On The Complex-Type Catalan Transform of the k-Fibonacci Numbers

Ömür Deveci Department of Mathematics Faculty of Science and Letters Kafkas University 36100, Turkey

                 odeveci36@hotmail.com

Anthony G. Shannon Honorary Fellows Warrane College University of New South Wales 2033, Australia

                 tshannon38@gmail.com

Abstract

We define a type of complex Catalan number and find some its properties. We also produce a complex Catalan transform and its inverse, together with associated generating functions and related matrices. These lead to connections with complex Catalan transforms of the k-Fibonacci numbers and the determinants of their Hankel matrices. The paper finishes with a conjecture.

1 Introduction and preliminaries

For n ≥ 1, the nth Catalan number C_{n} is described [13] by

C(n) = \frac{1}{n + 1}\binom{2n}{n}

with generating function given by

c(x) = \frac{1 − \sqrt{1 − 4x}}{2x}.

Barry [13] gave the Catalan transform for a given sequence (a_{n}) and its inverse as follows, respectively:

b_{n} = ∑_{k=0}^{n} \frac{k}{2n − k} \binom{2n − k}{n − k}a_{k}

and

a_{n} = ∑_{k=0}^{n} \binom{k}{n − k} (−1)^{n−k} b_{k}.

For a positive real integer k, the k-Fibonacci sequence (F_{k,n})_{n∈ℕ} is defined by the following homogeneous linear recurrence relation:

F_{k,n+1} = kF_{k,n} + F_{k,n−1}  (1)

for n ≥ 1, with initial conditions F_{k,0} = 0 and F_{k,1} = 0 = 1 (cf. [19, 20, 21]). When k = 1 and k = 2 in (1), the sequence (F_{k,n})_{n∈ℕ} reduces to the usual Fibonacci sequence and Pell sequence, respectively.

Number-theoretic properties such as these obtained from the Catalan transforms of the k-Fibonacci numbers relevant to this article have been studied by Falcon. Falcon [18] derived a number of closed-form formulas for the Catalan transform of the k-Fibonacci sequence (F_{k,n})_{n∈ℕ} by the matrix method as follows:

( CF_{k,1} ; CF_{k,2} ; CF_{k,3} ; CF_{k,4} ; CF_{k,5} ; CF_{k,6} ; ⋮ ) = ( \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 2 & 2 & 1 & 0 & 0 & 0 & 0 & ⋯ \\ 5 & 5 & 3 & 1 & 0 & 0 & 0 & ⋯ \\ 14 & 14 & 9 & 4 & 1 & 0 & 0 & ⋯ \\ 42 & 42 & 28 & 14 & 5 & 1 & 0 & ⋯ \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ( F_{k,1} ; F_{k,2} ; F_{k,3} ; F_{k,4} ; F_{k,5} ; F_{k,6} ; ⋮ )_{,}

where CF_{k,n} is the nth Catalan transform of the k-Fibonacci numbers. We denote the above lower-triangular matrix by C. Note that the first column of the matrix C is the sequence of the Catalan numbers. For more information on the matrix C, see [13, 18].

The Hankel matrix H_{n} associated with a given sequence of real numbers (a_{n}) is defined as follows:

H_{n} = ( \begin{matrix}a_{0} & a_{1} & a_{2} & ⋯ \\ a_{1} & a_{2} & a_{3} & ⋯ \\ a_{2} & a_{3} & a_{4} & ⋯ \\ a_{3} & a_{4} & a_{5} & ⋯ \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

The sequence (Ha_{n} = det(a_{i+j−2})_{1≤i,j≤n+1}) is called the Hankel matrix transform of the sequence (a_{n}). Namely, the Hankel determinant of order n of the sequence (a_{n}) is the upperleft n × n subdeterminant of the matrix H_{n} (cf. [1, 7, 10, 15, 17]). Falcon [18] gave the relationship between the Hankel matrix transform of Catalan transform of the k-Fibonacci sequence and the Fibonacci sequence by

(HCF_{k,n}) = (F_{2n+1}),

where F_{2n+1} is the (2n + 1)th Fibonacci number.

In the literature, many interesting properties and applications of the Catalan numbers relevant to this paper have been studied by many authors; for example, [2, 3, 4, 6, 8, 9, 16, 22]. In the first part of the paper, we consider a new generalization of the Catalan sequence that we call the complex-type Catalan sequence. Then we give some of its properties, such as the generating matrix, the generating function, the exponential representation, and some combinatorial representations.

Barry [13] defined the Catalan transform of a sequence and its inverse by the aid of the Riordan group and then gave many transformed sequences. Since then, the Catalan transforms of the linear recurrence sequences have been a topic of interest, and some new sequences have been derived by using these transforms; see [5, 11, 12, 14, 18]. Here we derive the complex-type Catalan transform and its inverse by the generating matrix of the complex-type Catalan sequence and then give the generating function of the complex-type Catalan transform of a given sequence. Finally, we obtain new sequences from the complextype Catalan transforms of the k-Fibonacci numbers and the determinants of their Hankel matrices as applications of the results produced.

2 The main results

Define the complex-type Catalan numbers as shown, for n ≥ 0

C_{n}^{(i)} = \frac{1}{n + 1}\binom{2n}{n}i^{n}  (2)

where \sqrt{−1} = i.

From Eq. (2), we can write the following lower-triangular matrix:

C^{(i)} = ( \begin{matrix}i \\ i & i^{2} \\ 2i & 2i^{2} & i^{3} \\ 5i & 5i^{2} & 3i^{3} & i^{4} \\ 14i & 14i^{2} & 9i^{3} & 4i^{4} & i^{5} \\ 42i & 42i^{2} & 28i^{3} & 14i^{4} & 5i^{5} & i^{6} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

The matrix C^{(i)} is said to be the complex-type Catalan matrix. Then we can write the following matrix relation:

( i^{3} i^{4} i^{5} i^{6} i^{7} i^{8} ⋯) ( i ; i ; 2i ; 5i ; 14i ; 42i ; ⋮ ) = ( C_{0}^{(i)} ; C_{1}^{(i)} ; C_{2}^{(i)} ; C_{3}^{(i)} ; C_{4}^{(i)} ; C_{5}^{(i)} ; ⋮ ) .

It can be readily established that

(C^{(i)})^{−1} = ( \begin{matrix}−i \\ 1 & −1 \\ 0 & −2i & i \\ 0 & 1 & −3 & 1 \\ 0 & 0 & −3i & 4i & −i \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

Let

M = ( \begin{matrix}i \\ i & i^{2} \\ i & i^{2} & i^{3} \\ i & i^{2} & i^{3} & i^{4} \\ i & i^{2} & i^{3} & i^{4} & i^{5} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

It is clear that C^{(i)} = C ∘ M, where C ∘ M denotes the Hadamard product C and M. Suppose that the terms of the matrices (C^{(i)})^{−1} and C^{−1} ∘ M are indicated by (c^{(i)})_{k,j}^{−1} and (c^{−1} ∘ m)_{k,j}^{−1}, respectively. Then, by a simple calculation, we obtain

((C^{(i)})_{k,j}^{−1})i^{k+j} = (c^{−1} ∘ m)_{k,j}^{−1}.

It is easy to see that the generating function of the complex-type Catalan sequence (C_{n}^{(i)}) is

c^{(i)}(x) = \frac{1 − \sqrt{1 − 4ix}}{2ix}
= 1 + ix − 2x^{2} − 5ix^{3} + 14x^{4} · · · .

Now we give an exponential representation for the complex-type Catalan numbers by the aid of the generating function with the following Proposition.

Proposition 1. The complex-type Catalan sequence (C_{n}^{(i)}) has the following exponential representation:

c^{(i)}(x) = \frac{1}{2ix exp ( ∞ ; ∑_{n=1} \frac{1}{n}(\sqrt{1 − 4ix})^{n})} .

Proof. By a simple calculation, we may write

ln(2ix · c^{(i)}(x)) = ln(1 − \sqrt{1 − 4ix} )

= −(\sqrt{1 − 4ix} + \frac{1}{2}(\sqrt{1 − 4ix} )^{2} + \frac{1}{3}(\sqrt{1 − 4ix} )^{3} + ⋯)

= − (∑_{n=1}^{∞} \frac{1}{n}(\sqrt{1 − 4ix} )^{n}) .

So we have the conclusion.

The following proposition gives an alternative version for the above expression.

Proposition 2. The complex-type Catalan sequence (C_{n}^{(i)}) has the following combinatorial representation:

c^{(i)}(x) = ∑_{n=0}^{∞} \binom{2n}{n}\frac{x^{n}}{n + 1}i^{n}.

Proof. By the generalized binomial theorem, it is readily seen that

\sqrt{1 − 4ix} = ∑_{n=0}^{∞} \frac{−1}{(2n − 1)}\binom{2n}{n}x^{n}i^{n}.

Then we have

\frac{1 − \sqrt{1 − 4ix}}{2ix} = \frac{1 − (1 − \binom{2}{1}xi − \frac{1}{3} \binom{4}{2}x^{2}i^{2} − \frac{1}{5} \binom{6}{3}x^{3}i^{3} − ⋯)}{2ix}

= 1 + \frac{1}{6}\binom{4}{2}xi + \frac{1}{10}\binom{6}{3}x^{2}i^{2} + ⋯

= ∑_{n=1}^{∞} \frac{1}{2 (2n − 1)}\binom{2n}{n}x^{n−1}i^{n−1}.

Substituting m for n − 1, in the equation, we may write

c^{(i)}(x) = ∑_{m=0}^{∞} \frac{1}{2(2m + 1)}\binom{2m + 2}{m + 1}x^{m}i^{m}

= ∑_{n=0}^{∞} \binom{2m}{m}\frac{x^{m}}{m + 1}i^{m}.

Thus the proof is complete.

Now we define a new sequence that we call the complex-type Catalan transform of a given sequence (a_{n}) as follows:

c^{(i)}[a_{n}] = ∑_{k=0}^{n} \frac{k}{2n − k} \binom{2n − k}{n − k}i^{k}a_{k}.  (3)

By Eq. (3), we can easily derive

( c^{(i)}[a_{1}] ; c^{(i)}[a_{2}] ; c^{(i)}[a_{3}] ; c^{(i)}[a_{4}] ; c^{(i)}[a_{5}] ; c^{(i)}[a_{6}] ; ⋮ ) = ( \begin{matrix}i \\ i & i^{2} \\ 2i & 2i^{2} & i^{3} \\ 5i & 5i^{2} & 3i^{3} & i^{4} \\ 14i & 14i^{2} & 9i^{3} & 4i^{4} & i^{5} \\ 42i & 42i^{2} & 28i^{3} & 14i^{4} & 5i^{5} & i^{6} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) ( a_{1} ; a_{2} ; a_{3} ; a_{4} ; a_{5} ; a_{6} ; ⋮ )

From the definition of the matrix (C^{(i)})^{−1}, we see that the inverse of the complex-type Catalan transform is

a_{n} = ∑_{k=0}^{n} \binom{k}{n − k}(−1)^{k}i^{n}c^{(i)}[a_{k}].

Let c(x) be the generating function of the Catalan sequence (C_{n}) and let g(x) be the generating function of the sequence (a_{n}). In [13], it is proved that the generating function of the Catalan transform of the sequence (a_{n}) is g(xc(x)). Since c^{(i)}(x) = c(xi), it can be clearly seen that g(xc^{(i)}(x)) is the generating function of the complex-type Catalan transform of the sequence (a_{n}).

Now we concentrate on the complex-type Catalan transform of the k-Fibonacci sequence (F_{k,n})_{n∈ℕ} Using Eq. (3), we define the following sequence:

c^{(i)}[F_{k,n}] = ∑_{j=1}^{n} \frac{j}{2n − j} \binom{2n − j}{n − j}i^{j}F_{k,j}

with c^{(i)}[F_{k,0}] = 0.

Since the generating functions of the k-Fibonacci sequence (F_{k,n})_{n∈ℕ} and the complextype Catalan sequence (C_{n}^{(i)}) are

g(x) = \frac{x}{1 − kx − x^{2}}

and

c^{(i)}(x) = \frac{1 − \sqrt{1 − 4ix}}{2ix},

respectively, we can give the generating function of the sequence (c^{(i)}[F_{k,n}]) as

g(xc^{(i)}(x)) = \frac{i − i\sqrt{1 − 4ix}}{−3 + i(2x − k) + (ik + 1)\sqrt{1 − 4ix}}.

The complex-type Catalan transform of the first k-Fibonacci numbers, that is, the first members of the sequence (c^{(i)}[F_{k,n}]) are the following polynomials in k:

c^{(i)}[F_{k,1}] = ∑_{j=1}^{1} \frac{j}{2 − j} \binom{2 − j}{1 − j}i^{j}F_{k,j} = i,

c^{(i)}[F_{k,2}] = ∑_{j=1}^{2} \frac{j}{4 − j} \binom{4 − j}{2 − j}i^{j}F_{k,j} = −k + i,

c^{(i)}[F_{k,3}] = ∑_{j=1}^{3} \frac{j}{6 − j} \binom{6 − j}{3 − j}i^{j}F_{k,j} = −ik^{2} − 2k + i,

c^{(i)}[F_{k,4}] = ∑_{j=1}^{4} \frac{j}{8 − j} \binom{8 − j}{4 − j}i^{j}F_{k,j} = k^{3} − 3ik^{2} − 3k + 2i,
c^{(i)}[F_{k,5}] = ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i,
c^{(i)}[F_{k,6}] = −k^{5} + 5ik^{4} + 10k^{3} − 13ik^{2} − 17k + 19i,
c^{(i)}[F_{k,7}] = −ik^{6} − 6k^{5} + 15ik^{4} + 24k^{3} − 36ik^{2} − 54k + 61i,
c^{(i)}[F_{k,8}] = k^{7} − 7ik^{6} − 21k^{5} + 40ik^{4} + 67k^{3} − 114ik^{2} − 176k + 200i,
c^{(i)}[F_{k,9}] = ik^{8} + 8k^{7} − 28ik^{6} − 62k^{5} + 115ik^{4} + 212k^{3} − 376ik^{2} − 584k + 670i,
c^{(i)}[F_{k,10}] = −k^{9} + 9ik^{8} + 36k^{7} − 91ik^{6} − 186k^{5} + 366ik^{4} + 706k^{3} − 1263ik^{2}
− 1974k + 2286i.

From the coefficients of the complex-type Catalan transform of the k-Fibonacci sequence (F_{k,n})_{n∈ℕ} we can produce the an infinite triangle, where the following are the first few rows:

c^{(i)}[F_{k,1}] i
c^{(i)}[F_{k,2}] −1 i
c^{(i)}[F_{k,3}] −i −2  i
c^{(i)}[F_{k,4}] 1 −3i −3  2i
c^{(i)}[F_{k,5}] i 4 −6i −6 6i
c^{(i)}[F_{k,6}] −1 5i 10 −13i −17  19i
c^{(i)}[F_{k,7}] −i −6 15i 24 −36i −54  61i
c^{(i)}[F_{k,8}] 1 −7i −21 40i  67 −114i −176 200i
c^{(i)}[F_{k,9}] i  8 −28i −62 115i 212 −376i −584 670i
c^{(i)}[F_{k,10}] −1 9i 36 −91i −186 366i 706 −1263i −1974 2286i

Table 1: The complex-type Catalan triangle of the k-Fibonacci sequence.

Now we give the following useful results by the aid of the above triangle:

• The first diagonal sequence (i, i, i, 2i, 6i, 19i, 61i, 200i, 670i, 2286i, …) is the complextype Catalan transform of the sequence (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, …).

• The second diagonal sequence (−1, −2, −3, −6, −17, −54, −176, −584, −1974 …) is the complex-type Catalan transform of the sequence (i, 1, i, 1, 2i, 2, 2i, 2, 3i, …), and so on.

It is well-known that the iteration of a function f is denoted by superscript; f^{n} means the nth iteration of function f, i.e.,

f^{n}(x) = f(f … f(x) …).

Clearly,

f^{0}(x) = x and f^{n}(x) = f^{n−1}(f(x)) for n ≥ 1.

We denote c^{(i)}[(c^{(i)})^{n−1}[a_{n}]] by (c^{(i)})^{n}[a_{n}]. The sequence (c^{(i)})^{n}[a_{n}] is said to be the nth iteration of the complex-type Catalan transform of the sequence (a_{n}).

The second iteration of the complex-type Catalan transform of the first k-Fibonacci numbers, that is, the first members of the sequence ((c^{(i)})^{2}[F_{k,n}]) are the following polynomials in k:

(c^{(i)})^{2} [F_{k,1}] = ∑_{j=1}^{1} \frac{j}{2 − j} \binom{2 − j}{1 − j}i^{j}c^{(i)} [F_{k,j}] = −1,

(c^{(i)})^{2}[F_{k,j}] = ∑_{j=1}^{2} \frac{j}{4 − j} \binom{4 − j}{2 − j}i^{j}c^{(i)}[F_{k,j}] = k − i − 1,

(c^{(i)})^{2}[F_{k,j}] = ∑_{j=1}^{3} \frac{j}{6 − j} \binom{6 − j}{3 − j}i^{j}c^{(i)}[F_{k,j}] = −k^{2} + k(2 + 2i) − 2i − 1,

(c^{(i)})^{2}[F_{k,4}] = ∑_{j=1}^{4} \frac{j}{8 − j} \binom{8 − j}{4 − j}i^{j}c^{(i)}[F_{k,j}] = k^{3} + k^{2}(−3i − 3) + k(6i + 2) − 3i − 2,
(c^{(i)})^{2}[F_{k,5}] = −k^{4} + k^{3}(4i + 4) + k^{2}(−12i − 3) + k(12i + 2) − 6i − 11,
(c^{(i)})^{2}[F_{k,6}] = k^{5} + k^{4}(−5i − 5) + k^{3}(20i + 4) + k^{2}(−29i + 2) + k(26i + 17) + −33i − 44,
(c^{(i)})^{2}[F_{k,7}] = −k^{6} + k^{5}(6i + 6) + k^{4}(−30i − 5) + k^{3}(56i − 12) + k^{2}(−66i − 6)
+ k(114i + 90) − 150i − 101.

From the coefficients of the second iteration of the complex-type Catalan transform of the k-Fibonacci sequence (F_{k,n})_{n∈ℕ}, we can produce an infinite triangle, of which the following are the first few rows:

(c^{(i)})^{2}[F_{k,1}] −1
(c^{(i)})^{2}[F_{k,2}] −1 −1 − i
(c^{(i)})^{2}[F_{k,3}] −1 2 + 2i −2i − 1
(c^{(i)})^{2}[F_{k,4}] 1 −3i − 3 6i + 2 −3i − 2
(c^{(i)})^{2}[F_{k,5}] −1 4i + 4 −12i − 3 12i + 2 −6i − 11
(c^{(i)})^{2}[F_{k,6}] 1 −5i − 5 20i + 4 −29i + 2 26i + 17 −33i − 44
(c^{(i)})^{2}[F_{k,7}] −1 6i + 6 −30i − 5 56i − 12 −66i − 6 114i + 90 −150i − 101
Table 2: The triangle of the sequence ((c^{(i)})^{2}[F_{k,n}]).

Thus, it can be easily seen that first diagonal sequence (−1, −1 − i, −2i − 1, −3i − 2, −6i − 11, −33i − 44, −150i − 101, …) is the complex-type Catalan transform of the sequence (i, i, i, 2i, 6i, 19i, 61i, 200i, 670i, 2286i, …), which is first diagonal sequence of the complex-type Catalan triangle of the k-Fibonacci sequence.

We will now address the Hankel matrix transform of the sequence (c^{(i)}[F_{k,n}]). Consider the following recursively defined sequence:

x_{n+2} = i · x_{n+1} + x_{n}

for n ≥ 0, with initial conditions x_{0} = 0 and x_{1} = 1.

Let the Hankel determinant of the complex-type Catalan transform of the mth term of the k-Fibonacci sequence (F_{k,n})_{n∈ℕ} denoted by Hc^{(i)}[F_{k,m}]. Then we obtain the early part of the sequence (x_{n}) as follows:

Hc^{(i)} [F_{k,0}] = det (0) = 0 = x_{0},
Hc^{(i)} [F_{k,1}] = | \begin{matrix}0 & i \\ i & −k + i \end{matrix} | = 1 = x_{1},

Hc^{(i)} [F_{k,2}] = | \begin{matrix}0 & i & −k + i \\ i & −k + i & −ik^{2} − 2k + i \\ −k + i & −ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i \end{matrix} | = i = x_{2},

Hc^{(i)} [F_{k,3}] = | \begin{matrix}0 & i & −k + i & −ik^{2} − 2k + i \\ i & −k + i & −ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i \\ −k + i & −ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i & ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i \\ −ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i & ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i & −k^{5} + 5k^{4}i + 10k^{3} − 13k^{2}i − 17k + 19i \end{matrix} |
= 0 = x_{3},

\begin{matrix}Hc^{i} [F_{k,4}] = | \begin{matrix}0 & i & −k + i \\ i & −k + i & −ik^{2} − 2k + i \\ −k + i & −ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i \\ −ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i & ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i \\ k^{3} − 3ik^{2} − 3k + 2i & ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i & −k^{5} + 5k^{4}i + 10k^{3} − 13k^{2}i − 17k + 19i \end{matrix} \\ \begin{matrix}−ik^{2} − 2k + i & k^{3} − 3ik^{2} − 3k + 2i \\ k^{3} − 3ik^{2} − 3k + 2i & ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i \\ ik^{4} + 4k^{3} − 6ik^{2} − 6k + 6i & −k^{5} + 5k^{4}i + 10k^{3} − 13k^{2}i − 17k + 19i \\ −k^{5} + 5k^{4}i + 10k^{3} − 13k^{2}i − 17k + 19i & −k^{6}i − 6k^{5} + 15k^{4}i + 24k^{3} − 36k^{2}i − 54k + 61i \\ −k^{6}i − 6k^{5} + 15k^{4}i + 24k^{3} − 36k^{2}i − 54k + 61i & k^{7} − 7ik^{6} − 21k^{5} + 40ik^{4} + 67k^{3} − 114ik^{2} − 176k + 200i \end{matrix} | = i = x_{4} . \end{matrix}

Thus we get immediately:

Conjecture 3. For n ≥ 0,

Hc^{(i)}[F_{k,n}] = x_{n}.

3 Acknowledgment

We would like to thank the editor and the referee for their valuable comments and suggestions.

References

[1] A. Cvetković, P. Rajković, and M. Ivković, Catalan numbers, the Hankel transform, and Fibonacci numbers, J. Integer Sequences 5 (2002), Article 02.1.3.

[2] F. Qi, X. T. Shi, M. Mahmoud, and F. F. Liu, The Catalan numbers: a generalization, an exponential representation, J. Comput. Anal. Appl. 23 (2017), 937–944.

[3] I. Gordon and S. Griffeth, Catalan numbers for complex reflection groups, Amer. J. Math. 134 (2012), 1491–1502.

[4] J. C. Aval, Multivariate Fuss-Catalan numbers, Discrete Math. 308 (2008), 4660–4669.

[5] J. Cigler, Some relations between generalized Fibonacci and Catalan numbers, Osterreich Akad. Wiss. Math.-Natur. Kl. Sitzungsber. II 211 (2002), 143–154.

[6] J. Fürlinger and J. Hofbauer, q-Catalan numbers, J. Combin. Theory Ser. A 40 (1985), 248–264.

[7] J. W. Layman, The Hankel transform and some of its properties, J. Integer Sequences 4 (2001), Article 01.1.5.

[8] M. Aigner, Catalan and other numbers: a recurrent theme, in H. Crapo and D. Senato, Eds., Algebraic Combinatorics and Computer Science: A Tribute to Gian-Carlo Rota, Springer, 2001, pp. 347–390.

[9] M. Chamberland and C. French, Generalized Catalan numbers and generalized Hankel transformations, J. Integer Sequences 10 (2007), Article 07.1.1.

[10] M. Dougherty, C. French, B. Saderholm, and W. Qian, Hankel transforms of linear combinations of Catalan numbers, J. Integer Sequences 14 (2011), Article 11.5.1.

[11] M. Tastan and E. Ozkan, Catalan transform of the k-Jacobsthal sequence, Electron. J. Math. Anal. Appl. 2 (2020), 70–74.

[12] M. Tastan and E. Ozkan, Catalan transform of the k-Pell, k-Pell-Lucas and modified k-Pell sequence, Notes Number Theory Discrete Math. 27 (2021), 198–207.

[13] P. Barry, A Catalan transform and related transformations of integer sequences, J. Integer Sequences 8 (2005), Article 05.4.5.

[14] P. Barry, Generalized Catalan numbers, Hankel transforms and Somos-4 sequences, J. Integer Sequences 13 (2010), Article 10.7.2.

[15] P. Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, preprint (2020). Available at http://arxiv.org/abs/2011.10827.

[16] P. Hilton and J. Pedersen, Catalan numbers, their generalization, and their uses, Math. Intelligencer 13 (1991), 64–75.

[17] P. M. Rajković, M. D. Petković, and P. Barry, The Hankel transform of the sum of consecutive generalized Catalan numbers, Integral Transforms Spec. Funct. 18 (2007), 285–296.

[18] S. Falcon, Catalan transform of the k-Fibonacci sequence, Commun. Korean Math. Soc. 28 (2013), 827–832.

[19] S. Falcon and A. Plaza, The k-Fibonacci hyperbolic functions, Chaos Solitons Fractals 38 (2008), 409–420.

[20] S. Falcon and A. Plaza, On the Fibonacci k-numbers, Chaos Solitons Fractals 32 (2007), 1615–1624.

[21] S. Falcon and A. Plaza, The k-Fibonacci sequence and the Pascal 2-triangle, Chaos Solitons Fractals 33 (2007), 38–49.

[22] T. Davis, Catalan numbers, Unpublished notes, 2006.

2020 Mathematics Subject Classification: Primary 11K31; Secondary 60E10, 39B32, 15A15, 11C20.

Keywords: Catalan number, Catalan transform, k-Fibonacci number, Hankel matrix, Hankel determinant.

(Concerned with sequence \underline{A000108}.)

Received February 21 2022; revised versions received March 10 2022; April 4 2022. Published in Journal of Integer Sequences, April 11 2022.

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