EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 1, 2024, 135-146
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Common Terms of k-Pell and Tribonacci Numbers

Hunar Sherzad Taher1, Saroj Kumar Dash2,∗

1 Mathematics Division, School of Advanced Science, Vellore Institute of Technology,
Chennai Campus, Chennai 600127, India

Abstract. Let Tm be a Tribonacci sequence, and let the k-Pell sequence be a generalization of
the Pell sequence for k ≥ 2. The first k terms are 0, 0, ..., 0, 1, and each term after the forewords is
defined by linear recurrence

P (k)
n = 2P

(k)
n−1 + P

(k)
n−2 + ...+ P

(k)
n−k.

We study the solution of the Diophantine equation P
(k)
n = Tm for the positive integer (n, k,m)

with k ≥ 2. We use the lower bound for linear forms in logarithms of algebraic numbers with the
theory of the continued fraction.

2020 Mathematics Subject Classifications: 11D61, 11J86, 11J70, 11B83

Key Words and Phrases: Exponential Diophantine equation, Linear forms in logarithms, k-Pell
numbers, Tribonacci numbers.

1. Introduction

The Pell sequence is defined by Pn = 2Pn−1 + Pn−2, for all n ≥ 3, where P0 = 0 and
P1 = 1.
Let an integer k ≥ 2. The generalization of the Pell sequence is a k-Pell sequence, denoted

by {P (k)
n }n≥−(k−2) given linear recurrence as:

P (k)
n = 2P

(k)
n−1 + P

(k)
n−2 + ...+ P

(k)
n−k for all n ≥ 2, (1)

with the initial conditions P
(k)
−(k−2) = P

(k)
−(k−3) = ... = P

(k)
0 = 0 and P

(k)
1 = 1. If k = 2 in

equation (1), it becomes a linear recurrence of the Pell sequence.
The Tribonacci sequence Tm is defined by

Tm = Tm−1 + Tm−2 + Tm−3 for each m ≥ 3 (2)

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i1.4989

Email addresses: sarojkumar.dash@vit.ac.in ( S. K. Dash),
hunarsherzad.taher2022@vitstudent.ac.in (H. S. Taher)

https://www.ejpam.com 135 © 2024 EJPAM All rights reserved.



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 136

with initial conditions T0 = 0, T1 = T2 = 1. It's first few terms are

0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, ...

The Online Encyclopedia of Integer (OEIS) of Pell and Tribonacci sequences are A000129
and A000073, respectively. Presently, researchers are finding the intersection between two
recurrences, and several studies have been published on k-Fibonacci, k-Pell, Tribonacci,
Padovan, and Perrin sequences related to other sequences. One can cite [1, 3, 7, 9, 10, 13].
Our aim is to show that there are common terms between k-generalized Pell numbers and
Tribonacci numbers. The earlier findings guided our completion of the investigation.

2. Auxiliary Results

2.1. Properties of Tribonacci sequence

The characteristic polynomial of the Tibonacci sequence is

f(x) = x3 − x2 − x− 1.

The Tribonacci sequence has one real root η1 with two complex roots η2 and η3.

η1 =
1 +

3
√

19 + 3
√
33 +

3
√

19− 3
√
33

3
,

η2 =
1 + ω

3
√

19 + 3
√
33 + ω2 3

√
19− 3

√
33

3
,

η3 =
1 + ω2 3

√
19 + 3

√
33 + ω

3
√

19− 3
√
33

3
,

where ω = −1+i
√
3

2 . Spickerman [12] found the Binet formula of the Tribonacci numbers
as

Tm =
ηm+1
1

(η1 − η2)(η1 − η3)
+

ηm+1
2

(η2 − η1)(η2 − η3)
+

ηm+1
3

(η3 − η1)(η3 − η2)
, for all m ≥ 0. (3)

The generating function of the Tribonacci sequence is:

g(x) =
x

1− x− x2 − x3
=

∞∑
m=0

Tmx
m.

Note that we have the following identities

η1 + η2 + η3 = 1,

η1η2 + η2η3 + η1η3 = −1,

η1η2η3 = 1.



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 137

Furthermore, Dresden and Du [6] presented a Binet-style formula for generating k-generalized
Fibonacci numbers. If k = 3, it follows that:

Tm =
(η1 − 1)ηm−1

1

2 + 4(η1 − 2)
+

(η2 − 1)ηm−1
2

2 + 4(η2 − 2)
+

(η3 − 1)ηm−1
3

2 + 4(η3 − 2)
, for all m ≥ 0. (4)

Moreover, Dresden and Du [6, Lemma 5] found that the Tribonacci numbers can be written
as

Tm = cηm−1
1 + dm with |dm| < 1

2
, for all m ≥ 1, (5)

where c = (η1 − 1)/(4η1 − 6) ≈ 0.61. For m ≥ 1, the inequality

ηm−2
1 ≤ Tm ≤ ηm−1

1 , (6)

hold.

2.2. Properties of k−generalized Pell sequence

We are aware that the characteristic polynomial of the k-generalized Pell sequence is

Ψk(x) = xk − 2xk−1 − xk−2 − ...− x− 1.

Bravo, Herrera and Luca [4] showed that Ψk(x) is irreducible over Q[x] and has one
positive real root α(k) outside the unit circle. The other roots were inside the unit circle.
Moreover, they showed the following:

ϕ2(1− ϕ−k) < α(k) < ϕ2, for all k ≥ 2, (7)

where ϕ = ((1 +
√
5)/2). To simplify the notation, we omit the dependence on k of α.

The authors found that the Binet formula for P
(k)
n is

P (k)
n =

k∑
i=1

gk(αi)(αi)
n, (8)

where ai represents the root of the characteristic polynomial Ψk(x) and gk is given by

gk(x) =
x− 1

(k + 1)x2 − 3kx+ k − 1
, for all k ≥ 2.

Bravo and Herrera [2, Lemma 1] proved that

0.276 < gk(α) < 0.5 and |gk(αi)| < 1, 2 ≤ i ≤ k,

where gk(α) is not an algebraic integer. Furthermore, they proved that the logarithmic
height of gk is

h(gk) < 4k log(ϕ) + k log(k + 1), for all k ≥ 2. (9)



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 138

According to the above notation, Bravo, Herrera and Luca [4] showed that formula (8),
given by the approximation∣∣∣P (k)

n − gk(α)α
n
∣∣∣ < 1

2
, for all n ≥ 2− k.

Therefore, for n ≥ 1 and k ≥ 2 , we have

P (k)
n = gk(α)α

n + ek(n), where |ek(n)| ≤
1

2
. (10)

Moreover, the inequality
αn−2 ≤ P (k)

n ≤ αn−1 (11)

holds for all n ≥ 1 and k ≥ 2.

Lemma 1. ([2, Lemma 2]) If k ≥ 30 and n ≥ 1 are integers satisfying n < ϕk/2, then

gk(α)α
n =

ϕ2n

ϕ+ 2
(1 + ζ) , where |ζ| < 4

ϕk/2
, ϕ =

1 +
√
5

2
. (12)

Lemma 2. ([14, Lemma 2.2]) Let v, x ∈ R and 0 < v < 1. If |x| < v, then

| log(1 + x)| < − log(1− v)

v
|x|.

2.3. Linear forms in logarithms

Let γ be an algebraic number of degree d with minimal polynomial

c0x
d + c1x

d−1 + . . .+ cd = c0

d∏
i=1

(
x− γ(i)

)
∈ Z[x],

where the γ(i)’s are conjugates of γ, and the ci’s are relative primes to each other with
c0 > 0. Then the logarithmic height of γ is given by

h(γ) =
1

d

(
log c0 +

d∑
i=1

log
(
max

{∣∣∣γ(i)∣∣∣ , 1})) . (13)

If γ = a
b is rational number with gcd(a, b) = 1 and b > 0, then h(γ) = log(max{|a|, b}).

Some properties of the logarithmic height function are listed below, which will be used in
the next parts of this paper:

h(η ± γ) ≤ h(η) + h(γ) + log 2, (14)

h
(
ηγ±1

)
≤ h(η) + h(γ), (15)

h
(
ηk
)
= |k|h(η). (16)

We use the following [5, Theorem 9.4], which is a modified version of the Matveev result
[8]



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 139

Theorem 1. Let L be a real algebraic number field of degree D over Q. Let γ1, . . . , γt ∈ L
be a positive real algebraic number, and b1, b2, . . . , bt be nonzero integers such that

Λ := γb11 · · · γbtt − 1,

is not zero. Then

log |Λ| > (−1.4)
(
30t+3

) (
t4.5
) (
D2
)
(A1 . . . At) (1 + logD)(1 + logB),

where
B ≥ max {|b1| , . . . , |bt|} ,

and
Ai ≥ max {Dh (γi) , |log (γi)| , 0.16} , 1 ≤ i ≤ t.

2.4. De Weger reduction method

To reduce the upper bound, we present a variant of Baker and Davenport’s reduction
method [14]. Let ϑ1, ϑ2, β ∈ R be given, and let x1, x2 ∈ Z be unknowns. Let

Λ = β + x1ϑ1 + x2ϑ2. (17)

Let c, δ be positive constants. Set X = max {|x1| , |x2|}. Let X0, Y be positive. Assume
that

|Λ| < c · exp(−δ · Y ), (18)

Y ≤ X ≤ X0. (19)

When β = 0 in (17), we get
Λ = x1ϑ1 + x2ϑ2.

Put ϑ = −ϑ1/ϑ2. We assume that x1 and x2 are coprime. Let the continued fraction
expansion of ϑ be given by

[a0, a1, a2, . . .] ,

and let the k-th convergent of ϑ be pk/qk for k = 0, 1, 2, . . . We may assume without loss
of generality that |ϑ1| < |ϑ2| and that x1 > 0. We obtain the following results.

Lemma 3. ([14, Lemma 3.2]) Let

A = max
0≤k≤Y0

ak+1,

where

Y0 = −1 +
log
(√

5X0 + 1
)

log
(
1+

√
5

2

) .

If (18) and (19) hold for x1, x2 and β = 0, then

Y <
1

δ
log

(
c(A+ 2)X0

|ϑ2|

)
. (20)



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 140

When β ̸= 0 in (17), put ϑ = −ϑ1/ϑ2 and ψ = β/ϑ2. Then we have Λ
ϑ2

= ψ − x1ϑ + x2.
Let p/q be a convergent of ϑ with q > X0. Tthe distance between real number T and the
closest integer is expressed as ∥T∥ = min{|T −n| : n ∈ Z}. We obtain the following result.

Lemma 4. ([14, Lemma 3.3]) Suppose that

∥qψ∥ > 2X0

q
.

Then, the solutions of (18) and (19) satisfy

Y <
1

δ
log

(
q2c

|ϑ2|X0

)
. (21)

We need the following discovery to prove our theorem.

Lemma 5. ([11, Lemma 7]) If r ≥ 1 and S ≥ (4r2)r, and L
(logL)r < S, then

L < 2rS(logS)r.

3. Main Results

Theorem 2. The positive integer solutions of the Diophantine equation

P (k)
n = Tm, (22)

where k ≥ 2 are P
(k)
1 = T1 = T2, P

(k)
2 = T3, and P

(k)
4 = T6.

To prove Theorem 2 will be done in four steps.

3.1. Relation between n and m

For the Diophantine equation (22) in the range 1 ≤ n ≤ k + 1, we have P
(k)
n = F2n−1,

where Fn is a Fibonacci number, and we obtain the set of solutions in Theorem 2. For the
remaining possibility, we assumed that n ≥ k + 2 and k ≥ 2. By combining inequalities
(6) and (11) with equation (22), we obtain:

αn−2 ≤ P (k)
n = Tm ≤ ηm−1

1 and ηm−2
1 ≤ Tm = P (k)

n ≤ αn−1,

we conclude that

(n− 2)
log(α)

log(η1)
≤ m− 1 and m ≤ (n− 1)

log(α)

log(η1)
+ 2,

we obtain
0.79n− 1.58 < m− 1 < m < 1.58n+ 0.42,

because ϕ2(1− ϕ−k) < α(k) < ϕ2 for all k ≥ 2. We consider the following

0.79n− 1.58 < m− 1 < m < 2n. (23)



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 141

3.2. Bounding n in terms of k

In this step, we prove the following lemma to find an upper bound for n in terms of k.

Lemma 6. If (m,n, k) is a positive integers solution of equation (22) with k ≥ 2 and
n ≥ k + 2, then the inequalities

0.63m < n < 7.6 · 1016k5(log(k))3

hold.

Proof. Combining equation (22), (5), and (10), we obtain:

gk(α)α
n + ek(n) = cηm−1

1 + dm.

Taking absolute values for both sides, we get∣∣gk(α)αn − cηm−1
1

∣∣ < 1

2
+ |dm| < 1. (24)

Dividing both sides by cηm−1
1 , we deduce that∣∣∣(c−1gk(α))α

nη
−(m−1)
1 − 1

∣∣∣ < 1.6

ηm−1
1

. (25)

We apply Theorem 1 to the left-hand side inequality (25) with parameters t := 3, where
γ1 := c−1gk(α), γ2 := α, γ3 := η1, and b1 := 1, b2 := n, b3 = −(m − 1). So L :=
Q(γ1, γ2, γ3). Thus, D := [L,Q] = 3k. To show that Λ is nonzero, it is assumed that

Λ = 0, which implies that gk(α) = cη
(m−1)
1 θ−n

1 , we obtain gk(α) as an algebraic integer,
which is a contradiction. Hence Λ ̸= 0. Not that

h (γ1) < h(c) + h(gk(α)) <
log(44)

3
+ 4k log(ϕ) + k log(k + 1) < 5.3k log(k),

which holds for all k ≥ 2 and a minimal polynomial 44x3− 44x2+12x− 1 of c. Therefore,
h (γ2) =

log(α)
k < 2 log(ϕ)

k and h (γ3) =
log(η1)

3 . Thus, we obtained A1 := 15.9k2 log(k), A2 :=
6 log(ϕ), and A3 := k log(η1). In addition, taking B := 2n, since max{|1|, |n|, |−(m−1)|} ≤
2n. Thus, by Theorem 1, we get that

1.6

ηm−1
1

> |Λ| > exp{−G(1 + log(2n))(15.9k2 log(k))(6 log(ϕ))(k log(η1))},

where G = (1.4)
(
306
) (

34.5
)
(3k)2(1 + log(3k)). We get

(m− 1) log(η1)− log(1.6) < 3.61 · 1013k5 log(k)(1 + log(3k))(1 + log(2n)). (26)

Using the facts that (1 + log(3k)) < 4.1 log(k) for all k ≥ 2 and (1 + log(2n)) < 2.3 log(n)
for all n ≥ 4. Simplifying the calculation, we obtain

m− 1 < 5.6 · 1014k5(log(k))2 log(n),



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 142

By inequality (23), we deduce that

n

log(n)
< 7.1 · 1014k5(log(k))2,

Now we apply Lemma 5 take S := 7.1·1014k5(log(k))2, L := n, r := 1 with 34.2+5 log(k)+
2 log(log(k)) < 53.5 log(k) for all k ≥ 2, we get

n < 2(7.1 · 1014k5(log(k))2)(log(7.1 · 1014k5(log(k))2))
< (1.42 · 1015k5(log(k))2)(34.2 + 5 log(k) + 2 log(log(k)))

< 7.6 · 1016k5(log(k))3. (27)

3.3. The case 2 ≤ k ≤ 350

In the previous, we obtained a very large upper bound of n. We apply Lemma 4 to
reduce the upper bound. In this case, we will prove the following lemma.

Lemma 7. The only solution of the Diophantine equation (22) is P
(k)
4 = T6 where n ≥ k+2

and 2 ≤ k ≤ 350

Proof. To apply Lemma 4, let

v1 := n log(α)− (m− 1) log(η1) + log(c−1gk(α)).

Then we have, by inequality (25),

|ev1 − 1| < 1.6

ηm−1
1

.

We know v1 ̸= 0, since Λ ̸= 0. If m ≥ 2, we have

1.6

ηm−1
1

< 0.87.

By Lemma 2, we get

|v1| = | log(Λ + 1)| = − log(1− 0.87)

0.87
· 1.6

ηm−1
1

<
3.75

ηm−1
1

,

and

0 <
∣∣(m− 1)(− log(η1)) + n log(α) + log(c−1gk(α))

∣∣ < 3.75 · exp(−(m− 1) log(η1)). (28)

According to Lemma 4, we obtain

c := 3.75, δ := log(η1), ψ :=
log(c−1gk(α))

log(α)
,



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 143

ϑ :=
log(η1)

log(α)
, ϑ1 := − log(η1), ϑ2 := log(α), β := log(c−1gk(α)).

We are aware that ϑ is an irrational number. Taking X0 := 1.5 · 1017k5(log(k))3, which is
an upper bound of m− 1 and n. Using Maple program inspection, the maximum value of
1
δ log

(
q2c

|ϑ2|X0

)
for k ∈ [2, 350] is 143 . We get 1 ≤ m − 1 ≤ 143 and discover the possible

values of the Diophantine equation (22) for which k ∈ [2, 350] have 2 ≤ m ≤ 144, and
by inequality (23), we obtain 4 ≤ n ≤ 181. The only possible solution in this range was

P
(k)
4 = T6.

3.4. The case k > 350

In this case, we prove the following lemma

Lemma 8. The Diophantine equation (22) has no solution for n ≥ k + 2 and k > 350

Proof. For k > 350, as a result of Lemma 1, we have

n < 7.6 · 1016k5(log(k))3 < ϕk/2.

From (12),(22) and (24), we get∣∣∣∣ ϕ2nϕ+ 2
− cηm−1

1

∣∣∣∣ < ∣∣gk(α)αn − cηm−1
1

∣∣+ ϕ2n

ϕ+ 2
|ζ| < 1 +

4ϕ2n

(ϕ+ 2)ϕk/2
.

Dividing both sides by ϕ2n

ϕ+2 , it becomes

|Λ1| <
7.6

ϕk/2
, where Λ1 := c(ϕ+ 2)ϕ−2nηm−1

1 − 1. (29)

Using the fact that 1
ϕ2n <

1
ϕk/2 yield for n ≥ k + 2. It is known that Λ1 is nonzero. If Λ1

is zero, then ϕ2n

ηm−1
1

= c(ϕ + 2), and we get the left-hand side as an algebraic integer, but

the right-hand side is not an algebraic integer, which is impossible, hence, Λ1 ̸= 0. We
apply Theorem 1, we take parameters t := 3, and γ1 := c(ϕ + 2), γ2 := ϕ, γ3 := η1, and
b1 := 1, b2 := −2n, b3 = (m− 1). So L := Q(γ1, γ2, γ3). Thus D := [L,Q] = 6. Moreover,

h(η2) =
log(ϕ)

2 , h(η3) =
log(η1)

3 and

h(η1) ≤ h(c) + h(ϕ) + 2 log(2) < 2.9,

it follows that A1 := 17.4, A2 := 1.45 and A3 := 1.22. Since max{|1|, | − 2n|, |(m− 1)|} ≤
2n, we can take B := 2n. Thus, by Theorem 6 , we get

k

2
log(ϕ)− log(7.6) < 4.43 · 1014 · (1 + log(2n)).



H. S. Taher, S. K. Dash / Eur. J. Pure Appl. Math, 17 (1) (2024), 135-146 144

Using fact that 1 + log(2n) < 1.3 log(n) for all n ≥ k + 2 > 352, which implies that

k < 2.4 · 1015 log(n).

We have an upper bound of n in inequality(27), then 38.87 + 5 log(k) + 3 log(log(k)) <
13 log(k) for all k > 350, we get

k < 2.4 · 1015 log(7.6 · 1016k5(log(k))3)
< 2.4 · 1015(38.87 + 5 log(k) + 3 log(log(k)))

< 3.12 · 1016 log(k).

The above inequality gives
k < 1.3 · 1018.

Thus, we get

n < 7.6 · 1016(1.3 · 1018)5(log(1.3 · 1018))3 < 2.1 · 10112

m < 2(2.1 · 10112) < 4.2 · 10112.

Let
v2 := (m− 1) log(η1)− (2n) log(α) + log(c(ϕ+ 2)).

Then we have, by inequality (29),

|ev2 − 1| < 7.6

ϕk/2
.

We know v2 ̸= 0, since Λ1 ̸= 0. If k ≥ 350, we get

7.6

ϕk/2
< 0.1.

By Lemma 2, we obtain the inequality

|v2| = | log(Λ1 + 1)| = − log(1− 0.1)

0.1
· 7.6

ϕk/2
<

8.1

ϕk/2
.

Thus, we get

0 < |(m− 1) log(η1)− 2n log(ϕ) + log(c(ϕ+ 2))| < 8.1 · exp(−0.24 · k). (30)

Applying lemma 4, we can take

c := 8.1, δ := 0.24, ψ := − log(c(ϕ+ 2))

log(ϕ)
,

ϑ :=
log(η1)

log(ϕ)
, ϑ1 := log(η1), ϑ2 := − log(ϕ), β := log(c(ϕ+ 2)).



REFERENCES 145

We take M := 4.2 ·10112, which is the upper bound for m−1. A quick inspection with the
help of Maple programming found that q211 is convergent of ϑ. By Lemma 4, we obtain

k <
1

0.24

(
q2211 · 8.1

4.2 · 10122 · | − log(ϕ)|

)
< 1105. (31)

By inequalities of (27) and (23) we have

n < 4.3 · 1034 and m < 8.6 · 1034.

Again we apply Lemma 4 for (30) with M := 8.6 · 1034, we found that q72 is a convergent
of ϑ, and k < 389. Hence

n < 1.4 · 1032 and m < 2.8 · 1032.

Third time applying Lemma 4 for (30) with M := 2.8 · 1032, we found that q65 is a
convergent of ϑ, and k < 342, we get contradiction by our assumption that k > 350.
Theorem 2 is proved.

4. Conclusion

We found all solutions of the Diophantine equation (22), where P
(k)
n is a k-generalized

Pell number and Tm is a Tribonacci number, for each positive integer n,m and k. We
used a lower bound for linear forms in logarithms of algebraic numbers to get an upper
bound for n. Then, we used a variation of the Baker-Davenport reduction method called
the De Weger reduction method to reduce the upper bound.

Acknowledgements

The authors express their gratitude to the anonymous reviewers for the instructive
suggestions.

References

[1] A. Acikel and N. Irmak. Common terms of Tribonacci and Perrin sequences. Miskolc
Mathematical Notes, 23(1):5–11, 2022.

[2] J. J. Bravo and J. L. Herrera. Repdigits in generalized Pell sequences. Archivum
Mathematicum, 56(4):249–262, 2020.

[3] J. J. Bravo, J. L. Herrera, and F. Luca. Common values of generalized Fibonacci and
Pell sequences. Journal of Number Theory, 226:51–71, 9 2021.

[4] J. J. Bravo, J. L. Herrera, and F. Luca. On a generalization of the Pell sequence.
Mathematica Bohemica, 146(2):199–213, 2021.



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