The (s, t)-Padovan and (s, t)-Perrin matrix sequences Gamaliel Cerda-Morales∗†
Abstract In this study, we first define new sequences named (s, t)-Padovan and (s, t)-Perrin sequences. After that, by using these sequences, we establish (s, t)-Padovan and (s, t)-Perrin matrix sequences. Finally we present some important relationships between these matrix sequences.
Keywords: Matrix sequence, Padovan-like sequence, Pell-Padovan number, Perrin sequence. 2000 AMS Classification: 15A24, 11B39.
1. Introduction The study of Perrin numbers started in the beginning of nineteenth century under different names, the master study was published in 2006 by Shannon et al. [13]. The authors defined the Padovan {Pn }n≥0 and Perrin {Rn }n≥0 sequences as in the forms (1.1)
Pn+3 = Pn+1 + Pn , P0 = P1 = P2 = 1,
and (1.2)
Rn+3 = Rn+1 + Rn , R0 = 3, R1 = 0, R2 = 2,
respectively. Furthermore, the ratio of two consecutive Padovan and Perrin numbers converges to s s r r 1 23 1 23 3 1 3 1 ρ= + + − , 2 6 3 2 6 3 that is named as Plastic constant and was first defined in 1924 by G´erard Cordonnier. He described applications to architecture and illustrated the use of the Plastic constant in many buildings. On the other hand, the matrix sequences have taken so much interest for different type of numbers [2]. For instance, in [2], authors defined new matrix generalizations for Fibonacci and Lucas numbers, and using essentially a matrix approach they showed some properties of these matrix sequences. In [4], Gulec and Taskara gave new generalizations for (s, t)-Pell and (s, t)-Pell-Lucas sequences for Pell and Pell-Lucas numbers. Considering these sequences, they defined the matrix sequences which have elements of (s, t)-Pell and (s, t)-Pell-Lucas sequences. ∗Instituto de Matem´aticas, Pontificia Universidada Cat´olica de Valpara´ıso, Cerro Bar´on, Valpara´ıso, Chile. Email:
[email protected] †Corresponding Author.
2
Also, they investigated their properties. In [7], authors defined a new sequence in which it generalizes (s, t)-Fibonacci and (s, t)-Lucas sequences at the same time. After that, using it, they established generalized (s, t)-matrix sequence. Finally, they presented some important relationships among this new generalization, (s, t)Fibonacci and (s, t)-Lucas sequences and their matrix sequences. Moreover, in [6], authors develop the matrix sequences that represent Padovan and Perrin numbers and examined their properties. In [6], for n ≥ 0, authors defined Padovan and Perrin matrix sequences as in the forms Pn+3 = Pn+1 + Pn , 1 0 0 0 where P0 = 0 1 0 , P1 = 0 0 0 1 1 (1.3)
(1.4)
1 0 1
0 0 1 , P2 = 1 0 0
0 1 1
1 0 , and 1
Rn+3 = Rn+1 + Rn ,
where
4 R0 = −3 2
2 1 −1
−3 −3 1 , R1 = 2 1 1
1 −1 3
2 2 1 , R2 = 1 −1 −1
−1 3 0
1 −1 . 3
Our goal in this paper is to generalize several results about the Padovan and Perrin matrix sequences and sequences such as Padovan, Perrin, Pell-Padovan and Pell-Perrin numbers. The purpose of this paper is to obtain some properties of the (s, t)-Padovan matrix sequence and to demonstrate that some properties of the (s, t)-Padovan numbers and the matrices defined in [6] are valid for a more general sequence of matrices, introduced in the next section. In the rest of the paper, the matrix sequence in which it generalizes the Padovan and Perrin matrix sequences will be defined, the relationship between these matrix sequences will be presented, by giving the Binet formulas and summation formulas over these new matrix sequence, some fundamental properties on the (s, t)-Padovan numbers will be obtained. Therefore, by the (s, t)-Padovan matrix sequence defined in Section 2, we have a great opportunity to obtain some new properties over the (s, t)-Padovan numbers.
2. The (s, t)-Padovan and (s, t)-Perrin matrix sequence In this section, we will mainly focus on the (s, t)-Padovan sequence and the (s, t)-Padovan matrix sequence to get some important results. 2.1. Definition. For any integer numbers s > 0 and t 6= 0 and 27t2 − 4s3 6= 0, the nth (s, t)-Padovan sequence, say {Pn (s, t)}n≥0 is defined recurrently by (2.1)
Pn+3 (s, t) = sPn+1 (s, t) + tPn (s, t), n ≥ 0
where P0 (s, t) = 0, P1 (s, t) = 1 and P2 (s, t) = 0. To simplify notation, take Pn (s, t) = Pn . 2.2. Definition. For any integer numbers s > 0 and t 6= 0 and 27t2 − 4s3 6= 0, the nth (s, t)-Padovan matrix sequence, say {Pn (s, t)}n≥0 is defined recurrently by (2.2)
Pn+3 (s, t) = sPn+1 (s, t) + tPn (s, t), n ≥ 0
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0 0 , P1 (s, t) = 1 Pn+1 2.3. Theorem. n ≥ 0, Pn (s, t) = Pn Pn−1
1 with P0 (s, t) = 0 0
0 1 0
0 1 0
s 0 1
Pn+2 Pn+1 Pn
s t 0 , P2 (s, t) = 0 0 1 tPn tPn−1 . tPn−2
t s 0
0 t . 0
Proof. We use the second principle of finite induction on n to prove this theorem. When n = 0, using (2.1) we have P−1 (s, t) = 0 and P−2 (s, t) = t−1 , then the result is true. Now assume that Pn (s, t) is true for all 1 ≤ n ≤ N. Then, PN +1
= sPN −1 + tPN −2 PN PN +1 tPN −1 PN −1 PN tPN −2 + t PN −2 = s PN −1 PN −2 PN −1 tPN −3 PN −3 PN +2 PN +3 tPN +1 tPN . = PN +1 PN +2 PN PN +1 tPN −1
PN PN −1 PN −2
tPN −2 tPN −3 tPN −4
Thus it is true for every nonnegative integer n.
2.4. Theorem. Pm+n = Pm Pn , for any integers m, n ≥ 0. Proof. By induction over n. Because P0 = I3 is the identity matrix of order 3, we have Pm = Pm P0 and the assertion is true for n = 0. Now, assume that Pm+n = Pm Pn for 0 ≤ n ≤ N, then Pm+N +1
= sPm+N −1 + tPm+N −2 = sPm PN −1 + tPm PN −2 = Pm (sPN −1 + tPN −2 ) = Pm PN +1 ,
for m ≥ 0. Then, proof of the theorem is completed.
2.5. Definition. For any integer numbers s > 0 and t 6= 0 and 27t2 − 4s3 6= 0, the nth (s, t)-Perrin sequence, say {Rn (s, t)}n≥0 is defined recurrently by (2.3)
Rn+3 (s, t) = sRn+1 (s, t) + tRn (s, t), n ≥ 0
where R0 (s, t) = 3, R1 (s, t) = 0 and R2 (s, t) = 2s. To simplify notation, take Rn (s, t) = Rn . 2.6. Definition. For any integer numbers s > 0 and t 6= 0 and 27t2 − 4s3 6= 0, the nth (s, t)-Perrin matrix sequence, say {Rn (s, t)}n≥0 is defined recurrently by (2.4)
Rn+3 (s, t) = sRn+1 (s, t) + tRn (s, t), n ≥ 0
with
0 R0 = 3 − st
2s 3t 2s 3t 0 3t 2s2 0 −s , R1 = 0 2s 3t , R2 = 2s 3t 2 3 0 −s 0 2s 3 st
2s 0 . 3t
4
The following table summarizes special cases of Pn (s, t) and Rn (s, t): (s, t) (1, 1) (2, 1) (1, 2)
Pn Padovan numbers Pell-Padovan numbers Jacobsthal-Padovan numbers
Rn Perrin numbers Pell-Perrin numbers Jacobsthal-Perrin numbers
In the literature, Pn (1, 1) and Rn (1, 1) was studied in [13] for Shannon et. al. and the case Pn (2, 1) was studied in [15]. Reciently, in [3], Gogin studied a class of Padovan-like sequences Pn (α, 1) (α ∈ R) that can be generated using special matrices of the third order. Rn+1 Rn+2 tRn Rn+1 tRn−1 . 2.7. Theorem. n ≥ 0, Rn (s, t) = Rn Rn−1 Rn tRn−2 Proof. The proof follows from the induction method.
2.8. Theorem. Rn+1 = R1 Pn , for any integer n ≥ 0. Proof. By induction over n. Because P0 = I3 is the identity matrix of order 3, we have R1 = R1 P0 and the assertion is true for n = 0. Now, 0 s t 3t 2s2 2st 2s 3t 0 0 R1 P1 = 0 2s 3t 1 0 0 = 2s 3t 0 1 0 0 2s 3t 3 0 −s R3 R4 tR2 = R2 R3 tR1 = R2 , R1 R2 tR0 is true for n = 1. Assume that Rn+1 = P1 Rn for 0 ≤ n ≤ N, have R1 PN +1 = R1 PN P1 = RN +1 P1 and we can write RN +2 RN +3 tRN +1 0 tRN 1 R1 PN +1 = RN +1 RN +2 0 RN RN +1 tRN −1 RN +3 RN +4 tRN +2 = RN +2 RN +3 tRN +1 RN +1 RN +2 tRN = RN +2 .
then if n = N + 1 we s 0 1
t 0 0
Then, proof of the theorem is completed.
2.9. Corollary. For n ≥ 1, Rn+1 (s, t) = 2sPn (s, t) + 3tPn−1 (s, t). Proof. Using above theorem, we have Rn+1 = R1 Pn . Then, comparing the second row and first column in Rn+1 and R1 Pn , the corollary is completed. 2.10. Theorem. For m, n ≥ 0, we have (2.5)
Pm Rn+1 = Rn+1 Pm .
5
Proof. From next matrix 1 0 R1 = 2s 0 1 0 0
identity on R1 , 0 1 0 0 + 3t 0 0 1 1 0 t
0 1 = 2sP0 + 3tP−1 , − st
and equation Pn+m = Pn Pm , we can write Pm Rn+1
=
Pm R1 Pn
=
Pm (2sP0 + 3tP−1 ) Pn
=
2sPm+n + 3tPm+n−1
=
(2sP0 + 3tP−1 ) Pm+n
=
R1 Pn Pm
=
Rn+1 Pm .
This completes the proof.
2.11. Theorem. For n ≥ 0, we have Rn+1 = 2sPn + 3tPn−1 . Proof. For n = 1, it is obvious R1 = 2sP0 + 3tP−1 if we take for convenience P−2 = 1t and P−3 = − ts2 . For n ≤ N , assume that RN +1 = 2sPN + 3tPN −1 . We will show that the case also holds for n = N + 1: RN +2
= sRN + tRN −1 = s (2sPN −1 + 3tPN −2 ) + t (2sPN −2 + 3tPN −3 ) =
2s (sPN −1 + tPN −2 ) + 3t (sPN −2 + tPN −3 )
=
2sPN +1 + 3tPN .
2.12. Theorem. For n ≥ 0, (2.6)
tRn2 + 2Rn+1 Rn+2 = s2 P2n + 6tP2n+1 + 3P2n+4 .
Proof. For n ≥ 0, we obtain from above theorems that (2.7)
R2n+1 = Rn+1 Rn+1 = (R1 Pn ) (R1 Pn ) = R21 P2n .
Then, R2n+1 = R21 P2n and 2 2 Rn+2 Rn+3 tRn+1 2s 3t 0 P2n+1 Rn+1 Rn+2 tRn = 0 2s 3t P2n Rn Rn+1 tRn−1 3 0 −s P2n−1
P2n+2 P2n+1 P2n
tP2n tP2n−1 . tP2n−2
From the equality of the second row and first column in both of sides of the equation, we write tRn2 + 2Rn+1 Rn+2
This proof is completed.
=
4s2 P2n + 3stP2n−1 + 9tP2n+1
=
s2 P2n + 3s (sP2n + tP2n−1 ) + 9tP2n+1
=
s2 P2n + 3 (sP2n+2 + tP2n+1 ) + 6tP2n+1
=
s2 P2n + 6tP2n+1 + 3P2n+4 .
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2.13. Theorem. For n ≥ 0, we have (2.8)
tRn2 + 2Rn+1 Rn+2 = 2R2n+3 + tR2n .
Proof. For n ≥ 0, we obtain from above theorem that R2n+1 = R21 P2n . Then, we obtain the next identity R2n+1 = R1 R1 P2n = R1 R2n+1 . So, 2 Rn+2 Rn+3 tRn+1 2s 3t 0 R2n+2 R2n+3 tR2n+1 Rn+1 Rn+2 tRn = 0 2s 3t R2n+1 R2n+2 tR2n . Rn Rn+1 tRn−1 3 0 −s R2n R2n+1 tR2n−1 From the equality of the second row and first column in both of sides of the equation, we can write tRn2 + 2Rn+1 Rn+2
=
2sR2n+1 + 3tR2n
=
2 (sR2n+1 + tR2n ) + tR2n
=
2R2n+3 + tR2n .
This proof is completed.
Furthermore, R2n+1 = R1 P2n = Pn R1 Pn = Pn Rn+1 . Then, R2n+2 R2n+3 tR2n+1 Pn+1 Pn+2 tPn Rn+2 R2n+1 R2n+2 tR2n = Pn Pn+1 tPn−1 Rn+1 R2n R2n+1 tR2n−1 Pn−1 Pn tPn−2 Rn
Rn+3 Rn+2 Rn+1
tRn+1 tRn tRn−1
and we obtain R2n+1 =
Pn
Pn+1
tPn−1
Rn+2 Rn+1 = Pn Rn+2 +Pn+1 Rn+1 +tRn Pn−1 . Rn
3. Binet’s formula for the (s, t)-Padovan matrix sequence Binet’s formulas are well known in the Padovan and Perrin numbers theory [6]. In our case, Binet’s formula allows us to express the (s, t)-Padovan matrix sequences in function of the roots of the characteristic equation x3 = sx + t, associated to the recurrence relation (2.1), or (2.3). The roots of the incomplete cubic equation x3 − sx − t = 0 are given by √ i 3 1 (A − B), α = A + B, ω1,2 = − (A + B) ± 2 2 √ 31 √ 13 2 3 where A = 2t + ∆ , B = 2t − ∆ , ∆ = 2t − 3s . Note that αω1 ω2 = t and α + ω1 + ω2 = 0. Furthermore, A and B being any of the values of the respective cubic roots such that AB = 3s . There are three real roots of the cubic equation x3 = sx + t if the discriminant ∆ < 0, or one real and two complex conjugate roots if ∆ > 0. 3.1. Theorem. For every n ∈ N, one can write the Binet formulas for the (s, t)Padovan and (s, t)-Perrin matrix sequences as the form (3.1)
Pn = A1 αn + A2 ω1n + A3 ω2n ,
7
and (3.2)
Rn = B1 αn + B2 ω1n + B3 ω2n ,
where R2 +R1 α+R0 ω1 ω2 P2 +P1 α+P0 ω1 ω2 B1 A1 (α−ω2 )(α−ω1 ) (α−ω )(α−ω ) − R2 +R21 ω1 +R01αω2 P2 +P1 ω1 +P0 αω2 A2 = B , = − (ω 2 (ω1 −ω2 )(α−ω1 ) , 1 −ω2 )(α−ω1 ) P2 +P1 ω2 +P0 αω1 R2 +R1 ω2 +R0 αω1 B3 A3
(ω1 −ω2 )(α−ω2 )
(ω1 −ω2 )(α−ω2 )
such that α, ω1 and ω2 are roots of characteristic equations of (2.1) and (2.3). Proof. We note that the proof will be based on the recurrence relations(2.1) and (2.3). Now, we will show the Binet formula for (s, t)-Padovan matrix sequence. By the assumption, the roots of the characteristic equation of (2.1) are α, ω1 and ω2 . Hence its general solution of it is given by Pn = A1 αn + A2 ω1n + A3 ω2n , n ≥ 0. Using initial conditions of {Pn }n≥0 and also applying fundamental linear algebra operations, we clearly get the matrices At (t = 1, 2, 3), as desired. This implies the formula for Pn . In [6], the authors obtained the Binet formulas for Padovan and Perrin numbers by matrix methods. In the following corollary, we will present the formulas for (s, t)-Padovan and (s, t)-Perrin numbers numbers via related matrix sequences. In fact, in the proof of the next corollary, we will just compare the linear combination of the second row and first column entries of the matrices A1 , A2 and A3 with the matrix Pn in 3.1 and, similarly, B1 , B2 and B3 with the matrix Rn in 3.2. 3.2. Corollary. For n ≥ 0, the Binet Formulas for nth (s, t)-Padovan number and nth (s, t)-Perrin number are given by Pn (s, t) =
αn+1 ω1n+1 ω2n+1 − + , (α − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω2 )
and Rn (s, t) = αn + ω1n + ω2n , respectively. Proof. By taking into account Definition 2.2 and Theorem 3.1, we can write Pn
= A1 αn + A2 ω1n + A3 ω2n P2 + P1 α + P0 ω1 ω2 P2 + P1 ω1 + P0 αω2 = αn − ω1n (α − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω1 ) P2 + P1 ω2 + P0 αω1 + ω2n . (ω1 − ω2 ) (α − ω2 )
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Herein, since α, ω1 and ω2 are roots of the characteristic equation x3 − sx − t = 0, we clearly have sα+t t + sα tα α αn sα+t α t Pn = α (α − ω2 ) (α − ω1 ) t 1 α α sω1 +t t + sω1 tω1 ω1 ω1n sω1 +t ω1 t − ω1 (ω1 − ω2 ) (α − ω1 ) t 1 ω1 ω1 sω2 +t t + sω2 tω2 n ω 2 ω2 sω2 +t ω2 t . + ω2 (ω1 − ω2 ) (α − ω2 ) t 1 ω2 ω2 From the equality of the second row and first column in both of sides of the equation, we can write Pn (s, t) =
ω1n+1 ω2n+1 αn+1 − + . (α − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω2 )
Using Corollary 2.9, we have Rn+1 (s, t) = 2sPn (s, t) + 3tPn−1 (s, t). Then, Rn+1 (s, t)
= =
2sPn (s, t) + 3tPn−1 (s, t) (2sα + 3t) αn (2sω1 + 3t) ω1n (2sω2 + 3t) ω2n − + (α − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω2 )
=
(2s + 3ω1 ω2 ) αn+1 (2s + 3αω2 ) ω1n+1 (2s + 3αω1 ) ω2n+1 − + (α − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω1 ) (ω1 − ω2 ) (α − ω2 )
=
αn+1 + ω1n+1 + ω2n+1 .
since αω1 ω2 = t. The proof is completed.
Pn 3.3. Theorem. Let s + t − 1 6= 0 and Sn = k=1 Pk . For n ≥ 1, 1 Pn+2 − t − s + Pn+3 + tPn+1 . tPn + Pn+1 + Pn+2 − 1 Sn 1 = 1 Pn + Pn+1 + tPn−1 − 1 Pn Proof. Let Sn = k=1 Pk . In this case, Sn P1 = P2 + P3 + · · · + Pn+1 . From the above equation, Sn (P1 − P0 ) = Pn+1 − P1 . Furthermore, we can write Sn = (Pn+1 − P1 ) (P1 − P0 )−1 since det(P1 − P0 ) = s + t − 1 6= 0. Then, Pn+2 Pn+3 − s tPn+1 − t Pn+2 tPn Pn+1 − P1 = Pn+1 − 1 Pn Pn+1 − 1 tPn−1 1 s+t t −1 1 1 1 t . and (P1 − P0 ) = s+t−1 1 1 1−s
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Finally, we obtain Pn+2 1 Pn+1 − 1 Sn = s+t−1 Pn
Pn+3 − s Pn+2 Pn+1 − 1
tPn+1 − t 1 1 tPn tPn−1 1
s+t t 1 t 1 1−s
and Pn+2 − t − s + Pn+3 + tPn+1 1 1 . tPn + Pn+1 + Pn+2 − 1 Sn 1 = s+t−1 Pn + Pn+1 + tPn−1 − 1 1
3.4. Corollary. For n ≥ 1, s + t 6= 1. We have Pn 1 (tPn + Pn+1 + Pn+2 − 1) , P (s, t) = s+t−1 Pn k=1 k 1 k=1 Pk+1 (s, t) = s+t−1 (Pn+2 − t − s + Pn+3 + tPn+1 ) .
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