EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6452 ISSN 1307-5543 – ejpam.com Published by New York Business Global Integral Representations of Generalizations of Pell Numbers and Their Companion Numbers Weerayuth Nilsrakoo1, Achariya Nilsrakoo2,∗ 1 Department of Mathematics, Statistics and Computer, Faculty of Science, Ubon Ratchathani University, Ubon Ratchathani, 34190, Thailand 2 Department of Mathematics, Faculty of Science, Ubon Ratchathani Rajabhat University, Ubon Ratchathani, 34000, Thailand Abstract. This paper discusses a one-parameter generalization of Pell numbers that preserves the recurrence relation with arbitrary initial conditions. We introduce generalized Pell-Lucas-like numbers, which are simple associations of generalized Pell numbers. Consequently, we give some new and well-known identities. Furthermore, we propose integral representations of these numbers associated with generalized Pell and Pell-Lucas-like numbers. Our results not only generalize the integral representations of the Pell and Pell-Lucas numbers but also apply to all the companion numbers of generalized Pell numbers. 2020 Mathematics Subject Classifications: 11B39, 11B37 Key Words and Phrases: Generalized Pell number, generalized Pell-Lucas-like number, integral representation 1. Introduction Recall that Pell numbers Pn are defined by the recurrence relations P0 = 0, P1 = 1, and Pn = 2Pn−1 + Pn−2, n ≥ 2, and its associated numbers or Pell-Lucas numbers Qn are defined by the recurrence rela- tions Q0 = 2, Q1 = 1, and Qn = 2Qn−1 + Qn−2, n ≥ 2. The Binet’s formulas for the Pell and Pell-Lucas numbers are related to the silver ratio φ = 1 + √ 2. There are some generalizations of Pell and Pell-Lucas numbers defined in different ways. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6452 Email addresses: weerayuth.ni@ubu.ac.th (W. Nilsrakoo), achariya.n@ubru.ac.th (A. Nilsrakoo) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 2 of 18 In 2007, Falcón and Plaza [1] introduced the first kind of one-parameter generalization of Fibonacci and Pell numbers as follows: The k-Fibonacci numbers Fk,n are defined by the recurrence relations Fk,0 = 0, Fk,1 = 1, and Fk,n = kFk,n−1 + Fk,n−2, n ≥ 2, where k and n are non-negative integers with k ̸= 0. The associated numbers of k- Fibonacci numbers introduced in 2011 by Falcón [2] as follows: The k-Lucas numbers Lk,n are defined by the recurrence relations Lk,0 = 2, Lk,1 = k, and Lk,n = kLk,n−1 + Lk,n−2, n ≥ 2. In 2013, Catarino [3] introduced the second kind of one-parameter generalization of Pell numbers as follows: The k-Pell numbers Pk,n are defined by the recurrence relations Pk,0 = 0, Pk,1 = 1, and Pk,n = 2Pk,n−1 + kPk,n−2, n ≥ 2, Subsequently, Catarino and Vasco [4] introduced the association of Pell numbers as follows: The k-Pell-Lucas numbers Qk,n are defined by the recurrence relations Qk,0 = 2, Qk,1 = 2, and Qk,n = 2Qk,n−1 + kQk,n−2, n ≥ 2, In 2019, Trojnar-Spenlina and W loch [5] introduced the third kind of one-parameter generalization of Pell numbers as follows: The generalized Pell numbers Pk,n are defined by the recurrence relations Pk,0 = 0,Pk,1 = 1, and Pk,n = kPk,n−1 + (k − 1)Pk,n−2, n ≥ 2, (1) Subsequently, the association of generalized Pell numbers Qk,n, so-called generalized Pell- Lucas numbers, are defined by the recurrence relations Qk,0 = 2,Qk,1 = 2, and Qk,n = kQk,n−1 + (k − 1)Qk,n−2, n ≥ 2, In the paper [6], the other associated numbers of generalized Pell numbers, which are the so-called generalized modified Pell numbers qk,n, are defined by the recurrence relations qk,0 = 1, qk,1 = 1, and qk,n = kqk,n−1 + (k − 1)qk,n−2, n ≥ 2, It is known that Qk,n = 2qk,n. Recall that a pair of Lucas sequences ({Un} , {Vn}) [7] is defined by the formulas U0 = 0, U1 = 1, and Un = αUn−1 − βUn−2, n ≥ 2, V0 = 2, V1 = α, and Vn = αVn−1 − βVn−2, n ≥ 2. where α and β are integers such that the discriminant ∆ = α2 +4β ̸= 0. In this case, {Un} and {Vn} are called the Lucas sequences of the first and second kinds, respectively. Note that ({Fk,n} , {Lk,n}) and ({Pk,n} , {Qk,n}) are included in the more general definition by W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 3 of 18 assuming (α, β) = (k,−1) and (α, β) = (2,−k), respectively. However, the third kind of one-parameter generalization of Pell numbers ({Pk,n} , {Qk,n}) and ({Pk,n} , {qk,n}) are not a pair of Lucas sequences. If we define Lk,n by Lk,0 = 2,Lk,1 = k, and Lk,n = kLk,n−1 + (k − 1)Lk,n−2, n ≥ 2, then it is a convenient Lucas sequence of the second kind such that ({Pk,n} , {Lk,n}) is a pair of Lucas sequences with (α, β) = (k, 1− k) to consider it to be an associated number of Pk,n. Here Lk,n is called generalized Pell-Lucas-like. The tables presented below contain initial terms of the sequences {Lk,n} for selected values k (Table 1). Table 1: Initial terms of the generalized Pell-Lucas-like numbers {Lk,n}. n 0 1 2 3 4 5 6 7 8 9 L2,n 2 2 6 14 34 82 198 478 1154 2786 L3,n 2 3 13 45 161 573 2041 7269 25889 92205 L4,n 2 4 22 100 466 2164 10054 46708 216994 1008100 L5,n 2 5 33 185 1057 6025 34353 195865 1116737 6367145 L6,n 2 6 46 306 2066 13926 93886 632946 4267106 28767366 For k = 2, we can see that the classical Pell–Lucas numbers are obtained. Moreover, sequences {L2,n}, {L3,n}, and {L4,n} are listed in The Online Encyclopaedia of Integer Sequences [8] under the symbols A002203, A206776, and A080042, respectively. In this paper, we study all the third kind of one-parameter generalization of Pell numbers that preserve the recurrence relation (1) with arbitrary initial conditions. We see that the generalized Pell-Lucas-like is a simple association of generalized Pell numbers. Consequently, we give some new and well-known identities. Furthermore, we propose integral representations of these numbers associated with generalized Pell and Pell-Lucas- like numbers. 2. The companion numbers of generalized Pell numbers In this section, we point out the third kind of generalized Pell numbers to study a generalization of the Pell numbers with one parameter positive integer k ≥ 2 which is called the companion generalized Pell number, denoted by GPk,n = GPk,n(a, b), defined by a recurrence relation Gk,0 = a,GPk,1 = b, and GPk,n = kGPk,n−1 + (k − 1)GPk,n−2, n ≥ 2, (2) where a and b are arbitrary non-negative integers such that a + b ̸= 0. Note that GPk,n correspond to special cases of Horadam numbers [9]. The first terms GPk,n are: GPk,0 = a GPk,1 = b W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 4 of 18 GPk,2 = (a + b)k − a GPk,3 = (a + b)k2 − (a− b)k − b GPk,4 = (a + b)k3 + 2bk2 − 2(a + b)k + a GPk,5 = (a + b)k4 + (a + 3b)k3 − 2(2a + b)k2 + (a− 2b)k + b GPk,6 = (a + b)k5 + 2(a + 2b)k4 − (5a + b)k3 − (a + 6b)k2 + 3(a + b)k − a. Some particular cases of the previous definition are (i) Pk,n = GPk,n(0, 1), (ii) Lk,n = GPk,n(2, k), (iii) Qk,n = GPk,n(2, 2), (iv) qk,n = GPk,n(1, 1), (v) generalized Pell numbers introduced in [10], Ha,b n = GP2,n(a, b), (vi) Pn = P2,n = GP2,n(0, 1), (vii) Qn = L2,n = Q2,n = GP2,n(2, 2), and (viii) qn = GP2,n(1, 1). The Binet’s formula for GPk,n is given in the following theorem. Theorem 1 (Binet’s formulas). Let k and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then GPk,n = 2b− ak + a∆k 2∆k σn k + ak − 2b + a∆k 2∆k (1 − k)n σn k , (3) where σk = k+∆k 2 . Proof. The recurrence relation (2) generates a characteristic equation of the form r2 − kr + 1 − k = 0. (4) Since ∆2 k = k2 + 4k − 4 = k2 + 4(k − 1) > 0 for k ≥ 2, this equation has two roots, r1 = k + ∆k 2 = σk and r2 = k − ∆k 2 = (k − ∆k) (k + ∆k) 2 (k + ∆k) = 2(1 − k) (k + ∆k) = 1 − k σk . W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 5 of 18 Note that r1 + r2 = k, r1 − r2 = ∆k, and r1r2 = 1− k. Therefore, the general term GPk,n can be expressed as GPk,n = αrn1 + βrn2 = ασn k + β (1 − k)n σn k for some coefficients α and β. Since GPk,0 = a and GPk,1 = b, we get α + β = a and αr1 + βr2 = b. It can be shown that α = 2b− ak + a∆k 2∆k and β = ak − 2b + a∆k 2∆k . Therefore, (3) has been proved. If (a, b) ∈ {(0, 1), (2, k), (2, 2), (1, 1)}, then we have the following: Corollary 1. Let k and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then (i) Pk,n = 1 ∆k ( σn k − (1−k)n σn k ) , (ii) Lk,n = σn k + (1−k)n σn k , (iii) Qk,n = ( 2−k+∆k ∆k ) σn k + ( k−2∆k ∆k ) (1−k)n σn k , and (iv) qk,n = ( 2−k+∆k 2∆k ) σn k + ( k−2∆k 2∆k ) (1−k)n σn k . Remark 1. As in Corollary 1, we get the following: • (i) and (iii) are presented in [5, Corollary 2.3]; • (iv) is presented in [6, Theorem 2.3]; • the Binet’s formula of Lk,n is simpler than that of Qk,n and qk,n. If k = 2, then we have the following: Corollary 2. Let n be a non-negative integer. Then Ha,b n = b− a + √ 2a 2 √ 2 (1 + √ 2)n + a− b− √ 2a 2 √ 2 (1 − √ 2)n. (5) Proof. Notice that GP2,n = Ha,b n , ∆2 = 2 √ 2 and σ2 = 1 + √ 2. Setting k = 2 in (3), we get (5) which completes the proof. Theorem 2. Let k and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then the following hold: W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 6 of 18 (i) Lk,n + ∆k Pk,n = 2σn k ; (ii) Lk,n − ∆k Pk,n = 2 (1−k)n σn k ; (iii) L2 k,n − ∆2 kP2 k,n = 4(1 − k)n. Proof. The conclusions follow from (i) and (ii) of Corollary 1. Next, we present that the companion generalized Pell numbers are associated with the generalized Pell and generalized Pell-Lucas-like numbers in the following results. Theorem 3. Let k and n be non-negative integers with k ≥ 2. Then GPk,n = a 2 Lk,n + 2b− ak 2 Pk,n. Proof. It follows from (3), (i) and (ii) of Theorem 2 that GPk,n = ( 2b− ak + a∆k 2∆k ) σn k + ( ak − 2b + a∆k 2∆k ) (1 − k)n σn k = ( 2b− ak + a∆k 2∆k )( Lk,n + ∆k Pk,n 2 ) + ( ak − 2b + a∆k 2∆k )( Lk,n − ∆k Pk,n 2 ) = a 2 Lk,n + 2b− ak 2 Pk,n. This completes the proof. If (a, b) ∈ {(2, 2), (1, 1)}, then we have the following: Corollary 3. Let k and n be non-negative integers with k ≥ 2. Then (i) Qk,n = Lk,n − (k − 2)Pk,n; (ii) qk,n = 1 2Lk,n − (k−2) 2 Pk,n. If k = 2, then we have the following: Corollary 4. Let n be a non-negative integer. Then Ha,b n = a 2Qn + (b− a)Pn. From Theorem 2 and Corollary 3, we have the following: Corollary 5 ([11, Lemma 2.1]). Let k and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then the following hold: (i) Qk,n + (k − 2 + ∆k) Pk,n = 2σn k ; (ii) Qk,n + (k − 2 − ∆k) Pk,n = 2 (1−k)n σn k ; (iii) (Qk,n + (k − 2)Pk,n)2 − ∆kP2 k,n = 4(1 − k)n. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 7 of 18 Theorem 4. Let k, m and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then the following hold: (i) 2Pk,m+n = Pk,mLk,n + Pk,nLk,m; (ii) 2Lk,m+n = Lk,mLk,n + ∆2 kPk,mPk,n. Proof. (i) Using (i) and (ii) of Corollary 1, we have Pk,mLk,n = [ 1 ∆k ( σm k − (1 − k)m σm k )]( σn k + (1 − k)n σn k ) = 1 ∆k ( σm+n k + (1 − k)nσm k σn k − (1 − k)mσn k σm k − (1 − k)m+n σm+n k ) and Pk,nLk,m = [ 1 ∆k ( σn k − (1 − k)n σn k )]( σm k + (1 − k)m σm k ) = 1 ∆k ( σm+n k + (1 − k)mσn k σn k − (1 − k)nσm k σn k − (1 − k)m+n σm+n k ) . So, we get Pk,mLk,n + Pk,nLk,m = 2 ∆k ( σm+n k − (1 − k)m+n σm+n k ) = 2Pk,m+n. (ii) Using (i) of Corollary 1, we have Pk,mPk,n = [ 1 ∆k ( σm k − (1 − k)m σm k )][ 1 ∆k ( σn k − (1 − k)n σn k )] = 1 ∆2 k ( σm+n k − (1 − k)nσm k σn k − (1 − k)mσn k σm k + (1 − k)m+n σm+n k ) . By using (ii) of Corollary 1, we have Lk,mLk,n = ( σm k + (1 − k)m σm k )( σn k + (1 − k)n σn k ) = σm+n k + (1 − k)nσm k σn k + (1 − k)mσn k σm k + (1 − k)m+n σm+n k . This implies that Lk,mLk,n + (k2 + 4k − 4)Pk,mPk,n = 2 ( σm+n k + (1 − k)m+n σm+n k ) = 2Lk,m+n. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 8 of 18 Hence, (i) and (ii) complete the proof. Corollary 6 ([11, Lemma 2.2]). Let k, m and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then the following hold: (i) 2Pk,m+n = Pk,mQk,n + Pk,nQk,m + 2(k − 2)Pk,mPk,n; (ii) 2Qk,m+n = Qk,mQk,n + 8(k − 1)Pk,mPk,n. Proof. It follows from (i) of Theorem 4 and (i) of Corollary 3 that 2Pk,m+n = Pk,mLk,n + Pk,nLk,m = Pk,m (Qk,n + (k − 2)Pk,n) + Pk,n (QLk,m + (k − 2)Pk,m) = Pk,mQk,n + Pk,nQk,m + 2(k − 2)Pk,mPk,n. Since Lk,mLk,n = (Qk,m + (k − 2)Pk,m) (Qk,n + (k − 2)Pk,n) = Qk,mQk,n + (k − 2)Pk,mQk,n + (k − 2)Pk,nQk,m + (k − 2)2Pk,mPk,n = Qk,mQk,n + (k − 2) [Pk,mQk,n + Pk,nQk,m + 2(k − 2)Pk,mPk,n] − (k − 2)2Pk,mPk,n = Qk,mQk,n + 2(k − 2)Pk,m+n − (k − 2)2Pk,mPk,n, we get 2Qk,m+n = 2Lk,m+n − 2(k − 2)Pk,m+n = Lk,mLk,n + ( k2 + 4k − 4 ) Pk,mPk,n − 2(k − 2)Pk,m+n = Qk,mQk,n − (k − 2)2Pk,mPk,n + (k2 + 4k − 4)Pk,mPk,n = Qk,mQk,n + 8(k − 1)Pk,mPk,n. Hence, (i) and (ii) complete the proof. Theorem 5 (Asymptotic behavior). Let k be a positive integer with k ≥ 2. Then lim n→∞ GPk,n+1 GPk,n = σk. Proof. By using (3), we have lim n→∞ GPk,n+1 GPk,n = lim n→∞ ( 2b−ak+a∆k 2∆k ) σn+1 k + ( ak−2b+a∆k 2∆k ) (1−k)n+1 σn+1 k( 2b−ak+a∆k 2∆k ) σn k + ( ak−2b+a∆k 2∆k ) (1−k)n σn k = lim n→∞ ( 2b−ak+a∆k 2∆k ) σk + ( ak−2b+a∆k 2∆k ) (1−k)n σ2n k · (1−k) σk( 2b−ak+a∆k 2∆k ) + ( ak−2b+a∆k 2∆k ) (1−k)n σ2n k . (6) W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 9 of 18 Since σk is the root of (4), we have σ2 k = kσk + (k − 1) > k − 1 and so ∣∣∣1−k σ2 k ∣∣∣ < 1. Then lim n→∞ (1 − k)n σ2n k = lim n→∞ ( 1 − k σ2 k )n = 0. This together with (6) gives lim n→∞ GPk,n+1 GPk,n = σk. This completes the proof. If (a, b) ∈ {(2, k), (0, 1), (2, 2), (1, 1)}, then we have the following: Corollary 7 ([5, Lemma 2.4]). Let k be a positive integer with k ≥ 2. Then lim n→∞ Lk,n+1 Lk,n = lim n→∞ Pk,n+1 Pk,n = lim n→∞ Qk,n+1 Qk,n = lim n→∞ qk,n+1 qk,n = σk. If k = 2, then we have the following: Corollary 8. limn→∞ Ha,b n+1 Ha,b n = 1 + √ 2. Theorem 6 (Catalan’s identities). Let k, n, and r be non-negative integers with k ≥ 2, ∆k = √ k2 + 4k − 4, and n ≥ r. Then GPk,n−rGPk,n+r − GP2 k,n = 1 4 (2b− ak + a∆k)(ak − 2b + a∆k)(1 − k)n−rP2 k,r. Proof. Let α = 2b−ak+a∆k 2∆k and β = ak−2b+a∆k 2∆k . By using (3), we have GPk,n−rGPk,n+r = ( ασn−r k + β (1 − k)n−r σn−r k )( ασn+r k + β (1 − k)n+r σn+r k ) = α2σ2n k + β2 (1 − k)2n σ2n k + αβ(1 − k)n−r ( σ2r k + (1 − k)2r σ2r k ) and GP2 k,n = ( ασn k + β (1 − k)n σn k )2 = α2σ2n k + β2 (1 − k)2n σ2n k + αβ(1 − k)n−r (2(1 − k)r) . Then GPk,n−rGPk,n+r − GP2 k,n = αβ(1 − k)n−r ( σ2r k + (1 − k)2r σ2r k − 2(1 − k)r ) = ( 2b− ak + a∆k 2∆k )( ak − 2b + a∆k 2∆k ) (1 − k)n−r ( σr k − (1 − k)r σr k )2 W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 10 of 18 = 1 4 (2b− ak + a∆k)(ak − 2b + a∆k) (1 − k)n−r [ 1 ∆k ( σr k + (1 − k)r σr k )]2 = 1 4 (2b− ak + a∆k)(ak − 2b + a∆k)(1 − k)n−rP2 k,r. This completes the proof. If (a, b) ∈ {(2, k), (0, 1), (2, 2), (1, 1)}, then we have the following: Corollary 9 ([5, Theorem 2.6]). Let k, n, and r be non-negative integers with k ≥ 2, ∆k = √ k2 + 4k − 4, and n ≥ r. Then (i) Lk,n−rLk,n+r − L2 k,n = ∆2 k(1 − k)n−rP2 k,r; (ii) Pk,n−rPk,n+r − P2 k,n = −(1 − k)n−rP2 k,r; (iii) Qk,n−rQk,n+r −Q2 k,n = −8(1 − k)n−r+1P2 k,r; (iv) qk,n−rqk,n+r − q2k,n = −2(1 − k)n−r+1P2 k,r. If k = 2, then we have the following: Corollary 10. Let n and r be non-negative integers with n ≥ r. Then Ha,b n−rH a,b n+r − (Ha,b n )2 = (b− a + √ 2a)(a− b− √ 2a)(−1)n−rP 2 r . Note that r = 1 in Theorem 6, the Catalan’s identities give Cassini’s identities as follows: Theorem 7 (Cassini’s identities). Let k, n, and r be non-negative integers with k ≥ 2, ∆k = √ k2 + 4k − 4, and n ≥ r. Then GPk,n−1GPk,n+1 − GP2 k,n = 1 4 (2b− ak + a∆k)(ak − 2b + a∆k)(1 − k)n−1. If (a, b) ∈ {(2, k), (0, 1), (2, 2), (1, 1)}, then we have the following: Corollary 11. Let k, n, and r be non-negative integers with k ≥ 2, ∆k = √ k2 + 4k − 4, and n ≥ r. Then (i) Lk,n−1Lk,n+1 − L2 k,n = ∆2 k(1 − k)n−1; (ii) Pk,n−1Pk,n+1 − P2 k,n = −(1 − k)n−1; (iii) Qk,n−1Qk,n+1 −Q2 k,n = −8(1 − k)n; (iv) qk,n−1qk,n+1 − q2k,n = −2(1 − k)n. If k = 2, then we have the following: Corollary 12. Let n be a non-negative integer. Then Ha,b n−1H a,b n+1 − (Ha,b n )2 = (b− a + √ 2a)(a− b− √ 2a)(−1)n−1. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 11 of 18 3. The integral representation of the generalized Pell numbers There are several ways to represent the special numbers. One of them is the integral representation; see, for example, [11–22]. The integral representations for Pell and Pell- Lucas numbers are studied by the second author in [17] as follows: Pℓn = nPℓ 2n ∫ 1 −1 (Qℓ + 2 √ 2Pℓx)n−1dx and Qℓn = 1 2n ∫ 1 −1 (Qℓ + 2 √ 2(n + 1)Pℓx)(Qℓ + 2 √ 2Pℓx)n−1dx, where ℓ and n are non-negative integers. Subsequently, the authors [11, 18, 19] give new integral representations for the general of Pell and Pell-Lucas numbers such as the k-Fibonacci, k-Lucas, k-Pell, k-Pell-Lucas, and the k-Pell-Lucas-like numbers. In this section, we obtain new integral representations for the companion generalized Pell numbers. We start with the following theorem for the generalized Pell number Pk,ℓn by employing other known relations between the two numbers Pk,ℓ and Lk,ℓ. Theorem 8. Let k, ℓ and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then Pk,ℓn = nPk,ℓ 2n∆k ∫ ∆k −∆k (Lk,ℓ + Pk,ℓ x)n−1 dx. (7) Proof. For n = 0 or ℓ = 0, we have done. Let us assume that ℓ, n > 0. Using integration by substitution, we get∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n−1 dx = 1 nPk,ℓ [ (Lk,ℓ + Pk,ℓx)n ]∆k −∆k = 1 nPk,ℓ [(Lk,ℓ + ∆kPk,ℓ) n − (Lk,ℓ − ∆kPk,ℓ) n] . It follows from (i) and (ii) of Theorem 2 with n replaced with ℓ that∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n−1 dx = 1 nPk,ℓ [( 2σℓ k )n − ( 2 (1 − k)ℓ σℓ k )n] = 2n∆k nPk,ℓ [ 1 ∆k ( σℓn k − (1 − k)ℓn σℓn k )] . By using (i) of Corollary 1 with replace n by ℓn, we have∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n−1 dx = 2n∆k Pk,ℓn nPk,ℓ . Then (7) which completes the proof. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 12 of 18 Remark 2. As in Theorem 8, equation (7) is equivalent to Pk,ℓn = nPk,ℓ 2n ∫ 1 −1 (Lk,ℓ + ∆kPk,ℓ t) n−1 dt. In fact, substituting t = x ∆k produces dx = ∆kdt and the integration limits are changed to −1 and 1, respectively. The generalized Pell number Pk,ℓn by employing the two numbers Pk,ℓ and Qk,ℓ is presented as follows: Corollary 13 ([11, Theorem 2.3]). Let k, ℓ and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then Pk,ℓn = nPk,ℓ 2n∆k ∫ ∆k −∆k (Qk,ℓ + (k − 2 + x)Pk,ℓ) n−1 dx. (8) Proof. From (i) of Corollary 3, we have Lk,ℓ = Qk,ℓ + (k − 2)Pk,ℓ. This together with (7) that (8) holds. The integral representations of the generalized Pell number for even and odd orders are shown as follows: Theorem 9. Let k and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. (i) The generalized Pell number Pk,2n can be represented by Pk,2n = nk 2n∆k ∫ ∆k −∆k ( k2 + 2k − 2 + kx )n−1 dx. (9) (ii) The generalized Pell number Pk,2n+1 can be represented by Pk,2n+1 = 1 2n+1∆k ∫ ∆k −∆k ( 2k − 2 + (n + 1)(k2 + kx) ) ( k2 + 2k − 2 + k x )n−1 dx. Proof. (i) Notice that Pk,2 = k and Lk,2 = k2 + 2k − 2. Setting ℓ = 2 in (7), we have Pk,2n = nPk,2 2n∆k ∫ ∆k −∆k (Lk,2 + Pk,2x)n−1 dx = nk 2n∆k ∫ ∆k −∆k ( k2 + 2k − 2 + kx )n−1 dx. (ii) Re-indexing n by n + 1 in (9), we get Pk,2n+2 = (n + 1)k 2n+1∆k ∫ ∆k −∆k ( k2 + 2k − 2 + kx )n dx. (10) W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 13 of 18 Using Pk,2n+2 = kPk,2n+1 + (k − 1)Pk,2n with (9) and (10), we obtain Pk,2n+1 = 1 k Pk,2n+2 − k − 1 k Pk,2n = (n + 1) 2n+1∆k ∫ ∆k −∆k ( k2 + 2k − 2 + kx )n dx− n(k − 1) 2n∆k ∫ ∆k −∆k ( k2 + 2k − 2 + kx )n−1 dx = 1 2n+1∆k ∫ ∆k −∆k ( 2k − 2 + (n + 1)(k2 + kx) ) ( k2 + 2k − 2 + k x )n−1 dx. This completes the proof. Setting k = 2 in Theorem 8 and Remark 2, we have the following corollary. Corollary 14 ([17, Theorem 3.1]). Let ℓ and n be non-negative integers. Then Pℓn = nPℓ 2n+1 √ 2 ∫ 2 √ 2 −2 √ 2 (Qℓ + Pℓ x)n−1dx = nPℓ 2n ∫ 1 −1 (Qℓ + 2 √ 2Pℓ x)n−1dx. Next, we provide integral representations for the generalized Pell-Lucas-like number Lk,ℓn based on the two numbers Pk,ℓ and Lk,ℓ. Theorem 10. Let k, ℓ and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then Lk,ℓn = 1 2n∆k ∫ ∆k −∆k (Lk,ℓ + (n + 1)Pk,ℓx) (Lk,ℓ + Pk,ℓx)n−1 dx. (11) Proof. For n = 0 or ℓ = 0, it is easy to see that (11) holds. We assume now that ℓ, n > 0. We will solve (11) using integration by parts. Let u and v be such that u(x) = Lk,ℓ + (n + 1)Pk,ℓx and dv = (Lk,ℓ + Pk,ℓx)n−1 dx. Then I = 1 2n∆k ∫ ∆k −∆k (Lk,ℓ + (n + 1)Pk,ℓx) (Lk,ℓ + Pk,ℓx)n−1 dx = 1 n2n∆k Pk,ℓ [ (Lk,ℓ + (n + 1)Pk,ℓx) (Lk,ℓ + Pk,ℓx)n ]∆k −∆k − (n + 1) n2n∆k ∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n dx. (12) Replacing n by n + 1 in (7) becomes Pk,ℓn+ℓ = (n + 1)Pk,ℓ 2n+1∆k ∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n dx and so 2Pk,ℓn+ℓ nPk,ℓ = (n + 1) n2n∆k ∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n dx. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 14 of 18 This together with (12) gives I = 1 n2n∆k Pk,ℓ [ (Lk,ℓ + (n + 1)∆kPk,ℓ) (Lk,ℓ + ∆kPk,ℓ) n ] − 1 n2n∆k Pk,ℓ [ (Lk,ℓ − (n + 1)∆kPk,ℓ) (Lk,ℓ − ∆kPk,ℓ) n ] − 2Pk,ℓn+ℓ nPk,ℓ . (13) Applying (i) and (ii) of Theorem 2 to (13) gives I = 1 n2n∆k Pk,ℓ [ 2nσℓn k (Lk,ℓ + (n + 1)∆kPk,ℓ) ] − 1 n2n∆k Pk,ℓ [ 2n (1 − k)ℓn σℓn k (Lk,ℓ − (n + 1)∆kPk,ℓ) ] − 2Pk,ℓn+ℓ nPk,ℓ = 1 nPk,ℓ [ 1 ∆k ( σℓn k − (1 − k)ℓn σℓn k ) Lk,ℓ + ( σℓn k + (1 − k)ℓn σℓn k ) (n + 1)Pk,ℓ − 2Pk,ℓn+ℓ ] . Using (i) and (ii) of Corollary 1, and (i) of Theorem 4, it follows that I = 1 nPk,ℓ [Pk,ℓnLk,ℓ + (n + 1)Pk,ℓLk,ℓn − 2Pk,ℓn+ℓ] = 1 nPk,ℓ [Pk,ℓnLk,ℓ + Pk,ℓLk,ℓn − 2Pk,ℓn+ℓ] + Lk,ℓn = Lk,ℓn. This completes the proof. Now, new integral representations for the companion generalized Pell numbers asso- ciated with the generalized Pell and generalized Pell-Lucas-like numbers are presented as follows: Theorem 11. Let k, ℓ and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. The companion generalized Pell numbers GPk,ℓn are represented by GPk,ℓn = 1 2n+1∆k ∫ ∆k −∆k (aLk,ℓ + (2b− ak)nPk,ℓ + a(n + 1)Pk,ℓx) (Lk,ℓ + Pk,ℓx)n−1 dx. Proof. From Theorem 3, we obtain GPk,ℓn = a 2 Lk,ℓn + 2b− ak 2 Pk,ℓn. (14) Applying the integral representations of Pk,ℓn and Lk,ℓn from Theorems 8 and 10 to (14), this completes the proof. Remark 3. As in Theorems 3 and 11, we have the following results. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 15 of 18 (i) If a = 0, then GPk,n = bPk,n and GPk,ℓn = bnPk,ℓ 2n ∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n−1dx. (ii) If ak = 2b, then GPk,n = a 2Lk,n and GPk,ℓn = a 2n+1 ∫ ∆k −∆k (Lk,ℓ + (n + 1)Pk,ℓx)(Lk,ℓ + (k + 1)Pk,ℓx)n−1dx. Setting (a, b) = (2, k) in Theorem 11 and using (i) of Corollary 3, we have the following corollary. Corollary 15 ([11, Theorem 2.6]). Let k, ℓ and n be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then Qk,ℓn = 1 2n∆k ∫ ∆k −∆k (Qk,ℓ + (k − 2 + x− n(k − 2x))Pk,ℓ)(Qk,ℓ + (k − 2 + x)Pk,ℓ) n−1dx. Setting k = 2 in Theorem 11, we have the following corollary. Corollary 16. Let ℓ and n be non-negative integers. Then Ha,b ℓn = 1 2n+2 √ 2 ∫ 2 √ 2 −2 √ 2 (aQℓ + 2(b− a)nPℓ + a(n + 1)Pℓx) (Qℓ + Pℓx)n−1 dx. Setting k = 2 in Theorem 10 or (a, b) = (2, 2) in Corollary 16, we have the following corollary. Corollary 17 ([17, Theorem 3.4]). Let ℓ and n be non-negative integers. Then Qℓn = 1 2n+1 √ 2 ∫ 2 √ 2 −2 √ 2 (Qℓ + (n + 1)Pℓx) (Qℓ + Pℓx)n−1 dx. Finally, both Pk,ℓn and Lk,ℓn are then used to establish integral representations for Pk,ℓn+r and Lk,ℓn+r as the following theorems. Theorem 12. Let k, ℓ, n and r be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then Pk,ℓn+r = 1 2n+1∆k ∫ ∆k −∆k (nPk,ℓLk,r + Pk,rLk,ℓ + (n + 1)Pk,ℓPk,rx) (Lk,ℓ + Pk,ℓx)n−1dx. Proof. Using (i) of Theorem 4 with m and n replaced by ℓn and r respectively, we get Pk,ℓn+r = 1 2 Pk,ℓnLk,r + 1 2 Pk,rLk,ℓn. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 16 of 18 Applying the integral representations of Pk,ℓn and Lk,ℓn from Theorems 8 and 10, we obtain Pk,ℓn+r = 1 2 ( nPk,ℓ 2n∆k ∫ ∆k −∆k (Lk,ℓ + Pk,ℓx)n−1dx ) Lk,r + 1 2 Pk,r ( 1 2n∆k ∫ ∆k −∆k (Lk,ℓ + (n + 1)Pk,ℓx)(Lk,ℓ + Pk,ℓx)n−1dx ) = 1 2n+1∆k ∫ ∆k −∆k (nPk,ℓLk,r + Pk,rLk,ℓ + (n + 1)Pk,ℓPk,rx) (Lk,ℓ + Pk,ℓx)n−1dx. This completes the proof. Setting k = 2 in Theorem 12, we have the following corollary. Corollary 18 ([17], Theorem 3.5). Let ℓ, n and r be non-negative integers. Then Pℓn+r = 1 2n+2 √ 2 ∫ 2 √ 2 −2 √ 2 (nPℓQr + PrQℓ + (n + 1)PℓPrx) (Qℓ + Pℓx)n−1dx. Theorem 13. Let k, ℓ, n and r be non-negative integers with k ≥ 2 and ∆k = √ k2 + 4k − 4. Then Lk,ℓn+r = 1 2n+1∆k ∫ ∆k −∆k ( n∆2 kPk,ℓPk,r + Lk,ℓLk,r + (n + 1)Pk,ℓLk,rx ) (Lk,ℓ + Pk,ℓx)n−1dx. Proof. Using (ii) of Theorem 4 with m and n replaced by ℓn and r respectively, we get Lk,ℓn+r = 1 2 Lk,ℓnLk,r + ∆2 k 2 Pk,ℓnPk,r. This together with Theorems 8 and 10 gives that the proof is finish. Setting k = 2 in Theorem 13, we have the following corollary. Corollary 19 ([17], Theorem 3.6). Let ℓ, n and r be non-negative integers. Then Qℓn+r = 1 2n+2 √ 2 ∫ 2 √ 2 −2 √ 2 (8nPℓPr + QℓQr + (n + 1)PℓQrx) (Qℓ + Pℓx)n−1dx. Remark 4. The integral representations for the companion generalized Pell numbers GPk,ℓn+r are established by applying Theorems 3, 12 and 13. W. Nilsrakoo, A. Nilsrakoo / Eur. J. Pure Appl. Math, 18 (3) (2025), 6452 17 of 18 4. Conclusions This paper presents a comprehensive study on one-parameter generalizations of Pell numbers and their associated sequences, introducing generalized Pell-Lucas-like numbers and their integral representations. The paper further extends known identities, derives Binet-type formulas, and proposes several new integral formulations that encompass and generalize classical results. 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