On the Integral Representations of the k-Fibonacci and k-Lucas
Numbers
WEERAYUTH NILSRAKOO1, ACHARIYA NILSRAKOO2
1Department of Mathematics, Faculty of Science,
Ubon Ratchathani University, Ubon Ratchathani, 34190
THAILAND
2Department of Mathematics, Faculty of Science,
Ubon Ratchathani Rajabhat University, Ubon Ratchathani, 34000
THAILAND
Abstract: In this paper, we propose the integral representations of the k-Fibonacci and k-Lucas numbers. We use
the Binet’s formulas to establish some identities and use simple integral calculus to prove them. Our results are
also deduced with the Fibonacci, Lucas, Pell, and Pell-Lucas numbers.
Key-Words: k-Fibonacci number, k-Lucas number, Fibonacci number, Lucas number, integral representation
Received: May 14, 2024. Revised: September 28, 2024. Accepted: October 21, 2024. Published: November 28, 2024.
1 Introduction
Sequences of special numbers have been studied over
several years, with the greatest numbers on studies of
well-known Fibonacci and Lucas sequences that are
related to the golden ratio; for instance, see, [1], [2],
[3], [4], [5], [6], [7], [8], [9], [10].
Recall that the Fibonacci numbers Fnare defined
via the recurrence relation
Fn=Fn−1+Fn−2
for n≥2with F0= 0 and F1= 1. The Lucas
numbers Lnare defined via the recurrence relation
Ln=Ln−1+Ln−2
for n≥2with L0= 2 and L1= 1.
Like Fibonacci and Lucas numbers, the Pell family
is widely used. Recall that Pell number Pnis defined
by the recurrence relation
Pn= 2Pn−1+Pn−2
for n≥2with P0= 0 and P1= 1. The Pell-Lucas
number Qnis defined by the recurrence relation
Qn= 2Qn−1+Qn−2
for n≥2with Q0= 2 and Q1= 2. The Binet’s
formulas for the Pell and Pell-Lucas numbers are
related to the silver ratio φ= 1 + √2.
The generalization of Fibonacci and Pell numbers
were introduced by [11], in 2007 as follows:
the k-Fibonacci numbers Fk,n is defined by the
recurrence relation
Fk,n =kFk,n−1+Fk,n−2(1)
for n≥2with Fk,0= 0 and Fk,1= 1, where kand n
are non-negative integers with k= 0. In 2011, [12],
introduced and studied a generalization of Lucas and
Pell-Lucas numbers as follows: the k-Lucas numbers
Lk,n is defined by the recurrence relation
Lk,n =kLk,n−1+Lk,n−2,(2)
for n≥2with Lk,0= 2 and Lk,1=k. The initial
terms of the k-Fibonacci numbers Fk,n and k-Lucas
numbers Lk,n for selected values of kpresented as in
Table 1 and Table 2.
Table 1.The initial terms of the k-Fibonacci numbers
n01 2 3 4 5 6 7
F1,n 01 1 2 3 5 8 13
F2,n 0 1 2 5 12 29 70 169
F3,n 0 1 3 10 33 109 360 1189
F4,n 0 1 4 17 72 305 1292 5473
F5,n 0 1 5 26 135 701 3640 18901
F6,n 0 1 6 37 228 1405 8658 53353
F7,n 0 1 7 50 357 2549 18200 129949
Table 2.The initial terms of the k-Lucas numbers
n01 2 3 4 5 6 7
L1,n 21 3 4 7 11 18 29
L2,n 2 2 6 14 34 82 198 478
L3,n 2 3 11 36 119 393 1298 4287
L4,n 2 4 18 76 322 1364 5778 24476
L5,n 2 5 27 140 727 3775 19602 101785
L6,n 2 6 38 234 1442 8886 54758 337434
L7,n 2 7 51 364 2599 18557 132498 946043
We can see that the classical Fibonacci and
classical Lucas numbers are obtained for k= 1.
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And then the classical Pell and classical Pell–Lucas
numbers are appeared if k= 2. Moreover,
sequences {F3,n},{F4,n}and {F6,n}are listed in
The Online Encyclopaedia of Integer Sequences,
[13], under the symbols A006190, A001076 and
A005668, respectively, while sequences {L3,n},
{L4,n},{L5,n},{L6,n}and {L7,n}under the
symbols A006497, A014448, A087130, A085447
and A086902, respectively.
The recurrence relations (1) and (2) generate
characteristic equation of the form
r2−kr −1 = 0
Since k≥1, this equation has two roots r1=
1
2k+√k2+ 4and r2=1
2k−√k2+ 4.
Therefore, the Binet’s formulas for the k-Fibonacci
numbers {Fk,n}and the k-Lucas numbers {Lk,n}are
Fk,n =1
∆kφn
k−(−1)n
φn
k(3)
and
Lk,n =φn
k+(−1)n
φn
k
(4)
where ∆k=√k2+ 4 and φk=1
2(k+ ∆k), see
also, [12, Theorem 2.2], [14, Proposition 2].
Some identities have been proposed to represent
and extend of spacial numbers in recent years, [15],
[16], [17], [18], [19], [20], [21], [22], [23], [24].
Integral representations are important tools available
in their analysis (see, for example, [1], [3], [4], [8],
[9], [25], [26], [27], [28], [29], [30], [31], [32], [33],
[34]).
The first example for the integral representations
of the Fibonacci numbers upon even orders by using
the hypergeometric function showed in 2000 by [3],
as follows:
F2n=n
23
2n−1Zπ
01 + √5
3cos xn−1
sin xdx.
In 2015, [4], worked out an explicit integral
representation for Fninvolving trigonometric
functions. Indeed, the main result in their paper is
the representation of the form
Fn=αn
√5−2
πZ∞
0sin(x/2)
x
×cos(2nx)−2sin(nx)sin x
5sin2x+cos2xdx
where α=1+√5
2is the golden ratio. Another
representation is given by [1].
In a recent year, [8], derived some appealing
integral representations for Fibonacci numbers Fn
and Lucas numbers Ln. For instance, he proved the
representations
Fℓn =nFℓ
2nZ1
−1
(Lℓ+√5Fℓx)n−1dx
and
Lℓn =1
2nZ1
−1
(Lℓ+√5Fℓx)n−1
×(Lℓ+√5(n+ 1)Fℓx)dx,
where ℓand nare non-negative integers. The special
case of this identity for ℓ= 1 is also discussed in [9],
from 2023.
In this paper, we give new integral representations
of the k-Fibonacci and the k-Lucas numbers. To
prove them, we propose some identities relied on the
Binet’s formulas and simple integral calculus.
2 Preliminaries
We employ the technique of [8], to obtain new integral
representations for the k-Fibonacci numbers and the
k-Lucas numbers. We start with the following some
identities relied on the Binet’s formulas that we will
require.
Lemma 1. Let kand nbe non-negative integers with
k= 0,∆k=√k2+ 4 and φk=1
2(k+ ∆k). Then
the following hold:
(i) Lk,n + ∆kFk,n = 2φn
k;
(ii) Lk,n −∆kFk,n = 2(−1)n
φn
k;
(iii) L2
k,n −∆2
kF2
k,n = 4(−1)n.
Proof. (i) Combining Binet’s formulas (3) and (4)
gives
Lk,n + ∆kFk,n
=φn
k+(−1)n
φn
k+φn
k−(−1)n
φn
k
= 2φn
k.
(ii) Subtracting Binet’s formulas (3) and (4) gives
Lk,n −∆kFk,n
=φn
k+(−1)n
φn
k−φn
k−(−1)n
φn
k
= 2(−1)n
φn
k
.
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(iii) From (i) and (ii), we have
L2
k,n −∆2
kF2
k,n
=L2
k,n −(∆kFk,n)2
= (Lk,n + ∆kFk,n) (Lk,n −∆kFk,n)
= (2φn
k)2(−1)n
φn
k
= 4(−1)n.
This completes the proof.
Remark 2.Lemma 1 (iii) is presented in [12, Theorem
2.3].
Lemma 3. Let k,mand nbe non-negative integers
k= 0 and ∆k=√k2+ 4. Then the following hold:
(i) 2Fk,m+n=Fk,mLk,n +Fk,nLk,m;
(ii) 2Lk,m+n=Lk,mLk,n + ∆2
kFk,mFk,n.
Proof. (i) Using Binet’s formulas (3) and (4), we
obtain
Fk,mLk,n
=1
∆kφm
k−(−1)m
φm
kφn
k+(−1)n
φn
k
=1
∆kφm+n
k−(−1)mφn
k
φm
k
+1
∆k(−1)nφm
k
φn
k−(−1)m+n
φm+n
k
and
Fk,nLk,m
=1
∆kφn
k−(−1)n
φn
kφm
k+(−1)m
φm
k
=1
∆kφm+n
k−(−1)nφm
k
φn
k
+1
∆k(−1)mφn
k
φm
k−(−1)m+n
φm+n
k.
So, we get
Fk,mLk,n +Fk,nLk,m
=2
∆kφm+n
k−(−1)m+n
φm+n
k
= 2Fk,m+n.
(ii) Using Binet’s formulas (4), we obtain
Lk,mLk,n
=φm
k+(−1)m
φm
kφn
k+(−1)n
φn
k
=φm+n
k+(−1)mφn
k
φm
k
+(−1)nφm
k
φn
k
+(−1)m+n
φm+n
k
.
Using Binet’s formulas (3), we obtain
∆2
kFk,mFk,n
= ∆2
k1
∆kφm
k−(−1)m
φm
k
×1
∆kφn
k−(−1)n
φn
k
=φm
k−(−1)m
φm
kφn
k−(−1)n
φn
k
=φm+n
k−(−1)mφn
k
φm
k−(−1)nφm
k
φn
k
+(−1)m+n
φm+n
k
.
This implies that
Lk,mLk,n + ∆2
kFk,mFk,n
= 2 φm+n
k+(−1)m+n
φm+n
k
= 2Lk,m+n.
Hence, (i) and (ii) complete the proof.
Setting k= 1 and k= 2 in Lemma 3, we have the
following.
Remark 4.Let mand nbe non-negative integers.
Then the following hold:
(i) 2Fm+n=FmLn+FnLm;
(ii) 2Lm+n=LmLn+ 5FmFn;
(iii) 2Pm+n=PmQn+PnQm;
(iv) 2Qm+n=QmQn+ 8PmPn.
3 Main Results
In this section, we now present that the integral
representation for the k-Fibonacci numbers Fk,ℓn
can be found by employing other known relations
between the two numbers Fk,ℓ and Lk,ℓ.
Theorem 5. For k,ℓand nare non-negative integers
with k= 0, the k-Fibonacci numbers Fk,ℓn can be
represented by the integral
Fk,ℓn =nFk,ℓ
2nZ1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1dx, (5)
where ∆k=√k2+ 4.
Proof. For n= 0 or ℓ= 0, it is obvious. Let
us assume that ℓ, n > 0. Let u(x) = Lk,ℓ +
∆kFk,ℓx. Then du = ∆kFk,ℓdx. Using integration
by substitution leads to
Z1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1dx
=1
∆kFk,ℓ Zu(1)
u(−1)
un−1du
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=1
n∆kFk,ℓ
(un)
u(1)
u(−1)
=1
n∆kFk,ℓ h(Lk,ℓ + ∆kFk,ℓx)ni1
−1
=1
n∆kFk,ℓ
(Lk,ℓ + ∆kFk,ℓ)n
−1
n∆kFk,ℓ
(Lk,ℓ −∆kFk,ℓ)n.(6)
Applying (i) and (ii) of Lemma 1 in (6) with n
replaced with ℓ, we get
Z1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1dx
=1
n∆kFk,ℓ 2φℓ
kn−2(−1)ℓ
φℓ
kn
=2n
nFk,ℓ 1
∆kφℓn
k−(−1)ℓn
φℓn
k.
It follows from (3) with replace nby ℓn that
Z1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1dx =2n
nFk,ℓ
Fk,ℓn.
Then (5) which completes the proof.
The integral representations of the k-Fibonacci
numbers for even and odd orders are shown as
follows:
Theorem 6. Let kand nbe non-negative integers
with k= 0 and ∆k=√k2+ 4.
(i) The k-Fibonacci numbers Fk,2ncan be
represented by the integral
Fk,2n=kn
2nZ1
−1
(k2+2+k∆kx)n−1dx. (7)
(ii) The k-Fibonacci numbers Fk,2n+1 can be
represented by the integral
Fk,2n+1 =1
2n+1 Z1
−1k2+2+k∆kxn−1
×k2n+k2+2+k(n+ 1)∆kx)dx.
Proof. (i) Notice that Fk,2=kand Lk,2=k2+ 2.
Setting ℓ= 2 in (5), we have
Fk,2n=kn
2nZ1
−1
(k2+2+k∆kx)n−1dx.
(ii) Re-indexing of nby n+ 1 in (7), we get
Fk,2n+2 =k(n+ 1)
2n+1 Z1
−1
(k2+2+k∆kx)ndx.
This together with (7) and
Fk,2n+2 =kFk,2n+1 +Fk,2n
gives
Fk,2n+1 =Fk,2n+2
k−Fk,2n
k
=(n+ 1)
2n+1 Z1
−1
(k2+2+k∆kx)ndx
−n
2nZ1
−1
(k2+2+k∆kx)n−1dx
=1
2n+1 Z1
−1k2+2+k∆kxn−1×
k2n+k2+2+k(n+ 1)∆kx)dx.
This completes the proof.
Setting k= 1 in Theorems 5 and 6, we have the
following corollaries.
Corollary 7 ([8], Theorem 2.1).For ℓand nare
non-negative integers, the Fibonacci numbers Fℓn
can be represented by the integral
Fℓn =nFℓ
2nZ1
−1
(Lℓ+√5Fℓx)n−1dx.
Proof. Notice that F1,ℓn =Fℓn,F1,ℓ =Fℓ,L1,ℓ =Lℓ
and ∆1=√5. Then, by Theorem 5, the conclusion
follows.
Corollary 8 ([8], Remark 2.1).Let nbe a
non-negative integer.
(i) The Fibonacci numbers F2ncan be represented
by the integral
F2n=n
2nZ1
−1
(3 + √5x)n−1dx.
(ii) The Fibonacci numbers F2n+1 can be
represented by the integral
F2n+1 =1
2n+1 Z1
−13 + √5xn−1
×n+3+√5(n+ 1)x)dx.
Setting k= 2 in Theorems 5 and 6, we have the
following corollaries.
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Corollary 9 ([30], Theorem 3.1).For ℓand nare
non-negative integers with k= 0, the Pell numbers
Pℓn can be represented by the integral
Pℓn =nPℓ
2nZ1
−1
(Qℓ+√8Pℓx)n−1dx.
Proof. Notice that F2,ℓn =Pℓn,F2,ℓ =Pℓ,L2,ℓ =
Qℓand ∆2=√8. Then, by Theorem 5, the
conclusion follows.
Corollary 10 ([30], Corollary 3.2).Let nbe a
non-negative integer.
(i) The Pell numbers P2ncan be represented by the
integral
P2n=nZ1
−1
(3 + √8x)n−1dx.
(ii) The Pell numbers P2n+1 can be represented by
the integral
P2n+1 =1
2Z1
−13 + √8xn−1
×2n+3+√8(n+ 1)x)dx.
Proof. By Theorem 6, we get
P2n=F2,2n=2n
2nZ1
−1
(6 + 2√8x)n−1dx
=nZ1
−1
(3 + √8x)n−1dx
and
P2n+1 =F2,2n+1
=1
2n+1 Z1
−16+2√8xn−1
×4n+ 6 + 2√8(n+ 1)x)dx
=1
2Z1
−13 + √8xn−1
×2n+3+√8(n+ 1)x)dx.
This completes the proof.
Setting k= 3 in Theorem 5, we have the following
numerical example.
Example 11.The 3-Fibonacci numbers F3,ℓn can be
represented by the integral
F3,ℓn =nF3,ℓ
2nZ1
−1
(L3,ℓ +√13F3,ℓx)n−1dx.
It is known that F3,2= 3 and L3,2= 11. Then we
can find F3,4and F3,6as follows:
F3,4=F3,2(2)
=2F3,2
22Z1
−1
(L3,2+√13F3,2x)2−1dx
=3
2Z1
−1
(11 + 3√13x)dx
=3
211x+3√13x2
2
1
−1
= 33
and
F3,6=F3,2(3)
=3F3,2
23Z1
−1
(L3,2+√13F3,2x)3−1dx
=9
8Z1
−1
(11 + 3√13x)2dx
=9
8Z1
−1
(121 + 66√13x+ 117x2)dx
=9
8121x+ 33√13x2+117x3
3
1
−1
= 360.
In another way, we can find F3,6when we known that
F3,3= 10 and L3,3= 36 as follows:
F3,6=F3,3(2)
=2F3,3
22Z1
−1
(L3,3+√13F3,3x)2−1dx
= 5 Z1
−1
(36 + 10√13x)dx
= 5 36x+ 5√13x2
1
−1
= 360.
Moreover, we obtain
F3,9=F3,3(3)
=3F3,3
23Z1
−1
(L3,3+√13F3,3x)3−1dx
=15
4Z1
−1
(36 + 10√13x)2dx
= 15 Z1
−1
(324 + 45√13x+ 325x2)dx
= 15 324x+45√13x2
2+325x3
3
1
−1
= 12970.
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Next, we provide the integral representations for
the k-Lucas numbers Lk,ℓn based on the two numbers
Fk,ℓ and Lk,ℓ.
Theorem 12. For k,ℓand nare non-negative
integers with k= 0, the k-Lucas numbers Lk,ℓn can
be represented by the integral
Lk,ℓn =1
2nZ1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1
×(Lk,ℓ + (n+ 1)∆kFk,ℓx)dx, (8)
where ∆k=√k2+ 4.
Proof. We will solve this integral (8) using the
integration by parts. Let uand vbe such that
u(x) = Lk,ℓ + (n+ 1)∆kFk,ℓx
and
dv = (Lk,ℓ + ∆kFk,ℓx)n−1dx.
Then du = (n+ 1)∆kFk,ℓdx and from (6) gives
v=Z(Lk,ℓ + ∆kFk,ℓx)n−1dx
=1
n∆kFk,ℓ
(Lk,ℓ + ∆kFk,ℓ)n.
It follows that
I=1
2nZ1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1
×(Lk,ℓ + (n+ 1)∆kFk,ℓx)dx
=1
n2n∆kFk,ℓ
×h(Lk,ℓ + ∆kFk,ℓx)n(Lk,ℓ + (n+ 1)∆kFk,ℓx)i1
−1
−n+ 1
n2nZ1
−1
(Lk,ℓ + ∆kPk,ℓx)ndx. (9)
Replacing nby n+ 1 in (5) becomes
Fk,ℓn+ℓ=(n+ 1)Fk,ℓ
2n+1 Z1
−1
(Lk,ℓ + ∆kFk,ℓx)ndx.
and so
2Fk,ℓn+ℓ
nFk,ℓ
=(n+ 1)
n2nZ1
−1
(Lk,ℓ + ∆kFk,ℓx)ndx.
This together with (9) gives
I=1
n2n∆kFk,ℓ
×h(Lk,ℓ + ∆kFk,ℓ)n(Lk,ℓ + (n+ 1)∆kFk,ℓ)
−(Lk,ℓ −∆kFk,ℓ)n(Lk,ℓ −(n+ 1)∆kFk,ℓ)i
−2Fk,ℓn+ℓ
nFk,ℓ
.
Applying (i) and (ii) of Lemma 1 with nreplaced by
ℓto the righthand side of the above equation gives
I=1
n2n∆kFk,ℓ h2nφℓn
k(Lk,ℓ + (n+ 1)∆kFk,ℓ)
−2n(−1)ℓn
φℓn
k
(Lk,ℓ −(n+ 1)∆kFk,ℓ)i
−2Fk,ℓn+ℓ
nFk,ℓ
=1
nFk,ℓ 1
∆kφℓn
k−(−1)ℓn
φℓn
kLk,ℓ
+ (n+ 1)Fk,ℓ φℓn
k+(−1)ℓn
φℓn
k
−2Fk,ℓn+ℓ
nFk,ℓ
.
Applying both Binet’s formulas (3) and (4) with n
replaced by ℓn and Lemma 3 (i) leads to
I=1
nFk,ℓ
[Fk,ℓnLk,ℓ + (n+ 1)Fk,ℓLk,ℓn]
−2Fk,ℓn+ℓ
nFk,ℓ
=1
nFk,ℓ
(Fk,ℓnLk,ℓ +Fk,ℓLk,ℓn) + Lk,ℓn
−2Fk,ℓn+ℓ
nFk,ℓ
=Lk,ℓn,
which completes the proof.
Setting k= 1 in Theorem 12, we have the
following corollary.
Corollary 13 ([8], Theorem 2.2).For ℓand nare
non-negative integers, the Lucas numbers Lℓn can be
represented by the integral
Lℓn =1
2nZ1
−1
(Lℓ+√5Fℓx)n−1
×(Lℓ+√5(n+ 1)Fℓx)dx.
Setting k= 2 in Theorem 12, we have the
following corollary.
Corollary 14 ([30], Theorem 3.4).For ℓand nare
non-negative integers, the Pell-Lucas numbers Qℓn
can be represented by the integral
Qℓn =1
2nZ1
−1×(Qℓ+√5Pℓx)n−1
(Qℓ+√8(n+ 1)Pℓx)dx.
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Setting k= 3 in Theorem 12, we have the
following numerical example.
Example 15.The 3-Lucas numbers L3,ℓn can be
represented by the integral
L3,ℓn =1
2nZ1
−1
(L3,ℓ +√13F3,ℓx)n−1
×(L3,ℓ +√13(n+ 1)F3,ℓx)dx.
Since F3,2= 3 and L3,2= 11, we can find L3,4and
L3,6as follows:
L3,4=L3,2(2)
=1
22Z1
−1
(L3,2+√13F3,2x)2−1
×(L3,2+ 3√13F3,2x)dx
=1
4Z1
−1
(11 + 3√13x)(11 + 9√13x)dx
=1
4Z1
−1
(121 + 132√13x+ 351x2)dx
=1
4121x+ 66√13x2+ 117x3
1
−1
= 119
and
L3,6=L3,2(3)
=1
23Z1
−1
(L3,2+√13F3,2x)3−1
×(L3,2+ 4√13F3,2x)dx
=1
8Z1
−1
(11 + 3√13x)2(11 + 12√13x)dx
=1
8Z1
−1
(1331 + 2178√13x
+ 11583x2+ 1404√13x3)dx
=1
8(1331x+ 1089√13x2+ 3861x3+ 351√13x4)
1
−1
= 1298.
Moreover, we can find L3,6when we known that
F3,3= 10 and L3,3= 36 as follows:
L3,6=F3,3(2)
=1
22Z1
−1
(L3,3+√13F3,3x)2−1
×(L3,3+ 3√13F3,3x)dx
=1
4Z1
−1
(36 + 10√13x)(36 + 30√13x)dx
=Z1
−1
(324 + 360√13x+ 975x2)dx
=324x+ 180√13x2+ 325x3
1
−1
= 1298.
Finally, both Fk,ℓn and Lk,ℓn are then used
to establish the generalized forms of integral
representations for the k-Fibonacci numbers Fk,ℓn+r
and k-Lucas numbers Lk,ℓn+ras the following
theorems.
Theorem 16. For k,ℓ,nand rare non-negative
integers with k= 0, the k-Fibonacci number Fk,ℓn+r
can be represented by the integral
Fk,ℓn+r=1
2n+1 Z1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1
×(nFk,ℓLk,r +Fk,rLk,ℓ + (n+ 1)∆kFk,ℓFk,rx)dx,
where ∆k=√k2+ 4.
Proof. Using Lemma 3 (i) with mand nreplaced by
ℓn and rrespectively, we get
Fk,ℓn+r=1
2Fk,ℓnLk,r +1
2Fk,r Lk,ℓn.
Applying Theorems 5 and 12 leads to
Fk,ℓn+r
=1
2nFk,ℓ
2nZ1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1dxLk,r
+1
2Fk,r 1
2nZ1
−1
(Lk,ℓ + (n+ 1)∆kFk,ℓx)
×(Lk,ℓ + ∆kFk,ℓx)n−1dx
=1
2n+1 Z1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1
×(nFk,ℓLk,r +Fk,rLk,ℓ + (n+ 1)∆kFk,ℓFk,rx)dx.
This completes the proof.
Remark 17.Notice that the results for the integral
representations of the even and odd k-Fibonacci
numbers given in Theorem 6 are recovered from
Theorem 16 on setting (ℓ, r) = (2,0) and (ℓ, r) =
(2,1), respectively.
Setting k= 1 in Theorem 16, we have the
following corollary.
Corollary 18 ([8], Theorem 2.3).For ℓ,nand rare
non-negative integers, the Fibonacci numbers Fℓn+r
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can be represented by the integral
Fℓn+r=1
2n+1 Z1
−1
(Lℓ+√5Fℓx)n−1
×nFℓLr+FrLℓ+√5(n+ 1) FℓFrxdx.
Setting k= 2 in Theorem 16, we have the
following corollary.
Corollary 19 ([30], Theorem 3.5).For ℓ,nand rare
non-negative integers, the Pell numbers Pℓn+rcan be
represented by the integral
Pℓn+r=1
2n+1 Z1
−1
(Qℓ+√8Pℓx)n−1
×nPℓQr+PrQℓ+√8(n+ 1) PℓPrxdx.
Theorem 20. For k,ℓ,nand rare non-negative
integers with k= 0, the k-Lucas numbers Lk,ℓn+r
can be represented by the integral
Lk,ℓn+r=1
2n+1 Z1
−1
(Lk,ℓ + ∆kFk,ℓx)n−1
×n∆2
kFk,ℓFk,r +Lk,ℓLk,r
+(n+ 1)∆kFk,ℓLk,rxdx,
where ∆k=√k2+ 4.
Proof. Using Lemma 3 (ii) with mand nreplaced by
ℓn and rrespectively, we get
Lk,ℓn+r=1
2Lk,ℓnLk,r +∆2
k
2Fk,ℓnFk,r.
This together with Theorems 5 and 12 gives that the
proof is finish.
Using the same idea as in Theorem 6, or setting
(ℓ, r) = (2,0) and (ℓ, r) = (2,1) in Theorem 20, we
also have the following integral representations of the
k-Lucas numbers for even and odd orders.
Theorem 21. Let kand nbe non-negative integers
with k= 0 and ∆k=√k2+ 4.
(i) The k-Lucas numbers Lk,2ncan be represented
by the integral
Lk,2n=1
2nZ1
−1
(k2+2+k∆kx)n−1
×(k2+2+k(n+ 1)∆kx)dx.
(ii) The k-Lucas numbers Lk,2n+1 can be
represented by the integral
Lk,2n+1 =k
2n+1 Z1
−1k2+2+k∆kxn−1
×n∆2
k+k2+2+k(n+ 1)∆kx)dx.
Setting k= 1 in Theorems 20 and 21, we have the
following corollaries.
Corollary 22 ([8], Theorem 2.4).For ℓ,nand rare
non-negative integers, the Lucas numbers Lℓn+rcan
be represented by the integral
Lℓn+r=1
2n+1 Z1
−1
(Lℓ+√5Fℓx)n−1
×5nFℓFr+LℓLr+√5(n+ 1) FrLrxdx.
Corollary 23 ([8], Remark 2.4).Let nbe a
non-negative integer.
(i) The Lucas numbers L2ncan be represented by
the integral
L2n=1
2nZ1
−1
(3+√5x)n−1(3+√5(n+1) x)dx.
(ii) The Lucas numbers L2n+1 can be represented by
the integral
L2n+1 =1
2n+1 Z1
−1
(3 + √5x)n−1
×(5n+3+√5(n+ 1)x)dx.
Setting k= 2 in Theorems 20 and 21, we have the
following corollaries.
Corollary 24 ([30], Theorem 3.6).For ℓ,nand rare
non-negative integers, the Pell-Lucas numbers Qℓn+r
can be represented by the integral
Qℓn+r=1
2n+1 Z1
−1
(Qℓ+√8Pℓx)n−1
×8nPℓPr+QℓQr+√8(n+ 1)PℓQrxdx.
Corollary 25. Let nbe a non-negative integer.
(i) The Pell-Lucas numbers Q2ncan be represented
by the integral
Q2n=Z1
−1
(3 + √8x)n−1(3 + √8(n+ 1)x)dx.
(ii) The Pell-Lucas numbers Q2n+1 can be
represented by the integral
Q2n+1 =Z1
−1
(3 + √8x)n−1
×(4n+3+√8(n+ 1)x)dx.
Setting k= 3 and r= 1 in Theorems 16 and 20,
we have the following numerical example.
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Example 26.The 3-Fibonacci numbers F3,ℓn+1 and
3-Lucas numbers L3,ℓn+1 can be represented by the
integral
F3,ℓn+1 =1
2n+1 Z1
−1
(L3,ℓ +√13F3,ℓx)n−1
×(3nF3,ℓ +L3,ℓ +√13(n+ 1)F3,ℓx)dx
and
L3,ℓn+r=1
2n+1 Z1
−1
(L3,ℓ +√13F3,ℓx)n−1
×13nF3,ℓ + 3L3,ℓ + 3√13(n+ 1) F3,ℓxdx.
It is known that F3,2= 3 and L3,2= 11. Then we
can find F3,3and L3,3as follows:
F3,3=F3,2(1)+1
=1
22Z1
−1
(3F3,2+L3,2+ 2√13F3,2x)dx
=1
2Z1
−1
(10 + 3√13x)dx
=1
210x+3√13x2
2
1
−1
= 10
and
L3,3=L3,2(1)+1
=1
22Z1
−113F3,2+ 3L3,2+ 6√13 F3,2xdx
=1
2Z1
−1
(36 + 9√13x)dx
=1
236x+9√13x2
2
1
−1
= 36.
Moreover, we obtain F3,5and L3,5as follows:
F3,5=F3,2(2)+1
=1
23Z1
−1
(L3,2+√13F3,2x)
×(6F3,2+L3,2+ 3√13F3,2x)dx
=1
8Z1
−1
(11 + 3√13x)(29 + 9√13x)dx
=1
8Z1
−1
(319 + 186√13x+ 351x2)dx
=1
8319x+ 93√13x2+ 117x3
1
−1
= 109
and
L3,5=L3,2(2)+1
=1
23Z1
−1
(L3,2+√13F3,2x)
×(26F3,2+ 3L3,2+ 9√13F3,2x)dx
=1
8Z1
−1
(11 + 3√13x)(111 + 27√13x)dx
=1
8Z1
−1
(1221 + 630√13x+ 1053x2)dx
=1
81221x+ 315√13x2+ 351x3
1
−1
= 393.
4 Conclusion
In this paper, new integral representations of the
k-Fibonacci numbers and the k-Lucas numbers have
been introduced and studied. Many of the properties
of these numbers are proved by using Binet’s
formulas. We also establish some identities and
simple integral calculus to prove them. The approach
primarily builds on mathematical skills for deriving
integral representations and provides formulas for
both even and odd terms in these sequences. Indeed,
we present that the integral representation for the
k-Fibonacci numbers Fk,ℓn and k-Lucas numbers
Lk,ℓn can be found by employing other known
relations between the two numbers Fk,ℓ and Lk,ℓ. And
then both Fk,ℓn and Lk,ℓn are used to establish the
generalized forms of integral representations for the
k-Fibonacci numbers Fk,ℓn+rand k-Lucas numbers
Lk,ℓn+r. Moreover, we deduce results applicable
to related number sequences like Fibonacci, Lucas,
Pell, and Pell-Lucas numbers. Finally, we give
some numerical examples of 3-Fibonacci and 3-Lucas
numbers.
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Contribution of individual authors to
the creation of a scientific article
(ghostwriting policy)
Weerayuth Nilsrakoo is responsible for the
conceptualization of the research problem, formal
analysis, and the supervision of the work. Achariya
Nilsrakoo is responsible for the formal analysis,
validation, and corresponding author.
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scientific article itself
No funding was received for conducting this study.
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Conflict of Interest
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that are relevant to the content of this article.
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