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1 Volume 18, pp , August GENERALIZED PASCAL TRIANGLES AND TOEPLITZ MATRICES A R MOGHADDAMFAR AND S M H POOYA Abstract The purpose of this article is to study determinants of matrices which are known as generalized Pascal triangles (see R Bacher Determinants of matrices related to the Pascal triangle J Théor Nombres Bordeaux, 14:19 41, 2002 This article presents a factorization by expressing such a matrix as a product of a unipotent lower triangular matrix, a Toeplitz matrix, and a unipotent upper triangular matrix The determinant of a generalized Pascal matrix equals thus the determinant of a Toeplitz matrix This equality allows for the evaluation of a few determinants of generalized Pascal matrices associated with certain sequences In particular, families of quasi-pascal matrices are obtained whose leading principal minors generate any arbitrary linear subsequences (F nr+s n 1 or (L nr+s n 1 of the Fibonacci or Lucas sequence New matrices are constructed whose entries are given by certain linear non-homogeneous recurrence relations, and the leading principal minors of which form the Fibonacci sequence Key words Determinant, Matrix factorization, Generalized Pascal triangle, Generalized symmetric (skymmetric Pascal triangle, Toeplitz matrix, Recursive relation, Fibonacci (Lucas, Catalan sequence, Golden ratio AMS subject classifications 15A15, 11C20 1 Introduction Let P ( be the infinite symmetric matrix with entries P i,j = ( i+j i for i, j 0 The matrix P ( is hence the famous Pascal triangle yielding the binomial coefficients The entries of P ( satisfy the recurrence relation P i,j = P i 1,j + P i,j 1 Indeed, this matrix has the following form: P ( = Received by the editors January 19, 2009 Accepted for publication August 1, 2009 Handling Editor: Michael J Tsatsomeros Department of Mathematics, Faculty of Science, K N Toosi University of Technology, P O Box , Tehran, Iran (moghadam@kntuacir School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics (IPM, PO Box , Tehran, Iran (moghadam@mailipmir This research was in part supported by a grant from IPM (No Department of Mathematics, Faculty of Science, K N Toosi University of Technology, P O Box , Tehran, Iran 564

2 Volume 18, pp , August One can easily verify that (see [2, 6]: Generalized Pascal Triangles and Toeplitz Matrices 565 (11 P ( =L( L( t, where L( is the infinite unipotent lower triangular matrix (12 L( =(L i,j i,j 0 = with entries L i,j = ( i j WedenotebyL(n the finite submatrix of L( withentries L i,j,0 i, j n 1 To introduce our result, we first present some notation and definitions We recall that a matrix T ( =(t i,j i,j 0 is said to be Toeplitz if t i,j = t k,l whenever i j = k l Let α =(α i i 0 and β =(β i i 0 be two sequences with α 0 = β 0 We shall denote by T α,β ( =(t i,j i,j 0 the Toeplitz matrix with t i,0 = α i and t 0,j = β j We also denote by T α,β (n the submatrix of T α,β ( consisting of the entries in its first n rows and columns We come now back to the (11 In fact, one can rewrite it as follows: P ( =L( I L( t, where matrix I (identity matrix is a particular case of a Toeplitz matrix In [1], Bacher considers determinants of matrices generalizing the Pascal triangle P ( He introduces generalized Pascal triangles as follows Let α =(α i i 0 and β = (β i i 0 be two sequences starting with a common first term α 0 = β 0 = γ Then, the generalized Pascal triangle associated with α and β, is the infinite matrix P α,β ( =(P i,j i,j 0 with entries P i,0 = α i, P 0,j = β j (i, j 0 and P i,j = P i 1,j + P i,j 1, for i, j 1 We denote by P α,β (n the finite submatrix of P α,β ( withentriesp i,j,0 i, j n 1 An explicit formula for entry P i,j of P α,β (n is also given by the following formula (see [1]: ( i + j ( i ( i + j s ( j ( i + j t P i,j = γ + (α s α s 1 + (β t β t 1 j j i s=1 t=1

3 Volume 18, pp , August ARMoghaddamfarandSMHPooya For an arbitrary sequence α =(α i i 0, we define the sequences ˆα =(ˆα i i 0 and ˇα =(ˇα i i 0 as follows: (13 ˆα i = i ( i ( 1 i+k α k and ˇα i = k k=0 i k=0 ( i α k k With these definitions we can now state our main result Indeed, the purpose of this article is to obtain a factorization of the generalized Pascal triangle P α,β (n associated with the arbitrary sequences α and β, as a product of a unipotent lower triangular matrix L(n, a Toeplitz matrix T ˆα, ˆβ(n and a unipotent upper triangular matrix U(n (see Theorem 31, that is Similarly, we show that P α,β (n =L(n T ˆα, ˆβ(n U(n T α,β (n =L(n 1 Pˇα, ˇβ(n U(n 1 In fact, we obtain a connection between generalized Pascal triangles and Toeplitz matrices In view of these factorizations, we can easily see that det(p α,β (n = det(t ˆα, ˆβ(n Finally, we present several applications of Theorem 31 to some other determinant evaluations We conclude the introduction with notation and terminology to be used throughout the article For convenience, we will let 1 denote the complex number i C By x we denote the integer part of x, ie, the greatest integer that is less than or equal to x Wealsodenoteby x the smallest integer greater than or equal to x Given a matrix A, wedenotebyr i (A andc j (A therowi and the column j of A, respectively We use the notation A t for the transpose of A In general, an n n matrix of the following form: [ ] [ A B ( A B resp C P α,β (n k C T α,β (n k ] where A, B and C are arbitrary matrices of order k k, k (n k and(n k k, respectively, is called a quasi-pascal (resp quasi-toeplitz matrix Throughout this article we assume that:

4 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 567 F =(F i i 0 =(0, 1, 1, 2, 3, 5, 8,,F i, F =(F i i 1 =(1, 1, 2, 3, 5, 8,,F i, L =(L i i 0 =(2, 1, 3, 4, 7, 11, 18,,L i, C =(C i i 0 =(1, 1, 2, 5, 14, 42, 132,,C i, I =(I i i 0 =(0!, 1!, 2!, 3!, 4!, 5!, 6!,,I i = i!, I =(I i i 1 =(1!, 2!, 3!, 4!, 5!, 6!,,I i = i!, (Fibonacci numbers, (Fibonacci numbers 0, (Lucas numbers, (Catalan numbers, The rest of this article is organized as follows In Section 2, we derive some preparatory results In Section 3, we prove the main result (Theorem 31 Section 4 deals with applications of Theorem 31 In Section 5, we present two matrices whose entries are recursively defined, and we show that the leading principal minor sequence of these matrices is the sequence F =(F i i 1 2 Preliminary Results As we mentioned in the Introduction, if α =(α i i 0 is an arbitrary sequence, then we define the sequences ˆα =(ˆα i i 0 and ˇα =(ˇα i i 0 as in (13 For certain sequences α the associated sequences ˆα and ˇα seem also to be of interest since they have appeared elsewhere In Tables 1 and 2, we have presented some sequences α and the associated sequences ˆα and ˇα Table 1 Some sequences α and associated sequences ˇα α Reference ˇα (0, 1, 1, 2, 3, 5, 8, A in [8] F (1, 0, 1, 1, 2, 3, 5, 8, A in [8] F (2, 1, 3, 4, 7, 11, 18, A in [8] L (1, 0, 1, 1, 3, 6, 15, A in [8] C (1, 0, 1, 2, 9, 44, 265, A in [8] I (1, 1, 3, 11, 53, 309, 2119, A in [8] I Table 2 Some sequences α and associated sequences ˆα α Reference ˆα (0, 1, 3, 8, 21, 55, 144, A in [8] F (1, 2, 5, 13, 34, 89, 233, A in [8] F (2, 3, 7, 18, 47, 123, 322, A in [8] L (1, 2, 5, 15, 51, 188, 731, A in [8] C (1, 2, 5, 16, 65, 326, 1957, A in [8] I (1, 3, 11, 49, 261, 1631, A in [8] I

5 Volume 18, pp , August ARMoghaddamfarandSMHPooya As another nice example, consider α =R i (L( where L is introduced as in (12 Then, we have ˇα =R i (P ( Lemma 21 Let α be an arbitrary sequence Then, we have ˆˇα = ˇˆα = α Proof Suppose α =(α i i 0 and ˆˇα =(ˆˇα i i 0 Then, we have ˆˇα i = i k=0 ( 1i+k( i kˇαk = i k=0 ( 1i+k( i k ( k k s=0 s αs = i k k=0 s=0 ( 1i+k( i k k( s αs = ( 1 i i l=0 α i l h=l ( 1h( i h h( l = ( 1 i i l=0 α i l h=l ( 1h( ( i i l l h l = ( 1 i i l=0 α i i l( l h=l ( 1h( i l h l But, if l<i,thenwehave i h=l ( 1h( i l h l = 0 Therefore, we obtain and hence ˆˇα = α ˆˇα i =( 1 i i ( i ( i i l α l ( 1 h l h l l=i h=l The proof of second part is similar to the previous case = α i, Lemma 22 Let i, j be positive integers Then, we have i j ( ( { i k + j 0 if i j, ( 1 k = k + j j 1 if i = j k=0 The proof follows from the easy identity ( ( i k + j = k + j j ( i j ( i j k The following Lemma is a special case of a general result due to Krattenthaler (see Theorem 1 in [9] Lemma 23 Let α =(α i i 0 and β =(β j j 0 be two geometric sequences with α i = ρ i and β j = σ j Then, we have det(p α,β (n = (ρ + σ ρσ

6 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices Main Result Now, we are in the position to state and prove the main result of this article Theorem 31 Let α =(α i i 0 and β =(β i i 0 be two sequences starting with a common first term α 0 = β 0 = γ Then, we have (31 P α,β (n =L(n T ˆα, ˆβ(n U(n, and (32 T α,β (n =L(n 1 Pˇα, ˇβ(n U(n 1, where L(n =(L i,j 0 i,j<n is a lower triangular matrix with L i,j = { 0 if i<j ( i j if i j, and U(n =L(n t In particular, we have det(p α,β (n = det(t ˆα, ˆβ(n Proof First, we claim that P α,β (n =L(n Q(n, where L(n =(L i,j 0 i,j<n is a lower triangular matrix with L i,j = { 0 if i<j ( i j if i j, and Q(n =(Q i,j 0 i,j<n with Q i,0 =ˆα i, Q 0,i = β i and (33 Q i,j = Q i 1,j 1 + Q i,j 1, 1 i, j < n For instance, when n = 4 the matrices L(4 and Q(4 are given by: and Q(4 = L(4 = , γ β 1 β 2 β 3 γ + α 1 α 1 β 1 + α 1 β 1 + β 2 + α 2 γ 2α 1 + α 2 α 1 + α 2 α 2 β 1 + α 1 + α 2 γ +3α 1 3α 2 + α 3 α 1 2α 2 + α 3 α 2 + α 3 α 3

7 Volume 18, pp , August ARMoghaddamfarandSMHPooya Note that the entries of L(n satisfying in the following recurrence (34 L i,j = L i 1,j 1 + L i 1,j, 1 i, j < n For the proof of the claimed factorization we compute the (i, j-th entry of L(n Q(n, that is n (L(n Q(n i,j = L i,k Q k,j k=1 In fact, it suffices to show that R 0 (L(n Q(n = R 0 (P α,β (n, C 0 (L(n Q(n = C 0 (P α,β (n and (35 (L(n Q(n i,j =(L(n Q(n i,j 1 +(L(n Q(n i 1,j, for 1 i, j < n First, suppose that i = 0 Then, we obtain (L(n Q(n 0,j = L 0,k Q k,j = L 0,0 Q 0,j = β j, k=0 and so R 0 (L(n Q(n = R 0 (P α,β (n = (β 0, β 1,, β Next, suppose that i 1andj = 0 In this case, we have (L(n Q(n i,0 = = i k=0 k=0 L i,kq k,0 { (i k = i j=0 β j k l=0 ( 1l+k( } k l βl { i j t=0 ( 1t( ( i t+j } t+j j = β i, (by Lemma 22 and so C 0 (L(n Q(n = C 0 (P α,β (n = (α 0, α 1,, α t

8 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 571 Finally, we must establish (35 Therefore, we assume that 1 i, j < n Inthis case, we have (L(n Q(n i,j = k=0 L i,kq k,j which is (35 Next, we claim that = L i,0 Q 0,j + k=1 L i,kq k,j = L i,0 Q 0,j + k=1 L i,k(q k 1,j 1 + Q k,j 1 (by (33 = L i,0 Q 0,j + k=1 L i,kq k 1,j 1 + k=1 L i,kq k,j 1 = L i,0 Q 0,j + k=1 (L i 1,k 1 + L i 1,k Q k 1,j 1 + k=0 L i,kq k,j 1 L i,0 Q 0,j 1 (by (34 = L i,0 Q 0,j + k=1 L i 1,k 1Q k 1,j 1 + k=1 L i 1,k(Q k,j Q k,j 1 +(L(n Q(n i,j 1 L i,0 Q 0,j 1 (by (33 = L i,0 Q 0,j + n 2 k=0 L i 1,kQ k,j 1 + k=1 L i 1,kQ k,j k=0 L i 1,kQ k,j 1 + L i 1,0 Q 0,j 1 +(L(n Q(n i,j 1 L i,0 Q 0,j 1 = L i,0 Q 0,j + k=0 L i 1,kQ k,j 1 + k=0 L i 1,kQ k,j L i 1,0 Q 0,j k=0 L i 1,kQ k,j 1 + L i 1,0 Q 0,j 1 +(L(n Q(n i,j 1 L i,0 Q 0,j 1 (note that L i 1, =0 = (L i,0 L i 1,0 Q 0,j +(L(n Q(n i 1,j +(L i 1,0 L i,0 Q 0,j 1 +(L(n Q(n i,j 1 = (L(n Q(n i 1,j +(L(n Q(n i,j 1, (note that L i,0 = L i 1,0 =1 Q(n =T ˆα, ˆβ(n U(n, where U(n =L(n t and T α,β (n =(T i,j 0 i,j<n with T i,0 =ˆα i, T 0,j = ˆβ j,and Note that, we have T i,j = T i 1,j 1, 1 i, j < n (36 U i,j = U i 1,j 1 + U i,j 1, 1 i, j < n

9 Volume 18, pp , August ARMoghaddamfarandSMHPooya For instance, when n = 4 the matrices T ˆα, ˆβ(4 and U(4 are given by: 2 γ γ + β 1 γ 2β 1 + β 2 γ +3β 1 3β 2 + β 3 Tˆα, ˆβ(4 γ + α 1 γ γ + β 1 γ 2β 1 + β 2 = 6 4 γ 2α 1 + α 2 γ + α 1 γ γ + β 1 γ +3α 1 3α 2 + α 3 γ 2α 1 + α 2 γ + α 1 γ and U(4 = As before, the proof of the claim requires some calculations If we have i =0, then (T ˆα, ˆβ(n U(n 0,j = = j k=0 = j i=0 β i k=0 T 0,kU k,j {( k (j l=0 ( 1l+k( k l } βl k { j i t=0 ( 1t( ( j t+i } t+i i = β j, (by Lemma 22 which implies that R 0 (T ˆα, ˆβ(n U(n = R 0 (Q(n = (β 0, β 1,, β If j =0, then we obtain (T ˆα, ˆβ(n U(n i,0 = T i,k U k,1 = T i,0 U 0,0 = ˆα i, k=0 and so C 0 (T ˆα, ˆβ(n U(n = C 0 (Q(n = (ˆα 0, ˆα 1,, ˆα t Finally, we assume that 1 i, j < n 1 and establish (33 Indeed, by calculations we observe that (T ˆα, ˆβ(n U(n i,j = k=0 T i,ku k,j = T i,0 U 0,j + k=1 T i,ku k,j = T i,0 U 0,j + k=1 T i,k(u k 1,j 1 + U k,j 1 (by (36 = T i,0 U 0,j + k=1 T i,ku k 1,j 1 + k=1 T i,ku k,j 1 = T i,0 U 0,j + k=1 T i 1,k 1U k 1,j 1 + k=0 T i,ku k,j 1 T i,0 U 0,j 1 (note that, T i,k = T i 1,k 1 = T i,0 (U 0,j U 0,j 1 + k=0 T i 1,kU k,j 1 +(T ˆα, ˆβ(n U(n i,j 1 (note that U,j 1 =0 = (T ˆα, ˆβ(n U(n i 1,j 1 +(T ˆα, ˆβ(n U(n i,j 1,

10 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 573 which is (33 The proof of (31 is now complete To prove of (32, we observe that L(n 1 Pˇα, ˇβ(n U(n 1 = L(n 1 L(n Tˆˇα,ˆˇβ(n U(n U(n 1 (by (31 = Tˆˇα,ˆˇβ(n = T α,β (n (by Lemma 21 The proof is complete 4 Some Applications Let A =(a i,j i,j 0 be an infinite matrix and let D n be the nth leading principal minor of A consisting of the entries in its first n rows and columns We will mainly be interested in the sequence of leading principal minors (D 1,D 2,D 3,, especially, in the case that it forms a Fibonacci or Lucas (subsequence 41 Generalized Pascal Triangle Associated With an Arithmetic or Geometric Sequence It is of interest to evaluate the determinant of generalized Pascal triangle P α,β (n, where one of the sequences α or β is an arithmetic or geometric sequence Corollary 41 Let α =(α i i 0 be an arithmetic sequence with α i = a + id, and let β =(β i i 0 be an arbitrary sequence with β 0 = a WesetD n =det(p α,β (n Then, we have with D 0 =1 D n = ( d k ˆβk D n k 1, k=0 Proof By Theorem 31, we deduce that D n =det(t ˆα, ˆβ(n, where ˆα =(ˆα i i 0 with ˆα i = i k=0 ( 1i+k( i k αk ( β0 + kd = i k=0 ( 1i+k( i k = β i 0 k=0 ( 1i+k( i k + d i k=0 ( 1i+k( i k k β 0 if i =0, = d if i =1, 0 if i>1 Now, expanding through the first row of T ˆα, ˆβ(n, we obtain the result

11 Volume 18, pp , August ARMoghaddamfarandSMHPooya Corollary 42 Let α =(α i i 0 be an arithmetic sequence with α i = a + id, and let β =(β i i 0 be an alternating sequence with β i =( 1 i a Then, we have det(p α,β (n = a(2d + a Proof LetD n =det(p α,β (n By Corollary 41, we have D n = ( d k ˆβk D n k 1 (n 1, k=0 with D 0 = 1 An easy calculation shows that ˆβ k = k l=0 ( 1k+l( k l βl Hence, we have = k l=0 ( 1k+l( k l ( 1 l a = a( 1 k k ( k l=0 l = a( 2 k D n = a (2d k D n k 1 k=0 Now, replacing n by n 1, we obtain n 2 D = a (2d k 1 D n k 2, k=0 and by calculation it follows that ( D n 2dD = a k=0 (2dk D n k 1 n 2 k=0 (2dk 1 D n k 2 ( = a k=0 (2dk D n k 1 k=1 (2dk D n k 1 = ad, or equivalently But, this implies that D n = a(2d + a D n =(2d + ad Corollary 43 Let α =(α i i 0 be an arithmetic sequence with α i = id, and let β =(β i i 0 be the square sequence, ie, β i = i 2 If D n =det(p α,β (n, thenwe have (41 D n = dd n 2 +2d 2 D n 3

12 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 575 Proof By Corollary 41, we get D n = ( d k ˆβk D n k 1 k=0 We claim that ˆβ 0 =0, ˆβ 1 =1, ˆβ 2 =2and ˆβ k =0fork 3 Now, it is obvious that our claim implies the validity of (41 Clearly ˆβ 0 =0, ˆβ 1 =1and ˆβ 2 = 2 Now, we assume that k 3 In this case we have k ( k k ( k ˆβ k = ( 1 k+j β j = ( 1 k+j j 2 j j j=0 We define the functions f and g as follows: k ( k f(x =(1 x k = ( x j j and g(x = xf (x j=0 Now, an easy calculation shows that and putting x =1,weget g (x =(1 x k 2 ( = k j=0 j=0 j=0 k j=0 ( k j 2 ( 1 j 1 =0 j Now, by multiplying both sides by ( 1 k+1,weobtain k ( k j 2 ( 1 k+j =0, j as claimed ( k j 2 ( x j 1, j Corollary 44 Let α =(α i i 0 and β =(β j j 0 be two geometric sequences with α i = ρ i and β j = σ j Then, we have det(t α,β (n = (1 ρσ Proof By Theorem 31, we have det(t α,β (n = det(pˇα, ˇβ(n On the other hand, straightforward computations show that ˇα =(ˇα i i 0 with ˇα i =(1+ρ i and similarly ˇβ =(ˇβ j j 0 with ˇβ j =(1+σ j By applying Lemma 23, we conclude the assertion

13 Volume 18, pp , August ARMoghaddamfarandSMHPooya 42 Certain Generalized Pascal Triangles Proposition 45 Let a, b, c C and let n be a positive integer Let α =(α i i 0 and β =(β j j 0 be two sequences with α i =(2 i 1a + c and β j =(2 j 1b + c Then, we have det(p α,β (n = 1 n c if a = b = c, [ ] c + a(n 1 (c a if a = b c, b b a (c an + a b a (c bn if a b Proof By Theorem 31, we have det(p α,β (n = det(t ˆα, ˆβ(n A straightforward computation shows that ˆα =(c,a,a,a, and ˆβ =(c,b,b,b, Therefore, in the notation of [7], we have T ˆα, ˆβ(n =M n (b, a, c, and since M n (a, b, c = M n (b, a, c t we have det(t ˆα, ˆβ(n = det(m n (a, b, c Now, the proof follows the lines in the proof of Theorem 2 in [7] Proposition 46 Let a, b, c C and let n be a positive integer Let α =(α i i 0 and β =(β j j 0 be two sequences with α i =2 i 1 (ia +2c and β j =2 j 1 (jb +2c Then, we have det(p α,β (n = ( 1 n+1 (a + b n 2[ c(a + b+(n 1ab ] Proof Again from Theorem 31, we have det(p α,β (n = det(t ˆα, ˆβ(n, where ˆα and ˆβ are two arithmetic sequences as ˆα =(ˆα i i 0 =(c, c + a, c +2a,,c+ ia, and ˆβ =(ˆβ j j 0 =(c, c + b, c +2b,,c+ jb, Now we compute the determinant of T ˆα, ˆβ(n elementary column operations: To do this, we apply the following C j C j C j 1, j = n 1,n 2,,2;

14 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 577 and we obtain the following quasi-toeplitz matrix: c b b b c + a c +2a T λ,µ (n 1 c +(a, where λ =( a, a, a,andµ =( a,b,b, Again, we subtract column j from column j +1, j = n 2,n 3,,2 It is easy to see that, step by step, the rows and columns are emptied until finally the determinant c b 0 0 c + a a det(t ˆα, ˆβ(n c +2a a (a + bi (n 2 (n 2 = det, c +(n 2a a c +(a a 0 0 is obtained The proposition follows now immediately, by expanding the determinant along the last row 43 Fibonacci and Lucas Numbers as Leading Principal Minors of a Quasi-Pascal Matrix There are several infinite matrices for which the leading principal minors form a Fibonacci or Lucas (subsequence For instance, in [10], we have presented a family of tridiagonal matrices with the following form: 1 λ λ λ λ F λ ( = λ 2 0 ( λ where λ =(λ i i 0 with λ i C = C\{0} Indeed, the leading principal minors of these matrices for every λ form the sequence ( F n+1 (Theorem 1 in [10] Also, n 1 for the special cases λ 0 = λ 1 = {1, 1}, see [3, 4] and [11] In ([12], pp , Strang presents the infinite tridiagonal (Toeplitz matrices: (43 P = T (3,t,0,0,,(3,t,0,0, (, where t = ±1, and it is easy to show that the leading principal minors of T form the subsequence ( F 2n+2 from the Fibonacci sequence As another example, the n 1

15 Volume 18, pp , August ARMoghaddamfarandSMHPooya leading principal minors of the Toeplitz matrices: (44 Q = T (2,1,1,1,,(2,t,0,0, (, where t = ±1, form the sequence ( F n+2 n 1 for t = 1 and the sequence ( F 2n+1 n 1 for t = 1 ([3], Examples 1, 2 We can summarize the above results in the following proposition Proposition 47 ([3, 4, 11, 12] Let n be a natural number, α = (α i i 0 and β =(β i i 0 be two sequences, and let D n be the nth leading principal minor of T α,β ( Then, the following hold (1 If α = β =(1, 1, 0, 0,, thend n = F n+1 (2 If α = β =(3,t,0, 0, 0, where t = ±1, thend n = F 2n+2 (3 If α =(1, 1, 0, 0, and β =(1, 1, 0, 0,, thend n = F n+1 (4 If α =(2, 1, 1, 1, and β =(2, 1, 0, 0,, thend n = F 2n+1 (5 If α =(2, 1, 1, 1, and β =(2, 1, 0, 0,, thend n = F n+2 Let φ = , the golden ratio, and Φ = 1 5 2, the golden ratio conjugate The recent article of Griffin, Stuart and Tsatsomeros [7] gives the following result: Proposition 48 ([7], Lemma 7 For each positive integer n, let P (n =T (1,Φ,Φ,,(1,φ,φ, (n, and Q(n =T (0, Φ, Φ,,(0, φ, φ, (n Then, we have det(p (n = F n+1, and det(q(n = F Using Propositions 47, 48 and Theorem 31, we immediately deduce the following corollary Corollary 49 Let n be a natural number, α =(α i i 0 and β =(β i i 0 be two sequences, and let D n be the nth leading principal minor of P α,β ( Then, the following hold (1 If α i = β i =1+i 1, thend n = F n+1 (2 If α i = β i =3 i, thend n = F 2n+2 (3 If α i = β i =3+i, thend n = F 2n+2 (4 If α i =1 i and β i =1+i, thend n = F n+1

16 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 579 (5 If α i =2 i +1 and β i =2 i, thend n = F 2n+1 (6 If α i =2 i +1 and β i =2+i, thend n = F n+2 (7 If α i =(2 i 1Φ + 1 and β i =(2 i 1φ +1,thenD n = F n+1 (8 If α i =(1 2 i Φ and β i =(1 2 i φ, thend n = F In the sequel, we study together the sequences F and L, and, in order to unify our treatment, we introduce the following useful notations For ε {+, } we let F ε n = F n if ε =+;andf ε n = L n if ε = Theorem 410 Let r be a non-negative integer and s be a positive integer Suppose that F ε φ r,s = 2r+s and ψ r,s = φ r,s Fr+s ε Fε 2r+s F ε r+s Then, the leading principal minors of the following infinite quasi-pascal matrix: F ε r+s ψ r,s 0 0 ψ r,s φ r,s ( 1 r ( 1 r P [r,s] ( = 0 ( 1 r 0 ( 1 r P α,α ( where α =(α i i 0 is an arithmetic sequence with α i = Fr subsequence {Fnr+s ε } n=1 from Fibonacci or Lucas sequences + i ( 1 r, form the Proof Cahill and Narayan in [5] introduce the following quasi-toeplitz matrices: T [r,s] ( = Fr+s ε ψ r,s 0 0 ψ r,s φ r,s ( 1 r 0 0 ( 1 r 0 0 T β,β ( where β =(F r, ( 1 r, 0, 0, Moreover, they show that (45 det(t [r,s] (n = F ε nr+s Now, we decompose the matrix T [r,s] (n as follows: (46 T [r,s] (n = L(nP [r,s] (n L(n t,

17 Volume 18, pp , August ARMoghaddamfarandSMHPooya where L(n =I 2 2 L 1 (n 2 The proof of (46 is similar to the proof of Theorem 31 and we omit it here Now, using (45 and (46, we easily see that det(p [r,s] (n = F ε nr+s, and the proof of theorem is complete 44 Generalized Pascal Triangle Associated With a Constant Sequence In this subsection, we study the generalized Pascal triangle P α,β (n, where one of the sequences α or β is a constant sequence The following result is another consequence of Theorem 31 Corollary 411 Let α = (α i i 0 and β = (β i i 0 be two sequences with α 0 = β 0 = γ Ifα or β is a constant sequence, then we have det(p α,β (n = γ n Proof By Theorem 31, we have det(p α,β (n = det(t ˆα, ˆβ(n But in both cases, thetoeplitzmatrixt ˆα, ˆβ(n is a lower triangular matrix or an upper triangular one with γ on its diagonal This implies the corollary The generalized Pascal triangle P α,α ( associated with the pair of identical sequences α and α, is called the generalized symmetric Pascal triangle associated with α and yields symmetric matrices P α,α (n by considering principal submatrices consisting of the first n rows and columns of P α,α ( For an arbitrary sequence α =(α i i 0 with α 0 = 0, we define α =( α i i 0 where α i =( 1 i α i for all i Then, the generalized Pascal triangle P α, α ( associated with the sequences α and α, is called the generalized skymmetric Pascal triangle associated with α and α, and yields skymmetric matrices P α, α (n by considering principal submatrices consisting of the first n rows and columns of P α, α ( Example 412 Let n 2 be a natural number Then, we have: (i The generalized symmetric Pascal triangle P F,F (n has determinant 2 n 2 (ii The generalized skymmetric Pascal triangle P F, F(n has determinant 2 n 2 All assertions in this example follow from Theorem 31 of [1] However, we will reprove them independently

18 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 581 Proof (i Consider the generalized symmetric Pascal triangle P F,F (n = F F n F F n+1 Now, we apply the following elementary row operations: R i R i R i 1 R i 2, i = n 1,n 2,,2 It is easy to see that F F n+1 det ( P F,F (n (F n 1 = det = det F F n P λ,µ (n 2 where λ =(2, 2, 2,andµ =(2(F 3 1, 2(F 4 1, 2(F 5 1, 2(F 6 1, Now, by Corollary 411, we get ( 0 1 det(p F,F (n = det 1 2 det ( P λ,µ (n 2 = 2 n 2, as desired

19 Volume 18, pp , August ARMoghaddamfarandSMHPooya (ii Here, we consider the generalized skymmetric Pascal triangle P F, F(n = ( 1 F F Similarly, we apply the following elementary column operations: C j C j +C j 1 C j 2, j = n 1,n 2,,2; and we obtain det(p F, F (n = det = det F F n P ν,λ (n 2 F F n+1 2 where λ =(2, 2, 2, Again, by Corollary 411, we get as desired ( 0 1 det(p F, F(n = det 1 0 det ( P ν,λ (n 2 =2 n 2, Example 413 Let n 2 be a natural number Then, the generalized Pascal triangle P F,I (n has determinant ( 1n

20 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 583 Proof Consider the following generalized Pascal triangle: P F,I (n = n! F F n+1 Again, we use the similar elementary row operations as Example 1(i: R i R i R i 1 R i 2, i = n 1,n 2,,2 Therefore, we deduce that n! det ( P F,I (n = det n! = det 0 0 P λ,µ (n where λ =( 1, 1, 1,andµ =( 1, 10, 73, Now, by Corollary 411, we get as desired det ( ( 1 2 P F,I (n =det 1 3 det ( P λ,µ (n 2 =( 1 n 2 =( 1 n, 5 New Matrices Whose Leading Principal Minors Form the Fibonacci Sequence In this section, we present two matrices whose entries are recursively

21 Volume 18, pp , August ARMoghaddamfarandSMHPooya defined, and we show that the leading principal minor sequence of these matrices is the sequence F =(F n n 1 It is worth mentioning that to construct these matrices we use non-homogeneous recurrence relations Theorem 51 Let a, b C and let n be a natural number Let (r i,j i,j 0 be the doubly indexed sequence given by the recurrence r i,j = r i,j 1 + r i 1,j + ai + bj, for i, j 1, and the initial conditions r i,0 = r 0,j =1, i, j 0 Let D n =det(r i,j 0 i,j Then, D n satisfies the following recursion: D 1 =1, (51 D 2 =1+a + b, D n =(1+a + bd abd n 2 (n 3 In particular, we have D n = F n if and only if (a, b {(1, 1, ( 1, 1} Proof LetR(n denotethematrix(r i,j 0 i,j First, we claim that where U(n =L(n t and 1 T (n = R(n =L(n T (n U(n, ω b a ω b a ω b a ω where ω =1+a + b Now it is easy to see that D n =det(t (n and by expansion through the last row of T (n, we obtain (51 In what follows, for convenience, we will let R = R(n, L = L(n, T = T (n and U = U(n Now, for the proof of the claimed factorization we compute the (i, j-entry of L T U, thatis (52 (L T U i,j = L i,r T r,s U s,j s=0 In fact, so as to prove the theorem, we should establish n n R 0 (L T U =R 0 (R =(1, 1,,1,,

22 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 585 C 0 (L T U =C 0 (A =(1, 1,,1, and (53 (L T U i,j =(L T U i 1,j 1 +(L T U i 1,j + ai + bj, for 1 i, j n 1 Let us do the required calculations First, suppose that i = 0 Then, we have (L T U 0,j = L 0,r T r,s U s,j = T 0,s U s,j = U 0,j =1, s=0 and so R 0 (L T U =R 0 (R =(1, 1,,1 Next, assume that j = 0 In this case, we obtain (L T U i,0 = L i,r T r,s U s,0 = L i,r T r,0 = L i,0 =1, s=0 s=0 and hence we have C 0 (L T U =C 0 (R =(1, 1,,1 Finally, we must establish (53 At the moment, let us assume that 1 i, j n 1 In this case we have (L T U i,j = s=0 = = = L i,r T r,s U s,j L i,r T r,0 U 0,j + L i,r T r,0 U 0,j + ( by (36 + = L i,r T r,0 U 0,j + s=1 + s=1 s=1 s=1 L i,r T r,s U s,j 1 L i,r T r,0 U 0,j + s=0 + s=1 r=1 s=1 L i,r T r,s U s,j 1 L i,0 T 0,s U s 1,j 1 L i,r T r,s U s,j L i,r T r,s ( Us 1,j 1 + U s,j 1 L i,r T r,s U s 1,j 1 L i,r T r,s U s 1,j 1 L i,r T r,0 U 0,j 1

23 Volume 18, pp , August ARMoghaddamfarandSMHPooya = + = = L i,r T r,0 U 0,j + s=1 r=1 s=1 L i,0 T 0,s U s 1,j 1 L i,r T r,0 U 0,j + r=1 s=1 +(L T U i,j s=1 L i,r T r,0 U 0,j + s=2 + = s=1 + = r=2 s=2 ( Li 1,r 1 + L i 1,r Tr,s U s 1,j 1 +(L T U i,j 1 L i,r T r,0 U 0,j 1 ( by (34 and (52 L i 1,r 1 T r,s U s 1,j 1 + L i,0 T 0,s U s 1,j 1 L i 1,0 T 1,s U s 1,j 1 + L i,0 T 0,s U s 1,j 1 L i,r T r,0 U 0,j + s=2 r=2 s=2 r=1 s=1 L i,r T r,0 U 0,j 1 L i 1,r 1 T r,s U s 1,j 1 + L i 1,0 T 1,s U s 1,j 1 + +(L T U i,j s=1 L i,r T r,0 U 0,j + r=1 + = r=1 s=1 r=2 s=2 r=1 s=1 r=1 L i 1,r T r,s U s 1,j 1 L i 1,r 1 T r,1 U 0,j 1 L i 1,r T r,s U s 1,j 1 +(L T U i,j 1 L i,r T r,0 U 0,j 1 L i 1,r 1 T r,s U s 1,j 1 + r=1 s=1 L i,0 T 0,s U s 1,j 1 L i 1,r 1 T r,1 U 0,j 1 + r=1 L i 1,r T r,s ( Us,j U s,j 1 L i 1,r 1 T r 1,s 1 U s 1,j 1 s=2 L i 1,r T r,s U s,j +(L T U i,j 1 + s=1 ( by the structure of T + L i,r T r,0 U 0,j + s=2 r=1 s=1 r=1 s=1 r=1 s=1 L i 1,0 T 1,s U s 1,j 1 L i 1,r T r,s U s,j 1 L i,0 T 0,s U s 1,j 1 L i 1,0 T 1,s U s 1,j 1 + L i 1,r T r,s U s,j 1 + r=1 s=0 L i 1,r 1 T r,1 U 0,j 1 L i,r T r,0 U 0,j 1 ( by (36 L i,r T r,0 U 0,j 1 r=1 L i 1,r T r,s U s,j L i 1,r T r,s U s,j 1 +(L T U i,j 1 + s=1 L i 1,r 1 T r,1 U 0,j 1 r=1 L i,r T r,0 U 0,j 1 ( note that Li 1, = U,j 1 =0 L i 1,r T r,0 U 0,j L i,0 T 0,s U s 1,j 1

24 Volume 18, pp , August Generalized Pascal Triangles and Toeplitz Matrices 587 = + = L i,r T r,0 U 0,j + s=0 r=1 L i 1,r T r,s U s,j +(L T U i,j 1 + s=1 L i,r T r,0 U 0,j + r=1 +(L T U i 1,j s=0 +(L T U i,j 1 + s=1 L i 1,r 1 T r,1 U 0,j 1 + s=0 s=2 L i 1,0 T 0,s U s,j L i,0 T 0,s U s 1,j 1 L i 1,r 1 T r,1 U 0,j 1 + L i 1,0 T 0,s U s,j L i,0 T 0,s U s 1,j 1 L i,r T r,0 U 0,j 1 ( by (52 = (L T U i 1,j +(L T U i,j 1 +Ψ i,j, r=1 s=2 L i 1,0 T 1,s U s 1,j 1 r=1 L i 1,r T r,0 U 0,j L i,r T r,0 U 0,j 1 L i 1,0 T 1,s U s 1,j 1 L i 1,r T r,0 U 0,j where Ψ i,j = L i,r T r,0 U 0,j + s=0 r=1 L i 1,0 T 0,s U s,j L i,r T r,0 U 0,j 1 L i 1,r 1 T r,1 U 0,j 1 + r=1 s=2 L i 1,r T r,0 U 0,j + s=1 L i 1,0 T 1,s U s 1,j 1 L i,0 T 0,s U s 1,j 1 But by an easy calculation one can show that Ψ i,j = ai + bj, which implies the first part of theorem The second part of theorem is now obvious Acknowledgments We would like to thank the referee for the careful reading of an earlier version of this article and for suggestions and comments The first author would like to thank IPM for their financial support

25 Volume 18, pp , August ARMoghaddamfarandSMHPooya REFERENCES [1] R Bacher Determinants of matrices related to the Pascal triangle J Théor Nombres Bordeaux, 14:19 41, 2002 [2] R Bacher and R Chapman Symmetric Pascal matrices modulo p European J Combin, 25: , 2004 [3] N D Cahill, J R D Errico, D A Narayan, and J Y Narayan Fibonacci determinants The College Math J, 33: , 2002 [4] N D Cahill, J R D Errico, and J P Spence Complex factorizations of the Fibonacci and Lucas numbers Fibonacci Quart, 41:13 19, 2003 [5] N D Cahill and D A Narayan Fibonacci and Lucas numbers as tridiagonal matrix determinants Fibonacci Quart, 42: , 2004 [6] A Edelman and G Strang Pascal matrices Amer Math Monthly, 111: , 2004 [7] KGriffin,JLStuart,andMJTsatsomeros NoncirculantToeplitzmatricesallofwhose powers are Toeplitz Czechoslovak Math J, 58: , 2008 [8] Integer-sequences, njas/sequences/indexhtml [9] C Krattenthaler Evaluations of some determinants of matrices related to the Pascal triangle Sém Lothar Combin, Article B47g, 19 pp, 2001/2002 [10] A R Moghaddamfar, S M H Pooya, S Navid Salehy, and S Nima Salehy Fibonacci and Lucas sequences as the principal minors of some infinite matrices J Algebra Appl, (to appear [11] G Strang Introduction to Linear Algebra Third Edition Wellesley-Cambridge Press, 2003 [12] G Strang and K Borre Linear Algebra Geodesy, and GPS Wellesley-Cambridge Press, 1997