FIBONACCI AND LUCAS SEQUENCES AS THE

December 15, 2009 15:7 WSPC/171-JAA

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Journal of Algebra and Its Applications Vol. 8, No. 6 (2009) 869–883 c World Scientific Publishing Company

FIBONACCI AND LUCAS SEQUENCES AS THE PRINCIPAL MINORS OF SOME INFINITE MATRICES∗

A. R. MOGHADDAMFAR†,‡ , S. M. H. POOYA, S. NAVID SALEHY and S. NIMA SALEHY Department of Mathematics, Faculty of Science K. N. Toosi University of Technology P. O. Box 16315-1618, Tehran, Iran †[email protected] ‡[email protected] Received 23 September 2008 Accepted 8 May 2009 Communicated by D. Passman In the literature one may encounter certain infinite tridiagonal matrices, the principal minors of which, constitute the Fibonacci or Lucas sequence. The major purpose of this article is to find new infinite matrices with this property. It is interesting to mention that the matrices found are not tridiagonal which have been investigated before. Furthermore, we introduce the sequences composed of Fibonacci and Lucas k-numbers for the positive integer k and we construct the infinite matrices the principal minors of which generate these sequences. Keywords: Fibonacci and Lucas k-numbers; determinant; recursive relation; matrix factorization; generalized Pascal triangle. Mathematics Subject Classification 2000: 11B39, 15A36, 15A15, 11C20

1. Introduction The problem of symbolic evaluation of determinants has rightly received considerable attention in the literature (see [4] and [5] for an excellent survey). Let A = (ai,j )i,j≥0 be an infinite matrix and let dn be the nth principal minor of A consisting of the entries in its first n rows and columns. We will mainly be interested in the sequence of principal minors (d1 , d2 , d3 , . . .), especially, in the case that it forms the Fibonacci or Lucas sequence. The goal of this article is to find new infinite matrices the sequence of their principal minors is one of these two sequences. ∗ This

work has been supported by the Research Institute for Fundamental Sciences, Tabriz, Iran. Institute for Fundamental Sciences (RIFS), Tabriz, Iran.

‡ Research

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A. R. Moghaddamfar et al.

Given G(1) and G(2), a Gibonacci sequence (or generalized Fibonacci sequence) {G(n)}∞ n=1 is recursively defined by G(n + 2) = G(n + 1) + G(n), for n ≥ 1. Two important special cases are the classical Fibonacci sequence {F (n)}∞ n=1 with with L(1) = 1 and F (1) = 1 and F (2) = 1, and the Lucas sequence {L(n)}∞ n=1 L(2) = 3. Binet, the French mathematician of the 19th century, devised two noticeable analytical formulas for the Fibonacci and Lucas numbers (see [9]): √ n √ n 1+ 5 1− 5 1 − F (n) = √ 2 2 5 and

L(n) =

√ n √ n 1+ 5 1− 5 + . 2 2

Definition 1. A lower Hessenberg matrix, H(n) = (hi,j )1≤i,j≤n , is an n × n matrix where hi,j = 0 whenever j > i + 1 and hi,i+1 = 0 for some i, 1 ≤ i ≤ n − 1, so we have   h1,1 h1,2 0 ... 0   ..  h2,1 h2,2 h2,3 . 0      .. . H(n) =  h3,1 h3,2 h3,3 . 0      . .. .. ..  .. . . hn−1,n  . hn,1 hn,2 . . . hn,n−1 hn,n In fact, all entries above the superdiagonal are 0 but the matrix is not lower triangular. For convenience, we set H(∞) = (hi,j )i,j≥1 . We recall that a lower Hessenberg matrix is called tridiagonal if all of its nonzero entries appear only on the main diagonal, superdiagonal or subdiagonal. In [2], the authors present the following    2 −1 0 1 −1 0 ... 0    ..  1 . 2 −1 2 −1 0   1    . ,  1 1 2 1 2 .. 0   1  .  .. . . . .. . . . . . . −1  ...  .. . . 1

1

...

1

2

1

1

...

n × n Hessenberg matrices:   2 1 0 ... 0   .. 1 2 1 . 0     .. and 1 1 2  . 0  . . .  .. ..  .. ..  . −1 1

2

1 1

... .. . .. . .. . ... 1

 0  0   , 0   1 2

whose determinants are F (2n), F (2n + 1) and F (n + 2), respectively. Therefore, if H(n) is one of these matrices and dn = det(H(n)), then the sequence {dn }∞ n=1 is a subsequence of classical Fibonacci sequence. In [3], the authors construct families of tridiagonal matrices whose determinants generate any linear subsequence {F (ak + b)} or {L(ak + b)}, k = 1, 2, 3, . . . of the Fibonacci or Lucas sequence.


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Fibonacci and Lucas Sequences as the Principal Minors of Some Infinite Matrices

871

One may occasionally face with some infinite matrices having the same principal minors. Such matrices are called equimodular (see [8]) and they are formally defined as follows. Definition 2. Infinite matrices A = (ai,j )i,j≥1 and B = (bi,j )i,j≥1 are equimodular if the sequences of their principal minors coincide, i.e., det(ai,j )1≤i,j≤n = det(bi,j )1≤i,j≤n for all n ≥ 1. For instance, the following infinite tridiagonal  1 λ1 0 −λ−1 1 λ2  1 −1  0 1 −λ 2 Fλ (∞) =    0 0 −λ−1 3  .. .. .. . . .

matrices are equimodular:  0 0 ··· 0 0 · · ·  λ3 0 · · ·   .. . · · · 1  .. .. . . ···

(1)

where λ = (λi )i≥1 with λi ∈ C∗ = C\{0}. Indeed, the principal minors of these matrices for every λ form the sequence {F (n + 1)}∞ n=1 (see Theorem 1). In this article, we will try to find out other infinite matrices which are equimodular with the matrices presented in Eq. (1). In [1], Bacher considers the determinants of matrices associated with the Pascal’s triangle. Furthermore, he introduces the generalized Pascal triangles as follows. Let α = (αi )i≥1 and β = (βi )i≥1 be two sequences starting with a common first term α1 = β1 . Define a matrix Pα,β (n) of order n whose (i, j)-entry pi,j obey the following rule: pi,1 = αi ,

p1,i = βi

for 1 ≤ i ≤ n,

and pi,j = pi−1,j + pi,j−1

for 2 ≤ i, j ≤ n.

It is easy to verify that Pβ,α (n) = Pα,β (n)T , and so det(Pα,β (n)) = det(Pβ,α (n)). Now assume that α and β are two arithmetic sequences with a common first term α1 = β1 = 1 and common differences c = 0 and −c−1 , respectively. Therefore, the matrix Pα,β (4), for instance, is given by   1 1 − c−1 1 − 2c−1 1 − 3c−1  1 + c 2 + c − c−1 3 + c − 3c−1 4 + c − 6c−1  . Pα,β (4) =  −1 1 + 2c 3 + 3c − c−1 6 + 4c − 4c 10 + 5c − 10c−1  1 + 3c 4 + 6c − c−1 10 + 10c − 5c−1 20 + 15c − 15c−1 In Corollary 1, we prove that det(Pα,β (n)) = F (n + 1) for all n ≥ 1. Hence, the generalized Pascal triangle Pα,β (∞) and Fλ (∞) are equimodular. Other infinite matrices having the same property are given in Corollary 2(i). They are as follows:     1 0 1 0 1 ··· 1 0 1 0 1 ··· 2  1 1 1 1 · · · 1 1 1 1 · · ·    0      3 2 2 2 · · · −1 1 2 2 2 · · · 3  ,   4 6 5 4 4 · · · −2 0 3 4 4 · · ·.     5 10 11 9 8 · · · −3 −2 3 7 8 · · ·     .. .. .. .. .. . . .. .. .. .. .. . . . . . . . . . . . . . .


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Actually, the principal minors of these matrices form the Fibonacci sequence {F (n)}∞ n=1 . These matrices are constructed as follows. The first row is the 2-periodic sequence (1, 0, 1, 0, . . .) and the first column is the arithmetic sequence with the first term 1 and common difference 1 (resp. −1). The remaining entries are obtained by taking the sum of the entry directly above it, and the entry to its upper left. Another family of equimodular infinite matrices are the following: 

3 −λ−1  1  0 Lλ (∞) =    0  .. .

λ1 1 −λ−1 2

0 λ2 1

0 0 λ3

0 .. .

−λ−1 3 .. .

1 .. .

0 0 0 .. . ..

.

 ··· · · ·  · · ·   · · · 

(2)

···

where λ = (λi )i≥1 with λi ∈ C∗ = C\{0}. Here, the principal minors of these matrices form the subsequence {L(n + 1)}∞ n=1 of the Lucas sequence (see Theorem 2). Again, in Corollary 2(ii), we show that the principal minors of the following √ infinite matrices, where i = −1, form the Lucas sequence and so they are equimodular: 

1 2i 1 2i  1 + i 1 + 2i 1 + 2i 1 + 2i  1 + 2i 2 + 3i 2 + 4i 2 + 4i  1 + 3i 3 + 5i 4 + 7i 4 + 8i  .. .. .. .. . . . .

 ··· · · ·  · · · , · · ·  .. .



1 −2i 1 −2i  1 − i 1 − 2i 1 − 2i 1 − 2i  1 − 2i 2 − 3i 2 − 4i 2 − 4i  1 − 3i 3 − 5i 4 − 7i 4 − 8i  .. .. .. .. . . . .

 ··· · · ·  · · · . · · ·  .. .

The above matrices are constructed as follows. Here, the first row is the 2-periodic sequence (1, 2i, 1, 2i, . . .) (resp. (1, −2i, 1, −2i, . . .)) and the first column is the arithmetic sequence with the first term 1 and common difference i (resp. −i). The remaining entries are calculated by taking the sum of the entry directly above it, and the entry to its upper left. Our notation and terminology are standard. Given a matrix A, we denote by Ri (A) and Cj (A) the row i and the column j of A, respectively. We also denote by [j ,j ,...,j ]

A[i11,i22,...,ilk]

the submatrix of A obtained by erasing rows i1 , i2 , . . . , il and columns j1 , j2 , . . . , jk . In our next calculations, when m, n are integers with m > n, we assume that n i=m

f (i) = 0 and

n

f (i) = 1.

i=m

(Note that, in general, f is a function defined at the integers m, m + 1, . . . , n.) By x we denote the integer part of x, i.e., the greatest integer that is less than or equal to x.


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2. Preliminary Results Before presenting the main results, we prefer to state the results to be used later. As in introduction, let H(n) denote the lower Hessenberg matrix of order n. We set h(0) = 1 and h(n) = det(H(n)) for n ≥ 1. In [2], the authors proved the following result by simply calculating determinants by expanding along entries in the nth row. Proposition 1. The principal minors of H(∞), i.e., the sequence {h(n)}∞ n=0 satifies: h(0) = 1, h(1) = h1,1 and for n ≥ 2,   n−1 n−1 (−1)n−r hn,r h(n) = hn,n · h(n − 1) + hj,j+1 · h(r − 1). r=1

j=r

In [6], we proved the following proposition (see Theorem 2.1 in [6]). Proposition 2. Let α = (αi )i≥1 be a given sequence and let A = (ai,j )1≤i,j≤n be the doubly indexed sequence given by the recurrence ai,j = ai−1,j−1 + ai−1,j ,

2 ≤ i, j ≤ n,

and the initial conditions ai,1 = α1 + (i − 1)d, a1,j = αj , 1 ≤ i, j ≤ n. Then we have A = L · B, where L = (Li,j )1≤i,j≤n is a lower triangular matrix given by the recurrence Li,j = Li−1,j−1 + Li−1,j ,

2 ≤ i, j ≤ n,

and the initial conditions L1,1 = 1, L1,j = 0, 2 ≤ j ≤ n, and Li,1 = 1, 2 ≤ i ≤ n, and B = (Bi,j )1≤i,j≤n is a matrix given by the recurrence Bi,j = Bi−1,j−1 ,

2 ≤ i, j ≤ n,

and the initial conditions B1,j = αj , 1 ≤ j ≤ n, B2,1 = d and Bi,1 = 0, 3 ≤ i ≤ n. In particular, det(A) = det(B). 3. Main Results In [7], the author introduced the so-called Fibonacci k-numbers. For a given positive integer k the Fibonacci k-numbers are given by the following recurrence relation: Fk (n) = Fk (n − 1) + Fk (n − k − 1) with n > k + 1, and the initial terms Fk (1) = Fk (2) = · · · = Fk (k) = Fk (k + 1) = 1. Note that the above recurrence relation generates numerous numerical sequences. In particular, for k = 1, it generates the classical Fibonacci sequence. Let F1 (n) = F (n) for every n ≥ 1.


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As mentioned in the introduction, the sequence of the principal minors of infinite matrices of the form (1) is {F (n + 1)}∞ n=1 . Here, we construct the families of lower Hessenberg matrices the determinants of which generate the sequence {Fk (n + 1)}∞ n=1 composed of the Fibonacci k-numbers. Let α = (αi )i≥1 be an arbitrary sequence with αi = 0 for every i. Let k be a natural number and let HFk (∞) = (hi,j )i,j≥1 be an infinite lower Hessenberg matrix having entries  1 if j = i,       if j = i + 1, αi k hi,j = (−1)  if i = j + k,    αj αj+1 · · · αj+k−1    0 otherwise. We set DFk (n) = det(HFk (n)), where HFk (n) = (hi,j )1≤i,j≤n . Theorem 1. With the above notation, we have DFk (n) = Fk (n + 1), n ≥ 1. Proof. When n ≤ k, the matrix HFk (n) is an upper triangular matrix with 1’s on the diagonal, and so DFk (n) = 1. In other words DFk (1) = DFk (2) = · · · = DFk (k) = 1. On the other hand, if n = k + 1, then by expanding in terms of the nth row we have DFk (k + 1) = 2. Next, we assume that n ≥ k + 2. Now, by Proposition 1, we deduce that DFk (n) = DFk (n − 1) + DFk (n − k − 1). So DFk (n) = Fk (n + 1). Similarly we can define the Lucas k-numbers as follows. Given positive integer k, the Lucas k-numbers are given by the following recurrence relation: Lk (n) = Lk (n − 1) + Lk (n − k − 1) with

n > k + 1,

and the initial terms Lk (1) = Lk (2) = · · · = Lk (k) = 1,

and Lk (k + 1) = 3.

As before in the case that k = 1, we have the classical Lucas sequence, i.e., 1, 3, 4, 7, 11, . . . . Here, we set L1 (n) = L(n) for every n ≥ 1. As we stated in the Introduction, the sequence of the principal minors of infinite matrices of the form (2) generates the Lucas subsequence {L(n + 1)}∞ n=1 . Here, we are going to construct a new family of lower Hessenberg matrices the determinants of which generate the sequence {Lk (n + 1)}∞ n=1 consists of the Lucas k-numbers.


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Therefore, we assume that α = (αi )i≥1 is an arbitrary sequence with αi = 0 for every i. Suppose that HLk (∞) = (hi,j )i,j≥1 is an infinite lower Hessenberg matrix having entries   1 if i = j = k,      3 if i = j = k,     αi if j = i + 1, hi,j = k  (−1)    if i = j + k,   α α  j j+1 · · · αj+k−1   0 otherwise. We set DLk (n) = det(HLk (n)), where HLk (n) = (hi,j )1≤i,j≤n . Theorem 2. With the above notation, we have DLk (n) = Lk (n + 1), n ≥ 1. Proof. The proof is similar to the proof of Theorem 1. Now, let us consider another theorem. Theorem 3. Let α = (αi )i≥1 be an arithmetic sequence with common difference d and let β = (βi )i≥1 be an arbitrary sequence with β1 = α1 . Then we have Pα,β (n) = L · B, where L = (Li,j )1≤i,j≤n is a lower triangular matrix given by the recurrence 2 ≤ i, j ≤ n,

Li,j = Li−1,j−1 + Li−1,j ,

(3)

and the initial conditions L1,j = 0, 2 ≤ j ≤ n, and Li,1 = 1, 1 ≤ i ≤ n, and B = (Bi,j )1≤i,j≤n is a matrix given by the recurrence Bi,j = Bi,j−1 + Bi−1,j−1 ,

2 ≤ i, j ≤ n,

(4)

and the initial conditions B1,j = βj , 1 ≤ j ≤ n, B2,1 = d and Bi,1 = 0, 3 ≤ i ≤ n. In particular, det(Pα,β (n)) = det(B). Proof. For the proof of the claimed factorization we compute the (i, j)−entry of L · B, that is (L · B)i,j =

n

Li,k Bk,j .

k=1

In fact, so as to prove the theorem, we should establish R1 (L · B) = R1 (Pα,β (n)) = (β1 , β2 , . . . , βn ), C1 (L · B) = C1 (Pα,β (n)) = (β1 , β1 + d, . . . , β1 + (n − 1)d),


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and (L · B)i,j = (L · B)i,j−1 + (L · B)i−1,j ,

(5)

for 2 ≤ i, j ≤ n. First, suppose that i = 1. Then (L · B)1,j =

n

L1,k Bk,j = L1,1 B1,j = βj ,

k=1

and so R1 (L · B) = R1 (Pα,β (n)) = (β1 , β2 , . . . , βn ). Next, we suppose that j = 1, and we obtain (L · B)i,1 =

n

Li,k Bk,1 = Li,1 B1,1 + Li,2 B2,1 = β1 + (i − 1)d,

k=1

which implies C1 (L · B) = C1 (Pα,β (n)) = (β1 , β1 + d, . . . , β1 + (n − 1)d). Finally, we must establish Eq. (5). At the moment, let us assume that 2 ≤ i, j ≤ n. In this case we have (L · B)i,j =

n

Li,k Bk,j

k=1

= Li,1 B1,j +

n

Li,k Bk,j

k=2

= Li,1 B1,j +

n

(Li−1,k−1 + Li−1,k )Bk,j

(by Eq. (3))

k=2

= Li,1 B1,j +

n

Li−1,k−1 Bk,j +

k=2

= Li,1 B1,j +

n

n

Li−1,k Bk,j

k=2

Li−1,k−1 Bk,j +

k=2

= (Li,1 − Li−1,1 )B1,j +

n

Li−1,k Bk,j − Li−1,1 B1,j

k=1 n

Li−1,k−1 Bk,j +

k=2

=

n

=

n k=2

Li−1,k Bk,j

k=1

Li−1,k−1 [Bk,j−1 + Bk−1,j−1 ] +

k=2

n

n

Li−1,k Bk,j

(by (4))

k=1

Li−1,k−1 Bk,j−1 +

n k=2

Li−1,k−1 Bk−1,j−1 +

n k=1

Li−1,k Bk,j


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=

n

(Li,k − Li−1,k )Bk,j−1 +

k=2

+

n

877

Li−1,k−1 Bk−1,j−1

k=2

n

Li−1,k Bk,j

(by Eq. (3))

k=1

=

n

(Li,k − Li−1,k )Bk,j−1 − (Li,1 − Li−1,1 )B1,j

k=1

+

n

Li−1,k Bk,j−1 +

k=1

=

n k=1

n

Li−1,k Bk,j

(notice Li−1,n = 0),

k=1

Li,k Bk,j−1 +

n

Li−1,k Bk,j

k=1

= (L · B)i,j−1 + (L · B)i−1,j , which is Eq. (5). Our proof is thus complete. In the following corollary we study the determinants of generalized Pascal triangles associated with two arithmetic sequences with a common first term. Corollary 1. Let α = (αi )i≥1 and β = (βi )i≥1 be two arithmetic sequences with a common first term α1 = β1 = a and common differences c and d, respectively. Then n+1 n+1 a+ω a−ω 1 − , det(Pα,β (n)) = ω 2 2 where ω =

√ a2 − 4cd. In particular, in the case that a = 1 and d = −c−1 , then det(Pα,β (n)) = F (n + 1).

Proof. Using Theorem 3, we deduce that det(Pα,β (n)) = det(B), where B = (Bi,j )1≤i,j≤n is a matrix given by the recurrence Bi,j = Bi,j−1 + Bi−1,j−1 ,

2 ≤ i, j ≤ n,

and the initial conditions B1,j = βj , 1 ≤ j ≤ n, B2,1 = c and Bi,1 = 0, 3 ≤ i ≤ n. Now, we claim that ˜ · U, B=B

(6)

where U = (Ui,j )1≤i,j≤n is an upper triangular matrix by the recurrence Ui,j = Ui−1,j−1 + Ui,j−1 ,

i, j ≥ 2,

(7)


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˜ = and the initial conditions U1,j = 1, j ≥ 1, and Ui,1 = 0, i ≥ 2, and where B ˜ (Bi,j )1≤i,j≤n is a matrix given by the recurrence ˜i,j = B ˜i−1,j−1 , B

2 ≤ i, j ≤ n,

(8)

˜1,2 = d, B ˜2,1 = c, B ˜1,j = B ˜i,1 = 0, 3 ≤ i, j ≤ n. ˜1,1 = a, B and the initial conditions B ˜ is a tridiagonal matrix where all of the terms on the main diagonal are In fact, B a, the terms on the superdiagonal are d and all terms on subdiagonal are c. For the proof of the claimed factorization we need some calculations. In fact, the (i, j)−entry ˜ · U is of B ˜ · U )i,j = (B

n

˜i,k Uk,j . B

k=1

˜ · U )1,j = βj , for Therefore, in order to prove the result it is enough to show that (B ˜ · U )i,1 = 0 for i ≥ 3 and ˜ · U )2,1 = c, (B j ≥ 1, (B ˜ · U )i,j−1 + (B ˜ · U )i−1,j−1 , ˜ · U )i,j = (B (B for 2 ≤ i, j ≤ n. First, suppose that i = 1. Then ˜ · U )1,j = (B

n

˜1,k Uk,j = B ˜1,1 U1,j + B ˜1,2 U2,j = a.1 + d.(j − 1) = βj . B

k=1

Next, suppose that j = 1. In this case we obtain ˜ · U )i,1 = (B

n

˜i,k Uk,1 = B ˜i,1 U1,1 = B

k=1

c

if i = 2,

0

if i ≥ 3.

Finally, we assume that 2 ≤ i, j ≤ n. Here, we have ˜ · U )i,j = (B

n

˜i,k Uk,j B

k=1

˜i,1 U1,j + =B ˜i,1 U1,j + =B ˜i,1 U1,j + =B ˜i,1 U1,j + =B

n k=2 n k=2 n k=2 n k=2

˜i,k Uk,j B ˜i,k (Uk−1,j−1 + Uk,j−1 ) (by (7)) B ˜i,k Uk−1,j−1 + B ˜i,k Uk−1,j−1 + B

n k=2 n k=1

˜i,k Uk,j−1 B ˜i,k Uk,j−1 − B ˜i,1 U1,j−1 B

(9)


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˜i,1 (U1,j − U1,j−1 ) + =B

n

˜i,k Uk−1,j−1 + B

k=2

=

=

n k=2 n

˜i−1,k−1 Uk−1,j−1 + B

n

879

˜i,k Uk,j−1 B

k=1

n

˜i,k Uk,j−1 , B

(by (8))

k=1

˜i−1,k Uk,j−1 + B

k=1

n

˜i,k Uk,j−1 , B

(notice Un,j−1 = 0)

k=1

˜ · U )i−1,j−1 + (B ˜ · U )i,j−1 , = (B which is Eq. (9), and the proof of our claim (6) is completed. ˜ To obtain det(B) ˜ we put By Eq. (6), it follows that det(Pα,β (n)) = det(B). ˜ Then by Proposition 1, we easily see that D(n) := det(B). D(1) = a,

D(2) = a2 − cd,

and D(n) = aD(n − 1) − cdD(n − 2) for n ≥ 3.

To solve this recurrence relation, we see that its characteristic equation as follows x2 − ax + cd = 0, and its characteristic roots are x1 = where ω =

a−ω 2

a+ω , 2

and x2 =

√ a2 − 4cd. Therefore, the general solution is given by D(n) = λ1

a−ω 2

n + λ2

a+ω 2

n ,

where λ1 and λ2 are constants. The initial conditions D(1) = a and D(2) = a2 − cd 1 a+ω imply that λ1 = − ω1 ( a−ω 2 ) and λ2 = ω ( 2 ), and so 1 D(n) = ω

a+ω 2

n+1 −

a−ω 2

n+1 .

This completes the proof of the corollary. Theorem 4. Let α = (a, a, . . . , a, b, b, . . . , b, . . .) be a 2k-periodic sequence and let k−times

k−times

(ai,j )i,j≥1 be the doubly indexed sequence given by the recurrence ai,j = ai−1,j−1 + ai−1,j ,

i, j ≥ 2,


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and the initial conditions ai,1 = a + (i − 1)c, a1,j = αj , 1 ≤ i, j ≤ n. We set D(n) = det(ai,j )1≤i,j≤n . Then we have D(n) = (−1)k+1 ck [(c − b)D(n − k − 1) + D(n − k)] + (a − c)D(n − 1),

n > k + 1.

Proof. Using Proposition 2, D(n) = det(B) where B = (Bi,j )1≤i,j≤n is a matrix given by the recurrence Bi,j = Bi−1,j−1 ,

2 ≤ i, j ≤ n,

and the initial conditions B1,j = αj , 1 ≤ j ≤ n, B2,1 = c and Bi,1 = 0, 3 ≤ i ≤ n. To obtain the result we thus need to compute det(B). Calculating the determinant D(n) = det(B) by expanding along the first column we obtain [1]

D(n) = aD(n − 1) − c · det(B[2] ). [1]

Now, we can consider the first row of B[2] as follows [1]

R1 (B[2] ) =

a, a, . . . , a , (b − a) + a, b, b, . . . , b , (a − b) + b, . . . ,

(k−1)−times

(k−1)−times

which is a 2k-periodic sequence, and hence it follows that [1] ˜ D(n) = aD(n − 1) − c · det(B[2] ) = aD(n − 1) − c · (D(n − 1) + det(B)),

(10)

˜ is a matrix of order n − 1 with the first row: where B ˜ = R1 (B)

0, 0, . . . , 0 , b − a, 0, 0, . . . , 0 , a − b, . . .

(k−1)−times

(k−1)−times

˜[1] = B [1] . Again, if we expand along the which is a 2k-periodic sequence, and B [1,2] ˜ then we get first row of B, n−1 k

˜ = det(B)

(−1)ik+1 (b − a)(−1)i−1 cik−1 D(n − ik − 1)

i=1 n−1 k

= (b − a)

(−1)i(k+1) cik−1 D(n − ik − 1).

i=1

Now, if this is substituted in Eq. (10), then we obtain n−1 k

D(n) = (a − c)D(n − 1) − c · (b − a)

i=1

(−1)i(k+1) cik−1 D(n − ik − 1).

(11)


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A similar argument, as mentioned above, with n − k instead of n shows that D(n − k) = (a − c)D(n − k − 1) n−k−1 k

− c · (b − a)

(−1)i(k+1) cik−1 D(n − k − ik − 1)

i=1

= (a − c)D(n − k − 1) n−1 k −1

− c · (b − a)

(−1)i(k+1) cik−1 D(n − k(i + 1) − 1).

(12)

i=1

Multiplying both sides of Eq. (12) by (−1)k+1 ck , we get (−1)k+1 ck D(n − k) = (−1)k+1 ck (a − c)D(n − k − 1) − c · (b − a) n−1 k −1

×

(−1)(i+1)(k+1) c(i+1)k−1 D(n − k(i + 1) − 1)

i=1 k+1 k

= (−1)

c (a − c)D(n − k − 1) n−1 k

− c · (b − a)

(−1)i(k+1) cik−1 D(n − ik − 1).

(13)

i=2

Therefore, it follows from Eqs. (11) and (13) that D(n) + (−c)k D(n − k) = a[D(n − 1) + (−c)k D(n − k − 1)] − c[D(n − 1) + (−c)k D(n − k − 1) + (b − a)(−c)k−1 D(n − k − 1)] = (a − c)D(n − 1) + (−c)k (b − c)D(n − k − 1). After some simplification this leads to D(n) = (−1)k+1 ck [(c − b)D(n − k − 1) + D(n − k)] + (a − c)D(n − 1), as desired. As a consequence of Theorem 4 we can deduce the following corollary. Corollary 2. In Theorem 4 we have: (i) D(n) = F (n) if and only if a = k = 1, b = 0 and c = ±1. In particular, in these cases the infinite matrices (ai,j )i,j≥1 and (bi,j )i,j≥1 , with  if j = i,  1   λ if j = i + 1, i > 1 i−1 bi,j = −1  −λj−1 if i = j + 1, j > 1     0 otherwise, where λk ∈ C∗ , (k = 1, 2, 3, . . .), are equimodular.


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√ √ (ii) D(n) = L(n) if and only if a = k = 1 and (b, c) ∈ {(2 −1, −1), √ √ (−2 −1, − −1)}. In particular, in these cases the infinite matrices (ai,j )i,j≥1 and (ci,j )i,j≥1 , with  3 if j = i = 2,      if j = i = 2,  1 ci,j = λi−1 if j = i + 1, i > 1   −1   −λj−1 if i = j + 1, j > 1    0 otherwise, where λk ∈ C∗ , (k = 1, 2, 3, . . .), are equimodular. Proof. (i) It is enough to prove the necessity. Let D(n) = F (n). Hence, D(n) is a linear homogeneous recurrence relation of order 2 with constant coefficients 1, and by Theorem 4, we must have c = 0. Clearly a = D(1) = F (1) = 1. If k ≥ 2, then 1 − c = D(2) = F (2) = 1 which implies that c = 0, a contradiction. Therefore k = 1, and we obtain 1 − bc = D(2) = F (2) = 1, which implies that bc = 0, and so b = 0. Now, according to Theorem 4, we obtain D(n) = c2 D(n − 2) + D(n − 1),

n > 2.

But, we must have c2 = 1, which implies that c = ±1. Now, assume that a = k = 1, b = 0, c = ±1. Let A(n) and B(n) denote the matrices (ai,j )1≤i,j≤n and (bi,j )1≤i,j≤n , respectively. By the previous part, it follows that det(A(n)) = F (n). On the other hand, we have B(n) = [1] ⊕ Fλ (n − 1). Now, by Theorem 1 (for k = 1), we conclude that det(B(n)) = det(Fλ (n − 1)) = F (n), which completes the proof of part (i). (ii) Again, it is enough to prove the necessity. Let D(n) = L(n). Then D(n) is a linear homogeneous recurrence relation of order 2 with constant coefficients 1. It is easy to see that a = D(1) = L(1) = 1. Now, we claim that k = 1. Assume, on the contrary, that k ≥ 2. From 1 − c = D(2) = L(2) = 3 we deduce that c = −2. If k ≥ 3, then we obtain 9 = D(3) = L(3), which is a contradiction. Therefore, we assume that k = 2. In this case, we obtain 5 + 4b = D(3) = L(3) = 4, which implies that b = −1/4. But then 10 = D(4) = L(4), a contradiction. Therefore k = 1, and we obtain 1 − bc = D(2) = L(2) = 3, which implies that bc = −2. Now, from Theorem 4, we obtain D(n) = c(c − b)D(n − 2) + D(n − 1),

n > 2.

√ But, we must have c(c − b) = 1, which implies that c2 = −1, and so c = ± −1 and √ b = ±2 −1.


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√ √ √ √ Now, assume that a = k = 1, (b, c) ∈ {(2 −1, −1), (−2 −1, − −1)}. Let A(n) and C(n) denote the matrices (ai,j )1≤i,j≤n and (ci,j )1≤i,j≤n , respectively. Then, by the previous paragraph, we obtain det(A(n)) = L(n). Moreover, since C(n) = [1] ⊕ Lλ (n − 1), by Theorem 2 (for k = 1), it yields that det(C(n)) = det(Lλ (n − 1)) = L(n). This completes the proof of part (ii) and the theorem. Acknowledgments We are very grateful to the referee for careful reading and helpful comments. Our special thanks go to the Research Institute for Fundamental Sciences, Tabriz, Iran, to sponsor this research. References [1] R. Bacher, Determinants of matrices related to the Pascal triangle, J. Théor. Nombres Bordeaux 14(1) (2002) 19–41. [2] N. D. Cahill, J. R. D’Errico, D. A. Narayan and J. Y. Narayan, Fibonacci determinants, College Math. J. 33(3) (2002) 221–225. [3] N. D. Cahill and D. A. Narayan, Fibonacci and Lucas numbers as tridiagonal matrix determinants, Fibonacci Quart. 42(3) (2004) 216–221. [4] C. Krattenthaler, Advanced determinant calculus, Sém. Lothar. Combin. 42 (1999) Art. B42q. [5] C. Krattenthaler, Advanced determinant calculus: a complement, Linear Algebra Appl. 411 (2005) 68–166. [6] A. R. Moghaddamfar, S. M. H. Pooya, S. Navid Salehy and S. Nima Salehy, More calculations on determinant evaluations, Electron. J. Linear Algebra 16 (2007) 19–29. [7] A. P. Stakhov, Introduction into algorithmic measurement theory, Moscow: Soviet Radio (1977) [in Russian]. [8] H. Zakrajˇsek and M. Petkovˇsek, Pascal-like determinants are recursive, Adv. Appl. Math. 33(3) (2004) 431–450. [9] S. Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications (John Wiley and Sons, New York, 1989).