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\begin{center}
\vskip 1cm{\LARGE\bf
On The Complex-Type Catalan Transform of the $k$-Fibonacci Numbers \\
}
\vskip 1cm
\large
\"{O}m\"{u}r Deveci \\
Department of Mathematics \\ Faculty of Science and Letters\\
Kafkas University \\ 36100, Turkey\\
\href{mailto:odeveci36@hotmail.com}{\tt odeveci36@hotmail.com} \\
\ \\
Anthony G. Shannon\\
Honorary Fellows \\ Warrane College\\
University of New South Wales \\ 2033, Australia\\
\href{mailto:tshannon38@gmail.com}{\tt tshannon38@gmail.com} 
\end{center}

\vskip .2 in

\begin{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.
\end{abstract}
	
\section{Introduction and preliminaries}\label{sec:intro}
	
For $n\geq 1$, the $n$th Catalan number $C_{n}$ is described \cite{Barry2005} by
\begin{equation*}
C( n) =\frac{1}{n+1}\binom{2n}{n}
\end{equation*}
with generating function given by
\begin{equation*}
c( x) =\frac{1-\sqrt{1-4x}}{2x}.
\end{equation*}
Barry \cite{Barry2005} gave the Catalan transform for a given sequence $(
a_{n}) $ and its inverse as follows, respectively:
\begin{equation*}
b_{n}=\sum_{k=0}^{n}\frac{k}{2n-k}\binom{2n-k}{n-k}a_{k}
\end{equation*}
and
\begin{equation*}
a_{n}=\sum_{k=0}^{n}\binom{k}{n-k}\left( -1\right) ^{n-k}b_{k}.
\end{equation*}

For a positive real integer $k$, the $k$-Fibonacci sequence $(
F_{k,n}) _{n\in \mathbb{N} }$
is defined by the following homogeneous linear recurrence relation:
\begin{equation}
F_{k,n+1}=kF_{k,n}+F_{k,n-1}  \label{1}
\end{equation}
for $n\geq 1$, with initial conditions $F_{k,0}=0$ and $F_{k,1}=0=1$
(cf.~\cite{Falcon2008,Falcon2007(5),Falcon2007(1)}). When $k=1$ and $k=2$ in ~\eqref{1}, the sequence $
( F_{k,n}) _{n\in \mathbb{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 \cite{Falcon2013} derived a number of closed-form formulas
for the Catalan transform of the $k$-Fibonacci sequence $(
F_{k,n}) _{n\in \mathbb{N}}$
by the matrix method as follows:
\begin{equation*}
\left(
\begin{array}{c}
CF_{k,1} \\
CF_{k,2} \\
CF_{k,3} \\
CF_{k,4} \\
CF_{k,5} \\
CF_{k,6} \\
\vdots
\end{array}
\right) =\left(
\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\
1 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\
2 & 2 & 1 & 0 & 0 & 0 & 0 & \cdots \\
5 & 5 & 3 & 1 & 0 & 0 & 0  & \cdots\\
14 & 14 & 9 & 4 & 1 & 0 & 0  & \cdots\\
42 & 42 & 28 & 14 & 5 & 1 & 0  & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\right) \left(
\begin{array}{c}
F_{k,1} \\
F_{k,2} \\
F_{k,3} \\
F_{k,4} \\
F_{k,5} \\
F_{k,6} \\
\vdots
\end{array}
\right) _{\text{,}}
\end{equation*}
where $CF_{k,n}$ is the $n$th 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 \cite{Barry2005,Falcon2013}.

The Hankel matrix $H_{n}$ associated with a given sequence of real numbers $ ( a_{n}) $ is defined as follows:
\begin{equation*}
H_{n}=\left(
\begin{array}{cccc}
a_{0} & a_{1} & a_{2} & \cdots \\
a_{1} & a_{2} & a_{3} & \cdots \\
a_{2} & a_{3} & a_{4} & \cdots \\
a_{3} & a_{4} & a_{5} & \cdots \\
\vdots & \vdots & \vdots & \ddots
\end{array}
\right ) 
\end{equation*}
The sequence $( Ha_{n}=\det ( a_{i+j-2}) _{1\leq i,j\leq
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 upper-left $n\times n$
subdeterminant of the matrix $H_{n}$ (cf.~\cite{Cvetkovic,Layman,Dougherty,Barry2020,Rajkovic}). Falcon \cite
{Falcon2013} gave the relationship between the Hankel matrix transform of Catalan
transform of the $k$-Fibonacci sequence and the Fibonacci sequence by
\begin{equation*}
( HCF_{k,n}) =( F_{2n+1})\text{,}
\end{equation*}
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, \cite{Qi,Gordon,Aval,Frlinger,Aigner,Chamberland,Hilton,Davis}. 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 \cite{Barry2005} 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 \cite{Cigler,Tastan2020,Tastan2021,Barry2010,Falcon2013}. 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 complex-type Catalan
transforms of the $k$-Fibonacci numbers and the determinants of their Hankel
matrices as applications of the results produced.
		
	\section{The main results} \label{sec:rook}
	
Define the complex-type Catalan numbers as shown, for $n\geq 0$
\begin{equation}
C_{n}^{(i)}=\frac{1}{n+1}\binom{2n}{n}i^{n}  \label{D2}
\end{equation}
where $\sqrt{-1}=i$.

From Eq.~\eqref{D2}, we can write the following lower-triangular
matrix:
\begin{equation*}
C^{(i)}=\left(
\begin{array}{ccccccc}
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} &  \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\right)  .
\end{equation*}

The matrix $C^{(i)}$ is said to be the complex-type Catalan matrix. Then we
can write the following matrix relation:
\begin{equation*}
\left(
\begin{array}{ccccccc}
i^{3} & i^{4} & i^{5} & i^{6} & i^{7} & i^{8} & \cdots
\end{array}
\right) \left(
\begin{array}{c}
i \\
i \\
2i \\
5i \\
14i \\
42i \\
\vdots
\end{array}
\right) =\left(
\begin{array}{c}
C_{0}^{(i)} \\
C_{1}^{(i)} \\
C_{2}^{(i)} \\
C_{3}^{(i)} \\
C_{4}^{(i)} \\
C_{5}^{(i)} \\
\vdots
\end{array}
\right) .
\end{equation*}
It can be readily established that

\begin{equation*}
\left( C^{(i)}\right) ^{-1}=\left(
\begin{array}{cccccc}
-i &  &  &  &  &  \\
1 & -1 &  &  &  &  \\
0 & -2i & i &  &  &  \\
0 & 1 & -3 & 1 &  &  \\
0 & 0 & -3i & 4i & -i &  \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\right) .
\end{equation*}

Let
\begin{equation*}
M=\left(
\begin{array}{cccccc}
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} &  \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\right)  .
\end{equation*}

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

It is easy to see that the generating function of the complex-type Catalan
sequence $(C_{n}^{(i)}) $ is
\begin{align*}
c^{(i)}(x) &=\frac{1-\sqrt{1-4ix}}{2ix} \\
&=1+ix-2x^{2}-5ix^{3}+14x^{4}\cdots .
\end{align*}

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

\begin{proposition}
The complex-type Catalan sequence $( C_{n}^{(i)}) $ has the
following exponential representation:
\begin{equation*}
c^{(i)}(x) =\frac{1}{2ix\exp \left( \overset{\infty }{\underset{n=1
}{\sum }}\frac{1}{n}( \sqrt{1-4ix}) ^{n} \right ) }  \ .
\end{equation*}

\begin{proof}
By a simple calculation, we may write
\begin{align*}
\ln ( 2ix\cdot 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}+\cdots ) \\
&=-\left( \sum_{n=1}^{\infty }\frac{1}{n}( \sqrt{1-4ix}~)
^{n}\right) .
\end{align*}
So we have the conclusion.
\end{proof}
\end{proposition}

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

\begin{proposition}
The complex-type Catalan sequence $( C_{n}^{(i)}) $ has the
following combinatorial representation:
\begin{equation*}
c^{(i)}( x) =\sum_{n=0}^{\infty }\binom{2n}{n}\frac{x^{n}}{n+1}i^{n}
.
\end{equation*}

\begin{proof}
By the generalized binomial theorem, it is readily seen that
\begin{equation*}
\sqrt{1-4ix}=\sum_{n=0}^{\infty }\frac{-1}{( 2n-1) }\binom{2n}{n}
x^{n}i^{n} .
\end{equation*}
Then we have
\begin{align*}
\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}-\cdots ) }{2ix
} \\[.1in]
&=1+\frac{1}{6}\binom{4}{2}xi+\frac{1}{10}\binom{6}{3}x^{2}i^{2}+\cdots \\[.1in]
&=\sum_{n=1}^{\infty }\frac{1}{2\left( 2n-1\right) }\binom{2n}{n}
x^{n-1}i^{n-1} .
\end{align*}
Substituting $m$ for $n-1$, in the equation, we may write
\begin{align*}
c^{(i)}( x) &=\sum_{m=0}^{\infty }\frac{1}{2( 2m+1) }
\binom{2m+2}{m+1}x^{m}i^{m} \\
&=\sum_{n=0}^{\infty }\binom{2m}{m}\frac{x^{m}}{m+1}i^{m} .
\end{align*}
Thus the proof is complete.
\end{proof}
\end{proposition}

\bigskip Now we define a new sequence that we call the complex-type Catalan
transform of a given sequence $( a_{n}) $ as follows:
\begin{equation}
c^{(i)}[ a_{n}] =\sum_{k=0}^{n}\frac{k}{2n-k}\binom{2n-k}{n-k}
i^{k}a_{k} .  \label{D3}
\end{equation}
By Eq.~\eqref{D3}, we can easily derive
\begin{equation*}
\left(
\begin{array}{c}
c^{(i)}[ a_{1}] \\
c^{(i)}[ a_{2}] \\
c^{(i)}[ a_{3}] \\
c^{(i)}[ a_{4}] \\
c^{(i)}[ a_{5}] \\
c^{(i)}[ a_{6}] \\
\vdots
\end{array}
\right) =\left(
\begin{array}{ccccccc}
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} &  \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
\right)\left (
\begin{array}{c}
a_{1} \\
a_{2} \\
a_{3} \\
a_{4} \\
a_{5} \\
a_{6} \\
\vdots
\end{array}
\right) 
\end{equation*}
From the definition of the matrix $( C^{(i)})^{-1}$, we see that
the inverse of the complex-type Catalan transform is
\begin{equation*}
a_{n}=\sum_{k=0}^{n}\binom{k}{n-k}( -1) ^{k}i^{n}c^{(i)}[ a_{k}
] .
\end{equation*}

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 \cite{Barry2005}, 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\in \mathbb{N}}$
Using Eq.~\eqref{D3}, we define the following sequence:
\begin{equation*}
c^{(i)}[ F_{k,n}] =\sum_{j=1}^{n}\frac{j}{2n-j}\binom{2n-j}{n-j}
i^{j}F_{k,j}
\end{equation*}
with $c^{(i)}[ F_{k,0}] =0$.

Since the generating functions of the $k$-Fibonacci sequence $(
F_{k,n}) _{n\in \mathbb{N} }$
and the complex-type Catalan sequence $( C_{n}^{(i)}) $ are
\begin{equation*}
g( x) =\frac{x}{1-kx-x^{2}}
\end{equation*}
and
\begin{equation*}
c^{(i)}( x) =\frac{1-\sqrt{1-4ix}}{2ix}\text{,}
\end{equation*}
respectively, we can give the generating function of the sequence $(
c^{(i)}[ F_{k,n}] ) $ as
\begin{equation*}
g( xc^{(i)}( x)) =\frac{i-i\sqrt{1-4ix}}{-3+i(
2x-k) +( ik+1) \sqrt{1-4ix}} .
\end{equation*}

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 $k$:
\begin{align*}
c^{(i)}[ F_{k,1}] &=\sum_{j=1}^{1}\frac{j}{2-j}\binom{2-j}{1-j}
i^{j}F_{k,j}=i\text{,} \\
c^{(i)}[ F_{k,2}] &=\sum_{j=1}^{2}\frac{j}{4-j}\binom{4-j}{2-j}
i^{j}F_{k,j}=-k+i\text{,} \\
c^{(i)}[ F_{k,3}] &=\sum_{j=1}^{3}\frac{j}{6-j}\binom{6-j}{3-j}
i^{j}F_{k,j}=-ik^{2}-2k+i\text{,} \\
c^{(i)}[ F_{k,4}] &=\sum_{j=1}^{4}\frac{j}{8-j}\binom{8-j}{4-j}
i^{j}F_{k,j}=k^{3}-3ik^{2}-3k+2i\text{,} \\
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} \\
& \quad -1974k+2286i .
\end{align*}

From the coefficients of the complex-type Catalan transform of the 
$k$-Fibonacci sequence $(F_{k,n}) _{n\in \mathbb{N}}$
we can produce the an infinite triangle, where the following
are the first few rows:
\begin{table}[H]
\begin{tabular}{c|ccccccccccc}
$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$ & $\ 706$ & $-1263i$ & $-1974$ & $2286i$ &  
\end{tabular}
\caption{The complex-type Catalan triangle of the  $k$-Fibonacci
sequence.} 
\end{table}

Now we give the following useful results by the aid of the above triangle:
\begin{itemize}
\item  The first diagonal sequence $(
i,i,i,2i,6i,19i,61i,200i,670i,2286i,\ldots ) $ is the complex-type
Catalan transform of the sequence $ ( 1,0,1,0,1,0,1,0,1,0,\ldots
) $.

\item The second diagonal sequence $(
-1,-2,-3,-6,-17,-54,-176,-584,-1974\ldots ) $ is the complex-type
Catalan transform of the sequence $( i,1,i,1,2i,2,2i,2,3i,\ldots
) $,
and so on.
\end{itemize}

It is well-known that the iteration of a function $f$ is denoted by
superscript; $f^{n}$ means the $n$th iteration of function $f$, i.e.,
\begin{equation*}
f^{n}( x) =f( f\ldots f( x) \ldots ) .
\end{equation*}
Clearly,
\begin{equation*}
f^{0}( x) =x\text{ and }f^{n}( x) =f^{n-1}(
f( x)) \text{ for }n\geq 1 .
\end{equation*}
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 $n$th 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 $k$:
\begin{align*}
( c^{(i)}) ^{2}\left[ F_{k,1}\right] &=\sum_{j=1}^{1}\frac{j}{2-j}
\binom{2-j}{1-j}i^{j}c^{(i)}\left[ F_{k,j}\right] =-1\text{,} \\
( c^{(i)}) ^{2}[ F_{k,j}] &=\sum_{j=1}^{2}\frac{j}{4-j}
\binom{4-j}{2-j}i^{j}c^{(i)}[ F_{k,j}] =k-i-1\text{,} \\
( c^{(i)}) ^{2}[ F_{k,j}] &=\sum_{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\text{,} \\
( c^{(i)}) ^{2}[ F_{k,4}] &=\sum_{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\text{,} \\
( 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) \\
& \quad +k( 114i+90) -150i-101. 
\end{align*}
From the coefficients of the second iteration of the complex-type Catalan
transform of the $k$-Fibonacci sequence $( F_{k,n}) _{n\in \mathbb{N}}$,
we can produce an infinite triangle, of which the following are 
the first few rows:
\begin{table}[H]
\begin{tabular}{c|ccccccc}
$( 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
$ & $\ 56i-12$ & $-66i-6$ & $114i+90$ & $-150i-101$ 
\end{tabular}
\caption{The triangle of the sequence  $( ( c^{(i)}) ^{2}[ F_{k,n}] ) $.}
\end{table}

Thus, it can be easily seen that first diagonal sequence { $(
-1,-1-i,-2i-1,-3i-2,-6i-11,-33i-44,-150i-101,\ldots ) $ } is the
complex-type Catalan transform of the sequence $(
i,i,i,2i,6i,19i, 61i, 200i, 670i, 2286i,\ldots ) $, 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:
\begin{equation*}
x_{n+2}=i\cdot x_{n+1}+x_{n}
\end{equation*}
for $n\geq 0$, with initial conditions $x_{0}=$ $0$ and $x_{1}=1$.

Let the Hankel determinant of the complex-type Catalan transform of the $m$th term of the $k$-Fibonacci sequence $( F_{k,n}) _{n\in \mathbb{N}}$
denoted by $Hc^{(i)}[ F_{k,m}] $. Then we obtain the early part
of the sequence $( x_{n}) $ as follows:
\begin{align*}
Hc^{(i)}\left[ F_{k,0}\right] &=\det \left( 0\right) =0=x_{0}\text{,} \\
Hc^{(i)}\left[ F_{k,1}\right] &=\left\vert
\begin{array}{cc}
0 & i \\
i & -k+i
\end{array}
\right\vert =1=x_{1}\text{,} \\
Hc^{(i)}\left[ F_{k,2}\right] &=\left\vert
\begin{array}{ccc}
0 & i & -k+i \\
i & -k+i & -ik^{2}-2k+i \\
-k+i & -ik^{2}-2k+i & k^{3}-3ik^{2}-3k+2i
\end{array}
\right\vert =i=x_{2}, \\
Hc^{(i)}\left[ F_{k,3}\right] &=\left\vert {\tiny
\begin{array}{cccc}
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{array}
}\right\vert \\ &=0=x_{3}\text{,}
\end{align*}
\begin{tabular}{l}
$Hc^{i}\left[ F_{k,4}\right] {\tiny =}\left\vert {\tiny
\begin{array}{ccc}
0 & i & -k+i \\
i & -k+i & -ik^{2}-2k+i \\
-k+i & -ik^{2}-2k+i & k^{3}-3ik^{2}-3k+2i \\
-ik^{2}-2k+i & k^{3}-3ik^{2}-3k+2i & ik^{4}+4k^{3}-6ik^{2}-6k+6i \\
k^{3}-3ik^{2}-3k+2i & ik^{4}+4k^{3}-6ik^{2}-6k+6i &
-k^{5}+5k^{4}i+10k^{3}-13k^{2}i-17k+19i%
\end{array}%
}\right. \text{\ }$ \\
\ \\
$\ \ \ \ \ \ \text{\ \ }\left. {\tiny
\begin{array}{cc}
-ik^{2}-2k+i & k^{3}-3ik^{2}-3k+2i \\
k^{3}-3ik^{2}-3k+2i & ik^{4}+4k^{3}-6ik^{2}-6k+6i \\
ik^{4}+4k^{3}-6ik^{2}-6k+6i & -k^{5}+5k^{4}i+10k^{3}-13k^{2}i-17k+19i \\
-k^{5}+5k^{4}i+10k^{3}-13k^{2}i-17k+19i &
-k^{6}i-6k^{5}+15k^{4}i+24k^{3}-36k^{2}i-54k+61i \\
-k^{6}i-6k^{5}+15k^{4}i+24k^{3}-36k^{2}i-54k+61i &
k^{7}-7ik^{6}-21k^{5}+40ik^{4}+67k^{3}-114ik^{2}-176k+200i%
\end{array}%
}\right\vert {\tiny =i=x_{4}\text{\ }} .$
\end{tabular}

\bigskip

Thus we get immediately:
\begin{conjecture}
For $n\geq 0$,
\begin{equation*}
Hc^{(i)}[ F_{k,n}] =x_{n} .
\end{equation*}
\end{conjecture}

\section{Acknowledgment}

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

	
	
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	\bigskip
	\hrule
	\bigskip
	
\noindent
2020 \emph{Mathematics Subject Classification}:~Primary 11K31; Secondary 60E10, 39B32, 15A15, 11C20.

\medskip
	
\noindent
\emph{Keywords}:~ Catalan number, Catalan transform,
$k$-Fibonacci number, Hankel matrix, Hankel determinant.
	
\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000108}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  February 21 2022;
revised versions received  March 10 2022; April 4 2022.
Published in {\it Journal of Integer Sequences}, April 11 2022.

\bigskip
\hrule
\bigskip

\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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