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\begin{center}
\vskip 1cm
{\LARGE\bf Some Properties of the Multiple Binomial\\
\vskip .04in
Transform and the Hankel Transform \\
\vskip .17in
of Shifted Sequences}
\vskip 1cm
\large
Jiaqiang Pan\\ 
School of Biomedical Engineering and Instrumental Science\\ 
Zhejiang University\\ 
Hangzhou 210027\\
China\\ 
\href{mailto:panshw@mail.hz.zj.cn} {\tt panshw@mail.hz.zj.cn}\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, we study the multiple binomial transform
and the Hankel transform of shifted sequences of an integer
sequence, particularly a linear homogeneous recurrence sequence,
and some of their properties.
\end{abstract}

\section{Notation}

In this paper, we generally use function symbols, like $a(t)$,
$b(t)$, etc., to express integer sequences, where
$t\in{\mathbb{N}}_0=\{0,1,2,\ldots\}$. However sometimes, to
employ matrix tools in deduction process, we also denote the
integer sequences by using (infinite-dimensional) vector symbols,
like $a=(a(0),a(1),a(2),a(3),\cdots,\cdots)^T$,
$b=(b(0),b(1),b(2),b(3),\cdots,\cdots)^T$, etc.

\section{Multiple binomial transforms of shifted sequences}

\begin{definition}[Shifting integer sequences]\label{d:ShiftingSequence}
Let $a(t)$ be an integer sequence and $\sigma$ be the shift
operator. Then we define \emph{the $p$th-order shifted sequence}
$a_{(p)}(t))$, ($p=0,1,2,\ldots$), of $a(t)$,  as follows:
\begin{equation}\label{e:ShiftingTrans}
a_{(p)}(t)=\sigma^{p}(a)=a(t+p), \qquad t=0,1,2,\ldots,
\end{equation}
Note that in the case $p=0$, $a_{(0)}(t)=\sigma^{0}(a)=a(t)$.
\end{definition}

\begin{definition}[Multiple binomial transforms]\label{d:BinomialTrans}
Let $a(t)$ be an integer sequence. Then according to
Pan \cite{ref1}, we define \emph{the $n$-fold binomial transform}
of $a(t)$, and denote its image sequence by ${\mathcal{B}}_n(a)$
or $a^{(n)}(t)$, as follows:
\begin{equation}\label{e:MultiTrans1}
a^{(1)}(t)={\mathcal{B}}_1(a)=\sum_{k=0}^{t} \binom{t}{k}a(k),
\qquad
a^{(n)}(t)={\mathcal{B}}_n(a)=\overbrace{{\mathcal{B}}_1({\mathcal{B}}_1
(\cdots({\mathcal{B}}_1}^{n-fold}(a)))),
\end{equation}
where $n=0,1,2,\ldots$. Note that in the case $n=0$,
${\mathcal{B}}_0(a)=a^{(0)}(t)=a(t)$, that is, the transform
${\mathcal{B}}_0$ just is the identity transform.
\end{definition}


\begin{definition}[Inverse multiple binomial transform]\label{d:InverseTrans}
Let $a(t)$ be an integer sequence. Then according to
Pan \cite{ref1}, we define \emph{the $m$-fold inverse binomial
transform} of $a(t)$, and denote its image sequence by
${\mathcal{B}}_{-m}(a)$ or $a^{(-m)}(t)$, as follows:
\begin{equation}\label{e:InverseMulti1}
a^{(-1)}(t)={\mathcal{B}}_{-1}(a)=\sum_{k=0}^{t}
(-1)^{t-k}\binom{t}{k}a(k), \quad
a^{(-m)}(t)={\mathcal{B}}_{-m}(a)=\overbrace{{\mathcal{B}}_{-1}({\mathcal{B}}_{-1}
(\cdots({\mathcal{B}}_{-1}}^{m-fold}(a)))),
\end{equation}
where $m=1,2,\ldots$.
\end{definition}

\begin{remark}\label{r:MatrixForm1}
We can express (\ref{e:MultiTrans1}) in the matrix form:
$a^{(1)}=B_1 a$, where the transform matrix $B_1$ is an infinite-order
lower-triangular matrix, as follows:
\begin{equation}\label{e:MultiBinom1}
B_1=\left(\begin{array}{ccccc} \binom{0}{0} & & & & \\
\binom{1}{0} & \binom{1}{1} & & & \\ \binom{2}{0} & \binom{2}{1} &
\binom{2}{2} & &
 \\ \binom{3}{0} & \binom{3}{1} & \binom{3}{2} & \binom{3}{3} & \\
  \vdots & \vdots & \vdots & \vdots
  & \ddots \end{array} \right)=\left(\begin{array}{ccccc} 1 & & &
& \\
 1 & 1 & & & \\ 1 & 2 & 1 & &
 \\ 1 & 3 & 3 & 1 & \\
  \vdots & \vdots & \vdots & \vdots
  & \ddots \end{array} \right),
\end{equation}
and
\begin{equation}\label{e:MultiBinom3}
a^{(n)}=(a^{(n)}(0),a^{(n)}(1),a^{(n)}(2),\cdots,\cdots)^T=B_n a=
B_1^n a,
\end{equation}
where $n=0,1, 2, 3, \ldots$. The transform matrix of the $n$-fold
binomial transform $B_n$ $(=B_1^n)$ is always a lower-triangular
transform matrix with each of the diagonal elements being one.
\end{remark}

\begin{remark}\label{r:MatrixForm2}
We can also express (\ref{e:InverseMulti1}) in matrix form, as
$a^{(-1)}=B_{-1} a$, where the transform matrix $B_{-1}$ is an
infinite-order lower-triangular matrix, as
\begin{equation}\label{e:InverseMulti2}
B^{-1}=\left(\begin{array}{rrrrr} \binom{0}{0} & & & & \\
-\binom{1}{0} & \binom{1}{1} & & & \\ \binom{2}{0} & -\binom{2}{1}
& \binom{2}{2} & &
 \\ -\binom{3}{0} & \binom{3}{1} & -\binom{3}{2} & \binom{3}{3} & \\
  \vdots & \vdots & \vdots & \vdots
  & \ddots \end{array} \right)=\left(\begin{array}{ccccc} 1 & &
& & \\
 -1 & 1 & & & \\ 1 & -2 & 1 & & \\ -1 & 3 & -3 & 1 &
 \\ \vdots & \vdots & \vdots & \vdots
  & \ddots \end{array} \right),
\end{equation}
and
\begin{equation}\label{e:InverseMulti3}
a^{(-m)}=(a^{(-m)}(0),a^{(-m)}(1),a^{(-m)}(2),\cdots,\cdots)^T=B_{-m}a=B_{-1}^{m}
a,
\end{equation}
where $m=1, 2, 3, \ldots$. The transform matrix $B_{-m}$
$(=B_{-1}^{m})$ is also always a lower-triangular transform matrix
with each of the diagonal elements being one. We see that
$B_1B_{-1}=B_{-1}B_1=E$, where $E$ is the infinite-order unit
matrix. It is the matrix form of well-known inversion relation:
$\sum_{k=i}^t (-1)^{t-k}\binom{t}{k}\binom{k}{i}=\sum_{k=i}^t
(-1)^{k-i}\binom{t}{k}\binom{k}{i}=\delta_{ti}$, where
$t,i=0,1,2,\ldots.$
\end{remark}

\begin{remark}\label{r:AbelianGroup}
We view the $n$-fold binomial or inverse binomial transform
${\mathcal{B}}_n$, $(n=0,\pm 1,\pm 2,\pm 3,\ldots)$, to be one
simple transform of integer sequences, because such inversion
relations as $B_2B_{-2}=B_{-2}B_2=E$, $B_3B_{-3}=B_{-3}B_3=E$
hold, and so forth. For example, for $2$-fold binomial and inverse
binomial transforms, the transform matrices are respectively
\begin{equation}\label{e:TwoBinom}
B_2=\left(\begin{array}{ccccc} 1 & & & & \\
 2 & 1 & & & \\ 4 & 4 & 1 & &
 \\ 8 & 12 & 6 & 1 & \\
  \vdots & \vdots & \vdots & \vdots
  & \vdots \end{array} \right), 
\quad B_{-2}=\left(\begin{array}{rrrrr} 1 & & & & \\
 -2 & 1 & & & \\ 4 & -4 & 1 & &
 \\ -8 & 12 & -6 & 1 & \\
  \vdots & \vdots & \vdots & \vdots
  & \vdots \end{array} \right),
\end{equation}
\end{remark}

Now, let us give the multiple binomial transforms of the shifting
sequences $a_{(p)}(t)$, $(p=0,1,2,\ldots)$, of an integer sequence
$a(t)$.

\begin{theorem}\label{t:BinomShift}
Let $a(t)$ be an integer sequence. Then
\begin{equation}\label{e:BinomShift1} % {\mathcal{B}}_n(a)
{\mathcal{B}}_n(a_{(p)})=(\sigma-n)^{p}({\mathcal{B}}_n(a))=
(\sigma-n)^{p}(a^{(n)})=\sum^{p}_{k=0}(-n)^{p-k}\binom{p}{k}\sigma^k(a^{(n)}),
\end{equation}
where $n=0,\pm 1,\pm 2,\ldots$.
\end{theorem}

\begin{proof}
Use the mathematical induction. When $n=\pm 1$ and $p=1$,
\begin{multline}
{\mathcal{B}}_1(\sigma(a))=\sum_{k=0}^{t}\binom{t}{k}a(k+1)=\sum_{k=1}^{t+1}\binom{t}{k-1}a(k)
=\sum_{k=1}^{t+1}\binom{t+1}{k}a(k)-\sum_{k=1}^{t+1}\binom{t}{k}a(k)
\\
=\sum_{k=0}^{t+1}\binom{t+1}{k}a(k)-a(0)-[\sum_{k=0}^{t}\binom{t}{k}a(k)-a(0)]
=\sigma({\mathcal{B}}_1(a))-{\mathcal{B}}_1(a)=(\sigma-1)({\mathcal{B}}_1(a)),
\nonumber
\end{multline}
and
\begin{multline}
{\mathcal{B}}_{-1}(\sigma(a))=\sum_{k=0}^{t}(-1)^{t-k}\binom{t}{k}a(k+1)
=\sum_{k=1}^{t+1}(-1)^{t+1-k}\binom{t}{k-1}a(k)
\\
=\sum_{k=0}^{t+1}(-1)^{t+1-k}\left[\binom{t+1}{k}-\binom{t}{k}\right]a(k)
=\sum_{k=0}^{t+1}(-1)^{t+1-k}\binom{t+1}{k}a(k)+\sum_{k=0}^{t}(-1)^{t-k}\binom{t}{k}a(k)
\\
=\sigma({\mathcal{B}}_{-1}(a))+{\mathcal{B}}_{-1}(a)=(\sigma+1)({\mathcal{B}}_{-1}(a)).
\nonumber
\end{multline}
If for $n=\pm k$($k$ is some positive integer),
${\mathcal{B}}_{\pm k}(\sigma(a))=(\sigma\mp k)({\mathcal{B}}_{\pm
k}(a))$ holds, then for $n=\pm(k+1)$,
${\mathcal{B}}_{\pm(k+1)}(\sigma(a))={\mathcal{B}}_{\pm
1}(\sigma({\mathcal{B}}_{\pm k}(a)))\mp k{\mathcal{B}}_{\pm
1}({\mathcal{B}}_{\pm k}(a))= (\sigma\mp
1)({\mathcal{B}}_{\pm(k+1)}(a))\mp
k{\mathcal{B}}_{\pm(k+1)}(a)=(\sigma\mp(k+1))({\mathcal{B}}_{\pm(k+1)}(a))$
also holds. Hence, for any integer $n$,
${\mathcal{B}}_n(\sigma(a))=(\sigma-n)({\mathcal{B}}_n(a))$ holds.
On the other hand, if for $p=m$($m$ is some positive integer) that
${\mathcal{B}}_n(\sigma^m(a))=(\sigma-n)^{m}({\mathcal{B}}_n(a))$
holds, then when $p=m+1$, we get that
${\mathcal{B}}_n(\sigma^{m+1}(a))=(\sigma-n)^{m}({\mathcal{B}}_n(\sigma(a)))=
(\sigma-n)^{m}((\sigma-n)({\mathcal{B}}_n(a)))=(\sigma-n)^{m+1}({\mathcal{B}}_n(a))$.
Hence, for any positive integer $n$ and $p$,
${\mathcal{B}}_n(\sigma^p(a))=(\sigma-n)^{p}({\mathcal{B}}_n(a))$.
Special cases that $n=0$ and/or $p=0$ are trivial.
\end{proof}

\begin{corollary}\label{c:BinomShift}
Let $a(t)$ be an integer sequence, and $P(\sigma)$ be an
integer-coefficient polynomial in $\sigma$. Then
\begin{equation}\label{e:BinomShift2}
{\mathcal{B}}_n(P(\sigma)(a))=P(\sigma-n)({\mathcal{B}}_n(a))=P(\sigma-n)(a^{(n)}),
\end{equation}
where $n=0,\pm 1,\pm 2,\ldots.$
\end{corollary}

\begin{proof}
Let $P(\sigma)$ be a integer-coefficient polynomial of degree $p$
($p=0,1,2,\ldots$) in $\sigma$:
$P(\sigma)=\sum_{k=0}^{p}c_k\sigma^{k}$, where $c_k$s are ($p+1$)
integers. From Theorem \ref{t:BinomShift}, we have that
${\mathcal{B}}_n(P(\sigma)(a))={\mathcal{B}}_n(\sum_{k=0}^{p}c_k\sigma^{k}(a))=
\sum_{k=0}^{p}c_k{\mathcal{B}}_n(\sigma^{k}(a))=\sum_{k=0}^{p}c_k(\sigma-n)^{k}({\mathcal{B}}_n(a))=
P(\sigma-n)({\mathcal{B}}_n(a))=P(\sigma-n)(a^{(n)})$.
\end{proof}

\begin{remark}\label{r:EigenValue}
By using Corollary \ref{c:BinomShift}, we can more succinctly
prove the following known property of recurrence sequences (see
\cite[Thm.\ 17]{ref1}). Let $a(t)$ be a linear homogeneous
recurrence sequence of order $q$ with the recurrence equation
\begin{equation}\label{e:RecurSeq}
P(\sigma)(a)=\sum_{k=0}^{q}b_k\sigma^{q-k}(a)=0,
\end{equation}
where $b_0=1$, $b_1,b_2,\ldots,b_q$ are $q$ given integers. Then
its $q$ complex characteristic values $\lambda_k$,
$k=1,2,\ldots,q$, are the roots of polynomial (algebraic)
equation:
\begin{equation}\label{e:EigenValue}
P(\lambda)=\sum_{k=0}^{q}b_k\lambda^{q-k}=0.
\end{equation}
On the other hand, by taking transformation ${\mathcal{B}}_n$ of
the two sides of (\ref{e:RecurSeq}), and then employing Corollary
\ref{c:BinomShift}, we find that sequences $a^{(n)}(t)$, ($n=0,\pm
1,\pm 2,\ldots$), satisfy recurrence equation:
\begin{equation}\label{e:EigenValue1}
P(\sigma-n)(a^{(n)})=0.
\end{equation}
This implies that $q$ complex characteristic values
$\lambda^{(n)}_k$, $(k=1,2,\ldots,q)$, of $a^{(n)}(t)$ are the
roots of the algebraic equation:
\begin{equation}\label{e:EigenValue2}
P(\lambda^{(n)}-n)=\sum_{k=0}^{q}b_k(\lambda^{(n)}-n)^{q-k}=0.
\end{equation}
Comparing (\ref{e:EigenValue}) with (\ref{e:EigenValue2}), we find
that $\lambda^{(n)}_k-n=\lambda_k$, namely
\begin{equation}\label{e:EigenValue3}
\lambda^{(n)}_k=\lambda_k+n, \quad (k=1,2,\ldots,q).
\end{equation}
\end{remark}


\section{Shifted sequences and the Hankel transform}


Layman proved the invariance of the Hankel
transform under applications of the binomial transform or its
inverse transform (see \cite{ref2}). For an integer
sequence, the $n$-fold binomial (or inverse binomial) transform 
is the same as the $n$ times successive binomial (or inverse binomial) transform
operation, Pan \cite{ref1} pointed out that the invariance of
the Hankel transform holds under applications of the $n$-fold
binomial (or $n$-fold invert binomial) transform. Now by using
Theorem \ref{t:BinomShift}, we give a more direct and succinct
proof of the invariance, as follows.

\begin{remark}\label{r:InvarianceHankel}
By using Definition \ref{d:ShiftingSequence}, we express the
Hankel matrix $H_a$ of sequence $a(t)$ as
\begin{equation}\label{e:HabkelMatri1}
H_a=\left(\begin{array}{ccccc}
  a & \sigma(a) & \sigma^2(a) & \sigma^3(a)
  & \cdots \end{array} \right) \\
  =\left(\begin{array}{ccccc}
  a & a_{(1)} & a_{(2)} & a_{(3)}
  & \cdots \end{array} \right),
\end{equation}
and Hankel matrix $H_{a^{(n)}}$ of integer sequence $a^{(n)}(t)$
as
\begin{equation}\label{e:HabkelMatri2}
H_{a^{(n)}}=\left(\begin{array}{ccccc}
  a^{(n)} & \sigma(a^{(n)}) & \sigma^2(a^{(n)}) & \sigma^3(a^{(n)})
  & \cdots \end{array} \right),
\end{equation}
According to Theorem \ref{t:BinomShift}, we have that
\begin{multline}\label{e:HabkelMatri3}
B_n H_a=\left(\begin{array}{ccccc}
  B_n a & B_n a_{(1)} & B_n a_{(2)} & B_n a_{(3)}
  & \cdots \end{array} \right) \\
  =\left(\begin{array}{ccccc}
  a^{(n)} & (\sigma-n)(a^{(n)}) & (\sigma-n)^2(a^{(n)}) & (\sigma-n)^3(a^{(n)})
  & \cdots \end{array} \right).
\end{multline}
Comparing (\ref{e:HabkelMatri3}) with (\ref{e:HabkelMatri2}), we
see that the upper-left $(t+1)\times(t+1)$ ($t=0,1,2,\ldots$)
sub-matrix of $B_nH_a$ has the same determinant to the upper-left
sub-matrix of the Hankel matrix $H_{a^{(n)}}$ of sequence
$a^{(n)}(t)$. On the other hand, the determinant of the upper-left
$(t+1)\times(t+1)$ ($t=0,1,2,\ldots$) sub-matrix of matrix $B_n
H_a$ is equal to the determinant of the upper-left
$(t+1)\times(t+1)$ ($t=0,1,2,\ldots$) sub-matrix of matrix $H_a$,
because the determinant of any upper-left sub-matrices of matrix
$B_n$ ($n=\pm 1,\pm 2,\pm 3,\ldots$) is always equal to one. In
other words, the sequences $a$ and $a^{(n)}$ both have the same Hankel
transform, for any integer $n$.
\end{remark}

\begin{remark}\label{r:InvarianceMulti}
This result gives an affirmative answer to one of Layman's two
questions raised in \cite{ref2}: \emph{Are there other interesting
transforms, $T$, of an integer sequence $S$, in addition to the
Binomial and Invert transforms, with the property that the Hankel
transform of $S$ is the same as the Hankel transform of the $T$
transform of $S$?} For example, $\mathcal{T}={\mathcal{B}}_2$ or
${\mathcal{B}}_{-2}$, which have transform matrices listed in
(\ref{e:TwoBinom}).
\end{remark}

Next, we investigate the Hankel transform of recurrence
sequences. The following theorem gives a basic property of the Hankel
transform of recurrence sequences.

\begin{theorem}\label{t:RecurSeq1}
Let $a(t)$ be a linear homogeneous recurrence sequence of order
$q$, with recurrence equation (\ref{e:RecurSeq}). Then the Hankel
transform $h_a(t)$ of sequence $a(t)$ is a finite sequence with
length $q$, that is, for $t\geq q$, $h_a(t)\equiv 0$.
\end{theorem}

\begin{proof}
We see from (\ref{e:HabkelMatri1}) and (\ref{e:RecurSeq}) that if
multiplying the first, the second, $\ldots$, the $q$-th column
vectors of the Hankel matrix $H_a$ by $b_q$, $b_{q-1}$, $\ldots$,
$b_1$ respectively, and then adding them to the $(q+1)$th column
$\sigma^q(a)$, we cause the $(q+1)$-th column to be a zero-column.
This operation does not change the determinants of principal
sub-matrices of $H_a$. On the other hand, for a infinite-order
square matrix with its $(q+1)$-th column being a zero-column,
determinants of the principal sub-matrices of order $q+1$, $q+2$,
$q+3$, $\ldots$, namely $h(q)$, $h(q+1)$, $h(q+2)$, $\ldots$, are
always equal to zeros. That is, the Hankel transform $h(t)$ is a
finite integer sequence with the length of $q$.
\end{proof}

\begin{corollary}\label{c:RecurSeq1}
All of the $n$-fold binomial transforms $a^{(n)}(t)$ ($n=0,\pm
1,\pm 2,\pm 3,\ldots$) of a $q$-order recurrence sequence $a(t)$
have identical Hankel transform with the length of $q$.
\end{corollary}

\begin{remark}\label{r:Examples1}
For example, as recurrence sequences of order $2$ and $3$, the
Fibonacci sequence $F(t)$ (\seqnum{A000045} in \cite{ref3}) and its
multiple binomial transforms \seqnum{A001906}, 
\seqnum{A093131}, \seqnum{A039834}, etc. (see
Pan \cite{ref1}) all have the same Hankel transform with length $2$:
$h_F(0)=1$, $h_F(1)=1$, and the Tribonacci sequence $T(t)$
(\seqnum{A000073} in \cite{ref3}) and its multiple binomial transforms
\seqnum{A115390}, etc. (see Pan \cite{ref1}) all have the same Hankel transform
with length $3$: $h_T(0)=3$, $h_T(1)=8$, $h_T(2)=-44$.
\end{remark}

Finally, we give special relations of the Hankel transforms of
$a^{(n)}(t)$, $(n=0,\pm 1,\pm 2,\ldots)$, and $a_{(p)}(t)$,
$(p=0,1,2,\ldots)$, with the general term formula of the recurrent
sequences $a(t)$, respectively.

\begin{theorem}\label{t:HankelRecur1}
Let $a(t)$ be a linear homogeneous recurrence sequence of order
$q$, with the general-term formula:
$a(t)=\sum^{q}_{i=1}c_i\lambda_i^t$, $t\in{\mathbb{N}}_0$. Then
the Hankel transforms $h_{a^{(n)}}(t)$, ($n=0,\pm 1,\pm
2,\ldots$), are such that
\begin{equation}\label{e:HankelRecur1}
h_{a^{(n)}}(t)=\sum_{(i_1,i_2,\cdots,i_{t+1})}\prod_{k=1}^{t+1}(c_{i_{k}}
\lambda_{i_{k}}^{k-1})\prod_{1\leq
k<m\leq(t+1)}(\lambda_{i_{k}}-\lambda_{i_{m}}), \quad
t=0,1,\ldots,q-1,
\end{equation}
where the summation is over the $q!/(q-t-1)!$ different
$(t+1)$-permutations $(i_1,i_2,\cdots,i_{t+1})$ of set
$\{1,2,\ldots,q\}$. Particularly, the first term
$h_{a^{(n)}}(0)=\sum^{q}_{i=1}c_i=a(0)$, and the $q$th (last) term
$h_{a^{(n)}}(q-1)=\prod_{i=1}^q c_i\prod_{1\leq i<j\leq
q}(\lambda_i-\lambda_j)^2$.
\end{theorem}

\begin{proof}
Denoting $j$-order vectors
$(1,\lambda_i,\lambda_i^2,\cdots,\lambda_i^{j-1})$ by
$\lambda(i,j)$, and $(j\times j)$ Vandermonde square-matrices
$\big(\lambda(i_1,j),\lambda(i_2,j),\ldots,\lambda(i_j,j)\big)$ by
$\mathbb{V}(i_1,i_2,\cdots,i_j)$ respectively, where
$i\in\{1,2,\ldots,q\}$, and $(i_1,i_2,\cdots,i_j)$ is a
$j$-permutation of set $\{1,2,\ldots,q\}$, $(1\leq j\leq q)$, we
find that the $t$-th term of Hankel transform $h_a(t)$ of $a(t)$,
that is, the determinant of upper-left $(t+1)\times(t+1)$
sub-matrix of Hankel matrix (\ref{e:HabkelMatri1}), is
\begin{equation}
h_a(t)=\det \left[
\begin{array}{cccc}\sum^{q}_{i=1}c_i\lambda(i,t+1) &
\sum^{q}_{i=1}c_i\lambda_i\lambda(i,t+1) & \cdots &
\sum^{q}_{i=1}c_i\lambda_i^{t}\lambda(i,t+1)
\end{array}\right]\nonumber
\end{equation}
\begin{equation}
=\sum_{(i_1,i_2,\cdots,i_{(t+1)})}\big(\prod_{k=1}^{t+1}(c_{i_{k}}
\lambda_{i_{k}}^{k-1})\big)\det
\mathbb{V}(i_1,i_2,\cdots,i_{t+1}), \nonumber
\end{equation}
where the summation is over $q!/(q-t-1)!$ different
$(t+1)$-permutations $(i_1,i_2,\cdots,i_{t+1})$ of set
$\{1,2,\ldots,q\}$. The Vandermonde determinant $\det
\mathbb{V}(i_1,i_2,\cdots,i_{t+1})$ equals $\prod_{1\leq
k<m\leq(t+1)}(\lambda_{i_{k}}-\lambda_{i_{m}})$. Because
$h_{a^{(n)}}(t)=h_a(t)$, (\ref{e:HankelRecur1}) holds. In case
$t=0$, we see that $h_{a^{(n)}}(0)=h_a(0)=\sum^{q}_{i=1}c_i=a(0)$;
in the case $t=q-1$, we have that
\begin{equation}
h_{a^{(n)}}(q-1)=h_a(q-1)=\det\left[
\begin{array}{cccc}\sum^{q}_{i=1}c_i\lambda(i,q) &
\sum^{q}_{i=1}c_i\lambda_i\lambda(i,q) & \cdots &
\sum^{q}_{i=1}c_i\lambda_i^{q-1}\lambda(i,q)
\end{array}\right], \nonumber
\end{equation}
The matrix in the right side of the above equality just equals a
product of three square matrices: $\mathbb{V}(1,2,\cdots,q)\cdot
{\rm diag}\{c_1,c_2,\ldots,c_q\}\cdot{\mathbb{V}}^T(1,2,\cdots,q)$.
Hence, we have that
$$h_{a^{(n)}}(q-1)=\det\mathbb{V}(1,\cdots,q)\times\det
{\rm diag}\{c_1,\ldots,c_q\}\times\det{\mathbb{V}}^T(1,\cdots,q)=\prod_{i=1}^q
c_i\prod_{1\leq i<j\leq q}(\lambda_i-\lambda_j)^2$$
\end{proof}

\begin{theorem}\label{t:HankelRecur2}
Let $a(t)$ be a linear homogeneous recurrence sequence of order
$q$, with a general-term formula:
$a(t)=\sum^{q}_{i=1}c_i\lambda_i^t$, $t\in{\mathbb{N}}_0$. Then
the Hankel transform $h_{a_{(p)}}(t)$ of the shifted sequence
$a_{(p)}$, $(p=0,1,2,\ldots)$, of sequence $a(t)$ are given by
\begin{equation}\label{e:HankelRecur2}
h_{a_{(p)}}(t)=\sum_{(i_1,i_2,\cdots,i_{t+1})}\prod_{k=1}^{t+1}(c_{i_{k}}
\lambda_{i_{k}}^{k-1+p})\prod_{1\leq
k<m\leq(t+1)}(\lambda_{i_{k}}-\lambda_{i_{m}}),\quad
t=0,1,\ldots,q-1,
\end{equation}
where summarizing is over $q!/(q-t-1)!$ different
$(t+1)$-permutations $(i_1,i_2,\cdots,i_{t+1})$ of set
$\{1,2,\ldots,q\}$. Particularly, the first term
$h_{a_{(p)}}(0)=\sum^{q}_{i=1}c_i\lambda_i^p$, and the $q$-th
(last) term $h_{a_{(p)}}(q-1)=\prod_{i=1}^q
(c_i\lambda_i^p)\prod_{1\leq i<j\leq q}(\lambda_i-\lambda_j)^2$.
\end{theorem}

\begin{proof}
The general term of $a_{(p)}(t)$ is 
$a_{(p)}(t)=\sum^{q}_{i=1}c_i\lambda_i^{t+p}=\sum^{q}_{i=1}d_i\lambda_i^t$,
$t\in{\mathbb{N}}_0$, where $d_i=c_i\lambda_i^p$
($i=1,2,\ldots,q$). We see from Theorem \ref{t:HankelRecur1} that
the Hankel transform $h_{a_{(p)}}(t)$ of sequence $a_{(p)}$ (note
that it is also a recurrence sequence of order $q$) is $$
h_{a_{(p)}}(t)=\sum_{(i_1,i_2,\cdots,i_{t+1})}\prod_{k=1}^{t+1}(d_{i_{k}}
\lambda_{i_{k}}^{k-1})\prod_{1\leq
k<m\leq(t+1)}(\lambda_{i_{k}}-\lambda_{i_{m}}), $$ where
summarizing is over $q!/(q-t-1)!$ different $(t+1)$-permutations
$(i_1,i_2,\cdots,i_{t+1})$ of set $\{1,2,\ldots,q\}$. Replacing
$d_1,d_2,\ldots,d_q$ by
$c_1\lambda_1^p,c_2\lambda_2^p,\ldots,c_1\lambda_q^p$
respectively, we obtain (\ref{e:HankelRecur2}). From Theorem
\ref{t:HankelRecur1}, we obtain that
$h_{a_{(p)}}(0)=\sum^{q}_{i=1}d_i=\sum^{q}_{i=1}c_i\lambda_i^p$,
and $$h_{a_{(p)}}(q-1)=\prod_{i=1}^q d_i\prod_{1\leq i<j\leq
q}(\lambda_i-\lambda_j)^2=\prod_{i=1}^q
(c_i\lambda_i^p)\prod_{1\leq i<j\leq q}(\lambda_i-\lambda_j)^2$$.
\end{proof}

\begin{remark}
We take the generalized Lucas sequence
$s(t)=3,1,3,7,11,21,39,\ldots$ (sequence \seqnum{A001644} in \cite{ref3})
as an example used for verification. The third order recurrent
sequence has a general term formula that
$s(t)=\lambda_1^t+\lambda_2^t+\lambda_3^t$ (Note that
$c_1=c_2=c_3=1$), where three characteristic values $\lambda_i$
$(i=1,2,3)$ are the roots of algebraic equation
$\lambda^3-\lambda^2-\lambda-1=0$. They are that $$
\lambda_1=\frac{1}{3}(1+\alpha+\beta),\quad
\lambda_2=\frac{1}{3}(1+\omega_1\alpha+\omega_2\beta),\quad
\lambda_2=\frac{1}{3}(1+\omega_2\alpha+\omega_1\beta).$$ where two
real numbers $\alpha=\sqrt[3]{19+\sqrt{297}}$,
$\beta=\sqrt[3]{19-\sqrt{297}}$; and $1,\omega_1,\omega_2$ are
three complex cubic roots of $1$.  Hence, noting that
$\omega_1+\omega_2=-1$ and $\omega_1\omega_2=1$, we get that the
Hankel transform of $s(t)$ (and any of its multiple binomial
transforms) has the three terms: $$h_s(0)=c_1+c_2+c_3=1+1+1=3,$$
$$h_s(1)=c_1c_2\lambda_2(\lambda_2-\lambda_1)+c_2c_1\lambda_1(\lambda_1-\lambda_2)
+c_1c_3\lambda_3(\lambda_3-\lambda_1)+c_3c_1\lambda_1(\lambda_1-\lambda_3)
+c_2c_3\lambda_3(\lambda_3-\lambda_2) $$ $$
+c_3c_2\lambda_2(\lambda_2-\lambda_3)=(\lambda_1-\lambda_2)^2+(\lambda_1-\lambda_3)^2
+(\lambda_2-\lambda_3)^2=2\alpha\beta=8,
 $$ $$
h_s(2)=c_1c_2c_3(\lambda_1-\lambda_2)^2(\lambda_1-\lambda_3)^2(\lambda_2-\lambda_3)^2
=(\lambda_1-\lambda_2)^2(\lambda_1-\lambda_3)^2(\lambda_2-\lambda_3)^2
$$ $$
=-\frac{1}{27}(\alpha^2+\beta^2+\alpha\beta)^2(\alpha-\beta)^2=-\frac{1}{27}
(\alpha^3+\beta^3+16)(\alpha^3+\beta^3-16)=-44. $$
\end{remark}


\section{Acknowledgement}

I would like to thank the referee for his/her useful suggestions
and improvements.


\begin{thebibliography}{9}

\bibitem{ref1}
J.-Q.~Pan,
\newblock Multiple binomial transforms and families of integer sequences,
\newblock {\em J. Integer Seq.}, {\bf 13} (2010),
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL13/Pan/pan8.html}{Article
10.4.2}.

\bibitem{ref2}
J.~W.~Layman,
\newblock The Hankel transform and some of its properties,
\newblock {\em J. Integer Seq.}, {\bf 4} (2001), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL4/LAYMAN/hankel.html}{Article 01.1.5}.

\bibitem{ref3}
N. J.~A. Sloane,
\newblock {\em The On-Line Encyclopedia of Integer Sequences},
\newblock published electronically at
\href{http://oeis.org/}{\tt http://oeis.org/}.


\end{thebibliography}

\bigskip
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\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B65; Secondary 11B75.

\noindent {\it Keywords}:  shifted sequences, multiple binomial
transforms, Hankel transform.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000045},
\seqnum{A000073},
\seqnum{A001644},
\seqnum{A001906},
\seqnum{A039834},
\seqnum{A093131}, and
\seqnum{A115390}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 22 2011;
revised version received February 22 2011. 
Published in {\it Journal of Integer Sequences},
March 25 2011.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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