q-Riordan representation

Accepted Manuscript q-Riordan representation

Fatma Yesil Baran, Naim Tuglu

PII: DOI: Reference:

S0024-3795(17)30180-5 http://dx.doi.org/10.1016/j.laa.2017.03.018 LAA 14094

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Linear Algebra and its Applications

Received date: Accepted date:

24 June 2016 19 March 2017

Please cite this article in press as: F. Yesil Baran, N. Tuglu, q-Riordan representation, Linear Algebra Appl. (2017), http://dx.doi.org/10.1016/j.laa.2017.03.018

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q−Riordan Representation Fatma YESIL BARANa,∗, Naim TUGLUb a Amasya University, Faculty of Technology, Department of Computer Engineering, 05100, Amasya-Turkey b Gazi University, Faculty of Science, Department of Mathematics, 06500 Teknikokullar, Ankara-Turkey

Abstract The aim of this study is to establish further theory of q−Riordan representation. We emphasize that any q−matrices can be obtained by q−Riordan representation. Then we analyse the structure of the entries in the matrix product of general q−Riordan matrices. Keywords: Riordan representation, Pascal matrices, q-calculus 2010 MSC: 11C20, 05A30, 15B65

1. Introduction and Preliminaries q−calculus plays an important role in elliptic and hypergeometric functions, partition theory and other branches of mathematics. Also, q−calculus plays an important role in analytic number theory and combinatorics, and it has wide applications in computer sciences and the theory of special functions. If one takes the limit as q → 1− , in a simple q−calculus identity, one gets an ordinary identity, so q−calculus may be viewed as a generalization of ordinary calculus. Accordingly, several mathematical concepts are generalized thanks to q−analysis. q−calculus was first introduced by Euler [1]. In references [2, 3, 4], the authors provide some transformational formulas for q−hypergeometric functions and also present some interesting results on q−generating functions. Carlitz defined q−binomial coefficients in the following way. Let n be a nonnegative integer and q a real number with 0 < q < 1. Then the quantity [n]q is defined as  1, n=0 [n]q = 1−q n , n >0 1−q [n]q is called the q−analogue of n or the q−number of n. The q−analogue of n! is denoted by [n]q ! and it is defined as  1, n=0 [n]q ! = [n]q [n − 1]q · · · [1]q , n = 1, 2, . . . Let n and k be nonnegative integers. Then q−binomial coefficient is defined as   [n]q ! n = k q [n − k]q ! [k]q !     for n ≥ k ≥ 1. Also n0 q = 1 and nk q = 0 for n < k [5]. Also, q−analogues of several polynomials are defined in an analogous way. For instance, the q−analogue n of (x + a) is defined by  1,  n  n=0 (x + a)q = (1.1) (x + a) (x + qa) · · · x + q n−1 a , n ≥ 1 ∗ Corresponding

author Email addresses: [email protected] (Fatma YESIL BARAN), [email protected] (Naim TUGLU )

Preprint submitted to Elsevier

March 21, 2017

and this polynomial satisfies the formula (x +

n a)q

n    n

=

j

j=0

j

q (2) aj xn−j . q

This formula is called the Gauss Binomial formula [6]. Heine’s Binomial formula is:  ∞   1 n+j−1 = xj n j (1 − x)q q j=0

(1.2)

[7]. Also, the following identity holds xk =



(−1)

n−k

n≥k





n+m n−k

n−k 2

q(

)

q

xn n+m+1

(1 − x)q

(1.3)

where k ≥ 0 [8]. Pascal matrices have significant influence on combinatorics and matrix theory. Many mathematicians deal with Pascal matrices for solution of matrix problems [9, 10]. Pascal matrices are defined by binomial coefficients. Using q−binomial coefficients, analogues of Pascal matrices can be defined. The q−Pascal matrix P = Pn,q is the (n + 1) × (n + 1) matrix whose (i, j) entry pn,q (i, j) (0 ≤ i, j ≤ n) is given by the formula  j+1   q ( 2 ) ji , if i ≥ j q (1.4) pn,q (i, j) = 0, otherwise Its inverse matrix P −1 = (aij ) where  aij =

(−1)

i−j

q(

j+1 2

)−i(j+1)  i  , j q

0,

if i ≥ j otherwise.

(1.5)

The combinatorial properties of q−Pascal matrices are studied by Ernst in [11]. Riordan arrays can be used to prove some combinatorial identities, investigate sequences of numbers and solve number theoretic problems (see [12, 13, 14, 15]). Now we recall the properties of Riordan arrays and q−composition which will be used in this paper. The Riordan group is a set of lower triangular matrices each of which is defined by two generating functions g (x) = g0 + g1 x + g2 x2 + · · · with g0 = 0 and f (x) = f1 x + f2 x2 + f3 x3 + · · · . The j−th column of matrix is generated by g(x)f (x)j for j = 0, 1, 2, . . . and the resulting matrix is denoted by (g, f ). The group law is (g, f ) ∗ (u, v) = (g (u ◦ f ) , v ◦ f ) and the identity element of this group is I = (1, x). The inverse of (g, f ) −1 is (g, f ) = g◦1f¯, f¯ [16]. The following theorem is keystone in the study of Riordan arrays and it is called the Fundamental Theorem of Riordan Arrays (see [17]). ∞ ∞ Theorem 1. Let A (z) = k=0 ak z k and B (z) = k=0 bk z k be any formal power series and g, f a pair of generating functions as above for which there is a corresponding Riordan array (g, f ). Then we have (g, f ) A = B ⇔ B (z) = g (z) A (f (z)) . ∞ Let F (x) = k=0 Fk xk and f (x) = k=0 fk xk be any formal series. Garsia [18] defined the q−analogue of composition of functions as    fn F (x) F (qx) . . . F q n−1 x f ◦F = ∞

n≥0

and F ◦f =



Fn f (x)f (x/q) . . . f (x/q n−1 ).

n≥0

2

Cheon et al. [19], combined the study of q−binomial coefficients and Riordan arrays and they showed that any q−Riordan matrix may be represented using Eulerian generating functions and they also studied combinatorial properties of these matrices [20]. Also, Tuglu et al. discovered a connection between Riordan arrays and fibonomial coefficients [21]. In [22], Tuglu et al. obtained q−Riordan representation of q−Pascal matrices. This paper mainly concerns the following objects: • defining the q−analogue of the power of any function • defining the q−analogue of a Riordan representation • obtaining new formulas for the matrix product of q−Riordan matrices. 2. q−Analogue of Riordan Representation Tuglu et al. [22] introduced the q−analogue of the m−th power of a function. For m ≥ 1, they defined −

−

f [m] (x) = f (x) ∗q f [m−1] (x) and

[m]

f

−

(x) = f (x) ∗1/q f

[m−1]

−

(x)

def

def

where ∗q and ∗1/q operations is defined as g(x)∗q f (x) := g(x)·f (qx) and g(x)∗1/q f (x) := g(x)·f (x/q). For −

j ≥ 0, the authors showed that if j−th column of any Riordan matrix has generating function g(x)∗q f [j] (x) then the matrix may be shown as (g, f )q . Similarly, if j−th column of any Riordan matrix has generating [j]

function g(x) ∗1/q f − (x) then the matrix may be shown as (g, f )1/q [22]. The next lemma proves that any power of a function may be calculated in an easy way.  fn xn be any function. For m ≥ 2, the q−analogues of the m th power of f (x) Lemma 1. Let f (x) = n≥0

are given by 



f [m] (x) =

fk1

f [m] (x) =





fki+1 −ki q i(ki+1 −ki ) xkm

fk1

m−1

fki+1 −ki

i=1

km ≥0 k1 +k2 +···km−1 =km

(2.1)

i=1

km ≥0 k1 +k2 +···km−1 =km

and

m−1

 i(ki+1 −ki ) 1 xkm . q

(2.2)

Proof. The proof is by induction on m. We have f [2] (x) = f (x) ∗q f (x) = f (x) f (qx)   = fn xn fn q n x n n≥0

=

n≥0

n 

fk fn−k q n−k xn

n≥0 k=0

which shows that the statement is true for m = 2. Assume the formula of (2.1) holds for t. We will prove it t + 1. Thus we get f

[t]

(x) =





fk1

kt ≥0 k1 +k2 +···kt−1 =kt

3

t−1 i=1

fki+1 −ki q i(ki+1 −ki ) xkt .

(2.3)

Then we write

  f [t+1] (x) = f [t] (x) f q t x

and using (2.3) we obtain f [t+1] (x) =





fk1

=

 

fki+1 −ki q i(ki+1 −ki ) xkm−1

i=1

kt ≥0 k1 +k2 +···kt−1 =kt kt+1

t−1



fk 1

=



fk 1

fk q tk xk

k≥0 t−1

fki+1 −ki q i(ki+1 −ki ) xkt fkt+1 −kt q t(kt+1 −kt ) xkt+1

i=1

kt+1 ≥0 kt =0 k1 +k2 +···kt−1 =kt





t

fki+1 −ki q i(ki+1 −ki ) xkt+1 .

i=1

kt+1 ≥0 k1 +k2 +···kt =kt+1

So the proof is completed. (2.2) follows by the same method as in proof of (2.1). Let L = (ln,k ) = (g, f ) be any Riordan matrix. Obviously the following formula holds  ln,k xn = g (x) f k (x) n≥k

(see [16]). Let L = (ln,k ) = (g, f )q be any q−Riordan matrix. Using definition of (g, f )q , it can be written as  ln,k xn = g (x) ∗q f [k] (x) . (2.4) n≥k

Similarly for L = (ln,k ) = (g, f )1/q , it can be obtained 

ln,k xn = g (x) ∗1/q f [k] (x) .

(2.5)

n≥k

Lemma 2. Let L = (ln,k ) = (g, f )q be any q−Riordan matrix, then we have ln,k = [xn ]





g k1

n≥0 t1 +t2 +···tk =n

Proof. From (2.4), we have



k−1

fki+1 −ki q i(ki+1 −ki ) xn .

(2.6)

i=1

ln,k xn = g (x) ∗q f [k] (x) .

(2.7)

n≥k

Using the method of proof of (2.1), we get g (x) ∗q f

[k]





(x) =

g k1

tk+1 ≥0 t1 +t2 +···tk =tk+1

k−1

fki+1 −ki q i(ki+1 −ki ) xtk+1

i=1

Then by (2.7) and (2.8), we immediately obtain equation (2.6). Lemma 3. Let L = (ln,k ) = (g, f )1/q be any q−Riordan matrix. Then ln,k = [xn ]

∞ 



g k1

n=0 k1 +k2 +···kr =n

r i=1

Proof. It is easily seen that observing proof of Lemma 2. 4

fki+1 −ki

 i(ki+1 −ki ) 1 xn . q

(2.8)

The following theorems are the q−analogues of the FTRA [22].   Theorem 2. Let g(x) ∈ Fq (0), f (x) ∈ Fq (1), A(x) = k≥0 ak (q) xk and B(x) = k≥0 bk (q) xk . Then we have ⎤ ⎡ ⎤ ⎡ a0 (q) b0 (q) ⎢ a1 (q) ⎥ ⎢ b1 (q) ⎥ ⎥ ⎢ ⎥ ⎢ (g(x), f (x))q ⎢ a2 (q) ⎥ = ⎢ b2 (q) ⎥ , (2.9) ⎦ ⎣ ⎦ ⎣ .. .. . . if and only if the following equation holds (2.10) g(x) ∗q (A◦f ) (x) = B(x).   Theorem 3. Let g(x) ∈ Fq (0), f (x) ∈ Fq (1), A(x) = k≥0 ak (q) xk and B(x) = k≥0 bk (q) xk . Then we get ⎤ ⎡ ⎤ ⎡ b0 (q) a0 (q) ⎢ a1 (q) ⎥ ⎢ b1 (q) ⎥ ⎥ ⎢ ⎥ ⎢ (2.11) (g(x), f (x))1/q ⎢ a2 (q) ⎥ = ⎢ b2 (q) ⎥ ⎦ ⎣ ⎦ ⎣ .. .. . . if and only if the following equation holds g(x) ∗1/q (A◦f ) (x) = B(x).

(2.12)

3. The Structure of q−Riordan Matrix In [23, 24, 25], the authors show that D is a Riordan matrix if and only if there exist A and Z sequences. In this chapter, we show that the matrix product of any q−Riordan matrices can be represented by a special matrix. The following theorems yield information on the matrix product of q−Riordan matrices defined by q−operations. The special matrix need not be an ordinary Riordan matrix. This matrix is constructed by q−operations and q−powers of functions. We demonstrate that the product of any two q−Riordan matrices A = (g, f )q1 , B = (u, v)q2 (q1 , q2 ∈ {q, 1/q}) can be expressed by a representation. The following four theorems give use a new method for obtaining the matrix product of any q−matrices. We can also find a formula for the inverse of a q−matrix. [j] j For the next theorems, let hj (x) = u (x) .v (x) . Then using operations of ∗1/q and ∗q , we have h (x) = u (x) ∗1/q v

[j]

−

−

(x) and h[j] (x) = u (x) ∗q v [j] (x).

Theorem 4. Let (g, f )q and (u, v)1/q be any q−Riordan matrices. Then we have    (g, f )q . (u, v)1/q = g (x) , h◦f q 2 x 1/q .

(3.1) −

Proof. Note that for the matrix (g, f )q , j−th column of (g, f )q is g(x) ∗q f [j] (x) (j = 0, 1, 2, . . .). (g, f )q can be written as ⎡ ⎤ g |xo 0 0 0 0 ··· ⎢ g |x g ∗q f | x 0 0 0 ··· ⎥ ⎢ ⎥ ⎢ ⎥ g |x2 g ∗q f |x2 g ∗q f [2] |x2 0 0 ··· ⎥ (g, f )q = ⎢ ⎢ ⎥ ⎢ g |x3 g ∗q f |x3 g ∗q f [2] |x3 g ∗q f [3] |x3 0 · · · ⎥ ⎣ ⎦ .. .. .. .. .. . . . . .

5

where |xn is coefficient of term of xn . Suppose that the generic element of this matrix is ai,j . Then we can write ai,j = g ∗q f [j] |xi . (u, v)1/q can be written as ⎡ (u, v)1/q

⎢ ⎢ ⎢ =⎢ ⎢ ⎣

u | xo u |x u | x2 u | x3 .. .

0 u ∗1/q v |x u ∗1/q v |x2 u ∗1/q v |x3 .. .

0 0 u ∗1/q v [2] |x2 u ∗1/q v [2] |x3 .. .

0 0 0 u ∗1/q v [3] |x3 .. .

··· ··· ··· ..

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎦

.

Similarly, assume that the generic element of this matrix is bi,j , then we can write bi,j = u ∗1/q v [j] |xi . Multiplying these matrices together gives us the matrix D = (di,j )i,j≥0 . Then ij−th element of D is di,j = =

∞ 

ai,k bk,j

k=1 ∞ 

g ∗q f [k] |xi



u ∗1/q v [j] |xk

k=1

where di,j = 0 for i < j. j−th column of D is T

[0, 0, . . . , 0, dj,j , dj+1,j , dj+2,j , . . .] . Let L denote the product of the above vector and the row vector [1, x, x2 , . . .], i.e. L = dj,j xj + dj+1,j xj+1 + dj+2,j xj+2 + · · ·   ∞ ∞





  [j] [j] [k] j [k] g ∗q f |xj u ∗1/q v |xk x + g ∗q f |xj+1 u ∗1/q v |xk xj+1 + · · · = k=1



=





k=1



 g ∗q f [1] |xj u ∗1/q v |x1 + g ∗q f [2] |xj u ∗1/q v [j] |x2 + · · · xj







  [j] + g ∗q f [1] |xj+1 u ∗1/q v |x1 + g ∗q f [2] |xj+1 u ∗1/q v [j] |x2 + · · · xj+1







  [j] + g ∗q f [1] |xj+2 u ∗1/q v |x1 + g ∗q f [2] |xj+2 u ∗1/q v [j] |x2 + · · · xj+2 + · · · . [j]



Here and subsequently, after several arrangements, we get 



 g ∗q f [1] |xj xj + g ∗q f [1] |xj+1 xj+1 + g ∗q f [1] |xj+2 xj+2 + · · · L = u ∗1/q v [j] |x1  



g ∗q f [2] |xj xj + g ∗q f [2] |xj+1 xj+1 + g ∗q f [2] |xj+2 xj+2 + · · · + · · · + u ∗1/q v [j] |x2



= u ∗1/q v [j] |x1 g ∗q f [1] + u ∗1/q v [j] |x2 g ∗q f [2] + · · ·   = g ∗q u ∗1/q v [j] |x1 f [1] + u ∗1/q v [j] |x2 f [2] + · · · 

 = g ∗q u ∗1/q v [j] ◦f

= g (x) ∗q h[j] ◦f (x) Thus we obtain



  g (x) ∗q h[j] ◦f (x) = g (x) ∗1/q h[j] ◦f q 2 x

and it shows that generating function of j−th column of D is

  g (x) ∗1/q h[j] ◦f q 2 x . 6

Finally, Riordan representation of D is

   g (x) , (h◦f ) q 2 x 1/q ,

which completes the proof.



     u ∗1/q v [j] ◦f q 2 x . Note that, for the matrix g (x) , (h◦f ) q 2 x 1/q , j−th column is g (x) ∗1/q

    Specifically, the 0−th column is g (x) ∗1/q u ∗1/q v [0] ◦f q 2 x = g (x) ∗1/q (u◦f ) q 2 x . Theorem 5. Let (g, f )q and (u, v)q be any q−Riordan matrices. Then we have (g, f )q . (u, v)q = (g (x) , h◦f (x))q . Proof. (g, f )q can be written as ⎡ ⎢ ⎢ ⎢ (g, f )q = ⎢ ⎢ ⎢ ⎣

g g g g .. .

|xo |x |x2 | x3

0 g ∗q f | x g ∗ q f | x2 g ∗ q f | x3 .. .

0 0 g ∗q f [2] |x2 g ∗q f [2] |x3 .. .

0 0 0 g ∗q f [3] |x3 .. .

0 0 0 0

··· ··· ··· ··· .. .

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

where |xn is coefficient of the term in xn . Suppose that the generic element of this matrix is ai,j , then we can write ai,j = g ∗q f [j] |xi . (u, v)q can be written as ⎡ ⎤ u | xo 0 0 0 ··· ⎢ u | x u ∗q v | x 0 0 ··· ⎥ ⎢ ⎥ ⎢ ⎥ [2] 2 2 2 u | u ∗ v | u ∗ v | 0 · · · ⎢ ⎥. q q x x x (u, v)q = ⎢ ⎥ ⎢ u |x3 u ∗q v |x2 u ∗q v [2] |x3 u ∗q v [3] |x3 ⎥ ⎣ ⎦ .. .. .. .. .. . . . . . Similarly, assume that the generic element of this matrix is bi,j . Then we can write bi,j = u ∗q v [j] |xi . Multiplying these matrices together gives us the D = (di,j )i,j≥0 . Then the generic element of D is di,j = =

∞ 

ai,k bk,j

k=1 ∞ 

g ∗q f [k] |xi



u ∗q v [j] |xk



k=1

where di,j = 0 for i < j. j−th column of D is T

[0, 0, . . . , 0, dj,j , dj+1,j , dj+2,j , . . .] . Let L denote the product of the above vector and the row vector [1, x, x2 , . . .], i.e. L = dj,j xj + dj+1,j xj+1 + dj+2,j xj+2 + · · · ∞ ∞  





  [k] [j] j [k] [j] = g ∗q f |xj u ∗q v |xk x + g ∗q f |xj+1 u ∗q v |xk xj+1 + · · · k=1

k=1







 = g ∗q f |xj u ∗q v [j] |x1 + g ∗q f [2] |xj u ∗q v [j] |x2 + · · · xj







  + g ∗q f [1] |xj+1 u ∗q v [j] |x1 + g ∗q f [2] |xj+1 u ∗q v [j] |x2 + · · · xj+1







  + g ∗q f [1] |xj+2 u ∗q v [j] |x1 + g ∗q f [2] |xj+2 u ∗q v [j] |x2 + · · · xj+2 + · · · 

[1]

7

After some rearrangements, we have  



L = u ∗q v [j] |x1 g ∗q f [1] |xj xj + g ∗q f [1] |xj+1 xj+1 + g ∗q f [1] |xj+2 xj+2 + · · · 



 + u ∗q v [j] |x2 g ∗q f [2] |xj xj + g ∗q f [2] |xj+1 xj+1 + g ∗q f [2] |xj+2 xj+2 + · · · + · · ·



= u ∗q v [j] |x1 g ∗q f [1] + u ∗q v [j] |x2 g ∗q f [2] + · · ·   = g ∗q u ∗q v [j] |x1 f [1] + u ∗q v [j] |x2 q f [2] + · · · 

 = g ∗q u ∗q v [j] ◦f

= g (x) ∗q h[j] ◦f (x) . We see that generating function of j−th column of D is

g (x) ∗q h[j] ◦f (x) . Therefore, Riordan representation of D is (g (x) , (h◦f ) (x))q . Thus we have proved the theorem. Theorem 6. Let (g, f )1/q and (u, v)q be any q−Riordan matrices. Then we have (g, f )1/q . (u, v)q = (g (x) , h◦f (x))q . Proof. Using definitions of ◦ , (g, f )1/q , (u, v)q and similar observations like previous theorems, we prove the theorem. Theorem 7. Let (g, f )1/q and (u, v)1/q be any q−Riordan matrices. Then we have    (g, f )1/q . (u, v)1/q = g (x) , h◦f q 2 x 1/q . Proof. The result follows on using similar considerations to those used in the proofs of the previous theorems. Using the usual operations on matrices, we do not obtain a Riordan representation for a q−Pascal matrix and its inverse matrix. In [22], show respectively

that q−Riordan representations of the authors 1 x 1 x . q−Pascal and its inverse matrix are 1−x , 1−x and 1+x/q , 1+x/q q

1/q

The next example shows how to use these results to obtain that the product of a q−Pascal matrix and its inverse is the unit matrix.



1 x 1 x Example 1. Let (g, f )q = 1−x , 1−x and (u, v)1/q = 1+x/q , 1+x/q . From (3.1), we get q

 (g, f )q . (u, v)1/q =

x 1 , 1−x 1−x

1/q



 .

q

x 1 , 1 + x/q 1 + x/q

 1/q

    = g (x) , hj ◦f q 2 x 1/q .



  The generating function of j−th column of the matrix is g (x) ∗1/q h[j] ◦f q 2 x . We have

   j   2  h ◦f q x = u ∗1/q v [j] ◦f q 2 x

8

and using Heine’s Binomial Formula, we get x/q 1 x/q j · · · · 1 + x/q 1 + x/q 2 1 + x/q j+1   n−j  (−1) n = xn . n(j+1)−(j+1 ) j 2 q n≥j q

u ∗1/q v [j] =

Changing x to f (x) gives us

u ∗1/q v

[j]



◦f =

 n≥j

  n n (f (x)) . n(j+1)−(j+1 j q 2 ) q (−1)

n−j

(3.2)

Then we obtain g (x) ∗1/q



u ∗1/q v

[j]



◦f

 (−1)n−j n   n 1 ∗1/q f q2 x q x = n(j+1)−(j+1 ) j 1−x 2 q n≥j q   n−j n 1  (−1) n = (f (qx)) n(j+1)−(j+1 ) 1−x j 2 q n≥j q   n−j  (−1) qx qn x n 1 ··· = j+1 n(j+1)−( 2 ) j 1 − qn x q 1 − x 1 − qx n≥j q  (−1)n−j n n+1 xn = q( 2 ) j+1 n+1 n(j+1)−( 2 ) j (1 − x)q q n≥j q    n−j xn n−j n = (−1) q( 2 ) n+1 . j q (1 − x)q n≥j



2



Putting k = j and m = 0 in (1.3), we obtain 

(−1)

n−j

n≥j

Then we get

  n−j n xn j q( 2 ) n+1 = x . j q (1 − x)q

  g (x) ∗1/q h[j] ◦f q 2 x = xj . T

Thus, j−th column is [0, 0, . . . , 0, 1, 0, . . .] . References [1] L. Euler, Introductio in analysin infinitorum, Marcun-Michaelem Bousquet, Lausanne, 1748. [2] H. Srivastava, A family of q−generating functions, Bull. Inst. Math. Acad. Sin. 12 (1984) 327–336. [3] A. Agarwal, H. Srivastava, Generating functions for a class of q−polynomials, Ann. Mat. Pura Appl. 154 (4) (1989) 99–109. ¨ [4] E. Heine, Uber die Z¨ahler und Nenner der N¨ aherungswerte von Kettenbr¨ uchen, J. Reine Angew. Math. 57 (1860) 231–247. [5] L. Carlitz, Sequences and inversions, Duke Math. J. 37 (1) (1970) 193–198.

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