On the Properties of Iterated Binomial Transforms for the Padovan and Perrin Matrix Sequences

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Rn+3 = Rn+1 + Rn, where R0 =3, R1 =0, R2 = 2 (1.1) and Pn+3 = Pn+1 + Pn, where P0 = P1 = P2 =1 , (1.2) respectively. On the other hand, the matrix sequences have taken so much interest for different type of numbers [7, 8, 9, 10]. For instance, in [7], 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 [8], 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. Also, they investigated their properties. In [9], authors defined a new sequence in which it generalizes (s,t)-Fibonacci and (s,t)-Lucas sequences at the same time. After that, by 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 [10], authors develop the matrix sequences that represent Padovan and Perrin numbers and examined their properties. In [10], for n > 0, authors defined Padovan and Perrin matrix sequences as in the forms Pn+3 = Pn+1 + Pn, (1.3) where 1 0 0 0 1 0 0 0 1 P0 = 0 1 0 , P1 = 0 0 1 , P2 = 1 1 0 ,  0 0 1   1 1 0   0 1 1        and Rn+3 = Rn+1 + Rn, (1.4) where 4 2 −3 −3 1 2 2 −1 1 R0 = −3 1 2 , R1 = 2 −1 1 , R2 = 1 3 −1 .  2 −1 1   1 3 −1   −1 0 3        Proposition 1.1 [10] Let us consider n,m > 0, the following properties are hold:

• PmPn = Pn+m,

• PmRn = RnPm = Rn+m.

2 In addition, some matrix based transforms can be introduced for a given sequence. Binomial transform is one of these transforms and there is also other ones such as rising and falling binomial transforms(see [11, 12, 16]). Given an integer sequence X = {x0, x1, x2,...} , the binomial transform B of the sequence X, B (X)= {bn} , is given by

n n b = x . n i i i=0 X In [13, 15], authors gave the application of the several class of transforms to the k-Fibonacci and k-Lucas sequence. In [14], the authors applied the binomial transforms to the Padovan (Pn) and Perrin matrix sequences (Rn). Proposition 1.2 [14] For n> 0, i) Recurrence relation of sequences {bn} is

bn+2 =3bn+1 − 2bn + bn 1, (1.5) − 1 0 0 1 1 0 with initial conditions b0 = 0 1 0 , b1 = 0 1 1 and b2 =  0 0 1   1 1 1  1 2 1     1 2 2 .  2 3 2    ii) Recurrence relation of sequences {cn} is

cn+2 =3cn+1 − 2cn + cn 1, (1.6) − 4 2 −3 1 3 −1 with initial conditions c0 = −3 1 2 , c1 = −10 3 and  2 −1 1   3 2 0  0 3 2     c2 = 2 2 3 .  3 5 2    iii) For n,m ≥ 0, we have bnbm = bn+m, where n ≤ m, iv) For n,m ≥ 0, we have bncm = cnbm = cn+m. Falcon [17] studied the iterated application of the some Binomial transforms to the k-Fibonacci sequence. For example, author obtained recurrence relation of the iterated binomial transform for k-Fibonacci sequence

(r) (r) 2 (r) (r) (r) ak,n+1 = (2r + k) ak,n − r + kr − 1 ak,n 1, ak,0 = 0 and ak,1 =1. − 3 Yilmaz and Taskara [18] studied the iterated application of the some Binomial transforms to the k-Lucas sequence. For example, author obtained recurrence relation of the iterated binomial transform for k-Lucas sequence

(r) (r) 2 (r) (r) (r) bk,n+1 = (2r + k) bk,n − r + kr − 1 bk,n 1, bk,0 = 2 and bk,1 =2r + k. − Motivated by [10,14,17,18], the goal of this paper is to apply iteratly the binomial transforms to the Padovan (Pn) and Perrin matrix sequences (Rn). Also, the binet formulas, summations, generating functions of these transforms are found by recurrence relations. Finally, it is illustrated the relations between of this transforms by deriving new formulas.

2 Iterated binomial transforms of the Padovan and Perrin matrix sequences

In this section, we will mainly focus on iterated binomial transforms of the Padovan and Perrin matrix sequences to get some important results. In fact, we will also present the recurrence relations, binet formulas, summations, generating functions of these transforms. The iterated binomial transforms of the Padovan (Pn) and Perrin matrix (r) (r) (r) (r) sequences (Rn) are demonstrated by B = bn and C = cn , where (r) (r) bn and cn are obtained by applying ”r” timesn theo binomial transformn o to the (r) (r) Padovan and Perrin matrix sequences. It are obvious that b0 = P0, b1 = (r) 2 (r) (r) (r) rP0 + P1, b2 = r P0 +2rP1 + P2 and c0 = R0, c1 = rR0 + R1, c2 = 2 r R0 +2rR1 + R2. The following lemma will be key of the proof of the next theorems.

Lemma 2.1 For n ≥ 0 and r ≥ 1, the following equalities are hold:

(r) n+1 n+1 (r 1) (r 1) i) bn+2 = j=0 j bj − + bj+1− , (r) P n+1 n+1 (r 1) (r 1) ii) cn+2 = j=0 j cj − + cj+1− . P Proof. Firstly, in here we will just prove i), since ii) can be thought in the same manner with i). i) By using deﬁnition of binomial transform, we obtain

n+2 (r) n +2 (r 1) b = b − n+2 j j j=0 X n+2 n +2 (r 1) (r 1) = b − + b − . j j 0 j=1 X

4 And by taking account the well known binomial equality n +2 n +1 n +1 = + , i i i − 1 we have

n+1 n+2 (r) n +1 (r 1) n +1 (r 1) (r 1) b = b − + b − + b − n+2 j j j − 1 j 0 j=1 j=1 X X n+1 n+1 n +1 (r 1) n +1 (r 1) = b − + b − j j j j+1 j=0 j=0 X X n+1 n +1 (r 1) (r 1) = b − + b − , j j j+1 j=0 X which is desired result.

From above Lemma, note that:

(r) (r) n n (r 1) (r 1) • bn+2 is also can be written as bn+2 = bn+1 + i=0 i bj+1− + bj+2− , (r) (r) Pn n (r 1) (r 1) • cn+2 is also can be written as cn+2 = cn+1 + i=0 i cj+1− + cj+2− . Theorem 2.1 For n ≥ 0 and r ≥ 1, the recurrenceP relations of sequences (r) (r) bn and cn are n o n o (r) (r) 2 (r) 3 (r) bn+2 =3rbn+1 − 3r − 1 bn + r − r +1 bn 1, (2.1) − (r) (r) 2 (r) 3 (r) cn+2 =3rcn+1 − 3r − 1 cn + r − r +1 cn 1, (2.2) − (r) (r) (r) 2 with initial conditions b0 = P0, b1 = rP0 + P1, b2 = r P0 +2rP1 + P2 and (r) (r) (r) 2 c0 = R0, c1 = rR0 + R1, c2 = r R0 +2rR1 + R2. Proof. Similarly the proof of the Lemma 2.1, only the first case, the equation (2.1) will be proved. We will omit the equation (2.2) since the proofs will not be different. The proof will be done by induction steps on r and n. First of all, for r = 1, from the i) condition of Proposition 1.2, it is true bn+2 =3bn+1 − 2bn + bn 1. Let us consider definition− of iterated binomial transform, then we have

(r) 3 2 b3 = r +1 P0 + 3r +1 P1 +3rP2 . The initial conditions are

(r) (r) (r) 2 b0 = P0,b1 = rP0 + P1, b2 = r P0 +2rP1 + P2.

5 (r) (r) 2 (r) Hence, for n =1, the equality (2.1) is true, that is b3 =3rb2 − 3r − 1 b1 + 3 (r) r − r +1 b0 . Actually, by assuming the equation in (2.1) holds for all (r −1,n) and (r, n− 1), that is,

(r 1) (r 1) 2 (r 1) 3 (r 1) bn+2− = 3 (r − 1) bn+1− − 3 (r − 1) − 1 bn − + (r − 1) − (r − 1)+1 bn −1 , − and (r) (r) 2 (r) 3 (r) bn+1 =3rbn − 3r − 1 bn 1 + r − r +1 bn 2. − − Then, we need to show that it is true for (r, n) . That is,

(r) (r) 2 (r) 3 (r) bn+2 =3rbn+1 − 3r − 1 bn + r − r +1 bn 1. − (r) (r) 2 (r) 3 (r) Let us label bn+2 =3rbn+1 − 3r − 1 bn + r − r +1 bn 1 by RHS. Hence, we can write − n+1 n n 1 n +1 (r 1) 2 n (r 1) 3 − n − 1 (r 1) RHS = 3r b − − 3r − 1 b − + r − r +1 b − j j j j j j j=0 j=0 j=0 X X X n+1 n+1 2 n +1 (r 1) 2 n +1 (r 1) = 3r − 3r +1 b − + 3r − 1 b − j j j j j=0 j=0 X X n n 1 − 2 n (r 1) 3 n − 1 (r 1) − 3r − 1 b − + r − r +1 b − j j j j j=0 j=0 X X n+1 n+1 2 n +1 (r 1) 2 n (r 1) = 3r − 3r +1 b − + 3r − 1 b − j j j − 1 j j=0 j=1 X X n 1 3 − n − 1 (r 1) + r − r +1 b − j j j=0 X n+1 n+1 2 n +1 (r 1) 2 n (r 1) = 3r − 3r +1 b − + 3r − 3r +1 b − j j j − 1 j j=0 j=1 X X n+1 n 1 2 n (r 1) 3 − n − 1 (r 1) + 6r − 3r − 2 b − + r − r +1 b − j − 1 j j j j=1 j=0 X X n+1 n+1 2 n +1 (r 1) 2 n +1 (r 1) = 3r − 3r +1 b − + 3r − 3r +1 b − j j j j+1 j=0 j=0 X X n+1 n 1 − 2 n (r 1) 3 n − 1 (r 1) + 6r − 3r − 2 b − + r − r +1 b − j − 1 j j j j=1 j=0 X X n+1 2 n (r 1) − 3r − 3r +1 b − . j − 1 j+1 j=1 X

6 From Lemma 2.1, we have n 2 (r) 2 n (r 1) RHS = 3r − 3r +1 b + 6r − 3r − 2 b − n+2 j j+1 j=0 X n 1 n − 3 n − 1 (r 1) 2 n (r 1) + r − r +1 b − + 3r − 3r − 1 b − j j j j+2 j=0 j=0 X X n 1 2 (r) 3 − n − 1 (r 1) = 3r − 3r +1 b + r − r +1 b − n+2 j j j=0 X n n 2 n (r 1) (r 1) 2 n (r 1) + 3r − 3r b − + b − + 3r − 2 b − j j+1 j+2 j j+1 j=0 j=0 n X X n (r 1) − b − j j+2 j=0 X

n 1 − (r) 2 (r) 3 n − 1 (r 1) RHS = b + 3r − 3r b + r − r +1 b − n+2 n+1 j j j=0 X n n 1 2 n − 1 (r 1) 2 − n − 1 (r 1) + 3r − 2 b − + 3r − 2 b − j − 1 j+1 j j+1 j=1 j=0 X X n n (r 1) − b − j j+2 j=0 X (r) 2 (r) 2 (r) = bn+2 + 3r − 3r bn+1 + 3r − 3r bn n 1 n 1 3 2 − n − 1 (r 1) − n − 1 (r 1) + r − 3r +2r +1 b − + (3r − 2) b − j j j j+1 j=0 j=0 X X n 1 n − 2 n − 1 (r 1) n (r 1) + 3r − 2 b − − b − j j+2 j j+2 j=0 j=0 X X (r) 2 (r) (r) = bn+2 + 3r − 3r bn+1 − bn

n 1 (r 1) (r 1) − 3 2 n − 1 r − 3r +2r +1 bj − + (3r − 2) bj+1− + (r 1) (r 1) . j ++ 3r2 − 3 b − − b − j=0 " j+2 j+3 # X Afterward, by taking account assumption and Lemma 2.1, we deduce

(r) 2 (r) (r) RHS = bn+2 + 3r − 3r bn+1 − bn n 1 2 − n − 1 (r 1) (r 1) − 3r − 3r b − + b − j j+1 j+2 j=0 X (r) = bn+2

7 which is completed the proof of this theorem. (r) (r) The characteristic equation of sequences bn and cn in (2.1) and (2.2) is n o n o 3 2 2 3 λ − 3rλ + 3r − 1 λ − r − r +1 = 0. Let be λ1, λ2 and λ3 the roots (r) (r) of this equation. Then, binet’s formulas of sequences bn and cn can be expressed as n o n o (r) n n n bn = X1λ1 + Y1λ2 + Z1λ3 , (2.3) and (r) n n n cn = X2λ1 + Y2λ2 + Z2λ3 , (2.4) where

(r) 2 (r) 3 (r) λ1b − 3rλ1 − λ b + r − r +1 b X = 2 1 1 0 , 1 λ λ λ λ λ 1 ( 1 − 2) ( 1 − 3) (r) 2 (r) 3 (r) λ2b − 3rλ2 − λ b + r − r +1 b Y = 2 2 1 0 , 1 λ λ λ λ λ 2 ( 2 − 1) ( 2 − 3) (r) 2 (r) 3 (r) λ3b − 3rλ3 − λ b + r − r +1 b Z = 2 3 1 0 , 1 λ λ λ λ λ 3 ( 3 − 1) ( 3 − 2) and

(r) 2 (r) 3 (r) λ1c − 3rλ1 − λ c + r − r +1 c X = 2 1 1 0 , 2 λ λ λ λ λ 1 ( 1 − 2) ( 1 − 3) (r) 2 (r) 3 (r) λ2c − 3rλ2 − λ c + r − r +1 c Y = 2 2 1 0 , 2 λ λ λ λ λ 2 ( 2 − 1) ( 2 − 3) (r) 2 (r) 3 (r) λ3c − 3rλ3 − λ c + r − r +1 c Z = 2 3 1 0 . 2 λ λ λ λ λ 3 ( 3 − 1) ( 3 − 2)

Now, we give the sums of iterated binomial transforms for the Padovan and Perrin matrix sequences.

(r) (r) Theorem 2.2 Sums of sequences bn and cn are n o n o (r) (r) 3 (r) (r) (r) n 1 (r) bn+1 + (1 − 3r) bn + r − r +1 bn 1 + (3r − 1) b0 − b1 i) − b = − i=0 i r3 r +2 P (r) (r) 3 (r) (r) (r) n 1 (r) cn+1 + (1 − 3r) cn + r − r +1 cn 1 + (3r − 1) c0 − c1 ii) − c = − . i=0 i r3 r +2 Proof.P We omit Padovan case since the proof be quite similar. By considering equation (2.4), we have

n 1 n 1 − (r) − n n n ci = (X2λ1 + Y2λ2 + Z2λ3 ) . i=0 i=0 X X

8 Then we obtain

n 1 n n n − λ − 1 λ − 1 λ − 1 c(r) = X 1 + Y 2 + Z 3 . i 2 λ − 1 2 λ − 1 2 λ − 1 i=0 1 2 3 X (r) 3 Afterward, by taking account equations c 1 = 0, λ1 · λ2 · λ3 = r − r + 1 and − λ1 + λ2 + λ3 =3r, we conclude

n 1 (r) (r) 3 (r) (r) (r) − (r) cn+1 + (1 − 3r) cn + r − r +1 cn 1 + (3r − 1) c0 − c1 c = − . i r3 +2r i=0 X

Theorem 2.3 The generating functions of the iterated binomial transforms for (Pn) and (Rn) are

(r) (r) (r) (r) (r) 2 (r) 2 b0 + b1 − 3rb0 x + b2 − 3rb1 + 3r − 1 b0 x i ∞ b(r)xi , ) i = 2 2 3 3 i=0 1 − 3rx + (3r −1) x − (r − r+ 1) x P (r) (r) (r) (r) (r) 2 (r) 2 c0 + c1 − 3rc0 x + c2 − 3rc1 + 3r − 1 c0 x ii ∞ c(r)xi . ) i = 2 2 3 3 i=0 1 − 3rx + (3r −1) x − (r − r+ 1) x P Proof.

∞ (r) i i) Assume that b (x, r) = bi x is the generating function of the iterated i=0 binomial transform forP (Pn). From Theorem 2.1, we obtain

(r) (r) (r) 2 b (x, r) = b0 + b1 x + b2 x ∞ (r) 2 (r) 3 (r) i + 3rbi 1 − 3r − 1 bi 2 + r − r +1 bi 3 x − − − i=3 X (r) (r) (r) 2 ∞ (r) i 1 = b0 + b1 x + b2 x +3rx bi 1x − − i=3 X ∞ ∞ 2 2 (r) i 2 3 3 (r) i 3 − 3r − 1 x bi 2x − + r − r +1 x bi 3x − − − i=3 i=3 X X ∞ (r) (r) (r) 2 (r) i (r) (r) = b0 + b1 x + b2 x +3rx bi x − 3rx b0 + b1 x i=0 X 2 2 ∞ (r) i 2 2 (r) 3 3 ∞ (r) i − 3r − 1 x bi x + 3r − 1 x b0 + r − r +1 x bi x . i=0 i=0 X X Now rearrangement the equation implies that

(r) (r) (r) (r) (r) 2 (r) 2 b0 + b1 − 3rb0 x + b2 − 3rb1 + 3r − 1 b0 x b (x, r)= , 1 − 3rx + (3r2 −1) x2 − (r3 − r+ 1) x3

9 ∞ (r) i which equal to the bi x in theorem. Hence the result. i=0 P ii) The proof of generating function of the iterated binomial transform for Per- rin matrix sequences can see by taking account proof of i).

3 The relationships between new iterated binomial transforms

In this section, we present the relationships between the iterated binomial transform of the Padovan matrix sequence and iterated binomial transform of the Perrin matrix sequence.

Theorem 3.1 For n,m ≥ 0, we have

(r) (r) (r) i) bn bm = bn+m, where n ≤ m,

(r) (r) (r) (r) (r) ii) bn cm = cm bn = cn+m,

Proof. i) The proof will be done by induction step on r. First of all, for r = 1, from the iii) condition of Proposition 1.2, it is true bnbm = bn+m. Actually, by assuming the equation in i) holds for all r, that is,

n m (r) (r) n (r 1) m (r 1) b b = b − + b − n m i i j j i=0 j=0 X X n+m n + m (r 1) (r) = b − = b . k k n+m k X=0 Then, we need to show that it is true for r +1. That is, From deﬁnition of iterated binomial transform, we have

n n m m b(r+1)b(r+1) = b(r) b(r) n m i i  j j  i=0 ! j=0 X X n i  m j n i (r 1) m j (r 1) = b − b − . i k k  j l l  i=0 k ! j=0 l X X=0 X X=0  

10 And, by considering assumption, we obtain

n m n m b(r+1)b(r+1) = b(r) n m i j i+j i=0 j=0 X X n m n m n m = b(r) + b(r) + ··· + b(r) 0 0 0 0 1 1 0 m m n m n m n m + b(r) + b(r) + ··· + b(r) + 1 0 1 1 1 2 1 m m+1 . . n m n m n m + b(r) + b(r) + ··· + b(r) n 0 n n 1 n+1 n m n+m n m n m n m = b(r) + + b(r) 0 0 0 0 1 1 0 1 n m n m n m + + + b(r) + ··· 0 2 1 1 2 0 2 n m n m n m + + + ··· + b(r) + ··· 0 k 1 k − 1 k 0 k n m + b(r) . n m n+m k x y x+y By taking account Vandermonde identity j k j = k , we get j=0 − P n + m n + m n + m b(r+1)b(r+1) = b(r) + b(r) + b(r) + ··· n m 0 0 1 1 2 2 n + m n + m + b(r) + ··· + b(r) k k n + m n+m n m + n + m = b(r) i i i=0 X (r+1) = bn+m . ii) The proof is similar proof of i).

(r) (r) Theorem 3.2 The properties of the transforms bn and cn would be illustrated by following way: n o n o

(r) (r) (r 1) (r) i) bn+1 − bn = b1 − bn ,

(r) (r) (r 1) (r) ii) cn+1 − cn = b1 − cn ,

11 (r) (r) (r 1) (r) iii) cn+1 − cn = c1 − bn .

Proof. We will omit the proof of ii) and iii), since it is quite similar with i). Therefore, by considering deﬁnition of iterated binomial transform and Lemma 2.1-i), we have n (r) (r) n (r 1) b − b = b − . n+1 n i i+1 i=0 X From Theorem 3.1-i), we get

n (r) (r) n (r 1) (r 1) (r 1) (r) b − b = b − b − = b − b . n+1 n i i 1 1 n i=0 X

(r) Theorem 3.3 For n,m ≥ 0, the relation between the transforms bn and (r) cn is n o (r 1) (r) (r 1) (r) n o cm− bn = bm− cn .

Proof. By considering deﬁnition of iterated binomial transform, we have

n (r 1) (r) (r 1) n (r 1) c b = c b − m− n m− i i i=0 n X n (r 1) (r 1) = c b − . i m− i i=0 X From Theorem 3.1-ii), we get

n (r 1) (r) n (r 1) c b = c − m− n i m+i i=0 Xn n (r 1) (r 1) = b − c − i m i i=0 X (r 1) (r) = bm− cn .

By choosing m = 0 in Theorem 3.3 and using the initial conditions of equations (2.1) and (2.2), we obtain the following corollary.

Corollary 3.1 The following equalities are hold:

(r) (r) i) cn = R0bn ,

(r) 1 (r) ii) bn = R0− cn .

12 Corollary 3.2 We should note that choosing r =1 in the all results of Section 2 and 3, it is actually obtained some properties of the iterated binomial transforms for Padovan and Perrin matrix sequences such that the recurrence relations, Binet formulas, summations, generating functions and relationships of between binomial transforms for Padovan and Perrin matrix sequences.

Conclusion 3.1 In this paper, we deﬁne the iterated binomial transforms for Padovan and Perrin matrix sequences and present some properties of these transforms. By the results in Sections 2 and 3 of this paper, we have a great opportunity to compare and obtain some new properties over these transforms. This is the main aim of this paper. Thus, we extend some recent result in the literature. In the future studies on the iterated binomial transform for number sequences, we except that the following topics will bring a new insight.

(1) It would be interesting to study the iterated binomial transform for Fibonacci and Lucas matrix sequences, (2) Also, it would be interesting to study the iterated binomial transform for Pell and Pell-Lucas matrix sequences.

Acknowledgement 3.1 This research is supported by TUBITAK and Selcuk University Scientiﬁc Research Project Coordinatorship (BAP).

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