,1) , which 

proves the result for 𝑛 = 1. We next assume that it is true for all positive 
integers not exceeding 𝑛. Then the following holds: 

𝑥𝑛−1 =
{
 
 

 
 𝐹𝑛−1

(1,𝑞,𝑟,1)
𝑏−𝐹𝑛−1

(1,𝑞,𝑟,1)
𝑎{[(

1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

+𝑎{[(
1−𝑟

𝑞
)𝐹𝑛−3

(1,𝑞,𝑟,1)
+𝐹𝑛−2

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛−1)

[
1

𝑞
𝐹𝑛−2
(1,𝑞,𝑟,1)

]
𝜒(𝑛−1)

}𝐹
𝑘+1
(1,𝑞,𝑟,1)

}
 
 

 
 

𝐹
𝑘+1
(1,𝑞,𝑟,1)   

𝑥𝑛−2 =
{
 
 

 
 𝐹𝑛−2

(1,𝑞,𝑟,1)
𝑏−𝐹𝑛−2

(1,𝑞,𝑟,1)
𝑎{[(

1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

+𝑎{[(
1−𝑟

𝑞
)𝐹𝑛−4

(1,𝑞,𝑟,1)
+𝐹𝑛−3

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛−2)

[
1

𝑞
𝐹𝑛−3
(1,𝑞,𝑟,1)

]
𝜒(𝑛−2)

}𝐹
𝑘+1
(1,𝑞,𝑟,1)

}
 
 

 
 

𝐹
𝑘+1
(1,𝑞,𝑟,1)   

19



Khushbu J. Das and Devbhadra V. Shah 

 

 

Now, 𝑞1−𝜒(𝑛)𝑥𝑛−1 + 𝑟
𝜒(𝑛)𝑥𝑛−2 

=
{
 
 
 
 

 
 
 
 (𝑞

1−𝜒(𝑛)𝐹𝑛−1
(1,𝑞,𝑟,1)

+𝑟𝜒(𝑛)𝐹𝑛−2
(1,𝑞,𝑟,1)

)𝑏−𝑎(𝑞1−𝜒(𝑛)𝐹𝑛−1
(1,𝑞,𝑟,1)

+𝑟𝜒(𝑛)𝐹𝑛−2
(1,𝑞,𝑟,1)

)

{[(
1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

+𝑎

{
 

 𝑞1−𝜒(𝑛)[(
1−𝑟

𝑞
)𝐹𝑛−3

(1,𝑞,𝑟,1)
+𝐹𝑛−2

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛−1)

[
1

𝑞
𝐹𝑛−2
(1,𝑞,𝑟,1)

]
𝜒(𝑛−1)

+𝑟𝜒(𝑛)[(
1−𝑟

𝑞
)𝐹𝑛−4

(1,𝑞,𝑟,1)
+𝐹𝑛−3

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛−2)

[
1

𝑞
𝐹𝑛−3
(1,𝑞,𝑟,1)

]
𝜒(𝑛−2)

}
 

 
𝐹
𝑘+1
(1,𝑞,𝑟,1)

}
 
 
 
 

 
 
 
 

𝐹
𝑘+1
(1,𝑞,𝑟,1)   

=
{
  
 

  
 𝐹𝑛

(1,𝑞,𝑟,1)
𝑏−𝑎𝐹𝑛

(1,𝑞,𝑟,1)
{[(

1−𝑟

𝑞
)𝐹𝑘−1

(1,𝑞,𝑟,1)
+𝐹𝑘

(1,𝑞,𝑟,1)
]
𝜒(𝑘)

[
1

𝑞
𝐹𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

+𝑎

{
 

 𝑞1−𝜒(𝑛)[(
1−𝑟

𝑞
)𝐹𝑛−3

(1,𝑞,𝑟,1)
+𝐹𝑛−2

(1,𝑞,𝑟,1)
]
𝜒(𝑛)

[
1

𝑞
𝐹𝑛−2
(1,𝑞,𝑟,1)

]
1−𝜒(𝑛)

+𝑟𝜒(𝑛)[(
1−𝑟

𝑞
)𝐹𝑛−4

(1,𝑞,𝑟,1)
+𝐹𝑛−3

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛)

[
1

𝑞
𝐹𝑛−3
(1,𝑞,𝑟,1)

]
𝜒(𝑛)

}
 

 
𝐹
𝑘+1
(1,𝑞,𝑟,1)

}
  
 

  
 

𝐹
𝑘+1
(1,𝑞,𝑟,1)   

 We calculate the value of third term of numerator separately. Now, when 

𝑛 is odd, we get 

 
𝑞1−𝜒(𝑛) [(

1−𝑟

𝑞
)𝐹𝑛−3

(1,𝑞,𝑟,1)
+ 𝐹𝑛−2

(1,𝑞,𝑟,1)
]
𝜒(𝑛)

[
1

𝑞
𝐹𝑛−2
(1,𝑞,𝑟,1)

]
1−𝜒(𝑛)

+𝑟𝜒(𝑛) [(
1−𝑟

𝑞
)𝐹𝑛−4

(1,𝑞,𝑟,1)
+ 𝐹𝑛−3

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛)

[
1

𝑞
𝐹𝑛−3
(1,𝑞,𝑟,1)

]
𝜒(𝑛)

  

= 𝑞[
1

𝑞
𝐹𝑛−2
(1,𝑞,𝑟,1)

] + (
1−𝑟

𝑞
)𝐹𝑛−4

(1,𝑞,𝑟,1)
+ 𝐹𝑛−3

(1,𝑞,𝑟,1)
  

=
1

𝑞
[𝑞𝐹𝑛−2

(1,𝑞,𝑟,1)
− 𝑟𝐹𝑛−4

(1,𝑞,𝑟,1)
+ (𝑞𝐹𝑛−3

(1,𝑞,𝑟,1)
+ 𝐹𝑛−4

(1,𝑞,𝑟,1)
)]  

=
1

𝑞
[𝑞𝐹𝑛−2

(1,𝑞,𝑟,1)
− 𝑟𝐹𝑛−4

(1,𝑞,𝑟,1)
+ 𝐹𝑛−2

(1,𝑞,𝑟,1)
]  

=
1

𝑞
[(1 + 𝑞)𝐹𝑛−2

(1,𝑞,𝑟,1)
− 𝑟𝐹𝑛−4

(1,𝑞,𝑟,1)
]  

=
1

𝑞
[(1 − 𝑟)𝐹𝑛−2

(1,𝑞,𝑟,1)
+ 𝑞𝐹𝑛−2

(1,𝑞,𝑟,1)
+ 𝑟𝑞𝐹𝑛−3

(1,𝑞,𝑟,1)
]  

= (
1−𝑟

𝑞
)𝐹𝑛−2

(1,𝑞,𝑟,1)
+ 𝐹𝑛−1

(1,𝑞,𝑟,1)
  

= [(
1−𝑟

𝑞
)𝐹𝑛−2

(1,𝑞,𝑟,1)
+ 𝐹𝑛−1

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛)

. 

Also, when 𝑛 is even, we get 

𝑞1−𝜒(𝑛) [(
1 − 𝑟

𝑞
)𝐹𝑛−3

(1,𝑞,𝑟,1)
+ 𝐹𝑛−2

(1,𝑞,𝑟,1)
]
𝜒(𝑛)

[
1

𝑞
𝐹𝑛−2
(1,𝑞,𝑟,1)

]
1−𝜒(𝑛)

+𝑟𝜒(𝑛) [(
1 − 𝑟

𝑞
)𝐹𝑛−4

(1,𝑞,𝑟,1)
+ 𝐹𝑛−3

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛)

[
1

𝑞
𝐹𝑛−3
(1,𝑞,𝑟,1)

]
𝜒(𝑛)

 

= [(
1 − 𝑟

𝑞
)𝐹𝑛−3

(1,𝑞,𝑟,1)
+ 𝐹𝑛−2

(1,𝑞,𝑟,1)
] + 𝑟[

1

𝑞
𝐹𝑛−3
(1,𝑞,𝑟,1)

] 

=
1

𝑞
[𝑞𝐹𝑛−2

(1,𝑞,𝑟,1)
+ 𝐹𝑛−3

(1,𝑞,𝑟,1)
] =

1

𝑞
[𝐹𝑛−1

(1,𝑞,𝑟,1)
] = [

1

𝑞
𝐹𝑛−1
(1,𝑞,𝑟,1)

]
𝜒(𝑛)

 

20



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

 

Therefore, we have 

𝑥𝑛 = 𝑞
1−𝜒(𝑛)𝑥𝑛−1 + 𝑟

𝜒(𝑛)𝑥𝑛−2 

     = {
 
 

 
 𝐹𝑛

(1,𝑞,𝑟,1)
𝑏−𝐹𝑛

(1,𝑞,𝑟,1)
𝑎{[(

1−𝑟

𝑞
)𝐹
𝑘−1
(1,𝑞,𝑟,1)

+𝐹
𝑘
(1,𝑞,𝑟,1)

]
𝜒(𝑘)

[
1

𝑞
𝐹
𝑘
(1,𝑞,𝑟,1)

]
1−𝜒(𝑘)

}

+𝑎{[(
1−𝑟

𝑞
)𝐹𝑛−2

(1,𝑞,𝑟,1)
+𝐹𝑛−1

(1,𝑞,𝑟,1)
]
1−𝜒(𝑛)

[
1

𝑞
𝐹𝑛−1
(1,𝑞,𝑟,1)

]
𝜒(𝑛)

}𝐹
𝑘+1
(1,𝑞,𝑟,1)

}
 
 

 
 

𝐹
𝑘+1
(1,𝑞,𝑟,1)  , which 

proves the result for every positive integer 𝑛. 
 

6 Insertion of terms satisfying the recurrence 

relation of bifurcating sequence 𝑭𝒏
(𝟏,𝟏,𝒓,𝒔)

 
 

 If we consider 𝑝 = 𝑞 = 1 in (1.1), then we get the sequence {𝐹𝑛
(1,1,𝑟,𝑠)

} 

whose terms are defined by the recurrence relation 

𝐹𝑛
(1,1,𝑟,𝑠)

= 𝐹𝑛−1
(1,1,𝑟,𝑠)

+ 𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)𝐹𝑛−2
(1,1,𝑟,𝑠)

,  

where 𝐹0
(1,1,𝑟,𝑠)

= 0,𝐹1
(1,1,𝑟,𝑠)

= 1.This can be written in the form 
 

        𝐹𝑛
(1,1,𝑟,𝑠)

= {
𝐹𝑛−1
(1,1,𝑟,𝑠)

+ 𝑟𝐹𝑛−2
(1,1,𝑟,𝑠)

 ;when n is odd

 𝐹𝑛−1
(1,1,𝑟,𝑠)

+ 𝑠𝐹𝑛−2
(1,1,𝑟,𝑠)

 ;when n is even
(𝑛 ≥ 2)            (6.1) 

 

with the initial conditions 𝐹0
(1,1,𝑟,𝑠)

= 0,𝐹1
(1,1,𝑟,𝑠)

= 1. First few terms of this 
sequence are shown in Table 9. 
 

n 𝐹𝑛
(1,1,𝑟,𝑠)

 

0 0 

1 1 

2 1 

3 1 + 𝑟 

4 1 + 𝑟 + 𝑠 

5 1 + 2𝑟 + 𝑠 + 𝑟2 

6 1 + 𝑟𝑠 + 2𝑟 + 2𝑠 + 𝑟2 + 𝑠2 
 

Table 9 
 

Diwan, Shah [3] derived the Binet-type explicit formula for the terms of this 

sequence as 

21



Khushbu J. Das and Devbhadra V. Shah 

 

 

𝐹𝑛
(1,1,𝑟,𝑠)

=
(𝛼−𝑠)𝜒(𝑛)𝛼

[
𝑛
2
]
−(𝛽−𝑠)𝜒(𝑛)𝛽

[
𝑛
2
]

𝛼−𝛽
, 

where 𝛼 =
(𝑟+𝑠+1)+√(𝑟+𝑠+1)2−4𝑟𝑠

2
,𝛽 =

(𝑟+𝑠+1)−√(𝑟+𝑠+1)2−4𝑟𝑠

2
 with 𝛼 − 𝛽 =

√(𝑟 + 𝑠 + 1)2 − 4𝑟𝑠, 𝛼 + 𝛽 = 𝑟 + 𝑠 + 1 and 𝛼𝛽 = 𝑟𝑠. 
In this section, we find the formula for the inserted terms between the 

numbers 𝑎 and 𝑏 so that terms of the sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 satisfies the 
recurrence relation (6.1). 

When we consider the sequence 𝑎,𝑥1,𝑏, where 𝑥1 is the only inserted term 

between 𝑎 and 𝑏. Then using 𝐹2
(1,1,𝑟,𝑠)

= 𝐹1
(1,1,𝑟,𝑠)

+ 𝑠𝐹0
(1,1,𝑟,𝑠)

, we get                

𝑏 = 𝑥1 + 𝑠𝑎. Thus 𝑥1 = 𝑏 − 𝑠𝑎.  
When we consider the finite sequence as 𝑎,𝑥1,𝑥2,𝑏, so that it satisfies (6.1), 

we observe that 𝑥1 = 𝑥2 − 𝑠𝑎 and 𝑥2 = 𝑏 − 𝑟𝑥1. This gives 𝑥2 =
𝑏+𝑟𝑠𝑎

1+𝑟
 .This 

also gives 𝑥1 =
𝑏−𝑠𝑎

1+𝑟
.  

Further, considering the sequence𝑎,𝑥1,𝑥2,𝑥3,𝑏, it is now easy to observe 
that 𝑥1 = 𝑥2 − 𝑠𝑎, 𝑥2 = 𝑥3 − 𝑟𝑥1, 𝑥3 = 𝑏 − 𝑠𝑥2. Solving these three equations 

in three variables 𝑥1,𝑥2,𝑥3 we get 𝑥1 =
𝑏−𝑎𝑠(1+𝑠)

1+𝑟+𝑠
, 𝑥2 =

𝑏−𝑟𝑠𝑎

1+𝑟+𝑠
, 𝑥3 =

𝑏+𝑏𝑟+𝑎𝑟𝑠2

1+𝑟+𝑠
. 

If we continue extending the above finite sequence one more time, then we 

get the sequence 𝑎,𝑥1,𝑥2,𝑥3,𝑥4,𝑏. Using the similar approach as above, we 

obtain 𝑥1 =
𝑏−𝑠𝑎(1+𝑟+𝑠)

1+2𝑟+𝑠+𝑟2
,𝑥2 =

𝑏+𝑠𝑟𝑎+𝑠𝑎𝑟2

1+2𝑟+𝑠+𝑟2
,𝑥3 =

𝑏+𝑏𝑟−𝑠2𝑟𝑎

1+2𝑟+𝑠+𝑟2
 and                                           

𝑥4 =
𝑏+𝑏𝑟+𝑏𝑠+𝑟2𝑠2𝑎

1+2𝑟+𝑠+𝑟2
. 

 

We mention these results in the Table 10. 
 

Number 

of 

inserted 

term 

Formula 

𝑥1 𝑥2 𝑥3 𝑥4 

1 𝑏 − 𝑠𝑎 --- --- --- 

2 
𝑏 − 𝑠𝑎

1 + 𝑟
 

𝑏 + 𝑟𝑠𝑎

1 + 𝑟
 --- --- 

3 
𝑏 − 𝑠𝑎(1 + 𝑠)

1 + 𝑟 + 𝑠
 

𝑏 − 𝑟𝑠𝑎

1 + 𝑟 + 𝑠
 

𝑏 + 𝑏𝑟 + 𝑎𝑟𝑠2

1 + 𝑟 + 𝑠
 --- 

4 
𝑏 − 𝑠𝑎(1 + 𝑟 + 𝑠)

1 + 2𝑟 + 𝑠 + 𝑟2
 
𝑏 + 𝑠𝑟𝑎 + 𝑠𝑎𝑟2

1 + 2𝑟 + 𝑠 + 𝑟2
 
𝑏 + 𝑏𝑟 − 𝑠2𝑟𝑎

1 + 2𝑟 + 𝑠 + 𝑟2
 
𝑏 + 𝑏𝑟 + 𝑏𝑠 + 𝑟2𝑠2𝑎

1 + 2𝑟 + 𝑠 + 𝑟2
 

 

Table 10 
  

From this table, we observe that there is a similar pattern for the first 

inserted term in case of any number of inserted terms. We thus generalize it for 

22



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

the case of any number of inserted terms and when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 
are inserted between given two positive integers 𝑎 and 𝑏, we can now write the 
general formula for 𝑥1 as 

 

   𝑥1 =

𝑏−𝑠𝑎{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

𝐹
𝑘+1
(1,1,𝑟,𝑠) ,               (6.2) 

 

Next, when the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given two 
positive integers 𝑎 and 𝑏, we obtain the general formula for 𝑛th inserted term 
(1 ≤ 𝑛 ≤ 𝑘) using the recurrence relation (6.1).  

 

Theorem 6.1 When the terms 𝑥1,𝑥2,𝑥3,…,𝑥𝑘 are inserted between given 

integers 𝑎 and 𝑏, so that all 𝑥𝑖’s satisfy the recurrence relation of 𝐹𝑛
(1,1,𝑟,𝑠)

, the 

general formula for any arbitrary term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) is given by  

𝑥𝑛 =
{
  
 

  
 
𝐹𝑛
(1,1,𝑟,𝑠)

𝑏−𝐹𝑛
(1,1,𝑟,𝑠)

𝑠𝑎{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

+𝑠𝑎𝐹𝑘+1
(1,1,𝑟,𝑠)

(𝐹𝑛−1
(1,1,𝑟,𝑠)

)
𝜒(𝑛)

(∑ 𝑠
[
𝑛−2
2
]−𝑚[

𝑛−2
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−2
2
]
)

1−𝜒(𝑛)

}
  
 

  
 

𝐹
𝑘+1
(1,1,𝑟,𝑠) . 

 

Proof. We use the principle of mathematical induction to prove the result. Since 

by (6.2), we have 𝑥1 =

𝑏−𝑠𝑎{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

𝐹
𝑘+1
(1,1,𝑟,𝑠) , 

which proves the result for  𝑛 = 1. We next assume that it is true for all positive 
integers not exceeding 𝑛. Then the following holds: 

𝑥𝑛−1 =
{
  
 

  
 
𝐹𝑛−1
(1,1,𝑟,𝑠)

𝑏−𝐹𝑛−1
(1,1,𝑟,𝑠)

𝑠𝑎{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

+𝑠𝑎𝐹𝑘+1
(1,1,𝑟,𝑠)

(𝐹𝑛−2
(1,1,𝑟,𝑠)

)
𝜒(𝑛−1)

(∑ 𝑠
[
𝑛−3
2
]−𝑚[

𝑛−3
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−3
2
]
)

1−𝜒(𝑛−1)

}
  
 

  
 

𝐹
𝑘+1
(1,1,𝑟,𝑠) ,  

23



Khushbu J. Das and Devbhadra V. Shah 

 

 

𝑥𝑛−2 =
{
  
 

  
 
𝐹𝑛−2
(1,1,𝑟,𝑠)

𝑏−𝐹𝑛−2
(1,1,𝑟,𝑠)

𝑠𝑎{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

+𝑠𝑎𝐹𝑘+1
(1,1,𝑟,𝑠)

(𝐹𝑛−3
(1,1,𝑟,𝑠)

)
𝜒(𝑛−2)

(∑ 𝑠
[
𝑛−3
2
]−𝑚[

𝑛−3
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−3
2
]
)

1−𝜒(𝑛−2)

}
  
 

  
 

𝐹
𝑘+1
𝑅(1,1,𝑟,𝑠)  .   

Now, 𝑥𝑛−1 + 𝑟
𝜒(𝑛)𝑠1−𝜒(𝑛)𝑥𝑛−2 

=
{
 
 
 
 
 
 

 
 
 
 
 
 (𝐹𝑛−1

(1,1,𝑟,𝑠)
+𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)𝐹𝑛−2

(1,1,𝑟,𝑠)
)𝑏−(𝐹𝑛−1

(1,1,𝑟,𝑠)
+𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)𝐹𝑛−2

(1,1,𝑟,𝑠)
)𝑠𝑎

×{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

+𝑠𝑎𝐹𝑘+1
𝑅(1,1,𝑟,𝑠)

{
 
 
 

 
 
 
(𝐹𝑛−2

(1,1,𝑟,𝑠)
)
𝜒(𝑛−1)

(∑ 𝑠
[
𝑛−3
2
]−𝑚[

𝑛−3
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−3
2
]
)

1−𝜒(𝑛−1)

+𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)(𝐹𝑛−3
(1,1,𝑟,𝑠)

)
𝜒(𝑛−2)

×(∑ 𝑠
[
𝑛−4
2
]−𝑚[

𝑛−4
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−4
2
]
)

1−𝜒(𝑛−2)

}
 
 
 

 
 
 

}
 
 
 
 
 
 

 
 
 
 
 
 

𝐹
𝑘+1
(1,1,𝑟,𝑠)   

=
{
 
 
 
 
 

 
 
 
 
 

𝐹𝑛
(1,1,𝑟,𝑠)

𝑏−𝐹𝑛
(1,1,𝑟,𝑠)

𝑠𝑎×{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

+𝑠𝑎𝐹𝑘+1
(1,1,𝑟,𝑠)

{
 
 

 
 

(𝐹𝑛−2
(1,1,𝑟,𝑠)

)
1−𝜒(𝑛)

(∑ 𝑠
[
𝑛−3
2
]−𝑚[

𝑛−3
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−3
2
]
)

𝜒(𝑛)

+𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)(𝐹𝑛−3
(1,1,𝑟,𝑠)

)
𝜒(𝑛)

(∑ 𝑠
[
𝑛−4
2
]−𝑚[

𝑛−4
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−4
2
]
)

1−𝜒(𝑛)

}
 
 

 
 

}
 
 
 
 
 

 
 
 
 
 

𝐹
𝑘+1
(1,1,𝑟,𝑠)   

 

 We calculate the value of third term of numerator separately. Now, when 

𝑛 is odd, we get 

(𝐹𝑛−2
(1,1,𝑟,𝑠)

)
1−𝜒(𝑛)

(∑ 𝑠
[
𝑛−3

2
]−𝑚[

𝑛−3

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−3

2
]
)

𝜒(𝑛)

  

            +𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)(𝐹𝑛−3
(1,1,𝑟,𝑠)

)
𝜒(𝑛)

(∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
)

1−𝜒(𝑛)

  

= 𝐹𝑛−2
(1,1,𝑟,𝑠)

+ 𝑠(∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
)  

= 𝐹𝑛−2
(1,1,𝑟,𝑠)

+ 𝑠(
𝑠
[
𝑛−4

2
]
𝐹0
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−6

2
]
𝐹1
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−8

2
]
𝐹2
(1,1,𝑟,𝑠)

+⋯+ 𝐹𝑛−4
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]

)  

24



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

= 𝐹𝑛−2
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−2
2
]
𝐹0
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4
2
]
𝐹1
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−6
2
]
𝐹2
(1,1,𝑟,𝑠)

 

            +⋯+ 𝑠𝐹𝑛−4
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−2

2
]
 

            = ∑ 𝑠
[
𝑛−2

2
]−𝑚[

𝑛−2

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−2

2
]
  

            = (∑ 𝑠
[
𝑛−2

2
]−𝑚[

𝑛−2

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−2

2
]
)

1−𝜒(𝑛)

 . 

 

Also, when 𝑛 is even, we get 

(𝐹𝑛−2
(1,1,𝑟,𝑠)

)
1−𝜒(𝑛)

(∑ 𝑠
[
𝑛−3

2
]−𝑚[

𝑛−3

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−3

2
]
)

𝜒(𝑛)

   

            +𝑟𝜒(𝑛)𝑠1−𝜒(𝑛)(𝐹𝑛−3
(1,1,𝑟,𝑠)

)
𝜒(𝑛)

(∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
)

1−𝜒(𝑛)

   

= ∑ 𝑠
[
𝑛−3

2
]−𝑚[

𝑛−3

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−3

2
]
+ 𝑟𝐹𝑛−3

(1,1,𝑟,𝑠)
 . 

 

To prove the main result, we need to prove that this value should be 𝐹𝑛−1
(1,1,𝑟,𝑠)

. 

Clearly this result holds for 𝑛 = 3, since 𝐹2
(1,1,𝑟,𝑠)

= 1. Assume that this result 
holds for all positive integers not exceeding 𝑛. Now,   
 

𝐹𝑛−2
(1,1,𝑟,𝑠)

= ∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
+ 𝑟𝐹𝑛−4

(1,1,𝑟,𝑠)
and  

𝐹𝑛−3
(1,1,𝑟,𝑠)

= ∑ 𝑠
[
𝑛−5

2
]−𝑚[

𝑛−5

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−5

2
]
+ 𝑟𝐹𝑛−5

(1,1,𝑟,𝑠)
 . 

Now, 𝐹𝑛−2
(1,1,𝑟,𝑠)

+ 𝑠𝐹𝑛−3
(1,1,𝑟,𝑠)

 

= ∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
+ 𝑟𝐹𝑛−4

(1,1,𝑟,𝑠)
  

+𝑠(∑ 𝑠
[
𝑛−5

2
]−𝑚[

𝑛−5

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−5

2
]
+ 𝑟𝐹𝑛−5

(1,1,𝑟,𝑠)
)  

 = ∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
+ 𝑟𝐹𝑛−4

(1,1,𝑟,𝑠)
 

 +∑ 𝑠
[
𝑛−3

2
]−𝑚[

𝑛−5

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−3

2
]
+ 𝑟𝑠𝐹𝑛−5

(1,1,𝑟,𝑠)
 

= (∑ 𝑠
[
𝑛−4

2
]−𝑚[

𝑛−4

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−4

2
]
+ ∑ 𝑠

[
𝑛−3

2
]−𝑚[

𝑛−5

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−3

2
]
)   

 +𝑟(𝐹𝑛−4
(1,1,𝑟,𝑠)

+ 𝑠𝐹𝑛−5
(1,1,𝑟,𝑠)

) 

= ∑ 𝑠
[
𝑛−3

2
]−𝑚[

𝑛−3

2
]

𝑚=0
𝐹2𝑚
(1,1,𝑟,𝑠)

+ 𝑠
[
𝑛−3

2
]
+ 𝑟𝐹𝑛−3

(1,1,𝑟,𝑠)
= 𝐹𝑛−1

(1,1,𝑟,𝑠)
= (𝐹𝑛−1

(1,1,𝑟,𝑠)
)
𝜒(𝑛)

 . 

                                            . 

 Therefore, we have 

25



Khushbu J. Das and Devbhadra V. Shah 

 

 

𝑥𝑛 = 𝑥𝑛−1 + 𝑟
𝜒(𝑛)𝑠1−𝜒(𝑛)𝑥𝑛−2 

=
{
  
 

  
 
𝐹𝑛
(1,1,𝑟,𝑠)

𝑏−𝐹𝑛
(1,1,𝑟,𝑠)

𝑠𝑎{(𝐹𝑘
(1,1,𝑟,𝑠)

)
1−𝜒(𝑘)

(∑ 𝑠
[
𝑘−1
2
]−𝑚[

𝑘−1
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑘−1
2
]
)

𝜒(𝑘)

}

+𝑠𝑎𝐹𝑘+1
(1,1,𝑟,𝑠)

(𝐹𝑛−1
(1,1,𝑟,𝑠)

)
𝜒(𝑛)

(∑ 𝑠
[
𝑛−2
2
]−𝑚[

𝑛−2
2
]

𝑚=0 𝐹2𝑚
(1,1,𝑟,𝑠)

+𝑠
[
𝑛−2
2
]
)

1−𝜒(𝑛)

}
  
 

  
 

𝐹
𝑘+1
(1,1,𝑟,𝑠)   , which 

proves the result for every positive integer 𝑛. 
 

7 Conclusion 

 In this paper we derived the general formula which gives the value of any 

inserted term 𝑥𝑛 (1 ≤ 𝑛 ≤ 𝑘) between the given fixed positive integers 𝑎,𝑏 so 
that terms of the sequence 𝑎,𝑥1,𝑥2,𝑥3,…,𝑥𝑘,𝑏 satisfies the recurrence relation 

of 𝐹𝑛(𝑝,𝑞) and various bifurcating subsequence of {𝐹𝑛
(𝑝,𝑞,𝑟,𝑠)

} by considering 

some fixed values of 𝑝,𝑞,𝑟,𝑠. 
 

References 
 

[1]  Agnes M., Buenaventura N., Labao J. J., Soria C. K., Limbaco K. A., and 

L. Natividad, Inclusion of inserted terms (Fibonacci mean) in a Fibonacci 

sequence, Math Investigatory Project, Central Luzon State University, 

2010. 

[2]  Bilgici G., New generalizations of Fibonacci and Lucas Sequences, Applied 

Mathematical Sciences, Vol. 8, No. 29, 2014, 1429 – 1437. 

[3] Diwan D. M, Shah D. V., Extended Binet’s formula for the class of 

generalized Fibonacci sequences, Proceeding 19th Annual Cum 4th 

International Conference of Gwalior academy of Mathematical Sciences 

(GAMS), SVNIT, surat, Oct 3-6, 2014, 109 – 113. 

[4] Diwan D. M, Shah D. V., Extended Binet’s formula for the class of 

generalized Fibonacci sequences, VNSGU Journal of Science and 

Technology, Vol. 4, No. 1, 2015, 205 – 210. 

[5]  Diwan D. M, Shah D. V., Explicit and recursive formulae for the class of 

generalized Fibonacci sequence, International Journal of Advanced 

Research in Engineering, Science and Management, Vol. 1, Issue 10, July 

2015, 1 – 6. 

[6]  Edson M., Yayenie O., A new generalization of Fibonacci sequence and 

Extended binet’s formula, Integers, Vol. 9, 2009, 639 – 654.    

[7] Horadam A. F., Basic properties of certain generalized sequence of 

numbers, Fibonacci Quarterly, 3, 1965, 161-176. 

26



Insertion of terms satisfying the recurrence relations of Horadam sequence 

and Bifurcating Fibonacci sequences 
 

 

 

[8] Howell P., Nth term of the Fibonacci sequence, from Math Proofs: 

Interesting mathematical results and elegant solutions to various problems, 

http://mathproofs.blogspot.com/2005/04/nth-term-offibonacci-sequence. 

[9] Verma, Bala A., On Properties of generalized Bi-variate Bi-Periodic 

Fibonacci polynomials, International journal of Advanced science and 

Technology,29(3), 2020, 8065–8072.  

[10] Yayenie O., A note on generalized Fibonacci sequence, Applied   

Mathematics and computation, 217, 2011, 5603–5611. 

       

27