( ±1)-INVARIANT SEQUENCES AND TRUNCATED FIBONACCI SEQUENCES OF THE SECOND KIND
Gyoung-Sik Choi, Suk-Geun Hwang, and Ik-Pyo Kim
Abstract. In this paper we present another characterization of ( ±1)-invariant sequences. We also introduce truncated Fibonacci and Lucas sequences of the second kind and show that a sequence x ∈ R
∞is (−1)-invariant(1-invariant resp.) if and only if D ˆ
0x
˜ is perpendicular to every truncated Fibonacci(truncated Lucas resp.) sequence of the second kind where
D = diag(( −1)
0, ( −1)
1, ( −1)
2, . . .).
1. Introduction
Throughout this paper, let R ∞ denote the innite dimensional real vector space consisting of all real sequences (x 0 , x 1 , x 2 , . . .) T , and let P and D denote the Pascal matrix [ ( i
j
) ], (i, j = 0, 1, 2, . . .) and the diagonal matrix diag(( −1) 0 , ( −1) 1 , ( −1) 2 , . . .) respectively. We also assume that every matrix in this paper is assumed to have innitely many rows and columns numbered 0, 1, 2, . . . unless otherwise specied. The classical binomial inversion formula states that for x, y ∈ R ∞ , P Dx = y if and only if P Dy = x. Let F = (F 0 , F 1 , F 2 , . . .) T = (0, 1, 1, 2, 3, . . .) T and L = (L 0 , L 1 , L 2 , . . .) T = (2, 1, 3, 4, . . .) T be the Fibonacci sequence and the Lucas sequence respectively. Then P DF = −F and P DL = L.
Thus, as a linear transformation of R ∞ , P D has eigenvalues −1, 1. In fact,
−1 and 1 are the only eigenvalues of P D
because if λ ̸= ±1, then P Dx = λx has no nonzero solution. A sequence x ∈ R ∞ is called a 1-invariant sequence if P Dx = x, a ( −1)-invariant
Received March 30, 2004.
2000 Mathematics Subject Classication: Primary 11B39, 11B50.
Key words and phrases: ( ±1)-invariant sequence, truncated Fibonacci sequence of the second kind.
This research was supported by Kyungpook National University Research Team
Fund, 2003.
sequence if P Dx = −x ([1]). Associated with the Pascal matrix P , let P
→denote the matrix dened by
P
→=
1 0 0 0 0 0 0 0 · · · 0 1 1 0 0 0 0 0 · · · 0 0 1 2 1 0 0 0 · · · 0 0 0 1 3 3 1 0 · · · .. . .. . .. . .. . .. . .. . .. . .. . . ..
,
where the row i, (i = 0, 1, 2, . . .), is that of the Pascal matrix P preceded by 0 T i , the i-vector of zeros. Let
(1.1) Q = P +
[ 1 0 T 0 P ]
=
2 0 0 0 0 · · · 1 2 0 0 0 · · · 1 3 2 0 0 · · · 1 4 5 2 0 · · · 1 5 9 7 2 · · · .. . .. . .. . .. . .. . . ..
.
Associated with Q, let Q → denote the matrix dened by
Q
→=
2 0 0 0 0 0 0 0 0 · · · 0 1 2 0 0 0 0 0 0 · · · 0 0 1 3 2 0 0 0 0 · · · 0 0 0 1 4 5 2 0 0 · · · 0 0 0 0 1 5 9 7 2 · · · .. . .. . .. . .. . .. . .. . .. . .. . .. . . ..
,
where the row i is that of Q preceded by 0 T i , (i = 0, 1, 2, . . .). Then e T [0, P
→] = F and e T Q
→= L where e = (1, 1, . . .) T ∈ R ∞ . It is noted in [1] that
Q
→= P
→+
[ 1 0 0 T 0 0 P
→] .
For a matrix A whose row index set and column index set are J and K
respectively, and for J 0 ⊂ J, K 0 ⊂ K, let A(J 0 |K 0 ) denote the matrix
obtained from A by deleting rows in J 0 and columns in K 0 , and let
A[J 0 |K 0 ] denote the matrix A(J 0 |K 0 ) where J 0 = J − J 0 , K 0 = K − K 0 .
In [4], generating functions of ( ±1)-invariant sequences are investi-
gated. In [1], the authors found some characterization of ( ±1)-invariant
sequences in connection with the matrices P and Q. They also in-
troduced truncated Fibonacci and Lucas sequences and proved that a
sequence x ∈ R ∞ is ( −1)-invariant(1-invariant resp.) if and only if x
is expressible as a linear combination of truncated Fibonacci(truncated Lucas resp.) sequences.
In this paper we present another characterization of ( ±1)-invariant sequences. We also introduce truncated Fibonacci and Lucas sequences of the second kind and show that a sequence x ∈ R ∞ is ( −1)-invariant(1- invariant resp.) if and only if D [ 0
x
] is perpendicular to every truncated Fibonacci(truncated Lucas resp.) sequence of the second kind.
2. Relationship between ( ±1)-invariant sequences
In this section we rst investigate the relationship between ( ±1)- invariant sequences.
Lemma 2.1. [1] For a sequence x ∈ R ∞ , the following hold.
(a) x is ( −1)-invariant if and only if P
→Dx = 0, (b) x is 1-invariant if and only if Q
→(0 |0)Dx = 0.
The matrices P
→and Q
→are related as Q
→= P
→+
[ 1 0 0 T 0 0 P
→] .
Yet, there is another relation between P
→and Q
→. In what follows let
N denote the matrix diag(1, 2, 3, 4, . . .).
Lemma 2.2. The matrices P
→and Q
→are related as Q
→=
[ 2 0 T 0 −1 N
] P
→[ 1 0 T
0 N
] .
Proof. Let P = [β ij ], Q = [γ ij ], (i, j = 0, 1, 2, . . .). Then β ij = ( i
j
) and, by (1.1),
(2.1) γ ij = ( i
j )
+
( i − 1 j − 1 )
= i + j i
( i j
)
= i + j i β ij
for i, j ≥ 1. Thus we see that Q is obtained from P by multiplying the row i by 1 i , (i = 1, 2, . . .) and multiplying each of the entries in the line i + j = k by k, (k = 1, 2, . . .) and then multiplying the row 0 by 2. Let
P
→= [p 0 , p 1 , p 2 , . . .], Q
→= [q 0 , q 1 , q 2 , . . .].
Then
p k = (β 0,k , β 1,k −1 , . . . , β k −1,1 , β k,0 , 0, 0, . . .) T ,
q k = (γ 0,k , γ 1,k −1 , . . . , γ k −1,1 , γ k,0 , 0, 0, . . .) T ,
for k = 1, 2, . . ., from which we see that q k = (2kβ 0,k , kβ 1,k −1 , 1
2 kβ 2,k −2 , 1
3 kβ 3,k −3 , . . . , 1
k kβ k,0 , 0, 0, . . .) T , for k = 1, 2, . . ., by (2.1). Thus we have the equality in the Lemma.
The following theorem is a characterization of (±1)-invariant sequences which looks similar to but is dierent from Lemma 2 .1.
Theorem 2.3. Let x = (x 0 , x 1 , x 2 , . . .) T ∈ R ∞ . Then
(a) x is ( −1)-invariant if and only if Q
→(x 0 , −x 1 , 1 2 x 2 , − 1 3 x 3 , . . .) T = 0.
(b) x is 1-invariant if and only if
P
→(0 |0)(x 0 , −2x 1 , 3x 2 , −4x 3 , . . .) T = 0.
Proof. Since P
→Dx = 0 is equivalent to Q
→[ 1 0T
0
−1N] Dx = 0 by Lemma 2.2, (a) follows from Lemma 2.1(a). By Lemma 2.2, we also have Q
→(0 |0) = −1 N P
→(0 |0) N . So, Q
→(0 |0)Dx = 0 is equivalent to P
→(0|0) N Dx = 0, and (b) follows from Lemma 2.1(b).
We now present the relationship between the ( −1)-invariant sequences and the 1-invariant sequences.
Lemma 2.4. [1] Let (x 0 , x 1 , x 2 , . . .) T ∈ R ∞ and let s k = ∑ k
i=0 x i , (k = 0, 1, 2, . . .). Then (x 0 , x 1 , x 2 , . . .) T is λ-invariant if and only if (0, 0, s 0 , s 1 , s 2 , . . .) T is λ-invariant, for λ ∈ {1, −1}.
Theorem 2.5. Let x = (x 0 , x 1 , x 2 , . . .) T ∈ R ∞ . Then
(a) x is ( −1)-invariant if and only if x 0 = 0 and (x 1 , 1 2 x 2 , 1 3 x 3 , 1 4 x 4 , . . .) T is 1-invariant.
(b) x is 1-invariant if and only if (0, x 0 , 2x 1 , 3x 2 , . . .) T is ( −1)-invariant.
Proof. (a) Let y = (x 1 , x 2 , . . .) T so that x = [ x0
y
] and Dx = [ x0
−Dy ] . Suppose that x is (−1)-invariant. Then, by Lemma 2.1, P→Dx = 0,
i.e., [
1 0 T
0 P
→(0 |0) ] [ x 0
−Dy ]
= [ 0
0 ]
. Thus x 0 = 0 and
(2.2) P
→(0 |0)Dy = 0.
Now (2.2) gives us that
−1 N P→(0 |0) N −1
N Dy = 0, i.e., Q→(0 |0) −1 N Dy = 0.
So, by Lemma 2.1 again we see that −1 N y = (x 1 , 1 2 x 2 , 1 3 x 3 , 1 4 x 4 , . . .) T is 1-invariant. Reversing the above argument we can show that if −1 N y is 1-invariant, then x is ( −1)-invariant.
(b) Suppose that x is 1-invariant. Then, by Lemma 2.1, Q
→(0|0)Dx = 0 from which we get P
→(0 |0) N Dx = 0 so that
P
→[ 0
N Dx ]
=
[ 1 0 T 0 P
→(0 |0)
] [ 0
N Dx ]
= [ 0
0 ]
.
Since [
0
N Dx ]
= −D [ 0
N x ]
, we see, by Lemma 2.1, that [ 0
N
x
] = (0, x 0 , 2x 1 , 3x 2 , 4x 3 , . . .) T is ( −1)- invariant. Reversing the above argument, it can be shown that if [ 0
N