The elements of finite order in the Riordan group over the complex field

Linear Algebra and its Applications 439 (2013) 4032–4046

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Linear Algebra and its Applications www.elsevier.com/locate/laa

The elements of finite order in the Riordan group over the complex field ✩ Gi-Sang Cheon ∗ , Hana Kim 1 Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Republic of Korea

a r t i c l e

i n f o

Article history: Received 9 May 2013 Accepted 18 September 2013 Available online 23 October 2013 Submitted by R. Brualdi MSC: primary 05A30 secondary 05A15 Keywords: Riordan group Riordan matrix Pseudo order k-pseudo involution

a b s t r a c t An element of finite order in the Riordan group over the real field must have order 1 or 2. If we extend all the entries to be complex numbers then it may have any finite order. In the present paper, we investigate the elements of finite order of the Riordan group over the complex field. This notion leads us to define an element of pseudo order k  2 and k-pseudo involution, respectively. It turns out that the inverse of a k-pseudo involution only differs from it in signs. We clarify some relationship between the elements of pseudo order k and k-pseudo involutions. In particular, k-pseudo involutions for k ≡ 0 (mod 4) are characterized by a single sequence. The subgroups of the Riordan group formed by the elements having pseudo order of a prime power p r are also introduced. © 2013 Elsevier Inc. All rights reserved.

1. Introduction For brevity, let F0 and F1 be defined as

      F0 = g ( z) ∈ C[[ z]]  g (0) = 1 and F1 = f ( z) ∈ C[[ z]]  f (0) = 0, f  (0) = 0 where C[[ z]] is the ring of formal power series over the complex field.

✩ This work was supported by the National Research Foundation of Korea Grant funded by the Korean Government (NRF-2012-007648). Corresponding author. E-mail addresses: [email protected] (G.-S. Cheon), [email protected], [email protected] (H. Kim). 1 Current address: Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA.

*

0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.09.029

G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

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The Riordan group [10] over the complex field, R(C), is the set of infinite lower triangular matrices of the form



X = g ( z)

g ( z) f ( z)

g ( z) f ( z)2

···



(1)

where g ( z) f ( z)k is its kth column generating function for which g ( z) ∈ F0 and f ( z) ∈ F1 . An element of the form (1) is called a Riordan matrix and denoted as X = ( g ( z), f ( z)). As is customary in this setting, we index our rows and columns starting with the index 0. The binary operation of the Riordan group is the usual matrix multiplication described in terms of generating functions:



 







g ( z), f ( z) · h( z), ( z) = g ( z)(h ◦ f )( z), ( ◦ f )( z)

where ◦ denotes the composition of functions. The identity for this operation is I = (1, z) the infinite identity matrix, and the inverse of ( g ( z), f ( z)) is given by



−1

g ( z), f ( z)

  = 1/( g ◦ ¯f )( z), ¯f ( z)

where ¯f ( z) is the compositional inverse of f ( z). Many properties of the Riordan matrices have been studied in connection with counting problems, especially for combinatorial sums, identities and inversion (see [8,14]). While, algebraic properties of the Riordan group have not much studied. In particular, Riordan group elements of finite order over the complex field are of our interest in this paper. As noted in [1,12], any Riordan matrix over the real field having finite order must have order 1 or 2. However, Riordan matrices over the complex field may have any finite order. For instance, let

⎡

1 ⎢

⎢1 ζ ⎢ 1 2ζ ζz 1 =⎢ P k (ζ ) := , ⎢ 1 3ζ 1−z 1−z ⎢ ⎣ 1 4ζ

⎤ ⎥ ⎥ ⎥ ζ2 ⎥. ⎥ 3ζ 2 ζ 3 ⎥ 6ζ 2 4ζ 3 ζ 4 ⎦ ···

(2)

As we shall see in Theorem 4.3, P k (ζ ) has finite order k whenever ζ is a primitive kth root of unity for each k  2 i.e., ζ k = 1 and ζ j = 1 for 0 < j < k. Let X = ( g ( z), f ( z)) ∈ R(C). Then X may be factorized as





X = g ( z), f ( z)/ζ · (1, ζ z) for any nonzero complex number ζ . From now on, let D ζ = (1, ζ z), the diagonal matrix with diagonal entries 1, ζ, ζ 2 , . . . . As we shall see in Lemma 2.1, if X = ( g ( z), f ( z)) is of order k  2 then f  (0) = ζ and it must be a primitive kth root of unity. Note that Y := ( g ( z), f ( z)/ζ ) is a Riordan matrix whose diagonal entries are all ones. In this paper, we define an element Y = I in the Riordan group R(C) to have pseudo order k if k is the smallest integer  2 such that (Y D ζ )k = I for which ζ is a primitive kth root of unity. If such k does not exist, Y is said to have infinite pseudo order. It can be easily seen that Y has pseudo order k if and only if Y −1 has pseudo order k. Though Y D ζ is of finite order k  3, Y need not be of pseudo 1 order k. For instance, P k (ζ ) = P · (1, ζ z) in (2) has order k where P = ( 1− , z ) is the Pascal matrix, z 1− z but P has pseudo order 2 since P (1, − z) has order 2. It is well-known that Y has pseudo order 2 if and only if Y −1 = MY M where M = (1, − z). The inverse Y −1 coincides with Y except for alternating signs of its diagonals. In this sense, Y is called a pseudo involution [1]. More generally, in this paper we call Y a k-pseudo involution if

Y −1 = D ζ Y D ζ −1 .

(3)

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In particular, if k = 2 then ζ = −1 and so it is clear that Y is an element of pseudo order 2 if and only if Y is a 2-pseudo involution. We will show later that if k  3 then these two notions are different. The purpose of this paper is to investigate the elements of finite order of the Riordan group over the complex field and to classify k-pseudo involutions. Specifically, in Section 2 we present a large class of Riordan matrices of finite order and pose an open question. In Section 3, we investigate a structure of k-pseudo involutions and verify a relationship between k-pseudo involutions and Riordan matrices of finite pseudo order. Section 4 is devoted to the special subgroups of the Riordan group whose elements have pseudo order of a prime power. Finally in Section 5, we classify k-pseudo involutions with k ≡ 0 (mod 2) according to change of sign patterns of their inverses. As a result, we characterize k-pseudo involutions with k ≡ 0 (mod 4) by means of a single sequence. 2. Riordan matrices of finite order over the complex field In this section, we investigate the elements of finite order k of the Riordan group R(C) and prove some results concerning them. The section concludes with a demonstration of Riordan matrices of order k by means of k-cyclic symmetric functions. We assume that k  2. Let X = ( g ( z), f ( z)) ∈ R(C) be an element of finite order k. Then g ( z) and f ( z) satisfy the following two functional equations:

f (k) ( z) = z

and

k −1



 g ◦ f ( j ) ( z ) = 1,

(4)

j =0

where f ( j ) ( z) = ( f ◦ · · · ◦ f )( z) and f (0) ( z) := z.







j

Along the similar lines of the results in [7], we obtain the following lemma: Lemma 2.1. Let f ( z) ∈ F1 . Then f (k) ( z) = z and f ( j ) ( z) = z for 0 < j < k if and only if f ( z) is of the form





f ( z) = σ¯ ζ σ ( z)

(5)

where σ ( z) ∈ F1 , and ζ is a primitive kth root of unity. Proof. Let f ( z) be given by (5). Since (ζ σ ◦ σ¯ )( z) = ζ z for any





σ (z) ∈ F1 , we have



 f (k) ( z) = σ¯ ◦ ζ k σ ( z) = σ¯ σ ( z) = z. If f ( j ) ( z) = z for some 0 < j < k then it follows from f ( j ) ( z) = (σ¯ ◦ ζ j σ )( z) = z that ζ j σ ( z) = σ ( z). Thus we have ζ j = 1 which contradicts that ζ is a primitive kth root of unity. Conversely, let k be the smallest integer such that f (k) ( z) = z. From now on, we write an = [ zn ] A ( z)  j if A ( z) = n0 an zn . Let f  (0) = f 1 . Since [ z] f (k) ( z) = f 1k = 1, f 1 is a kth root of unity. If f 1 = 1 for

some 0 < j < k then f (k) ( z) = z implies f ( j ) ( z) = z which contradicts the minimality of k. Thus f 1 must be a primitive kth root of unity, say ζ . Define

σ (z) :=

k −1  1 ( j) f ( z). ζj j =0

Then

σ (z) ∈ F1 . Since f (k) (z) = z and ζ k = 1, it follows that 1 

ζ



σ f ( z) =

k −1  1 f ( j +1) ( z) = σ ( z), j ζ +1 j =0

which implies f ( z) = σ¯ (ζ σ ( z)) as required.

2

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We note that the Lagrange inversion formula gives the coefficients of the compositional inverse of

σ ( z ):



 n

1  n −1  z σ¯ ( z) = z n

z



σ ( z)

. f (z)

Theorem 2.2. Let ζ be a primitive kth root of unity. Then ( ζ z , f ( z)) has order k if and only if f ( z) is of the form f ( z) = σ¯ (ζ σ ( z)) where σ ( z) ∈ F1 . f (z)

Proof. Let g ( z) = ζ z . Since k −1



 g ◦ f ( j ) ( z) =

j =0

k −1 j =0

f ( j +1 ) ( z )

ζ f ( j ) ( z)

=

f (k) ( z)

ζkz

=

f (k) ( z) z

,

f (z)

we see that ( ζ z , f ( z)) has order k if and only if (1, f ( z)) has order k. Hence the result follows from (4) and Lemma 2.1. 2 Let us define a k-cyclic symmetric function to be a function

ϕ in k variables z1 , . . . , zk such that

ϕ (z1 , z2 , . . . , zk ) + ϕ (z2 , . . . , zk , z1 ) + · · · + ϕ (zk , z1 , . . . , zk−1 ) = 0. Following the underlying idea in [9], we write τ ϕ ( z1 , z2 , . . . , zk ) = ϕ (τ z1 , τ z2 , . . . , τ zk ) if τ is a permutation on the set { z1 , z2 , . . . , zk } mapping zi to τ zi . Let τ be the permutation defined as τ z1 = z2 , τ z2 = z3 , . . . , τ zk = z1 . Then ϕ is k-cyclic symmetric if and only if

ϕ + τ ϕ + · · · + τ k −1 ϕ = 0.

(6)

We note that a 2-cyclic symmetric function coincides with an anti-symmetric function in 2 variables. In [2,3], the involutions in R(C) have been completely characterized by means of antisymmetric functions. Theorem 2.3. Let f ( z) = σ¯ (ζ σ ( z)) for a σ ( z) ∈ F1 where ζ is a primitive kth root of unity, and let

g ( z) = exp

 



ϕ z, f (1) (z), . . . , f (k−1) (z)

(7)

for a k-cyclic symmetric function ϕ . Then ( g ( z), f ( z)) is an element of order k in the Riordan group R(C). Proof. Since k is the smallest positive integer such that ζ k = 1, it suffices to show that the two functional equations in (4) hold for our f ( z) and g ( z). The first equation f (k) ( z) = z follows from Lemma 2.1. Next, with f denoting f ( z), since f (k) = z and f (k+ j ) = f ( j ) we obtain



       g ◦ f ( j ) ( z) = exp ϕ f ( j ) , f ( j +1) , . . . , f (k+ j −1) = exp ϕ f ( j ) , f ( j +1) , . . . , f ( j −1)      = exp τ j ϕ z, f (1) ( z), . . . , f (k−1) ( z) =: exp τ j ϕ ,

where

τ is the same permutation as one in (6). Thus the second equation follows from k −1





g ◦ f ( j ) ( z) = exp





ϕ + τ ϕ + · · · + τ k−1 ϕ = exp(0) = 1.

j =0

Hence the proof is complete.

2

As also noted in [9], it is easy to give examples of k-cyclic symmetric functions. If h( z1 , . . . , zk ) is any function, then the function ϕ = h( z1 , . . . , zk ) − τ j h( z1 , . . . , zk ) is k-cyclic symmetric for j = 1, . . . , k − 1.

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Example 2.4. Let

σ ( z) = z





3n+1 1 n0 3n+1 n

which is the analytic solution of



zn be the generating function for the ternary numbers,

σ (z) = 1 + zσ (z)3 . Then σ¯ (z) =

z C (z)

√

where C ( z) = 1−

generating function for the Catalan numbers. Set k = 3. Let f ( z) = σ¯ (ζ σ ( z)) = primitive 3rd root of unity. Explicitly, we obtain

f ( z) = ζ z



1−4z 2z

is the

ζ σ (z) where C (ζ σ ( z))

ζ is a

pn (ζ ) zn

n0

3n

2k−23n−2k

n

+1 where pn (x) = 2n1+1 n − k=1 k(2nk− xk . By Lemma 2.1, f (3) ( z) = z. n−k k+1) k−1 Using h( z1 , z2 , z3 ) = z1 z2 we have the 3-cyclic symmetric function given by ϕ = h( z1 , z2 , z3 ) − h( z3 , z1 , z2 ) = z1 ( z2 − z3 ). From (7) we then obtain

g ( z) = exp

 

 





ϕ z, f (z), f (2) (z) = exp z f (z) − f (2) (z)

2

3

= 1 + (1 + 2ζ ) z + (2 + 4ζ ) z +

7 2



+ 10ζ z4 + (11 + 34ζ ) z5 + · · · .

By Theorem 2.3, we obtain an element of order 3:

⎡

1 0

⎢ ⎢ ⎢   ⎢ 1 + 2ζ g ( z), f ( z) = ⎢ ⎢ 72 + 4ζ ⎢ + 10ζ ⎢ 2 ⎣ 11 + 34ζ

ζ (1 − ζ )ζ (5 + ζ )ζ (18 + 3ζ )ζ ( 149 + 2ζ )ζ 2

ζ2 (2 − 2ζ )ζ 2 1 (9 − 3ζ )ζ 2 3 − 3ζ (40 − 10ζ )ζ 2 13 − 10ζ ··· ···

⎤

ζ (4 − 4ζ )ζ

⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ 2⎦ ζ

We end this section with an open question: Open question. Does the converse of Theorem 2.3 hold? That is, if ( g ( z), f ( z)) has order k, should g ( z) be of the form (7) for some k-cyclic symmetric function ϕ ? This is true for k = 2, see [2]. 3. k-pseudo involutions In this section, we investigate k-pseudo involutions (k  2) in the Riordan group R(C), and then we obtain some relationship between k-pseudo involutions and Riordan matrices of finite pseudo order. We first observe a Riordan matrix X whose inverse is X −1 = D 1 X D 2 for some diagonal matrices D 1 , D 2 ∈ R(C). If D 1 = D 2 say D, then X D is involution. Now assume that D 1 = D 2 . To eliminate trivial cases, we assume that X is not a diagonal matrix. Theorem 3.1. Let X = ( g ( z), f ( z)) ∈ R(C). Then X −1 = D 1 X D 2 if and only if Y −1 = D ω Y D ω−1 where Y = ( g (z), f (z)/ f  (0)) and ω is a primitive mth root of −1 for the smallest integer m  2 such that [ zm ] g (z) = 0.





Proof. Let g ( z) = 1 + n1 gn zn and f ( z) = n1 f n zn with f 1 = 0. Assume that X −1 = D 1 X D 2 for D 1 = (1, d1 z) and D 2 = (1, d2 z) where d1 = d2 . It follows from X D 1 X D 2 = (1, z) that





d2 f d1 f ( z ) = z .

(8)

We first show that D 1 = (1, ω z) and D 2 = (1, ω1f z), where ω is a primitive mth root of −1 for f1 1 the smallest integer m  2 such that gm = 0. Equating the first few coefficients on both sides of (8) yields

      [z]d2 f d1 f ( z) = d1 d2 f 12 = 1 and z2 d2 f d1 f ( z) = (d1 f 1 + 1)d1 d2 f 1 f 2 = 0.

G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

Since d1 = d2 , we have d1 = −

1 f1

4037

and f 2 = 0. Further, it follows from [ z] g ( z) g (d1 f ( z)) = (d1 f 1 + 1) g 1

that g 1 = 0. Let m  2 be the smallest integer such that gm = 0. Since [ zm ] g ( z) g (d1 f ( z)) = ω and d = 1 = 1 , where ω is a primitive mth root of −1. m (dm 2 2 1 f 1 + 1) gm , we obtain d1 = f ωf d1 f 1

1

1

Thus D 1 = (1, ω z) and D 2 = (1, ω1f z). Let Y = ( g ( z), f ( z)/ f 1 ). Since X = Y (1, f 1 z), we obtain f1 1

Y −1 = (1, f 1 z) X −1 = (1, f 1 z) D 1 X D 2 = D ω Y (1, f 1 z) D 2 = D ω Y D ω−1 . The converse is straightforward.

2

To characterize X = ( g ( z), f ( z)) ∈ R(C) satisfying X −1 = D 1 X D 2 for some diagonal matrices D 1 , D 2 ∈ R(C), it is enough to study Y = ( g ( z), f ( z)/ f  (0)) such that Y −1 = D ω Y D ω−1 by Theorem 3.1. More generally, we explore k-pseudo involutions in connection with the elements of finite pseudo order. It is obvious that I = (1, z) and M = (1, − z) are k-pseudo involutions for each k. From now on, we only consider the k-pseudo involutions other than I and M. We first note that if k = 2m (m  1) then Y in (3) is a k-pseudo involution if and only if

Y −1 = D ω Y D ω −1

(9)

where D ω = (1, ω z) for a primitive mth root

ω of −1.

Lemma 3.2. Let Y = ( g ( z), f ( z)) ∈ R(C) be a k-pseudo involution. Then we have: (i) If k is even then [ zn ] g ( z) = [ zn+1 ] f ( z) = 0 for n = 0 (mod 2k ). (ii) If k is odd then [ zn ] g ( z) = [ zn+1 ] f ( z) = 0 for n = 0 (mod k). Proof. (i) Let k = 2m. Then it follows from (9) that

g ( z) g



 1  f ω f ( z) = z



ω f (z) = 1 and

ω

(10)



where ω is a primitive mth root of −1. Let f ( z) = n1 f n zn . Equating the coefficients on both sides of the second equation of (10) yields the infinite system of equations

f 12 = 1,

( f 1 ω + 1) f 1 f 2 = 0,  2  f 1 f 3 ω2 + 2 f 22 ω + f 3 f 1 = 0, .. . From the first equation, f 1 = 1 or f 1 = −1. Assume f 1 = 1. Solving the second equation uniquely for f 2 , the third for f 3 , etc., we obtain



zn+1

1 

ω

f







ω f ( z ) = 1 + ω n f n +1 = 0.

Since ω = −1, f n+1 need not be zero if n ≡ 0 (mod m) and f n+1 must be zero, otherwise. The case for f 1 = −1 follows from that −ω is the primitive mth root of −1. Similarly, using the first equation of (10) we obtain m

 

0 = zn g ( z) g







 

ω f (z) = 1 + ωn zn g (z).

Again since ωm = −1, [ zn ] g ( z) need not be zero if n ≡ 0 (mod m) and [ zn ] g ( z) must be zero, otherwise. Thus (i) follows as required. (ii) may be similarly proved. 2

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Let Y = [ y i , j ] be a k-pseudo involution. By Lemma 3.2, if k is even (odd, resp.) then y i , j = 0 for k ) (i − j = 0 (mod k), resp.) i.e., Y consists of diagonals of nonzero (and 2

i , j with i − j = 0 (mod

possibly some zero) entries such that there are m := 2k − 1 (k − 1, resp.) zero diagonals between the adjacent nonzero diagonals. We call a Riordan matrix Y with such a zero–nonzero pattern of the entries an m-diagonally striped matrix. A 0-diagonally striped matrix is a full matrix. This kind of matrices (not necessarily Riordan matrices) arise in various areas such as sparse approximate inverse preconditioners for solving linear systems in Astrophysics [13] and in electronic engineering [6], etc. Further, if k = 2m then letting Y −1 = [ y i , j ] we obtain y i , j = ωi − j y i , j from (9). If i − j ≡ 0 (mod m) then ωi − j = ±1, and if i − j = 0 (mod m) then y i , j = 0. Thus †

†

y i , j = (−1)

i− j m

†

y i, j .

The following theorem is an immediate consequence of Lemma 3.2. Theorem 3.3. Let Y ∈ R(C) be a k-pseudo involution. Then we have: (i) If k is even then Y is ( 2k − 1)-diagonally striped. (ii) If k is odd then Y is (k − 1)-diagonally striped. Remark. Let us define Hm by the set of formal power series h( z) ∈ F1 of the form

h( z) =



hmn+1 zmn+1 .

n0

Then ( g ( z), f ( z)) is (m − 1)-diagonally striped if and only if zg ( z), f ( z) ∈ Hm . The centralizer of M denoted C ( M ) is a Riordan subgroup called the checkerboard subgroup [11]. This is the set of 1-diagonally striped matrices in R(C) i.e.,

C (M ) =







g ( z), f ( z)  zg ( z), f ( z) ∈ H2 .

In general, it can be shown that the centralizer of D ζ is the set of (k − 1)-diagonally striped matrices in R(C) i.e.,

C(Dζ ) =







g ( z), f ( z)  zg ( z), f ( z) ∈ Hk .

(11)

Theorem 3.4. Let Y ∈ R(C) be a k-pseudo involution. Then we have: (i) If k ≡ 1 or 3 (mod 4) then Y has order 2. (ii) If k ≡ 2 (mod 4) then Y has pseudo order 2. (iii) If k ≡ 0 (mod 4) then Y has pseudo order  k. Proof. Let Y = [ y i , j ] and Y −1 = [ y i , j ]. Since Y is a k-pseudo involution, y i j = ζ i − j y i j where ζ is a primitive kth root of unity. (i) If k ≡ 1 or 3 (mod 4) then k is odd. By (ii) of Lemma 3.2, Y is (k − 1)-diagonally striped. Thus from (11) we obtain Y −1 = D ζ Y D ζ −1 = Y D ζ D ζ −1 = Y . (ii) If k ≡ 2 (mod 4) then k = 2k for some odd k . By (i) of Lemma 3.2, Y is (k − 1)-diagonally †

†

i− j

† striped. If i − j ≡ 0 (mod k ) then y i , j = (−1) k y i , j . Since there are even number of zero diagonals † between the adjacent nonzero diagonals, y i , j = (−1)i − j y i , j for all i , j, which implies Y −1 = MY M. Thus Y has pseudo order 2. (iii) Let k ≡ 0 (mod 4). Using D ζ Y = Y −1 D ζ , it can be shown that

(Y D ζ )k = D kζ = I where ζ is a primitive kth root of unity. Thus Y has pseudo order less than or equal to k.

2

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We note that the converse of Theorem 3.4 does not hold in general. That is, not every element of pseudo order k is a k-pseudo involution. Theorem 3.5. Let Y ∈ R(C) be an element of pseudo order k. If k  3 is odd then Y cannot be a k-pseudo involution. Proof. Let Y = ( g ( z), f ( z)) and k = 2m + 1 (m  1). Suppose that Y is a (2m + 1)-pseudo involution. Since Y D ζ Y D ζ −1 = (1, z), we have

g ( z) · ( g ◦ ζ f )( z) = 1 and

( f ◦ ζ f )( z) = ζ z

(12)

where ζ is a primitive (2m + 1)th root of unity. Using (ζ f ◦ ζ f )( z) = ζ 2 z, we obtain (ζ f )(2m) ( z) = (ζ f ◦ ζ f )(m) (z) = ζ 2m z = ζ k−1 z. Thus

    (ζ f )(k) ( z) = ζ f ◦ (ζ f )(2m) ( z) = ζ f ζ k−1 z .

(13)

Since Y D ζ is of order k, it follows from (4) and (13) that z = (ζ f )(k) ( z) = ζ f (ζ k−1 z). Substituting ζ z into z yield ζ z = ζ f (ζ k z) = ζ f ( z). Thus f ( z) = z. Substituting z = (ζ f )(2 j −1) ( z) to both sides of the first equation of (12) leads to









g ◦ (ζ f )(2 j −1) ( z) · g ◦ (ζ f )(2 j ) ( z) = 1.

(14)

Since Y D ζ is of order k, it follows from (4) and (14) that

1 = g ( z)

2m  

 g ◦ f ( j ) ( z) = g ( z)

j =1

m  







g ◦ (ζ f )(2 j −1) ( z) · g ◦ (ζ f )(2 j ) ( z) = g ( z).

j =1

Hence Y = (1, z) = I which does not have pseudo order 2m + 1. Thus Y cannot be a (2m + 1)-pseudo involution. 2 Theorem 3.6. Let Y ∈ R(C) be an element of pseudo order k. If k = 2m (m  1) and Y is (m − 1)-diagonally striped then Y is a k-pseudo involution. Proof. We first show that (Y D ω )2 = D 2ω , where ω is a primitive mth root of −1. Since Y is (m − 1)-diagonally striped, zg (z), f (z) ∈ Hm . Then we obtain

 

D 2ω (Y D ω ) = g











ω2 z , ω f ω2 z = g (z), ω3 f (z) = (Y D ω ) D 2ω .

Thus D 2ω (Y D ω )2 = (Y D ω )2 D 2ω . Since (Y D ω )2m = I and D 2m ω = I , it follows that





2 2 O = (Y D ω )2m − D 2m ω = (Y D ω ) − D ω ·



(Y D ω )2 j D 2ω .

(15)

j ,1 j +=m−1

The second factor of the right-hand side of (15) is not a zero divisor since it is a lower triangular matrix whose diagonal entries are mω−2n for n  0. Thus (Y D ω )2 = D 2ω or equivalently Y D ω Y D ω−1 = I , which implies that Y −1 = D ω Y D ω−1 . 2 Remark. If Y ∈ R(C) has pseudo order 2m then (Y D ω )m is an involution where ω is a primitive mth root of −1. In particular, if Y is (2k − 1)-diagonally striped for integer k  1 such that 2k = m, then









MY D ω = g (− z), ω f (− z) = g ( z), −ω f ( z) = Y D ω M . Thus M (Y D ω )m = (Y D ω )m M. Since M 2 = I , we have



 



O = (Y D ω )2m − M 2 = (Y D ω )m − M · (Y D ω )m + M . It can be easily seen that (Y D ω )m + M is not a zero divisor. Thus (Y D ω )m is just the diagonal matrix M = (1, − z).

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G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

4. Riordan subgroups of the elements having pseudo order p r It is known [11] that there are several important subgroups of the Riordan group R(C). In this section, we investigate the Riordan subgroups whose nonidentity elements have the same pseudo order p r a prime power. We note that even though X ∈ R(C) has pseudo order k, a power of X need not have the same pseudo order k in general. We begin by observing the m-Bell subgroup B (m) , m ∈ Z+ and the hitting time subgroup H defined as

B(m) =



f ( z)/ z

m

  , f ( z)  f ( z) ∈ F1 , f  (0) = 1

and

H=







zf  ( z)/ f ( z), f ( z)  f ( z) ∈ F1 , f  (0) = 1 .

Lemma 4.1. The intersection of two subgroups B (m) and H is

 B

(m)

∩H=

1 1 − czm

   c∈C , m

z

, √ m

(16)

1 − cz

√

which contains the cyclic subgroup generated by (1/(1 − zm ), z/ m 1 − zm ). zf  ( z) f (z) − mf (1z)m =

Proof. It may be proved by solving the ordinary differential equation f  ( z)/ f ( z)m+1

m+1

= 1/ z

, integrating both sides with respect to z yields

C is a constant. By setting C =

c m

z . Thus 1−czm m z ). 2

we obtain f ( z) = √ m

the cyclic subgroup generated by (1/(1 − z ), z/ m

√ m

1−

= ( f (zz) )m . Since − mz1m + C where

we have (16). Clearly, it contains

We will now show that every element of B (m) ∩ H has pseudo order of a prime power. More generally, for m ∈ Z+ and β ∈ C let us consider the set Λm,β defined by

 Λm,β =

1

β

1 − czm

, √ m

   c∈C .  m

z

1 − cz

m,β It can be easily shown that Λ√ forms an abelian group, which contains the cyclic subgroup generated by λ := (1/(1 − zm )β , z/ m 1 − zm ). Thus λ is isomorphic to Z. Further, if we define H n,m,β for n, m ∈ Z+ , β ∈ C by

 H n,m,β =

1 1 − nzm

β , √ m

z



1 − nzm

then H n,m,β nZ. Theorem 4.2. Let m ∈ Z+ and β ∈ C. Then every nonidentity element in Λm,β has the same pseudo order p r for some prime power p r . In particular, we have: (i) If m is odd then every nonidentity element in Λm,β has pseudo order 2. (ii) If m is a power of 2 then every nonidentity element in Λm,β has pseudo order 3. (iii) If m is not a power of 2 but even then every nonidentity element in Λm,β has pseudo order k where k is a prime power in which k is the smallest integer such that k does not divide m. Proof. Let Y := (( 1−1czm )β , √ m

z ) 1−czm

∈ Λm,β . If c = 0 then Y = I which has pseudo order 1. Now as-

sume c = 0. Let ζ be a primitive kth root of unity. Then

G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

(Y D ζ ) =

β

1

k

k−1

1 − c(

j =0 ζ

jm ) zm

,  m



z

k−1

1 − c(

j =0 ζ

4041

. jm ) zm

Hence Y has pseudo order k  2 for any nonzero c ∈ C if and only if k is the smallest integer such k−1 jm = 0. It suffices to show the following. that j =0 ζ (i) Let m be odd. If k = 2 then ζ = −1 so 1 + (−1)m = 0. Thus Y has pseudo order 2. (ii) Let m = 2r for r  1. Similarly, if k = 2 then 1 + (−1)m = 2 = 0. Hence Y does not have pseudo order 2. Let ζ be a primitive 3rd root of unity. If r is odd then m ≡ 2 (mod 3) and so ζ m = ζ 2 . If r is even then m ≡ 1 (mod 3) and so ζ m = ζ . Thus 1 + ζ m + ζ 2m = 1 + ζ + ζ 2 = 0, which implies that Y has pseudo order 3. k−1 jm (iii) Let m = 2r be even. If k divides m then = k = 0. Assume that k does not divide m. j =0 ζ Since m  6, there must exist an integer j between 1 and m such that j does not divide m. Hence we may assume that 1 < k < m. Write m = kq + s (0 < s < k) in accord with the division algorithm. Then k −1 

ζ jm =

j =0

k −1 

ζ js =

j =0

(ζ s )k − 1 = 0. ζs − 1

It means that Y has pseudo order k if k is the smallest integer such that k does not divide m. Now we show that such k is a prime power p r for some r > 0. Suppose that k is not a prime power p r for any r > 0. Then k may be expressed as k = ab for a, b ∈ {2, 3, . . . , k − 1} where a and b are relatively prime. By the minimality of k, every integer less than k divides m. Thus a and b divide m. It follows from gcd(a, b) = 1 that k divides m, a contradiction. Hence the proof is complete. 2 Conversely, for any β ∈ C we can find the subgroups Λm,β consisting of the elements of pseudo order p r (= k) except I by choosing m ∈ Z+ to be the smallest integer k such that k does not divide m. For instance, if k = 23 then m = 22 · 3 · 5 · 7 = 420, m = 22 · 32 · 5 · 7 = 1260, etc. are integers such that k is the smallest integer that does not divide m. That is, Λ420,β and Λ1260,β are subgroups consisting of the elements of pseudo order 23 except I . We note that if i A : R(C) → R(C) is an inner automorphism defined by i A ( X ) = A X A −1 for a A ∈ R(C) then X and i A ( X ) have the same pseudo order k if and only if A is an element of the centralizer of D ζ given by (11). Further, Λm,β is invariant under the inner automorphism i D for a diagonal matrix D ∈ R(C) i.e.,











i D Λm,β = D X D −1  X ∈ Λm,β = Λm,β . Theorem 4.3. Let D ζ = (1, ζ z) where ζ is a primitive kth root of unity. Then Y D ζ ∈ R(C) has order k for any k  2 if and only if Y ∈ Λ1,β . Proof. Let Y = ( g ( z), f ( z)). Assume that for every k  2, X k := ( g ( z), ζ f ( z)) has order k. Let f ( z) =  n (k) k−1 = 0, it can be shown that f 1 = 1 and for n1 f n z . Using (ζ f ) ( z) = z and 1 + ζ + · · · + ζ k = 2, 3, . . .









zk+1 (ζ f )(k) ( z) = k f k+1 − f 2k + kk−2 1 + ζ + · · · + ζ k−1



  = k f k+1 − f 2k = 0.

Thus f 1 = 1 and f k+1 = f 2k for k = 2, 3, . . . . Letting f 2 = c ∈ C, we obtain

f ( z) =



z

f n zn =

n1

Now, let g ( z) = 1 +

(ζ f )( j ) ( z) =

1 − cz



n1

.

gn zn and f ( z) = 1−zcz . Using

ζ jz 1 − c (1 + ζ + · · · + ζ j −1 ) z

(17)

4042

where

G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

k−1

j =0 ζ

j

−1  k  k 

z

= 0, it can be shown that for k = 2, 3, . . . ( j)

g ◦ (ζ f )



( z) =

j =0

=



1

k −1

k ! gk −

(k − 1)!

k −1

k ! gk −

(k − 1)!

( g 1 + jc ) + k

j =0



1

k −2

k −1

( g 1 + jc )

j =0



k −1 

 ζ

j

j =0

( g 1 + jc ) .

(18)

j =0

k−1

( j) j =0 ( g ◦ (ζ f ) )( z) = 1 by (4), the left-hand side of (18) is equal to zero for k  2. Thus  k − 1 gk = k1! j =0 ( g 1 + jc ) for k = 2, 3, . . . . Letting g 1 = c β for β ∈ C, we obtain

Since

g ( z) = 1 +





gn zn =

n1

β

1

.

1 − cz

ζz

Conversely, assume that Y D ζ = (( 1−1cz )β , 1−cz ) = ( g ( z), ζ f ( z)) where ζz k  2, it immediately follows from (17) that ( 1−cz )(k) ( z) = ζ k z = z. Since





g ◦ (ζ f )( j ) ( z) =

k−1

1 − c (1 + ζ + · · · + ζ j −1 ) z 1 − c (1 + ζ + · · · + ζ j ) z

,

k−1

j =0 ζ

j

= 0. For every

j  2,

◦ (ζ f )( j) )(z) = 1 for every k  2. Thus for every k  2, Y D ζ has order k since k is the smallest positive integer such that ζ k = 1. Hence the proof is complete. 2

we also have

j =0 ( g

Remark. Every element of Λm,β is a 2m-pseudo involution, since for Y ∈ Λm,β



Y D ω Y D ω −1 = where

β

1 1 − c (1 + ωm ) zm

, √ m

z

1 − c (1 + ωm ) zm



= (1, z)

ω is a primitive mth root of −1.

5. Classification of k-pseudo involutions with k ≡ 0 (mod 2) By (i) of Theorem 3.4, if k is odd then every k-pseudo involution is an involution. Thus we focus on k-pseudo involutions for even k. Let k = 2m. By Theorem 3.3, if Y = [ y i , j ] is a 2m-pseudo involution then Y is (m − 1)-diagonally i− j

striped. Further, if Y −1 = [ y i , j ] then y i , j = (−1) m y i , j . i.e., the inverse of a 2m-pseudo involution coincides with itself except for alternating signs of its diagonals. In this section, we prove that 2m-pseudo involutions may be classified by diagonally striped sign patterns of their inverses. For instance, let us consider the 6-pseudo involution of the 2-diagonally striped form given by †

†

⎡

1 ⎢0 ⎢ ⎢0 ⎢ ⎢1 ⎢ ⎢0 ⎢ ⎢0 Y =⎢ ⎢2 ⎢ ⎢0 ⎢0 ⎢ ⎢6 ⎢ ⎣0

⎤ 1 0 0 2 0 0 5 0 0 16

1 0 0 3 0 0 9 0 0

1 0 1 0 0 4 0 0 5 0 0 14 0 0 20

1 0 0 6 0 0

···

1 0 1 0 0 7 0 0 8

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ 1 ⎥ ⎥ 0 1 ⎥ 0 0 1⎦

G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

In fact, Y −1 = D ω Y D ω−1 where

⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ Y −1 = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

+1 −0 +0 −1 +0 −0 +2 −0 +0 −6 +0

+1 −0 +0 −2 +0 −0 +5 −0 +0 −16

+1 −0 +0 −3 +0 −0 +9 −0 +0

ω is a primitive 3rd root of −1 i.e.,

+1 −0 +0 −4 +0 −0 +14 −0

+1 −0 +0 −5 +0 −0 +20

+1 −0 +0 −6 +0 −0 ···

+1 −0 +0 −7 +0

+1 −0 +0 −8

+1 −0 +0

4043

⎤

+1 −0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ +1 ⎦

Since the signs of diagonals in Y −1 do not affect zero entries, it can be also expressed as Y −1 = MY M where M = D −1 = (1, − z). It means that Y may be also viewed as a 2-pseudo involution. In general, a sign pattern of diagonals in Y −1 is determined by its first column. We call it the diagonal sign pattern (abbreviated d.s.p.) of Y −1 . In the above example, the d.s.p. of Y −1 has 1, 0, 0, −1, 0, 0, +1, 0, 0, −1, 0, . . . . To classify distinguished d.s.p.’s, we only consider 2m-pseudo involutions whose jth diagonal has at least one nonzero entry for every j with j ≡ 0 (mod m). For each m ∈ Z+ let Em be the set of 2m-pseudo involutions and let E = {E1 , E2 , . . .} be the collection of Em ’s. The d.s.p. of a 2m-pseudo involution may be expressed as a (0, 1, −1)-vector s m = (s0,m , s1,m , . . .) T defined by



sr ,m =

if r = 0 (mod m), (−1)r /m if r ≡ 0 (mod m).

0

Since every 2m-pseudo involution in Em has the same d.s.p., we assign the vector s m to the set Em for each m ∈ Z+ . Let us now define a relation ∼ on E by Em ∼ Em if and only if sr ,m × sr ,m = 1 for all r such that r ≡ 0 (mod ), where  = lcm(m, m ). Theorem 5.1. The relation ∼ is an equivalence relation on E . Proof. If r ≡ 0 (mod m) then sr2,m = 1. Thus Em ∼ Em . Obviously, the relation ∼ satisfies symmetry. To show transitivity, we first show that

Em ∼ Em



if and only if m ≡ m ≡ 2n mod 2n+1



for some n  0.

  Assume Em ∼ Em . Let  = lcm(m, m ). Since s,m × s,m = (−1)/m (−1)/m = 1, m + m is even. If both



m

 and m are even then m = 2k and m = 2k for some k, k  1. Hence it follows from  = 2km =

2k m that km 

= k m is the common multiple of m and m less than , a contradiction. Thus both m and m are odd. Now let n and j be the largest integers  0 such that 2n and 2 j divide m and m , respectively. Then m and m are written as m = 2n (2a + 1) and m = 2 j (2b + 1) for some a, b  0. Thus  the largest integer such that 2s divides  = lcm(m, m ) is s = max{n, j }. Since both m and m are odd, we have n = j which implies that m ≡ m ≡ 2n (mod 2n+1 ). The converse may be proved by chasing the reverse procedure. Suppose that Em ∼ Er and Er ∼ Et . Then



m ≡ r ≡ 2n mod 2n+1





and r ≡ t ≡ 2 j mod 2 j +1



for some n, j  0. Since r = 2n+1 a + 2n = 2 j +1 b + 2 j for some a, b  0, we have 2n (2a + 1) = 2 j (2b + 1). By the uniqueness of factorization, we obtain a = b and n = j. Hence m ≡ t ≡ 2n (mod 2n+1 ), which implies that Em ∼ Et . Therefore, ∼ is an equivalence relation on E . 2

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G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

By Theorem 5.1, E can be partitioned into equivalence classes as



E=

   Er  r ≡ 2n mod 2n+1 .

n0

Hence all 2m-pseudo involutions are classified by 2n -pseudo involutions for every n ∈ Z+ . We end this section by characterizing 2n -pseudo involutions by means of some sequences. A 2-pseudo involution in R(C) has been characterized by a single sequence called the -sequence [4]. We are now interested in a sequence characterization of 2n -pseudo involutions for n  2. In [5], Cheon et al. observed Riordan matrices whose element xn+1,k+1 except lying in column 0 can be expressed as a linear combination with coefficients in Ω = (c 0 , c 1 , . . .) of the elements lying on the slanting diagonal obtained by moving a units up or down and b units to the right starting from xn,k i.e., −k  na+  b

xn+1,k+1 =



c j xn−aj ,k+bj ,

c 0 = 0

j =0

where a and b are integers such that a + b  1 and b  0. The sequence Ω is called the (a, b)-ray sequence with its generating function Ω(a,b) ( z). In particular, they showed that X = ( f ( z)/ z, f ( z)) is a 22 -pseudo involution if and only if X has the (1, 1)-ray sequence such that Ω(1,1) ( z)Ω(1,1) (− z) = 1. More generally, we have the following theorem: Theorem 5.2. If X ∈ R(C) is a 2n -pseudo involution for n  2 then X has the (m, m)-ray sequence where m = 2n−2 such that

Ω(m,m) ( z)Ω(m,m) (− z) = 1.

(19)

In particular, if X = ( g ( z), f ( z)) satisfies g ( z) g (ω f ( z)) = 1 then the converse holds where 2m-th root of −1.

ω is a primitive

Proof. Suppose that X = ( g ( z), f ( z)) = [xn,k ] is a 4m-pseudo involution. If X −1 = [xn,k ] then xn,k = n−k (−1) 2m xn,k . By (i) of Lemma 3.2, f (z) ∈ H2m and hence by Theorem 2.1 in [5], X has the (m, m)-ray sequence i.e., †

†

−k  n2m 

xn+1,k+1 =



c j xn−mj ,k+mj .

(20)

j =0

Substituting (−1)

† xn+1,k+1

n−k 2m

†

xn,k into xn,k yields

−k  n2m 

=

 j =0

†

(−1) j c j xn−mj ,k+mj ,

(21)

which implies that X −1 has the (m, m)-ray sequence ((−1)n cn )n0 with the generating function Ω(m,m) (− z). On the other hand, by Theorem 2.5 in [5] X −1 has the (m, m)-ray sequence with the generating function 1/Ω(m,m) ( z). Thus Ω(m,m) ( z)Ω(m,m) (− z) = 1. Conversely, assume that X = ( g ( z), f ( z)) = [xn,k ] has the (m, m)-ray sequence (cn )n0 satisfying (19). Multiplying both sides of (20) by (−1)

(−1)

n−k 2m

n−k 2m

−k  n2m 

xn+1,k+1 =

 j =0

(−1) j c j (−1)

yields

n−k−2mj 2m

xn−mj ,k+mj .

(22)

G.-S. Cheon, H. Kim / Linear Algebra and its Applications 439 (2013) 4032–4046

4045

n−k

†

It follows from (21) and (22) that xn+1,k+1 and (−1) 2m xn+1,k+1 satisfy the same recurrence relation with the coefficients (−1) j c j . If g ( z) g (ω f ( z)) = 1 then for n with n ≡ 0 (mod 2m), we have †

 

xn,0 = zn

1 g ( ¯f ( z))

  n = zn g (ω z) = ωn xn,0 = (−1) 2m xn,0 , †

which means that the two recurrence relations have the same initial conditions. Hence xn,k = n−k (−1) 2m xn,k for every n, k  0 such that n ≡ k (mod 2m), which completes the proof. 2 σ (z)+ρ (z)

We note that for any even function σ ( z) and odd function ρ ( z), Ω(m,m) ( z) = σ (z)−ρ (z) and Ω(m,m) (z) = exp(ρ (z)) satisfy (19). Further, two simple examples of g (z) satisfying g (z) g (ω f (z)) = 1 are ( f ( z)/ z)k , k  0 and zf  ( z)/ f ( z). Example 5.3. Let n = 3 in Theorem 5.2 so that m = 2. By setting σ ( z) = 1 and ρ ( z) = z we obtain ρ (z) 1+ z 2 3 Ω(2,2) (z) = σσ ((zz)+ )−ρ (z) = 1−z = 1 + 2z + 2z + 2z + · · · . Then the following matrix is a 8-pseudo involution:

⎡

1 ⎢ 0 1 ⎢ ⎢ 0 0 1 ⎢ ⎢ 0 0 0 1 ⎢ ⎢ 2 0 0 0 ⎢ 0 4 0 0

⎢ ⎢ f ( z) ⎢ 0 0 6 0 , f ( z) = ⎢ ⎢ 0 0 0 8 z ⎢ ⎢ 10 0 0 0 ⎢ 0 24 0 0 ⎢ ⎢ 0 0 42 0 ⎢ ⎢ 0 0 0 64 ⎢ ⎣ 66 0 0 0

⎤

1 0 1 0 0 0 0 10 0 0 12 0 0 0 0 90 0

1 0 0 0 14 0 0

1 0 0 0 16 0

1 0 1 0 0 0 0 18 0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1 ⎥ ⎥ 0 1 ⎥ 0 0 1⎦

··· In fact, the (9, 1)-entry is obtained from 1 · 10 + 2 · 6 + 2 · 1 = 24. It can be also easily seen that the (i , j )-entry of the matrix counts the lattice paths from (0, 0) to (3i , 0) by using the steps (2, 1), (1, 2), and (1, −1) where the paths stay in the first quadrant and start with the j consecutive (2, 1) steps. References [1] N.T. Cameron, A. Nkwanta, On some (pseudo) involutions in the Riordan group, J. Integer Seq. 8 (2005), Article 05.3.7. [2] G.-S. Cheon, H. Kim, Simple proofs of open problems about the involutions in the Riordan group, Linear Algebra Appl. 428 (2008) 930–940. [3] G.-S. Cheon, H. Kim, L.W. Shapiro, Riordan group involutions, Linear Algebra Appl. 428 (2008) 941–952. [4] G.-S. Cheon, S.-T. Jin, H. Kim, L.W. Shapiro, Riordan group involutions and the -sequence, Discrete Appl. Math. 157 (2009) 1696–1701. [5] G.-S. Cheon, S.-T. Jin, Structural properties of Riordan matrices and extending the matrices, Linear Algebra Appl. 435 (2011) 2019–2032. [6] E. Chow, A priori sparsity patterns for parallel sparse approximate inverse preconditioners, in: Special Issue on Iterative Methods for Solving Systems of Algebraic Equations, SIAM J. Sci. Comput. 21 (2000) 1804–1822. [7] M. Kuczma, B. Choczewski, R. Ger, Iterative Functional Equations, Cambridge Univ. Press, 1990. [8] A. Luzón, D. Merlini, M.A. Morón, R. Sprugnoli, Identities induced by Riordan arrays, Linear Algebra Appl. 436 (2012) 631–647. [9] N. Metropolis, G.-C. Rota, Functions of three variables, Amer. Math. Monthly 98 (1991) 328–332. [10] L.W. Shapiro, S. Getu, W.-J. Woan, L. Woodson, The Riordan group, Discrete Appl. Math. 34 (1991) 229–239. [11] L.W. Shapiro, Bijections and the Riordan group, Theoret. Comput. Sci. 307 (2003) 403–413. [12] L.W. Shapiro, Some open questions about random walks, involutions, limiting distributions and generating functions, Adv. in Appl. Math. 27 (2001) 585–596.

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[13] D.C. Smolarski, Diagonally-striped matrices and approximate inverse preconditioners, J. Comput. Appl. Math. 186 (2006) 416–431. [14] R. Sprugnoli, Riordan arrays and combinatorial sums, Discrete Math. 132 (1994) 267–290.