On the pulsating (m,c)-Fibonacci sequence

Heliyon 7 (2021) e07883 Contents lists available at ScienceDirect Heliyon journal homepage: www.cell.com/heliyon Research article On the pulsating (𝑚, 𝑐)-Fibonacci sequence ✩ Kittipong Laipaporn, Kiattiyot Phibul, Prathomjit Khachorncharoenkul ∗ School of Science, Walailak University, Nakhon Si Thammarat, 80160, Thailand A R T I C L E I N F O Keywords: Fibonacci sequence Pulsating sequence Diagonalization Eigenvalue Matrix decomposition A B S T R A C T In this paper, we study new ideas in the generalization of additive and multiplicative pulsating Fibonacci sequences. Then, we construct two types of pulsating Fibonacci sequences of the 𝑚th order. Moreover, the closed forms of the two sequences are derived by basic linear algebra. 1. Introduction 𝛼0 = 𝑎,𝛽0 = 𝑏, 𝛾0 = 𝑐, K. T. Atanassov et al. [1, 2, 3] proposed the following four 2Fibonacci sequences in “additive and multiplicative” schemes: 𝛼1 = 𝑑,𝛽1 = 𝑒, 𝛾1 = 𝑓 , 𝛽𝑛+2 = 𝑥2𝑛+1 + 𝑦2𝑛 , 𝛼0 = 𝑎, 𝛽0 = 𝑏, 𝛼1 = 𝑐, 𝛽1 = 𝑑, 𝛼𝑛+2 = 𝛽𝑛+1 + 𝛽𝑛 , (1) 𝛽𝑛+2 = 𝛼𝑛+1 + 𝛼𝑛 𝛼𝑛+2 = 𝛼𝑛+1 + 𝛽𝑛 , (2) 𝛽𝑛+2 = 𝛽𝑛+1 + 𝛼𝑛 𝛼𝑛+2 = 𝛽𝑛+1 + 𝛼𝑛 , (3) 𝛽𝑛+2 = 𝛼𝑛+1 + 𝛽𝑛 𝛼𝑛+2 = 𝛼𝑛+1 + 𝛼𝑛 , 𝛼𝑛+2 = 𝑥1𝑛+1 + 𝑦1𝑛 , (4) 𝛾𝑛+2 = 𝑥3𝑛+1 + 𝑦3𝑛 (5) where (𝑥1𝑛+1 , 𝑥2𝑛+1 , 𝑥3𝑛+1 ) and (𝑦1𝑛 , 𝑦2𝑛 , 𝑦3𝑛 ) are any permutation of (𝛼𝑛+1 , 𝛽𝑛+1 , 𝛾𝑛+1 ) and (𝛼𝑛 , 𝛽𝑛 , 𝛾𝑛 ), respectively, for any non-negative integer 𝑛. In 2010, B. Singh and O. Sikhwal [6] studied other fundamental theorems for multiplicative coupled Fibonacci sequences. Moreover, some results of multiplicative triple Fibonacci sequences were presented by M. Singh et al. [7] in 2012, and later by S. Kumer et al. [8] in 2015. In 2013, K. T. Atanassov [9, 10] introduced two types of Fibonacci sequence. The first sequence is defined by 𝛽𝑛+2 = 𝛽𝑛+1 + 𝛽𝑛 for every non-negative integer 𝑛. Many researchers interested in the sequence (2), one of which is J. Z. Lee and J. S. Lee [4]. The multiplicative schemes are analogous to the additive schemes, with only the operators being interchanged. For example, the multiplicative scheme for the sequence (2) is: 𝛼𝑛+2 = 𝛼𝑛+1 ⋅ 𝛽𝑛 , 𝛼0 = 𝑎, 𝛽0 = 𝑏, 𝛼2𝑘+1 = 𝛽2𝑘+1 = 𝛼2𝑘 + 𝛽2𝑘 , 𝛼2𝑘+2 = 𝛼2𝑘+1 + 𝛽2𝑘 , 𝛽2𝑘+2 = 𝛽2𝑘+1 + 𝛼2𝑘 (6) for any 𝑎, 𝑏 ∈ ℝ and 𝑘 ∈ ℕ ∪ {0}, and the second sequence is defined by 𝛽𝑛+2 = 𝛽𝑛+1 ⋅ 𝛼𝑛 , which was introduced in [5]. Moreover, K. T. Atanassov et al. [3] constructed the following 3-Fibonacci sequence: ✩ * 𝛼0 = 𝑎, 𝛽0 = 𝑏, 𝛼1 = 𝛽1 = 𝑐, Fully documented templates are available in the elsarticle package on CTAN. Corresponding author. E-mail address: [email protected] (P. Khachorncharoenkul). https://doi.org/10.1016/j.heliyon.2021.e07883 Received 23 February 2021; Received in revised form 20 July 2021; Accepted 25 August 2021 2405-8440/ 2021 The Author(s). Published by Elsevier Ltd. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). K. Laipaporn, K. Phibul and P. Khachorncharoenkul Heliyon 7 (2021) e07883 Lemma 1. For any real number 𝑐 and integer 𝑚 ≥ 2, the eigenvalues of matrix 𝑈 = (𝑐 + 1)𝐽𝑚 − 𝐼𝑚 are (𝑐 + 1)𝑚 − 1 of multiplicity 1 and −1 of multiplicity 𝑚 − 1 and the corresponding eigenvectors are [1]𝑚×1 and [𝑣𝑗 ]𝑚×1 for 𝑗 = 1, 2, ..., 𝑚 − 1, where 𝛼2𝑘+1 = 𝛽2𝑘+1 = 𝛼2𝑘 + 𝛽2𝑘 , 𝛼2𝑘 = 𝛼2𝑘−1 + 𝛽2𝑘−2 , (7) 𝛽2𝑘 = 𝛽2𝑘−1 + 𝛼2𝑘−2 ⎧ −1 ; 𝑖 = 1 ⎪ 𝑣𝑗𝑖1 = ⎨ 1 ;𝑖 = 𝑗 + 1 ⎪0 ; otherwise . ⎩ for any 𝑎, 𝑏, 𝑐 ∈ ℝ and 𝑘 ∈ ℕ. These sequences are said to be the (𝑎, 𝑏)pulsated Fibonacci sequence and the (𝑎, 𝑏, 𝑐)-pulsated Fibonacci sequence, respectively. Note that the sequence (6) is the sequence (2) with regard to every even subscript. Moreover, by adding the initial conditions to (6), it becomes the sequence (7). In 2014, K. T. Atanassov [11] provided an extension of the sequence (6), namely, the (𝑎1 , 𝑎2 , … , 𝑎𝑚 )-pulsated Fibonacci sequence, which is defined as follows: Proof. Note that 𝑈 [1]𝑚×1 = [(𝑐 + 1)(𝑚 − 1) + 𝑐]𝑚×1 = ((𝑐 + 1)𝑚 − 1)[1]𝑚×1 . Then, (𝑐 + 1)𝑚 − 1 is the eigenvalue of the matrix 𝑈 with the corresponding eigenvector [1]𝑚×1 . Next, we will show that 𝑈 [𝑣𝑗 ] = −[𝑣𝑗 ] for all 𝑗 ∈ {1, 2, ..., 𝑚 − 1}. Let 𝑈 = [𝑢𝑖𝑙 ] and 𝑤𝑗𝑖 be the 𝑖th column of 𝑈 [𝑣𝑗 ]. Then, 𝛼1,0 = 𝑎1 , 𝛼2,0 = 𝑎2 , … , 𝛼𝑚,0 = 𝑎𝑚 , 𝛼1,2𝑘+1 = 𝛼2,2𝑘+1 = ⋯ = 𝛼𝑚,2𝑘+1 = 𝑚 ∑ 𝑤𝑗𝑖 = ⎧ −𝑐 + (𝑐 + 1) ⎪ = ⎨ −(𝑐 + 1) + 𝑐 ⎪ −(𝑐 + 1) + (𝑐 + 1) ⎩ (8) for any non-negative integers 𝑗, 𝑘, 𝑚 such that 𝑎1 , 𝑎2 , … , 𝑎𝑚 ∈ ℝ and 1 ≤ 𝑗 ≤ 𝑚. In 2015, A. Suvarnmani [12] transformed the sequence (6) into a multiplicative version, and in 2017, he [13] introduced a new idea, i.e., a generalization of pulsating Fibonacci sequences called multiplicative pulsating 𝑚-Fibonacci sequences. His inspiration may have been the 𝑚-fold multiplication of the term 𝛼 in (8). The sequence can be defined as follows: Let 𝑎1 , 𝑎2 , … , 𝑎𝑚 be real numbers, with 𝑚 ∏ 𝑖=1 𝛼𝑗,2𝑘+2 = 𝛼𝑗,2𝑘+1 𝑚 ∏ −𝑐 + 𝑢1(𝑗+1) −(𝑐 + 1) + 𝑢1(𝑗+1) ⎧1 ;𝑖 = 1 ⎪ ; 𝑖 = 𝑗 + 1 = ⎨ −1 ⎪0 ; otherwise ⎩ ;𝑖 = 1 ;𝑖 ≠ 1 ;𝑖 = 1 𝑗 ; 𝑖 = 𝑗 + 1 = −𝑣𝑖1 . ; otherwise Hence, we have the desired result. Lemma 2. Let 𝑈 = [𝑢𝑖𝑗 ] be the matrix in Lemma 1. Then, 𝑈 𝑛 = [𝑛 𝑢𝑖𝑗 ], where ⎧ ⎪ ⎪ 𝑛 𝑢𝑖𝑗 = ⎨ ⎪ ⎪ ⎩ 𝛼1,0 = 𝑎1 , 𝛼2,0 = 𝑎2 , … , 𝛼𝑚,0 = 𝑎𝑚 , 𝛼1,2𝑘+1 = 𝛼2,2𝑘+1 = ⋯ = 𝛼𝑚,2𝑘+1 = { 𝑢𝑖𝑙 𝑣𝑗𝑙1 = −𝑢𝑖1 + 𝑢𝑖(𝑗+1) = 𝑙=1 𝛼𝑖,2𝑘 , 𝑖=1 𝛼𝑗,2𝑘+2 = 𝛼𝑗,2𝑘+1 + 𝛼𝑚−𝑗+1,2𝑘 𝑚 ∑ 1 [((𝑐 + 1)𝑚 − 1)𝑛 + (−1)𝑛 (𝑚 − 1)] 𝑚 ;𝑖 = 𝑗 1 [((𝑐 + 1)𝑚 − 1)𝑛 − (−1)𝑛 ] 𝑚 ; otherwise for any positive integer 𝑛. That is, all the main diagonal elements have the same value, and all elements outside the main diagonal have the same value. 𝛼𝑖,2𝑘 , (9) Proof. By Lemma 1, we know that 𝑈 𝑛 = 𝑃 𝐷𝑛 𝑃 −1 , where 𝛼𝑖,2𝑘 [ 𝑃 = 1 𝑣1 𝑖=1 𝑖≠𝑗 for all 𝑗 ∈ {1, 2, … , 𝑚} and the non-negative integer 𝑘. All closed forms in [9, 10, 11, 12, 13] were proved by mathematical induction, which is not the only method for obtaining the closed forms. Furthermore, this method can be used when patterns of the closed forms are predicted. Nevertheless, we confront recurrence relation problems that require various methods to solve. Thus, basic linear algebra is used in our work. In this paper, we apply basic linear algebra, especially the matrix method, eigenvalues and eigenvectors, to find the closed form of two new types of pulsating 𝑚-Fibonacci sequence. We show some results in Section 2 which are essential to acquiring the main result. Section 3 is devoted to the main result. First, we study the sequence (9) in additive schemes in Theorem 4. The sequence in Theorem 4 is a generalization of the (𝑎, 𝑏)-pulsated Fibonacci sequence in (6). Next, we take the logarithm function of sequence (9) when all values of 𝑎𝑖,0 are positive and apply Theorem 4 to provide the closed form in [13] which is shown in Corollary 6. In the last section, we discuss our work and future research. 𝑣2 ⋯ 𝑣𝑚−1 ] and 𝐷 = diag((𝑐 + 1)𝑚 − 1, −1, … , −1). On the other hand, the matrices 𝑃 = [𝑝𝑖𝑗 ], 𝑃 −1 = [𝑝′𝑖𝑗 ] and 𝐷𝑛 = [𝑑𝑖𝑗 ] can be written in the form ⎧ ⎪1 ⎪1 𝑝𝑖𝑗 = ⎨ ⎪ −1 ⎪0 ⎩ ; 1 ≤ 𝑖 ≤ 𝑚, 𝑗 = 1 ; 𝑖 = 𝑗 ∈ {2, 3, ..., 𝑚} , ; 𝑖 = 1, 2 ≤ 𝑗 ≤ 𝑚 ; otherwise 𝑝′𝑖𝑗 ⎧ 1 ⎪ 𝑚 =⎨1− ⎪−1 ⎩ 𝑚 1 𝑚 ; 𝑖 = 1, 1 ≤ 𝑗 ≤ 𝑚 ; 𝑖 = 𝑗 ∈ {2, 3, ..., 𝑚} ; otherwise and ⎧ ((𝑐 + 1)𝑚 − 1)𝑛 ⎪ 𝑑𝑖𝑗 = ⎨ (−1)𝑛 ⎪0 ⎩ ;𝑖 = 𝑗 = 1 ; 𝑖 = 𝑗 ∈ {2, 3, ..., 𝑚} ; otherwise . Thus, for each 𝑖, 𝑗 ∈ {1, 2, … , 𝑚}, the (𝑖, 𝑗)th entry element of 𝑈 𝑛 is 𝑛 𝑢𝑖𝑗 = 𝑚 ∑ 𝑝𝑖𝑘 𝑑𝑘𝑘 𝑝′𝑘𝑗 = 𝑝𝑖1 ((𝑐 + 1)𝑚 − 1)𝑛 𝑝′1𝑗 + = ((𝑐 + 1)𝑚 − 1)𝑛 ( In this section, we establish some lemmas to obtain our main result, i.e., Theorem 4. First, we introduce some important notations. Throughout this paper, let 𝐼𝑚 be an 𝑚-by-𝑚 identity matrix and 𝐽𝑚 be an 𝑚-by-𝑚 matrix in which every entry is one. Let 𝐾𝑚 be an 𝑚-by-𝑚 reversal matrix that is a permutation matrix in which 𝑘𝑖,𝑚−𝑖+1 = 1 for 𝑖 = 1, 2, … , 𝑚 and all other entries are zero. Now, the following lemmas are used to find the closed form of the sequence (10) in Theorem 4. 𝑚 ∑ 1 𝑝𝑖𝑘 𝑝′𝑘𝑗 . ) + (−1)𝑛 𝑚 𝑘=2 Since 𝑃 𝑃 −1 = 𝐼𝑚 and 𝑝𝑖1 𝑝′1𝑗 = (1)( 𝑚1 ) = 𝑚 ∑ 𝑘=2 𝑝𝑖𝑘 𝑝′𝑘𝑗 = 𝑝𝑖𝑘 (−1)𝑛 𝑝′𝑘𝑗 𝑘=2 𝑘=1 2. Preliminaries 𝑚 ∑ 𝑚 ∑ 𝑘=1 { 𝑝𝑖𝑘 𝑝′𝑘𝑗 − 𝑝𝑖1 𝑝′1𝑗 = 1− − 𝑚1 1 , 𝑚 1 𝑚 it follows that ;𝑖 = 𝑗 ; 𝑖 ≠ 𝑗. Hence, we have the desired result. The structure of 𝑈 𝑛 directly affects the following lemma. 2 K. Laipaporn, K. Phibul and P. Khachorncharoenkul Heliyon 7 (2021) e07883 Lemma 3. For any real number 𝑐 and integer 𝑛 ≥ 3, let 𝐴 = 𝑈 𝑛−2 (𝐽𝑚 − 𝐼𝑚 ), 𝐵 = (𝐼𝑚 − 𝐽𝑚 )𝑈 𝑛−2 + 𝑈 𝑛−1 and 𝐶 = 𝐵(𝐽𝑚 − 𝐼𝑚 ) be given. Then, 𝛼1,2𝑘−1 = 𝛼2,2𝑘−1 = ⋯ = 𝛼𝑚,2𝑘−1 = 𝑐((𝑐 + 1)𝑚 − 1)𝑘−1 𝛼𝑗,2𝑘−2 = 𝑚 𝑚 ∑ ∑ (−1)𝑘 1 𝑎𝑖 . [−(𝑚 − 1)𝑎𝑗 + 𝑎𝑖 ] + ((𝑐 + 1)𝑚 − 1)𝑘−1 𝑚 𝑚 𝑖=1 𝑖=1 𝑖≠𝑗 1 [(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑛−2 + (−1)𝑛−1 (𝑚 − 1)] 𝑚 ;𝑖 = 𝑗 Proof. Define a linear transformation 𝑇 ∶ ℝ2𝑚 → ℝ2𝑚 by 1 [(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑛−2 − (−1)𝑛−1 ] 𝑚 ; otherwise, 𝑇 (𝑝1 , 𝑝2 , … , 𝑝𝑚 , 𝑞1 , 𝑞2 , … , 𝑞𝑚 ) = (𝑥1 , 𝑥2 , … , 𝑥𝑚 , 𝑦, … , 𝑦), where 𝑥𝑗 = 𝑞𝑗 + + 1)𝑚 − 1)𝑛−2 𝐽𝑚 , 2. 𝐵 = 𝑐((𝑐 3. 𝐶 = 𝑐(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑛−2 𝐽𝑚 . 𝑚 ∑ 𝑝𝑖 and 𝑦 = 𝑐(𝑚 − 1) 𝑖=1 𝑖≠𝑗 𝑚 ∑ 𝑝𝑖 + 𝑐 𝑚 ∑ 𝑞𝑖 for all 𝑗 = 1, 2, … , 𝑚. 𝑖=1 𝑖=1 Clearly, the matrix representation of 𝑇 with respect to the standard basis is [ ] 𝐽𝑚 − 𝐼𝑚 𝐼𝑚 𝑄= . 𝑐(𝑚 − 1)𝐽𝑚 𝑐𝐽𝑚 Proof. By Lemma 2, we have 1 [((𝑐 + 1)𝑚 − 1)𝑛−2 + (−1)𝑛−2 (𝑚 − 1)]𝐽𝑚 𝑚 𝑚−1 [((𝑐 + 1)𝑚 − 1)𝑛−2 − (−1)𝑛−2 ]𝐽𝑚 + 𝑚 ) ( 1 𝑚−1 𝐽𝑚 + = ((𝑐 + 1)𝑚 − 1)𝑛−2 𝑚 𝑚 𝑛−2 = ((𝑐 + 1)𝑚 − 1) 𝐽𝑚 . 𝑈 𝑛−2 𝐽𝑚 = [ ]𝑡 Let 𝑘 ∈ ℕ and 𝑋𝑘 = 𝛼1,2𝑘−2 ⋯ 𝛼𝑚,2𝑘−2 𝛼1,2𝑘−1 ⋯ 𝛼𝑚,2𝑘−1 . Observe that ]𝑡 [ 𝑚 𝑚 𝑚 ∑ ∑ ∑ 𝑎𝑖 and 𝑄𝑋𝑘 = 𝑋𝑘+1 . 𝑎𝑖 ⋯ 𝑐 𝑎𝑖 𝑐 𝑋1 = 𝑎1 𝑎2 ⋯ 𝑎𝑚 𝑐 𝑖=1 𝑖=1 𝑖=1 1. Since 𝑈 = (𝑐 + 1)𝐽𝑚 − 𝐼𝑚 , we determine that 𝐽𝑚 − 𝐼𝑚 = 𝑈 − 𝑐𝐽𝑚 . Then, we obtain 1×2𝑚 Hence, the main result will be obtained, that is, 𝑋𝑘 = 𝑄𝑘−1 𝑋1 where 𝑘 is greater than 1. To obtain the features of 𝑋𝑘 , we will find the structure of 𝑄𝑘−1 . For the matrix 𝑈 in Lemma 1, we know that 𝐴 = 𝑈 𝑛−2 (𝐽𝑚 − 𝐼𝑚 ) = 𝑈 𝑛−2 (𝑈 − 𝑐𝐽𝑚 ) = 𝑈 𝑛−1 − 𝑐𝑈 𝑛−2 𝐽𝑚 . Thus, by Lemma 2, we can write 𝐴 = [𝑎𝑖𝑗 ]𝑚 , where ⎧ ⎪ ⎪ 𝑎𝑖𝑗 = ⎨ ⎪ ⎪ ⎩ 𝑎𝑖 , 𝑖=1 1. 𝐴 = 𝑈 𝑛−1 − 𝑐𝑈 𝑛−2 𝐽𝑚 . In particular, we have 𝐴 = [𝑎𝑖𝑗 ], where ⎧ ⎪ ⎪ 𝑎𝑖𝑗 = ⎨ ⎪ ⎪ ⎩ 𝑚 ∑ and 𝐼𝑚 − 𝐽𝑚 + 𝑈 = 𝑐𝐽𝑚 (𝐼𝑚 − 𝐽𝑚 + 𝑈 )(𝐽𝑚 − 𝐼𝑚 ) = 𝑐(𝑚 − 1)𝐽𝑚 . 1 [(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑛−2 + (−1)𝑛−1 (𝑚 − 1)] 𝑚 ;𝑖 = 𝑗 In case 𝑘 = 2, it is easy to see that 𝑋2 = 𝑄𝑋1 and we have 1 [(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑛−2 − (−1)𝑛−1 ] 𝑚 ; otherwise . 𝛼𝑗,3 = 𝑐((𝑐 + 1)𝑚 − 1) [ 𝑄= = −𝐴 + 𝑈 𝑛−1 = 𝑐((𝑐 + 1)𝑚 − 1)𝑛−2 𝐽𝑚 . 𝑄𝑘−1 = In this section, we will prove the main result, i.e., Theorem 4. This theorem plays an important role in Corollary 5 and Corollary 6. Let 𝑎1 , 𝑎2 , … , 𝑎𝑚 and 𝑐 be real numbers, with 𝛼𝑗,2𝑘+2 = 𝛼𝑗,2𝑘+1 + 𝑚 ∑ 𝑎𝑖 where 1 ≤ 𝑗 ≤ 𝑚. 𝑖=1 𝑖≠𝑗 ][ 0𝑚 0𝑚 𝐼𝑚 𝑈 ][ 𝐼𝑚 𝐽𝑚 − 𝐼𝑚 ][ 𝐼𝑚 𝐼𝑚 − 𝐽𝑚 ] 0𝑚 . 𝐼𝑚 0𝑚 𝐼𝑚 ] [ is 𝐼𝑚 𝐽𝑚 − 𝐼𝑚 𝐼𝑚 𝐼𝑚 − 𝐽𝑚 0𝑚 𝐼𝑚 0𝑚 0𝑚 𝑈 𝑘−2 𝑈 𝑘−1 ] 0𝑚 , it im𝐼𝑚 ][ where 𝐴, 𝐵 and 𝐶 are defined as in Lemma 3. [ In order to make it easier to prove, we set 𝑋1 = 𝑌1 [ 𝑚 ] ∑ . Then, 𝑎𝑖 𝑌1 = [𝑎1 𝑎2 ⋯ 𝑎𝑚 ]𝑡 and 𝑌2 = 𝑐 𝛼1,0 = 𝑎1 , 𝛼2,0 = 𝑎2 , … , 𝛼𝑚,0 = 𝑎𝑚 , 𝑖=1 𝑚 ∑ ] 𝐼𝑚 0𝑚 𝐽𝑚 − 𝐼𝑚 𝐼𝑚 [ ][ ] 0𝑚 𝐼𝑚 0𝑚 𝑈 𝑘−2 = 𝑘−2 𝑘−1 𝐽𝑚 − 𝐼𝑚 𝐼𝑚 0𝑚 (𝐼𝑚 − 𝐽𝑚 )𝑈 +𝑈 [ 𝑘−2 ] ] [ 𝑘−2 𝐴 𝑈 𝑘−2 𝑈 (𝐽𝑚 − 𝐼𝑚 ) 𝑈 = , = 𝐶 𝐵 𝐵(𝐽𝑚 − 𝐼𝑚 ) 𝐵 [ 𝐽𝑚 . 3. Main results 𝛼1,2𝑘+1 = 𝛼2,2𝑘+1 = ⋯ = 𝛼𝑚,2𝑘+1 = 𝑐 0𝑚 𝐼𝑚 plies that − 𝐽𝑚 ) = 𝑐(𝑚 − 1)((𝑐 + 1)𝑚 − 1) 𝑚 ∑ 𝐼𝑚 𝐼𝑚 − 𝐽𝑚 Since the inverse of a matrix 𝐶 = 𝐵(𝐽𝑚 − 𝐼𝑚 ) = 𝑐((𝑐 + 1)𝑚 − 1)𝑛−2 𝐽𝑚 (𝐽𝑚 − 𝐼𝑚 ) = 𝑐((𝑐 + 1)𝑚 − 1) 𝑎𝑖 + 𝑖=1 [ 3. By the structure of 𝐽𝑚 , we can see that 𝐽𝑚2 = 𝑚𝐽𝑚 . We obtain 𝑛−2 𝑚 ∑ From now on, we consider 𝑘 ≥ 3. In facts, the matrix 𝑄 can be rewritten in the following form: 𝐵 = (𝐼𝑚 − 𝐽𝑚 )𝑈 𝑛−2 + 𝑈 𝑛−1 = 𝑈 𝑛−2 − 𝑈 𝑛−2 𝐽𝑚 + 𝑈 𝑛−1 (𝐽𝑚2 𝑎𝑖 and 𝛼𝑗,2 = 𝑐 𝑖=1 2. Since 𝑈 𝑛−2 is a symmetric matrix, it follows that 𝐽𝑚 𝑈 𝑛−2 = 𝑈 𝑛−2 𝐽𝑚 . By the feature of 𝑈 𝑛−2 𝐽𝑚 , we have 𝑛−2 𝑚 ∑ 𝛼𝑖,2𝑘 , (10) 𝑖=1 [ 𝛼𝑖,2𝑘 𝑋𝑘 = 𝑄𝑘−1 𝑋1 = 𝑖=1 𝑖≠𝑗 for any non-negative integers 𝑗, 𝑘, 𝑚 such that 1 ≤ 𝑗 ≤ 𝑚 and 𝑐 ≠ 0. This sequence is called an additive pulsating (𝑚, 𝑐)-Fibonacci sequence. 𝐴 𝐶 𝑈 ][ 𝑘−2 𝐵 ] 𝑌2 ]𝑡 where 𝑚×1 [ ] 𝑌1 𝐴𝑌1 + 𝑈 𝑘−2 𝑌2 = = [𝑠𝑗 ]2𝑚×1 , 𝑌2 𝐶𝑌1 + 𝐵𝑌2 where 𝑘 is greater than 1. After that, we will calculate the element 𝑠𝑗 by Lemma 2 and Lemma 3. By the features of 𝐴, we see that all its main diagonal entries are equal and the entries outside the main diagonal are equal. To write clear and concise content, we utilize 𝑎11 for the main diagonal entry and 𝑎12 for the other entry. Thus, we have Theorem 4. For all positive integers 𝑗, 𝑘, 𝑚 such that 𝑗 ≤ 𝑚, 𝑘 ≥ 2 and 𝑐 ∈ ℝ∖{0}, the closed form of the additive pulsating (𝑚, 𝑐)-Fibonacci sequence is 3 K. Laipaporn, K. Phibul and P. Khachorncharoenkul Heliyon 7 (2021) e07883 ⎡ 𝑎1 ⎤ ⎢ ⎥ 𝑎 𝐴𝑌1 = 𝐴 ⎢ 2 ⎥ ⎢ ⋮ ⎥ ⎢𝑎 ⎥ ⎣ 𝑚⎦ [ ]𝑡 ∑𝑚 ∑𝑚 ∑𝑚 = 𝑎11 𝑎1 + 𝑎12 𝑖=1 𝑎𝑖 𝑎11 𝑎2 + 𝑎12 𝑖=1 𝑎𝑖 ⋯ 𝑎11 𝑎𝑚 + 𝑎12 𝑖=1 𝑎𝑖 . 𝑖≠1 𝑖≠2 To illustrate Theorem 4, we provide the following example. Example 1. Let 𝑎, 𝑏, and 𝑐 be real numbers with 𝛼0 = 𝑎, 𝛽0 = 𝑏, 𝛾0 = 𝑐, 𝑖≠𝑚 𝛼2𝑛+1 = 𝛽2𝑛+1 = 𝛾2𝑛+1 = 𝛼2𝑛 + 𝛽2𝑛 + 𝛾2𝑛 , Similarly, under the features of 𝑈 𝑘−2 in Lemma 2, we utilize 𝑘−2 𝑢11 for the main diagonal entry and 𝑘−2 𝑢12 for the other entry. Then, 𝑈 𝑘−2 𝑌2 = (𝑐 [ 𝑚 ∑ 𝑎𝑖 )𝑈 𝑘−2 [1]𝑚×1 = (𝑐 𝑚 ∑ [ 𝑘−2 𝑎𝑖 ) ( 𝑖=1 𝑖=1 = (𝑘−2 𝑢11 + (𝑚 − 1)(𝑘−2 𝑢12 ))(𝑐 [ 𝑚 ∑ = 𝑐((𝑐 + 1)𝑚 − 1)𝑘−2 ( 𝑚 ∑ 𝑎𝑖 ) 𝑖=1 𝑢11 ) + (𝑚 − 1)( 𝛽2𝑛+2 = 𝛼2𝑛 + 𝛽2𝑛+1 + 𝛾2𝑛 , ] 𝑢12 ) 𝑚×1 𝛾2𝑛+2 = 𝛼2𝑛 + 𝛽2𝑛 + 𝛾2𝑛+1 ] for every non-negative integer 𝑛. We will find the closed form of this sequence using the process in Theorem 4. The above sequence leads us to define a linear transformation 𝑇 on ℝ6 by 𝑎𝑖 ) 𝑖=1 ] 𝑘−2 𝑚×1 . 𝑚×1 𝑇 (𝑝1 , 𝑝2 , 𝑝3 , 𝑞1 , 𝑞2 , 𝑞3 ) = (𝑞1 + 𝑝2 + 𝑝3 , 𝑞2 + 𝑝1 + 𝑝3 , 𝑞3 + 𝑝1 + 𝑝2 , 𝑦, 𝑦, 𝑦), Thus, we let 𝑗 ∈ {1, 2, … , 𝑚} and 𝑤𝑗 be the 𝑗th row of the matrix 𝐴𝑌1 + 𝑈 𝑘−2 𝑌2 . Then, by Lemma 3 (1), we have 𝑤𝑗 = 𝑎11 𝑎𝑗 + 𝑎12 𝑚 ∑ = where 𝑦 = 2(𝑝1 + 𝑝2 + 𝑝3 ) + (𝑞1 + 𝑞2 + 𝑞3 ). Then, the matrix 𝑄 in Theorem 4 is 𝑚 ∑ 𝑎𝑖 + 𝑐((𝑐 + 1)𝑚 − 1)𝑘−2 ( 𝑎𝑖 ) 𝑖=1 𝑖≠𝑗 (11) 𝛼2𝑛+2 = 𝛼2𝑛+1 + 𝛽2𝑛 + 𝛾2𝑛 , 0 ⎡ 1 ⎢ 0 1 ⎢ 0 ⎢ 0 ⎢ ⎢ 0 −1 ⎢ −1 0 ⎢ ⎣ −1 −1 𝑖=1 𝑚−1 [((𝑐 + 1)𝑚 − 1)𝑘−2 + (−1)𝑘−1 ]𝑎𝑗 𝑚 𝑚 ∑ 1 + [(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑘−2 + (−1)𝑘−2 ] 𝑎𝑖 𝑚 𝑖=1 0 0 1 0 0 0 0 0 0 −1 −1 0 1 0 0 0 1 0 0⎤ 0⎥ ⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎦ ⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎣0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0⎤ ⎡1 0 1 0⎥ ⎢0 ⎥ ⎢ 0 0 1⎥ ⎢0 ⎥ ⎢ 1 2 2⎥ ⎢0 2 1 2⎥ ⎢1 ⎥ ⎢ 2 2 1⎦ ⎣1 0 0 1 0 0 1 1 1 0 1 1 0 0 0 0⎤ 0 0 0⎥ ⎥ 0 0 0⎥ ⎥, 1 0 0⎥ 0 1 0⎥ ⎥ 0 0 1⎦ 𝑖≠𝑗 ⎡1 and we can see that 𝑈 = ⎢ 2 ⎢ ⎣2 that 𝑚 ∑ + 𝑐((𝑐 + 1)𝑚 − 1)𝑘−2 ( 𝑎𝑖 ) 𝑖=1 𝑚−1 = [((𝑐 + 1)𝑚 − 1)𝑘−2 + (−1)𝑘−1 ]𝑎𝑗 𝑚 𝑚−1 ((𝑐 + 1)𝑚 − 1)𝑘−2 (−𝑎𝑗 ) + 𝑚 𝑚 (−1)𝑘−2 ∑ 𝑎𝑖 + [𝑐((𝑐 + 1)𝑚 − 1)𝑘−2 + 𝑚 𝑖=1 ⎡1 𝑈 = ⎢1 ⎢ ⎣1 𝑛 ⎡ 1 (5𝑛 + 2(−1)𝑛 ) ⎢3 ⎢ = ⎢ 13 (5𝑛 − (−1)𝑛 ) ⎢ ⎢ ⎢ 1 (5𝑛 − (−1)𝑛 ) ⎣ 3 𝑖≠𝑗 + = 𝑚 ∑ 𝑚−1 ((𝑐 + 1)𝑚 − 1)𝑘−2 ]( 𝑎𝑖 ) 𝑚 𝑖=1 𝑚 𝑚 ∑ ∑ 1 [(𝑚 − 1)(−1)𝑘−1 𝑎𝑗 + (−1)𝑘−2 𝑎𝑖 ] 𝑎𝑖 + ((𝑐 + 1)𝑚 − 1)𝑘−1 𝑚 𝑖=1 𝑖=1 = 𝑚 [−(𝑚 − 1)𝑎𝑗 + 𝑚 ∑ 𝑎𝑖 ] + 𝑖=1 𝑖≠𝑗 𝑚 ∑ ⎡ 𝑎1 𝐴 = ⎢ 𝑎2 ⎢ ⎣ 𝑎2 Last but not least, since 𝑐𝐽𝑚 𝑌1 = 𝑌2 , 𝐽𝑚 𝑌2 = 𝑚𝑌2 and Lemma 3, we obtain 𝐶𝑌1 + 𝐵𝑌2 = 𝑐(𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑘−2 𝐽𝑚 𝑌1 + 𝑐((𝑐 + 1)𝑚 − 1)𝑘−2 𝐽𝑚 𝑌2 [ 𝑋1 = 𝑎 𝑚×1 ⎧ (−1)𝑘 1 𝑎𝑖 ] + ((𝑐 + 1)𝑚 − 1)𝑘−1 ⎪ 𝑚 [−(𝑚 − 1)𝑎𝑗 + 𝑚 𝑖=1 ⎪ 𝑖≠𝑗 𝑠𝑗 = ⎨ 𝑚 ∑ ⎪ 𝑎𝑖 ⎪ 𝑐((𝑐 + 1)𝑚 − 1)𝑘−1 ⎩ 𝑖=1 1 𝑛 (5 3 + 2(−1)𝑛 ) − (−1)𝑛 ) 1 3 − 13 2 3 ⎤ ⎥ ⎥ ⎥ ⎦ − (−1)𝑛 ) ⎤ ⎥ ⎥ 1 𝑛 𝑛 ⎥, (5 − (−1) ) 3 ⎥ ⎥ 1 𝑛 (5 + 2(−1)𝑛 ) ⎥⎦ 3 1 𝑛 (5 3 𝑎2 ⎤ 𝑎2 ⎥ , ⎥ 𝑎1 ⎦ 𝑎2 𝑎1 𝑎2 ⎡ 𝑏1 𝐵 = ⎢ 𝑏2 ⎢ ⎣ 𝑏2 𝑏2 𝑏1 𝑏2 𝑏2 ⎤ 𝑏2 ⎥ , ⎥ 𝑏1 ⎦ 𝐶 = 5𝑛−2 𝐽3 and 𝐷 = 2𝐶. 𝑏 𝑎+𝑏+𝑐 𝑐 𝑎+𝑏+𝑐 ]𝑡 𝑎+𝑏+𝑐 , which implies that Finally, we conclude that for any integer 𝑘 ≥ 2, the matrix 𝑋𝑘 is [𝑠𝑗 ]2𝑚×1 where 𝑚 ∑ 1 𝑛 (5 3 3 − (−1)𝑛 ) 1 3 2 3 − 13 From the initial conditions in (11), = (𝑚 − 1)((𝑐 + 1)𝑚 − 1)𝑘−2 𝑌2 + 𝑚𝑐((𝑐 + 1)𝑚 − 1)𝑘−2 𝑌2 [𝑚 ] ∑ . 𝑎𝑖 = 𝑐((𝑐 + 1)𝑚 − 1)𝑘−1 𝑚 ∑ 1 𝑛 (5 3 1 0 ⎤⎡ 3 ⎢ 1 ⎥ − 0 ⎥⎢ 3 (−1)𝑛 ⎦ ⎢⎣ − 1 (−1)𝑛−2 ), 𝑏1 = 23 (5𝑛−2 + (−1)𝑛−1 ) and 𝑏2 = 13 (2 ⋅ 5𝑛−2 + (−1)𝑛−2 ), then we have 1 𝑎𝑖 . ((𝑐 + 1)𝑚 − 1)𝑘−1 𝑚 𝑖=1 𝑖=1 0 (−1)𝑛 0 2⎤ 2 ⎥. In addition, by Lemma 2, it follows ⎥ 1⎦ and by Lemma 3, if we let 𝑎1 = 13 (5𝑛−2 + 2 ⋅ (−1)𝑛−2 ), 𝑎2 = 13 (5𝑛−2 − 𝑖≠𝑗 (−1)𝑘 −1 ⎤ ⎡ 5𝑛 0 ⎥⎢ 0 ⎥⎢ 1 ⎦⎣ 0 −1 1 0 2 1 2 [ 𝑎𝑖 ; 1 ≤ 𝑗 ≤ 𝑚 𝑋𝑛 = 𝑄𝑛−1 𝑋1 = 𝑖=1 ; 𝑚 + 1 ≤ 𝑗 ≤ 2𝑚. This completes the proof. 𝐵 𝐷 ⎡ 1 (−1)𝑛 (−2𝑎 + 𝑏 + 𝑐) + 1 5𝑛−1 (𝑎 + 𝑏 + 𝑐) ⎤ 3 ⎢ 31 ⎥ 1 𝑛−1 𝑛 ⎢ 3 (−1) (𝑎 − 2𝑏 + 𝑐) + 3 5 (𝑎 + 𝑏 + 𝑐) ⎥ ] 1 𝑛−1 𝑛 ⎢ 1 ⎥ 𝐴 𝑋1 = ⎢ 3 (−1) (𝑎 + 𝑏 − 2𝑐) + 3 5 (𝑎 + 𝑏 + 𝑐) ⎥ . 𝑛−1 (𝑎 + 𝑏 + 𝑐) 𝐶 5 ⎢ ⎥ ⎢ ⎥ 5𝑛−1 (𝑎 + 𝑏 + 𝑐) ⎢ ⎥ 5𝑛−1 (𝑎 + 𝑏 + 𝑐) ⎣ ⎦ Hence, the closed form of this sequence is 4 K. Laipaporn, K. Phibul and P. Khachorncharoenkul Heliyon 7 (2021) e07883 𝛼 2𝑛−1 = 𝛽2𝑛−1 = 𝛾2𝑛−1 = 5𝑛−1 (𝑎 + 𝑏 + 𝑐), 1 1 𝛼2𝑛−2 = (−1)𝑛 (−2𝑎 + 𝑏 + 𝑐) + 5𝑛−1 (𝑎 + 𝑏 + 𝑐), 3 3 1 1 𝛽2𝑛−2 = (−1)𝑛 (𝑎 − 2𝑏 + 𝑐) + 5𝑛−1 (𝑎 + 𝑏 + 𝑐), 3 3 1 1 𝛾2𝑛−2 = (−1)𝑛 (𝑎 + 𝑏 − 2𝑐) + 5𝑛−1 (𝑎 + 𝑏 + 𝑐) 3 3 where 𝑛 is any positive integer. And using the properties of the logarithmic function as the desired result. 4. Conclusion and discussion In the last ten years, many researchers have investigated abundant types of pulsating Fibonacci sequences, which are generalizations of the Fibonacci sequence. They use mathematical induction to verify the closed form of these sequences. This method is beautiful; however, it can only be used when patterns of the closed forms are predicted. In this paper, we use the matrix method to bridge the gap to find the closed forms. Moreover, we study new ideas in the generalization of additive and multiplicative pulsating Fibonacci sequences. Furthermore, we construct two types of pulsating Fibonacci sequences of the 𝑚th order. These types generalize other past efforts in pulsating Fibonacci sequences. One future research direction in this area is the extension and characterization of additive and multiplicative pulsating Fibonacci sequences by using permutation of some subscripts. Next, by replacing 𝑐 with 1 in the sequence (10), we obtain the following sequence: Let 𝑎1 , 𝑎2 , … , 𝑎𝑚 be any real numbers, with 𝛼1,0 = 𝑎1 , 𝛼2,0 = 𝑎2 , … , 𝛼𝑚,0 = 𝑎𝑚 , 𝛼1,2𝑘+1 = 𝛼2,2𝑘+1 = ⋯ = 𝛼𝑚,2𝑘+1 = 𝑚 ∑ 𝛼𝑖,2𝑘 , (12) 𝑖=1 𝛼𝑗,2𝑘+2 = 𝛼𝑗,2𝑘+1 + 𝑚 ∑ 𝛼𝑖,2𝑘 𝑖=1 𝑖≠𝑗 for any non-negative integers 𝑗, 𝑘, 𝑚 such that 1 ≤ 𝑗 ≤ 𝑚. Thus, the corresponding closed form of this sequence is provided by the following corollary. Declarations Author contribution statement Corollary 5. For all positive integers 𝑗, 𝑘, 𝑚 and 𝑗 ≤ 𝑚, the closed form of the pulsating 𝑚-Fibonacci sequence (12) is 𝛼1,2𝑘−1 = 𝛼2,2𝑘−1 = ⋯ = 𝛼𝑚,2𝑘−1 = (2𝑚 − 1)𝑘−1 𝑚 ∑ K. Laipaporn, K. Phibul, P. Khachorncharoenkul: Analyzed and interpreted the data; Contributed reagents, materials, analysis tools or data; Wrote the paper. 𝑎𝑖 , 𝑖=1 𝑚 𝑚 ∑ ∑ (−1)𝑘 1 𝛼𝑗,2𝑘−2 = 𝑎𝑖 . 𝑎𝑖 ] + (2𝑚 − 1)𝑘−1 [−(𝑚 − 1)𝑎𝑗 + 𝑚 𝑚 𝑖=1 𝑖=1 Funding statement 𝑖≠𝑗 This research was supported by Walailak University under Grant number WU63247 and partially supported by the new strategic research (P2P) project, Walailak University, Thailand. Proof. This follows directly by substituting 𝑐 = 1 in the result of Theorem 4. Data availability statement Moreover, by taking the logarithm of (9), we obtain 𝛽1,0 = ln 𝑎1 , 𝛽2,0 = ln 𝑎2 , … , 𝛽𝑚,0 = ln 𝑎𝑚 , 𝛽1,2𝑘+1 = 𝛽2,2𝑘+1 = ⋯ = 𝛽𝑚,2𝑘+1 = ln 𝛼𝑚,2𝑘+1 = ln 𝑚 ∏ 𝛼𝑖,2𝑘 = 𝛽𝑗,2𝑘+2 = ln 𝛼𝑗,2𝑘+2 = ln(𝛼𝑗,2𝑘+1 𝛼𝑖,2𝑘 ) = 𝛽𝑗,2𝑘+1 + 𝑖=1 𝑖≠𝑗 No data was used for the research described in the article. 𝛽𝑖,2𝑘 , 𝑖=1 𝑖=1 𝑚 ∏ 𝑚 ∑ 𝑚 ∑ Declaration of interests statement (13) The authors declare no conflict of interest. 𝛽𝑖,2𝑘 𝑖=1 𝑖≠𝑗 Additional information for positive real numbers 𝑎1 , 𝑎2 , … , 𝑎𝑚 and non-negative integers 𝑗, 𝑘, 𝑚 such that 1 ≤ 𝑗 ≤ 𝑚. If we consider the variable 𝛽 in terms of 𝛼 in (12), then the closed form of (9) is the following corollary. No additional information is available for this paper. Acknowledgements Corollary 6. For all positive integers 𝑗, 𝑘, 𝑚 and 1 ≤ 𝑗 ≤ 𝑚, the closed form of the multiplicative pulsating 𝑚-Fibonacci sequence (9) is The authors would like to thank the referees for useful comments that improve the final version of the manuscript. 𝑚 ∏ 𝑘−1 𝛼1,2𝑘−1 = 𝛼2,2𝑘−1 = ⋯ = 𝛼𝑚,2𝑘−1 = ( 𝑎𝑖 )(2𝑚−1) , 𝑖=1 𝛼𝑗,2𝑘−2 = (𝑎𝑗 ) (2𝑚−1)𝑘−1 −(−1)𝑘 (𝑚−1) 𝑚 ( 𝑚 ∏ 𝑎𝑖 ) References (2𝑚−1)𝑘−1 +(−1)𝑘 𝑚 , [1] K. Atanassov, L. Atanassova, D. Sasselov, A new perspective to the generalization of the fibonacci sequence, Fibonacci Q. 23 (1985) 21–28. [2] K. Atanassov, On a second new generalization of the fibonacci sequence, Fibonacci Q. 24 (1986) 362–365. [3] K. Atanassov, et al., New Visual Perpectives on Fibonacci Numbers, 2001. [4] J.Z. Lee, J.S. Lee, Some properties of the generalization of the fibonacci sequence, Fibonacci Q. 25 (1987) 111–117. [5] K. Atanassov, Remark on a new direction for a generalization of the fibonacci sequence, Fibonacci Q. 33 (1995) 249–250. [6] B. Singh, O. Sikhwal, Multiplicative coupled fibonacci sequences and some fundamental properties, Int. J. Contemp. Math. Sci. 5 (2010) 223–230. [7] M. Singh, S. Bhatnanar, O. Sikhwa, Multiplicative triple fibonacci sequences, Appl. Math. Sci. 6 (2012) 2567–2572. [8] S. Kumer, H. Kishan, D. Gupta, A note on multiplicative triple fibonacci sequences, Bull. Soc. Math. Serv. Stand. 13 (2015) 1–6. 𝑖=1 𝑖≠𝑗 which is the same as Suvarnmani’s result. Proof. Let each 𝛽𝑖,𝑗 in the sequence (13) be the 𝛼𝑖,𝑗 in the sequence (12). Applying Corollary 5, we get 𝛽1,2𝑘−1 = 𝛽2,2𝑘−1 = ⋯ = 𝛽𝑚,2𝑘−1 = (2𝑚 − 1)𝑘−1 𝑚 ∑ ln 𝑎𝑖 , 𝑖=1 𝛽𝑗,2𝑘−2 = 𝑚 𝑚 ∑ ∑ (−1)𝑘 1 [−(𝑚 − 1) ln 𝑎𝑗 + ln 𝑎𝑖 ] + (2𝑚 − 1)𝑘−1 ln 𝑎𝑖 . 𝑚 𝑚 𝑖=1 𝑖=1 𝑖≠𝑗 5 K. Laipaporn, K. Phibul and P. Khachorncharoenkul Heliyon 7 (2021) e07883 [9] K. Atanassov, Pulsating fibonacci sequence, Notes Number Theory Discrete Math. 19 (2013) 12–14. [10] K. Atanassov, Pulsating fibonacci sequence part 2, Notes Number Theory Discrete Math. 19 (2013) 33–36. [11] K. Atanassov, 𝑛-pulsating fibonacci sequence, Notes Number Theory Discrete Math. 20 (2014) 32–35. [12] A. Suvarnamani, S. Koyram, Multiplicative fibonacci sequence part 2, Univers. J. Appl. Math. 3 (2015) 89–93. [13] A. Suvarnamani, On the multiplicative pulsating 𝑛-fibonacci sequence part 2, SNRUJST 9 (2017) 502–508. 6