Notes on Number Theory and Discrete Mathematics Print ISSN 1310–5132, Online ISSN 2367–8275 Vol. 25, 2019, No. 3, 185–197 DOI: 10.7546/nntdm.2019.25.3.185-197 Sequences of Tridovan and their identities Renata Passos Machado Vieira 1 and Francisco Regis Vieira Alves 2 1 Department of Mathematics Institute Federal of Technology of the State of Ceará (IFCE) Fortaleza-CE, Brazil e-mail: [email protected] 2 Department of Mathematics Institute Federal of Technology of the State of Ceará (IFCE) Fortaleza-CE, Brazil e-mail: [email protected] Received: 19 December 2018 Revised: 20 June 2019 Accepted: 29 July 2019 Abstract: This work introduces the so-called Tridovan sequence which is an extended form of the Padovan sequence. In a general definition, this extension adds one more term to the Padovan recurrence relation, considering now the three terms preceding the penultimate one. Studies carried out on the proposed extension reveal properties of the positive and negative integer index, the sum of all, even and odd terms, the obtaining Tridovan Q-matrix and finally the Tridovan initial terms generalization. Keywords: Sequence of Tridovan, Positive indices, Negative indices, Generating matrix. 2010 Mathematics Subject Classification: 11B37, 11B39. 1 Introduction The Italian architect Richard Padovan (born 1935) discovered a kind of “cousin” of the well-known Fibonacci sequence, the Padovan numbers, which is also recursive, arithmetic and linear. Padovan was born in the city of Padua [12], a village a little far from the same birthplace of Fibonacci [1]. The Padovan sequence is defined by the recursive relation Pn = Pn2 + Pn3, n t 3, where Pn is the n-th term of the sequence. The first sequence terms are defined by P0 = P1 = P2 = 1, which may generate the following initial elements: 1, 1, 1, 2, 2, 3, 4, 5, 7, ... 185 An important feature of this sequence is the neighboring terms relationship Pn+1 / Pn, which converges to a value known as plastic number or plastic constant [8, 10]. The approximate value of the plastic number is ψ = 1.324718... and the ways of calculating this value are presented in the [6, 8]. Based on the Fibonacci sequence, [4, 9] proposed a sequence extension considering not only the sum of two previous elements but three or four previous elements. Following the same path it is possible to extend the Padovan numbers considering a greater number of terms to estimate the next one. The geometric representation of the Padovan sequence occurs through the Padovan spiral, as shown in Figure 1. This is composed by the juxtaposition of equilateral triangles respecting a characteristic construction rule. Consider the side 1 highlighted triangle in blue as the initial triangle. The spiral formation of the spiral is given by the addition of a new equilateral triangle to the largest side of the formed polygon, initially the blue triangle. After the addition of the other triangles, the new polygon is formed, known as plastic pentagon. The spiral presents itself by connecting with an arc the two corners of the newly added triangle. Figure 1. Padovan spiral. (Source: Prepared by the authors.) Although this sequence is recursive, its terms can be found through the Q-matrix. According to studies carried out by Alves [2] for the Fibonacci sequence, the same idea will be followed so that an application is made for the Tridovan numbers. Thus, it is possible to study these new identities in addition to providing the perception of a constantly evolving notion of this sequence. 186 2 Preliminary As defined in the previous section, any element of the Padovan sequence is calculated based on the sum of the last two terms ignoring the one immediately before. Following the same idea, one can define the Tridovan sequence as the sum of the tree last terms ignoring the one immediately before. The addition of this new element makes the Tridovan Sequence a 4th degree recurrent linear sequence, and its properties and definitions will be presented in the following sessions. Definition 1. The Tridovan sequence numbers recursive relation for positive indices is given by: TTnn 234 T n Tn n,4 t, where Tn is the n-th term of the sequence. The initial values of these numbers are given by: T0 = 0, T1 = 1, T2 = 0, T3 = 1. Thus, the list of terms in this sequence are (0, 1, 0, 1, 1, 2, 2, 4, 5, 8, 11, 17, ...). Definition 2. The Tridovan sequence recurrence relation for negative indices is given by: T–n = T–n + 4 – T–n + 2 – T–n + 1 or T–n = T–n + 4 – T–n + 2 + T–n + 1, with T0 = 0, T1 = 1, T2 = 0, T3 = 1. From Definition 1, we can naturally describe the Tridovan sequence. We can also observe a process of extension of integer indices, since, in the previous definition, its values are indicated only for n t 4 . So let us see that for nTTTT 3?310 1 and, thus, we will have the T1 0 . Similarly, we see that for nTTTT 2?20 1 2 and, we will have the value T2 0 . For a smaller index, we have nTTTT 1?1123 , getting T3 1. On a recurrent basis, we can further determine that TTTTTTTT4 1, 5 0, 6 1, 7 0, 8 2, 9 2, 10 1, 11 3 , etc. With this, from the previous values, we can systematize the description of the following sets with integers, as follows: (...,TTTTTnnnnn2 , 1 , , 1 , 2 ,... TTTTTTTTTTTTT 432101234 , , , , , , , , ,..., nnnn 3 , 2 , 1 , ,...) . When we want to determine any element for n t 4, we use the relationship Tn = Tn – 2 + Tn – 3 + Tn – 4, however, the elements T0 = 0, T1 = 1, T2 = 0, T3 = 1 are fixed preliminarily. Now, noticing that TTTT1310 or yet TTTT220 1 and also TTTT31 1 2, constructing: (..., 3, 1, 2, 2, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, ...) we can write the relation of recurrence to negative indexes as: TTnnnn 421 T T or TTnn 421 T n T n. This arrangement shows a different understanding of the one shown at the beginning, since before the development of the terms were found by performing a “movement” from left to right, these terms can now be found with right-to-left “movement”) of the sequence. Definition 3. From the Tridovan sequence, we use the following notations with initial values T0 = 0, T1 = 1, T2 = 0, T3 = 1, and S0 = 0, S1 = 1, S2 = 1, S3 = 2, and: n (a) Sum of positive terms: STTTnnni 012... T 1 T ¦ T, i 0 n (b) Sum of positive even-numbered terms: STTT2nnni 024... T 222 T ¦ T 2, i 0 187 n (c) Sum of positive odd-numbered terms: STTTTT21nnni 1 3 5... 21 21 ¦ T 21 , i 0 n (d) Sum of negative terms: STTTnnni 012 ... T 1 T ¦ T, i 0 n (e) Sum of negative even-numbered terms: STTT2nnni 024 ... T 222 T ¦ T 2, i 0 n (f) Sum of negative odd-numbered terms: STTTT21nni 1 3 5... 21 ¦ T 21. i 0 Theorem 1. For any integer n t 2 the following identities are verified: (i) SSnn13 T n1 ; (ii) SS22123nn T n1, (iii) SS21nn 22 T 22 n . Proof. Consider the following set of expressions: ­­T4 TTT 210 TT 42 TT 10 °°TTTT TTTT °°5 321 53 21 °°T6 TTT 432 TT 64 TT 32 °° ®®? °°TTTT TTTT °°nnnn1123 nnnn 1123 °°Tn212212 TTT nnn TTTT n nnn °° ¯¯Tn31 T n TT nn 1 T n 31 T n TT nn 1 We can effectively cancel out the similar terms in the above equations to determine the following sum nn1 TTTTnii323 ¦ ¦ T, ii 00 remembering that TT23 0, 1 and, therefore, we see that nn1 SSnn13 ¦¦ T i TT in 1, ii 00 or even more simplified as SSnn13 T n1, hence we prove (i). In a similar way, this time for the even indices, we have the following set of equations: ­TTTT5 321­ TTTT 53 21 °TTTT ° TTTT ° 7543° 7543 °TTTT9765 ° TTTT 9765 ° ° ®?® °TTTT ° TTTT ° 21nnnn 23 24 25° 21 nnnn 23 24 25 °TTTT21nnnn 21 22 23° TTTT 21 nnnn 21 22 23 ° ° ¯TTTT23nnnn 21 2 21 ¯ TT 23 n 21nnn TT 2 21 By means of cancelling out, we can obtain: 188 nn1 TTT23nn 3 231 ¦ T 2 ii¦ T 21 , ii 10 assuming T3 1, we find that TSS23nnn1 2 21, hence we prove (ii). We can also consider the following arrangement of the elements of the first list, eliminating the odd indexes as follows: ­­T4210 TTT TT 4210 TT °°TTTT TTTT °°6 432 64 32 °°TTTT8 654 TTTT 86 54 °° ®®TTTT10 8 7 6 ? TTTT 10 8 7 6 °° °° °°TTTT2nnnn 22 23 24 TTTT 2 nnnn 22 23 24 °° ¯¯T22n TTT 2 nnn 21 22 TTTT 22 n 2 nnn 21 22 Thus, considering the corresponding eliminations, we must find that nn11 TT22nnn 2 ¦ T 21¦ T 2, ii 00 therefore, we can write nn11 TTTSS22nnnnn ¦¦ 21 2 21 22, ii 00 which proves item (iii). Finally, let us note that nSSS 4? 1210 1110 3 S 4. Using another value for n, one can get nSSS 51?321 12115 S 5. Thus, through the inductive method, we can see that: SSSSnnnn 1234. For n = n + 1, we have SSTSSSTSSSTTTnnn1 1 (1 nnn 234 ) n 1 (1 nnn 234 ) ( nnn 123 ) SSTSTSTSSSnnnnnnnnnn1213243123 1( )( )( )1 Theorem 2. For any integer n t 2 the following identities are verified: (i) TSS22nnn 23 24 ; (ii) ()TTnn23 S n 4 S n 5. Proof. Let us consider the equations: ­TT0234 T T °TTTT ° 2456 °TTTT4678 ® °TTTT68910 ° ° ¯°TT2222324nn T n T n 189 Performing some algebraic manipulations, we have: ­TT02 T 34 T °TTTT ° 24 56 °TTTT46 78 ® .