Sequences of Tridovan and Their Identities

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 23 1 ¦ 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 ® . °TTTT68 910 ° ° ¯°TT2222324nn T n T n Adding and eliminating the corresponding terms, we find

nn11 nn 11 TT022niiniinnn ¦¦ T 212222 T T ¦¦ T 2122222324 T T S S , ii 11 ii 11 which proves item (i). Then, we will use the following set of relations

TT0234 T T °TTTT ° 1345

°TTTT2456 ° ®TTTT3567 ° ° °TTnn 234 T n T n ° ¯TTTTnnnn1345 Through simple algebraic manipulations, we obtain:

TT02 T 34 T °TT TT ° 13 45

°TTTT24 56 ° ®TT35 TT 67 ° ° °TTnn234 T n T n ° ¯TTnn13 TT n 45 n With this, after cancelling out, we can see that:

nn45 TTT01nn 2 T 3 ¦ T i ¦ T i. ii 34

Knowing that the values of TT01 0 , we have: ()TTnn23 S n 4 S n 5, which proves item (ii).

Definition 4. Given the arbitrary numbers G1, G2, G3, the Generalized Tridovan Sequence’s recurrence relation is defined as following:

Gn = Gn – 1 + Gn – 2 + Gn – 3.

190 Theorem 3. For any integer n t 3 will have the following general term

GTnn ()...3411223 TGTGTG n n n .

It is worth noting that Tn represents the n-th term of the Tridovan Sequence (see Definition 1) and Gn represents the n-th term of the Generalized Tridovan Sequence (see Definition 4). Proof. Consider

GGGGGG4 2 1 0 1 2 1. G 2 1 GTGTG 2 32 . 23 . . For

GGGG5123 111 G 1 G 2 G 3 ( TTGTGTG 1214233 ).. .

Admitting the inductive step GTnn ()...4311223 TGTGTG n n n , we will add the following terms GGGGn212341122312 nnn()()...() TTGTGTGGG nn n n nn . In the next step, we have: GTTGTGTG ()..., nnn14512233 n n GTTGTGTGnnn2 ().... 5 6 1 n 32 n 43 And then replacing the terms, using the hypothesis:

GGGGnnnn212 ()

()...().TTGTGTGTPGnn3411223451 n n n n

TSTGnn22..().. 3 3 T nnn 5 T 6 GTGTG 1 3 2 n 4 . 3

().().().TTTTTTGTTTGTTTGnnnnnn344556112322343 nnn nnn

().TTnn12GTGTG1123().().nn This completes the proof.

Now consider the sequence behavior for negative indices. To do so, we can describe properties involving the following set:

(,………Gnnnn11233210112 , G , G , G , G n ,, G , G , G , GGGG ,,, ,). Therefore, from Definition 3, we will take:

nGGGGGGGGGG 0?0 234? 4023 23

GGGTGTTGTG4232145243 ( 1). ( 1). . ( ). . To another value:

nGGGGGGGGGGGGGG ?1()1345513413 ? 23 12 Or, we can write

GGGGTGTTGTG51233156253 1. 1. 0. . ( ). . . For one more value, we have: nGGGGGGGGGGGGG ?2()() ? 245662452 23 12 GGG123 . Finally, we find: GGGG (1). 1. 1. 6123 TG41.( T 6 T 7 )... G 2 TG 63 From the previous recursive procedure, we must find the following set 191 GTGTTGTG42145243 .( ).. °GTGTTGTG .( ).. ° 53156253 °GTGTTGTG .( ).. ° 64167263 °GTGTTGTG75178273 .( ).. ® . °GTGTTGTG86189283 .( ).. °GTGTTGTG .( ).. ° 971910293 ° ° ¯GTGTTGTGnn 21.( nn 1 ).. 2 n 3 Now, from the previous relationships, we will formulate the following theorem.

Theorem 4. For any integer n t 0 we will have the following general term:

GTnn 21 GTT() nn 1 GTG 2 n 3. Proof. Some initial steps were indicated in the previous paragraphs, however, only to confirm our conjecture, we will perform the method of induction.

GTnn 21 GTT() nn 1 GTG 2 n 3 and let us notice the following relation

GGGGnnnn212 and replacing the index n with –n, it gives GGGGG ? nnnnn2122 GGGnnn21 .

In this way, we will add Gn to the expression described above, resulting in: ()()GGGGG G nnnnn2121 n ()GGnnn21 TGTTGTG 21 (). nn 12 n 3

On the left-hand side, we determine: GGGGnnnn22 1 while, on the right-hand side, we will develop the expression, noting that GTGTTGTGnn11112213 () nn n and GTGTTGTGnn24121223 (). nn n . Likewise,

GTGTTGTGnn24121223 () nn n ° ®GTGTTGTGnn11112213 () nn n ° ¯GTnn 21 GTT() nn 1 GTG 2 n 3 In the previous system, we evaluated:

GGGTTTGnnnnnn21 () 4121

()()TTTTTTGTTTGnnnnnn211122213 nnn Finally, noting the following relationships

TTnn 412 T n T n,

TTTTnnnn221 ,

TTTTnnnn3112 ,

192 we can replace and find: GGGG nnnn221 TGnnnnnn 123221()( T T G T T T ) G 3

Thus, we determine that GTGTTGTGnn2123223 () nn n . Corollary. For any integer, the following relations are valid:

nnn§·43 n 1 n 2 (i) ¦¦¦¦Giii ¨¸ t t G123 tG i ¦ tG i, iiii 4213©¹ i 2 nn11§· nn n (ii) ¦¦GtGttGtGii 123¨¸ ¦¦ ii ¦ i. ii 42 ©¹ ii 45 i 4

Proof. Consider the following identities and then add them up, so that the terms G1, G2, G3 are highlighted.

GTTGTGTG4 () 2 1 1 32 23 GTGTTGTG42145243 () ° ° GTTGTGTG () GTGTTGTG () ° 51214233 ° 53156253

°GTTGTGTG62315243 () °GTGTTGTG64167263 () ® , ® °GTTGTGTG73416253 () °GTGTTGTG75178273 () ° ° ° ° ¯°GTnn ().4311223 TGTGTG n n n ¯°GTnn 21 G () TT nn 1 GTG 2 n 3 We will immediately have the identities foreseen in Theorems 3 and 4.

According to the Padovan Q-matrix [11], we have the Tridovan Q-matrix, with T0 = 0, T1 = 1, T2 = 0, T3 = 1, ª0100º «1010» Q « » . «1001» « » ¬1000¼

Theorem 5. Any term of the Tridovan sequence can be obtained by calculating the n-th power n of Q, with TTTT0123 0, 1, 0, 1. This is represented by Q . n ªºª0 1 0 0QQ (1,1) (1, 2) QQ (1, 3) (1, 4) º «»«1 0 1 0QQ (2,1) (2,2) QQ (2,3) (2,4) » Qnn «»« »,3t «»«1 0 0 1QQ (3,1) (3,2) QQ (3,3) (3,4) » «»« » ¬¼¬1 0 0 0QQ (4,1) (4,2) QQ (4,3) (4,4) ¼ In order to simplify the terms visualization of the previous matrix, the calculation of each of these is presented in the following lines:

193 n1 n1 QT(1,1) ()n , QTT(3,1) (1)nn (2) , n1 n1 QT(1, 2) (1)n , QTT(3, 2) (2)nn (3) , n1 n1 QT(1, 3) (2)n , QTT(3,3) (3)nn (4) , n1 n1 QT(1, 4) (3)n , QTT(3, 4) (4)nn (5) ,

n1 n1 QTTT(2,1) (1)nnn (2) (3) , QT(4,1) (1)n , n1 n1 QTTT(2,2) (2)nnn (3)(4) , QT(4,2) (2)n , n1 n1 QTTT(2,3) (3)nnn (4) (5) , QT(4,3) (3)n , n1 n1 QTT(2,4) (4)nn (5)(6)T n , QT(4,4) (4)n . Proof. Using the principle of finite induction to prove the theorem. We have that for nn 1:

QQQnn1 .

n ªºªºQQ(1,1) (1, 2) QQ (1, 3) (1, 4) 0 1 0 0 «»«»QQ(2,1) (2,2) QQ (2,3) (2,4) 1 0 1 0 Qn «»«». «»«»QQ(3,1) (3,2) QQ (3,3) (3,4) 1 0 0 1 «»«» ¬¼¬¼QQ(4,1) (4,2) QQ (4,3) (4,4) 1 0 0 0 It is assumed that each element of the Tridovan Q-matrix is defined as: n1 QTTTT(1,1) (1)nn (2) (3) nn (1) , n1 QTT(1, 2) ()nn ( 1)1 , n1 QTT(1, 3) (1)nn (1)2 , QTTn1(1, 4) , (2)nn (1)3 n1 QTTTTTTTTT(2,1) (2)n (3)(4)(3) nnn (4)(5)(4) nnn (5)(6) nn

T(n)TT (nn 1) ( 2) T ( n 1) 1 T ( n 1) 2 T ( n 1) 3 , n1 QTTTTTT(2,2) =3(nnnn 1) 3( 2) ( 3) ( 1) 2 ( n 1) 3 ( n 1) 4 , n1 QTTTTTT(2,3) 3(2)nnnn (3) (4) (1)3 (1)4 n (1)5 n ,

n1 QTTTTTT(2,4) = (nnn 3) ( 4) 3( 5) 3( n 1) 4 3( n 1) 5 3( n 1) 6 , n1 Q(3,1) T(2)n TTTTTTTT (3)(3)(4)(4)(5) nnnnn () nnn (1) (1)1 T (1)2 n , n1 QTTTT(3, 2) (1)nn (2) (1)2 n (1)3 n , n1 Q (3,3) TTT(2)nnn (3) (1)3 T (1)4 n , n1 QTTTT(3, 4) (3)nn (4) (1)4 n (1)5 n , n1 QTTTTT(4,1) (nn 2) ( 3) 3( n 4) 3( nn ) 3( 1) 1 , n1 QTT(4,2) (1)nn (1)2 , n1 QTT(4,3) (2)nn (1)3 , n1 QTT(4,4) (3)nn (1)4.

According to Definition 5, we can observe the following matrices:

194 ªº4221 ªºTTTT6543 «»5422 «»TTTTTTTTTTTT Q6 «» «»543432321210, «»4321 «»TT54 TT 43 TT 32 TT 21 «»«» ¬¼2211 ¬¼TTTT5432

ªº5422 ªºTTTT7654 «»8542 «»TTTTTTTTTTTT Q7 «» «»654543432321. «»6432 «»TT65 TP 54 TT 43 TT 32 «»«» ¬¼4221 ¬¼TTTT6543 Similarly, we observe:

ªº000 1 ªºTTTT1234 «»100 0 «»TTTTTTTTTTTT Q1 «» «»234345456567, «»010 1 «»TT23 TT 34 TT 45 TT 56 «»«» ¬¼001 1 ¬¼TTTT2345

ªº00 1 1 ªºTTTT2345 «»00 0 1 «»TTTTTTTTTTTT Q2 «» «»345456567678. «»10 1 1 «»TT34 TT 45 TT 56 TT 67 «»«» ¬¼01 1 0 ¬¼TTTT3456

Theorem 6. The determinant calculation of the Tridovan Q-matrix powered to the n-th power is always equal to the number 1, namely, detQn 1. Proof. detQ 1, det(QQnn ) det( ) 1.

Theorem 7. For any natural numbers m and n, 3 d mn , we have the following relation:

TTTTn m..( nm m 11 T nm T nm 2 T nm 3 ).(). T m 21 T nm T nm 2 TT m 31 nm . Proof. QQQnmnm .

ªºTTnn123 T n T n «»TTTTTT TTTTTT Qn «»nnnnnn123234345456 nnnnnn «»TTnn12 TT nn 23 TT nn 34 TT nn 45 «» ¬¼TTTTnn1234 n n

ªºTTmm123 T m T m «»TTTTTT TTTTT T «»mm1232343454 m m mm mm m m mm56 «»TTmm12 TT m 23 m TT mm 34 TT m 45 m «» ¬¼TTTTmm1234 m m

ª TTTTnm nm123 nm nm º «TTTTTT TTTTTT » .« nm123234345456 nm nm nm nm nm nm nm m nm nm nm » « TTnm12 nm TT nm 23 nm Tnm34TTT nm nm 45 nm » « » ¬ TTTTnm1234 nm nm nm ¼

195 TTTTn m..( nm m 11 T nm T nm 2 T nm 3 ).(). T m 21 T nm T nm 2 TT m 31 nm .

If m = 3, TTTTT ..().(). T T TTT TT nn332 nnn 4 5 6 1 nn 4 5 04 n Pn2

TTTTTTTTTTnn 33221...(). n nn 4 5 04 n ∵TTTT01 1; 0; 23 1

TTnn 234 T n T n.

3 Conclusion

Based on the studies carried out on this Tribonacci Sequence, initially studied by Feinberg [5], and then continued by Koshy [7], we can investigate and establish new identities for the Tridovan numbers. In a similar way as for the Fibonacci extensions [2, 3], investigative and historical studies of this sequence are made so that it is possible to obtain the Tridovan Sequence, and thus discover new properties related to it. Based on the previous considerations, we emphasize that this work presents sufficient elements to characterize the general objective scope of the research. The new identities of the Tridovan Sequence, its generalization and the behavior of the terms for the positive and negative indices were presented. Besides that, it was still possible to obtain the Tridovan Q-matrix, so that it is possible to find the terms of the sequence without the use of the recurrence formula [12]. We can conclude that these Padovan numbers are in initial study and discoveries of relations and properties and more research and studies on the subject are required.

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