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\title{\bf Generalized Stirling Numbers and Hyper-Sums of Powers of Binomial
Coefficients}



\author{Claudio de J. Pita Ruiz V.\\
\small Department of Mathematics\\[-0.8ex]
\small Universidad Panamericana\\[-0.8ex]
\small Mexico City, Mexico\\
\small\tt cpita@up.edu.mx\\
}


\date{\dateline{Oct 9, 2013}{Dec 16, 2013}{Jan 12, 2014}\\
\small Mathematics Subject Classifications: 05A10, 11B73}

\begin{document}

\maketitle


\begin{abstract}
We work with a generalization of Stirling numbers of the second kind related
to the boson normal ordering problem (P. Blasiak \textit{et al.}). We show
that these numbers appear as part of the coefficients of expressions in
which certain sequences of products of binomials, together with their
partial sums, are written as linear combinations of some other binomials. We
show that the number arrays formed by these coefficients can be seen as
natural generalizations of Pascal and Lucas triangles, since many of the
known properties on rows, columns, falling diagonals and rising diagonals in
Pascal and Lucas triangles, are also valid (some natural generalizations of
them) in the arrays considered in this work. We also show that certain
closed formulas for hyper-sums of powers of binomial coefficients appear in
a natural way in these arrays.


  \bigskip\noindent \textbf{Keywords:} generalized Stirling numbers, powers of binomial coefficients\end{abstract}

\section{Introduction}\label{Sec.1}

Along the years, Stirling numbers have demonstrated to be a fundamental tool
(`of the greatest utility'\textit{\ }\cite{Jo}) for dealing with
combinatorial problems. We can find now many research works pursuing
generalizations (in some sense) of Stirling numbers of both kinds (some of
them accompanied with the corresponding combinatorial interpretations). We
mention some: L. Carlitz \cite{Car} studied a type of $\lambda $-weighted
Stirling numbers/polynomials which have demonstrated to be useful to
understand different mathematical problems (see for example \cite{Ga} and 
\cite{Ko}); W. Lang \cite{La2} considered a (one-parameter) generalization
of Stirling numbers, by using the language of `infinitesimal
transformations' (with some flavor from Lie groups theory). Lang himself 
\cite{La1} studied the corresponding combinatorial interpretation of his
generalized Stirling numbers. One more generalized version of Stirling
numbers appeared in the work of T. Mansour \textit{et al.} \cite{Ma}, and
one yet more in the work of M. Dziemia\'{n}czuk \cite{D}. We mention also
the work of Tian-Xiao He \cite{TXH}, which presents a generalization that
includes some other known generalizations of Stirling numbers. (See also
Remark \ref{Rem.1} at the end of this section.) Among all these works, there
is one concerning Stirling numbers of the second kind, which has a natural
connection with certain products of binomial coefficients. Having this
connection as starting point we construct some number arrays which are
natural generalizations of Pascal and Lucas triangles (and also of the
corresponding results of the previous work \cite{Pi}). To study these number
arrays is the main theme of this article. The mentioned generalization of
Stirling numbers of the second kind appears in the works of P. Blasiak 
\textit{et al.} \cite{Bl1, Bl3, Me} (see also \cite{Ar}). (Two works
relating this generalization with other combinatorial-physics objects are 
\cite{So1, So2}; two works about the combinatorial interpretations of these
generalized Stirling numbers are \cite{Bl2, Co}.) We devote the rest of this
section to present the basic facts of Blasiak's generalized Stirling numbers
of the second kind.

Throughout the work, $p$ denotes a given natural number. To keep the
original author's notation, we use $r\,$\ and $s$ to denote positive
integers such that $r\geq s$. The $\left( r,s\right) $\textit{-Stirling
numbers of the second kind} $S_{r,s}\left( p,k\right) $ (or simply \textit{%
generalized Stirling numbers of the second kind}), are introduced in the P.
Blasiak's main work \cite{Bl1} as certain coefficients appearing in the
so-called \textit{boson normal ordering problem}. Roughly speaking, the
numbers $S_{r,s}\left( p,k\right) $ are determined by the following
equation: 
\begin{equation}
\left( x^{r}\frac{d^{s}}{dx^{s}}\right) ^{p}=x^{p\left( r-s\right)
}\sum_{j=s}^{ps}S_{r,s}\left( p,j\right) x^{j}\frac{d^{j}}{dx^{j}}.
\label{1.1}
\end{equation}

For example, if we set $r=4$, $s=3$ and $p=2$, formula (\ref{1.1}) looks as%
\begin{equation}
\left( x^{4}\frac{d^{3}}{dx^{3}}\right)
^{2}=x^{2}\sum_{j=3}^{6}S_{4,3}\left( 2,j\right) x^{j}\frac{d^{j}}{dx^{j}}.
\label{1.1001}
\end{equation}

Letting act both sides of (\ref{1.1001}) on a $C^{6}\left( \mathbb{R}\right) 
$ function $y=y\left( x\right) $, the left-hand side becomes%
\begin{equation*}
x^{4}\frac{d^{3}}{dx^{3}}\left( x^{4}\frac{d^{3}y}{dx^{3}}\right)
=x^{2}\left( 24x^{3}\frac{d^{3}y}{dx^{3}}+36x^{4}\frac{d^{4}y}{dx^{4}}%
+12x^{5}\frac{d^{5}y}{dx^{5}}+x^{6}\frac{d^{6}y}{dx^{6}}\right) ,
\end{equation*}%
so we have that $S_{4,3}\left( 2,3\right) =24,S_{4,3}\left( 2,4\right)
=36,S_{4,3}\left( 2,5\right) =12$ and $S_{4,3}\left( 2,6\right) =1.$

By convention one takes $S_{r,s}\left( p,0\right) =\delta _{p,0}$. It can be
shown that $S_{r,s}\left( p,j\right) \neq 0$ if and only if $s\leq j\leq sp$%
, and that $S_{r,s}\left( p,ps\right) =1$ for $p\in \mathbb{N}$. The
standard Stirling numbers of the second kind $S\left( p,j\right) $
correspond to the case $r=s=1$. An explicit formula for $S_{r,s}\left(
p,j\right) $ is%
\begin{equation}
S_{r,s}\left( p,j\right) =\frac{\left( -1\right) ^{j}}{j!}%
\sum_{i=s}^{j}\left( -1\right) ^{i}\dbinom{j}{i}\prod_{l=1}^{p}\left(
i+\left( l-1\right) \left( r-s\right) \right) ^{\underline{\,s\,}},
\label{1.2}
\end{equation}%
where $m^{\underline{\,s\,}}=\prod_{l=0}^{s-1}\left( m-l\right) $ is the
falling factorial. From (\ref{1.2}) one can see at once that $S_{s,s}\left(
p,s\right) =\left( s!\right) ^{p-1}$. (Note that when $r=s=1$, formula (\ref%
{1.2}) is the known explicit formula for Stirling numbers of the second kind 
$S\left( p,k\right) =\frac{\left( -1\right) ^{k}}{k!}\sum_{i=1}^{k}\left(
-1\right) ^{i}\binom{k}{i}i^{p}$.) It is possible to see from (\ref{1.1})
that%
\begin{equation}
\prod_{j=1}^{p}\left( x+\left( j-1\right) \left( r-s\right) \right) ^{%
\underline{\,s\,}}=\sum_{k=s}^{ps}S_{r,s}\left( p,k\right) x^{\underline{%
\,k\,}}.  \label{1.3}
\end{equation}

Observe that the case $s=r$ of (\ref{1.3}) is%
\begin{equation}
\left( x^{\underline{\,r\,}}\right) ^{p}=\sum_{k=r}^{pr}S_{r,r}\left(
p,k\right) x^{\underline{\,k\,}},  \label{1.4}
\end{equation}%
and that the case $r=1$ of (\ref{1.4}) is the well-known expression 
\begin{equation}
x^{p}=\sum_{k=1}^{p}S\left( p,k\right) x^{\underline{\,k\,}},  \label{1.5}
\end{equation}
in which the Stirling numbers of the second kind $S\left( p,k\right) $
appear as the connecting coefficients when $x^{p}$ is written as a linear
combination of the falling factorials $x^{\underline{\,k\,}}$, $k=1,2,\ldots
,p$.

Formula (4.56) --corrected-- from Blasiak \cite{Bl1}, gives us the
recurrence relation%
\begin{equation}
S_{r,s}\left( p+1,k\right) =\sum_{i=0}^{s}\dbinom{p\left( r-s\right) +k+i-s}{%
i}s^{\underline{\,i\,}}S_{r,s}\left( p,k+i-s\right) .  \label{1.7}
\end{equation}

(When $r=s=1$ we have the known recurrence $S\left( p+1,k\right) =S\left(
p,k-1\right) +kS\left( p,k\right) $ for the usual Stirling numbers of the
second kind.)

An interesting particular case is when $r=2s$. We claim that%
\begin{equation}
S_{2s,s}\left( p,k\right) =\frac{\left( sp\right) !}{k!}\dbinom{s\left(
p-1\right) }{k-s}.  \label{1.9}
\end{equation}

Clearly this formula is true for $p=1$. Assuming the formula is true for a
given $p\in \mathbb{N}$, we have according to (\ref{1.7}) that%
\begin{equation}
S_{2s,s}\left( p+1,k\right) =\sum_{i=0}^{s}\dbinom{ps+k+i-s}{i}s^{\underline{%
\,i\,}}S_{2s,s}\left( p,k+i-s\right) .  \label{1.10}
\end{equation}

By using the induction hypothesis together with some algebraic
manipulations, we can write (\ref{1.10}) as%
\begin{equation}
S_{2s,s}\left( p+1,k\right) =\frac{\left( s\left( p+1\right) \right) !}{k!}%
\frac{1}{\binom{s\left( p-1\right) +k}{k}}\sum_{i=0}^{s}\binom{s}{i}\binom{sp%
}{k-i}\binom{sp+k-i}{k-i-s}.  \label{1.11}
\end{equation}

But identity (6.47) from Gould's book \cite{Go1} gives us that%
\begin{equation}
\sum_{i=0}^{s}\binom{s}{i}\binom{sp}{k-i}\binom{sp+k-i}{k-i-s}=\binom{sp}{k-s%
}\binom{s\left( p-1\right) +k}{k},  \label{1.12}
\end{equation}%
so we conclude finally from (\ref{1.11}) and (\ref{1.12}) that 
\begin{equation}
S_{2s,s}\left( p+1,k\right) =\frac{\left( s\left( p+1\right) \right) !}{k!}%
\binom{sp}{k-s},  \label{1.13}
\end{equation}%
proving our claim (\ref{1.9}). (The case $s=1$ of (\ref{1.9}) is $\frac{p!}{%
k!}\binom{p-1}{k-1}$, corresponding to the so-called (unsigned) Lah numbers 
\cite[A008297]{OEIS}.)

In \cite{Bl1} there are tables for $S_{r,s}\left( p,k\right) $ with $r=1,2,3$%
, $1\leq s\leq r$ and some small values of $p$ and the corresponding $s\leq
k\leq ps$. Some $S_{4,s}\left( p,k\right) $ for $1\leq s\leq 4$ are the
following:

\begin{equation*}
\begin{array}{ll}
\begin{array}{c}
S_{4,1}\!\left( p,k\right)  \\ 
1\!\!\leq \!k\!\leq \!p%
\end{array}%
\begin{tabular}{c||cccc}
$p\backslash k$ & $1$ & $2$ & $3$ & $\cdots $ \\ \hline\hline
$1$ & $1$ &  &  &  \\ 
$2$ & $4$ & $1$ &  &  \\ 
$3$ & $28$ & $12$ & $1$ & $\cdots $ \\ 
$\vdots $ &  &  & $\vdots $ & 
\end{tabular}
& 
\begin{array}{c}
S_{4,2}\!\left( p,k\right)  \\ 
2\!\leq \!k\!\leq \!2p%
\end{array}%
\begin{tabular}{c||cccccc}
$p\backslash k$ & $2$ & $3$ & $4$ & $5$ & $6$ & $\cdots $ \\ \hline\hline
$1$ & $1$ &  &  &  &  &  \\ 
$2$ & $12$ & $8$ & $1$ &  &  &  \\ 
$3$ & $360$ & $480$ & $180$ & $24$ & $1$ & $\cdots $ \\ 
$\vdots $ &  &  & $\vdots $ &  &  & 
\end{tabular}%
\end{array}%
\end{equation*}

\begin{equation*}
\begin{array}{c}
S_{4,3}\!\left( p,k\right)  \\ 
3\leq \!k\!\leq \!3p%
\end{array}%
\begin{tabular}{c||cccccccc}
$p\backslash k$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $\cdots $ \\ 
\hline\hline
$1$ & $1$ &  &  &  &  &  &  &  \\ 
$2$ & $24$ & $36$ & $12$ & $1$ &  &  &  &  \\ 
$3$ & $1440$ & $5760$ & $6120$ & $2520$ & $456$ & $36$ & $1$ & $\cdots $ \\ 
$\vdots $ &  &  & $\vdots $ &  &  & $\vdots $ &  & 
\end{tabular}%
\end{equation*}

\begin{center}
\begin{tabular}{ccc}
%\begin{array}{ccc}
$\begin{array}{c}
S_{4,4}\!\left( p,k\right) \\ 
4\leq \!k\!\leq \!4p%
\end{array}
$
&
%&  & 
\begin{tabular}{c||cccccccccc}
$p\backslash k$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$ & $%
\cdots $ \\ \hline\hline
$1$ & $1$ &  &  &  &  &  &  &  &  &  \\ 
$2$ & $24$ & $96$ & $72$ & $16$ & $1$ &  &  &  &  &  \\ 
$3$ & $576$ & $13824$ & $50688$ & $59904$ & $30024$ & $7200$ & $856$ & $48$
& $1$ & $\cdots $ \\ 
$\vdots $ &  &  & $\vdots $ &  &  & $\vdots $ &  &  & $\vdots $ & 
\end{tabular}%
%\end{array}%
\end{tabular}
\end{center}

Recall that the \textit{convolution} of the sequences $a_{n}$ and $b_{n}$ is
defined as the sequence $a_{n}\ast b_{n}=\sum_{t=0}^{n}a_{t}b_{n-t}$; in
particular $a_{n}\ast 1$ is the sequence of partial sums of $a_{n}$, and for
a non-negative integer $l$, the sequence $a_{n}\ast ^{l}1$ is defined
recursively as $a_{n}\ast ^{0}1=a_{n}$ and $a_{n}\ast ^{l}1=\left( a_{n}\ast
^{l-1}1\right) \ast 1$.

Beginning with the sequence $n^{p}$ (where $p\in \mathbb{N}$ is given),
written as 
\begin{equation}
n^{p}=\sum_{j=1}^{p}j!S\left( p,j\right) \binom{n}{j},  \label{1.01}
\end{equation}%
(see formula (6.10) in \cite{G-K-P}) and the partial sums of it, namely $%
n^{p}\ast ^{l}1$, we studied in \cite{Pi} the number arrays formed by the
coefficients resulting when we write these sequences as linear combinations
of binomial coefficients. We called them $p$-arrays. It turns out that these
arrays are natural generalizations of Pascal's triangle (case $p=1$) and
Lucas triangle (case $p=2$). We showed that many of the known properties of
Pascal and Lucas triangles (the expected generalizations of those
properties) are valid for the $p$-arrays, including the property on rising
diagonals (recall that the sums of rising diagonals of Pascal's triangle are
Fibonacci numbers, and in the Lucas triangle are Lucas numbers). We showed
that the sums of the rising diagonals in a $p$-array is equal to some
constant (the Stirling-Bernoulli transform of Fibonacci numbers, that
depends only on $p$; see \cite[A050946]{OEIS}) times a Fibonacci number,
when $p$ is odd, and times a Lucas number, when $p$ is even. In this work we
generalize the above results. Beginning with the sequence $\prod_{j=1}^{p}%
\binom{n+\left( j-1\right) \left( r-s\right) }{s}$, where $r,s\in \mathbb{N}$
are given, and $1\leq s\leq r$, we show that a similar formula to (\ref{1.01}%
) is valid, namely 
\begin{equation}
\prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}=\frac{1}{%
\left( s!\right) ^{p}}\sum_{j=s}^{ps}j!S_{r,s}\left( p,j\right) \dbinom{n}{j}%
.  \label{1.02}
\end{equation}

Observe that in the case $r=s=1$, formula (\ref{1.02}) is (\ref{1.01}), and
in the case $r=s$ formula (\ref{1.02}) looks as%
\begin{equation}
\dbinom{n}{s}^{p}=\frac{1}{\left( s!\right) ^{p}}\sum_{j=s}^{ps}j!S_{s,s}%
\left( p,j\right) \dbinom{n}{j},  \label{1.021}
\end{equation}%
which gives us the $p$-th power of the binomial coefficient $\binom{n}{s}$
written as a linear combination of the binomials $\binom{n}{j}$, $s\leq
j\leq sp$.

Moreover, we consider the partial sums of the sequence $\prod_{j=1}^{p}%
\binom{n+\left( j-1\right) \left( r-s\right) }{s}$ and show that they can be
written as linear combination of binomial coefficients according to

\begin{eqnarray}
&&\left( \prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}%
\right) \ast ^{k-pr+s-1}1  \label{1.03} \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{j=s}^{k-p\left( r-s\right)
+s-1}\sum_{i=s}^{j}\dbinom{k-pr+s-1}{j-i}i!S_{r,s}\left( p,i\right) \dbinom{n%
}{j},  \notag
\end{eqnarray}%
where $k\geq pr-s+1$. With the coefficients%
\begin{equation}
a_{k,j}^{\left( r,s\right) }=\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{j}%
\binom{k-pr+s-1}{j-i}i!S_{r,s}\left( p,i\right) ,  \label{1.04}
\end{equation}%
we form the $\left( r,s,p\right) $-arrays (with $k$ for lines and $j$ for
columns), which generalize the $p$-arrays of \cite{Pi}, and then also
generalize the Pascal's triangle (case $r=s=p=1$) and Lucas triangle (case $%
r=s=1$, $p=2$). Observe that the case $r=s$ of (\ref{1.03}) give us the
`hyper-sum' (partial sums) of powers of binomial coefficients expressed as a
linear combination of binomial coefficients, namely%
\begin{equation}
\dbinom{n}{s}^{p}\ast ^{k-pr+s-1}1=\frac{1}{\left( s!\right) ^{p}}%
\sum_{j=s}^{k+s-1}\sum_{i=s}^{j}\dbinom{k-p\left( s-1\right) -1}{j-i}%
i!S_{s,s}\left( p,i\right) \dbinom{n}{j}.  \label{1.05}
\end{equation}

Expressions like (\ref{1.05}) will appear several times in this work, with
variants in the right-hand side. What we have in the left-hand side of (\ref%
{1.05}) is a (partial) sum of (partial) sums of the sequence of $p$-th
powers of the binomial coefficients $\binom{n}{s}$, in which $s$ is fixed
and $n$ is the running index. To find closed formulas for sums of powers of
binomial coefficients of the type $\sum_{k=0}^{n}\binom{n}{k}^{p}$ are 
\textit{difficult problems} (it is known that in some cases these formulas
do not exist; see for example \cite{Ca}). Our sums of powers of binomial
coefficients do not belong to the difficult ones, but we believe that they
are new results.

In section \ref{Sec.2} we prove the main results of the work (formulas (\ref%
{1.02}) and (\ref{1.03}), in propositions \ref{Prop.2.1} and \ref{Prop.2.2},
respectively). We also exhibit in this section the first few rows and
columns of some $\left( r,s,p\right) $-arrays. In section \ref{Sec.3} we
mention some of the `easy' properties of the $\left( r,s,p\right) $-arrays,
related to their rows, columns, and falling diagonals. The proofs of most of
these properties are left as easy exercises to the reader. In section \ref%
{Sec.4} we consider the rising diagonals of the $\left( r,s,p\right) $%
-arrays. As in the previous work \cite{Pi}, this is the most important
section of the article, and justifies that our $\left( r,s,p\right) $-arrays
can be seen as generalizations of Pascal and Lucas triangles. In this
section we consider only $\left( r,s,p\right) $-arrays in which $r=s$ (see
Remark \ref{Remark4.1}). We prove that in the sums of rising diagonals in a $%
\left( r,r,p\right) $-array appear Fibonacci numbers when $r$ is even or $r$
and $p$ are odd (this includes the case $r=1$ and $p$ odd of \cite{Pi}), or
Lucas numbers when $r$ is odd and $p$ is even (this includes the case $r=1$
and $p$ even of \cite{Pi}), multiplied by a constant that depends only on $r$
and $p$. The corresponding sequences of these constants (with $p\in \mathbb{N%
}$ and $r$ is fixed) generalize the sequence of Stirling-Bernoulli transform
of Fibonacci numbers mentioned before. Finally, in section \ref{Sec.5} we
use a result of section \ref{Sec.4} to obtain a different version of (\ref%
{1.021}). It turns out that this new version can be simplified to a
different expression for $\binom{n}{s}^{p}$ similar to (\ref{1.021}) but
simpler than it.

\begin{remark}
\label{Rem.1}We would like to do some comments on the direction presented in 
\cite{Ch2} for a generalization of Stirling numbers (see also the previous
works related to this generalization \cite{Ch3,Kou}, and the later work \cite%
{Ch1} containing $q$-versions of it). Certainly what is presented in \cite%
{Ch2} is different to the direction contained in Blasiak's works, but we
think it is worth to mention it since we can see some similar mathematical
ingredients in both (besides the non-mathematical fact ---but a coincidence,
after all--- that the generalization presented in \cite{Ch2} is also a two
parameters one, and these parameters are also denoted by $r$ and $s$ ---with
nothing to do with Blasiak's parameters $r$ and $s$---).

The classical definition of Stirling numbers of the second kind $S\left(
n,k\right) $ in (\ref{1.5}) can be generalized if the `central' $x$ ($=x+0$)
of the left-hand side, is replaced by the `non-central' $x+r$, namely%
\begin{equation}
\left( x+r\right) ^{n}=\sum_{k=0}^{n}S_{r}\left( n,k\right) x^{\underline{%
\,k\,}}.  \label{1.20}
\end{equation}

The numbers $S_{r}\left( n,k\right) $ are called \underline{\textit{non-central Stirling numbers of the second kind}}. So\-me examples are the
following
\begin{equation*}
\begin{array}{lllll}
S_{r}\left( 1,0\right) =r &  & S_{r}\left( 2,0\right) =r^{2} &  & 
S_{r}\left( 3,0\right) =r^{3} \\ 
S_{r}\left( 1,1\right) =1 &  & S_{r}\left( 2,1\right) =2r+1 &  & S_{r}\left(
3,1\right) =3r^{2}+3r+1 \\ 
&  & S_{r}\left( 2,2\right) =1 &  & S_{r}\left( 3,2\right) =3r+3 \\ 
&  &  &  & S_{r}\left( 3,3\right) =1
\end{array}
\end{equation*}

If now the powers $\left( x+r\right) ^{n}$ of the left-hand side of (\ref{1.20}) are replaced by the non-central re-scaled falling factorials $\left(
sx+r\right) ^{\underline{\,n}}$, we obtain some new coefficients in the
linear combination of the right-hand side of (\ref{1.20}). We have
\begin{equation}
\left( sx+r\right) ^{\underline{\,n}}=\sum_{k=0}^{n}C_{r,s}\left( n,k\right)
x^{\underline{\,k\,}}.  \label{1.21}
\end{equation}

The numbers $C_{r,s}\left( n,k\right) $ are called \underline{\textit{non--central generalized factorial coefficients}} (or \linebreak[4] Gould-Hopper
numbers). Observe that (\ref{1.21}) share with (\ref{1.3}) the fact that
both formulas express certain falling factorials (those of the corresponding
left-hand sides) as linear combinations of some other falling factorials
(those of the corresponding right-hand sides). So we can see the non-central
generalized factorial coefficients $C_{r,s}\left( n,k\right) $ of (\ref{1.21}%
) as the objects playing a similar role to the generalized Stirling numbers
of the second kind $S_{r,s}\left( p,k\right) $ of (\ref{1.3}). Some examples
are the following%
\begin{equation*}
\begin{array}{lll}
& C_{r,s}\left( 1,0\right) =r &  \\ 
& C_{r,s}\left( 1,1\right) =s &  \\ 
&  &  \\ 
& C_{r,s}\left( 2,0\right) =r\left( r-1\right) &  \\ 
& C_{r,s}\left( 2,1\right) =s\left( 2r+s-1\right) &  \\ 
& C_{r,s}\left( 2,2\right) =s^{2} &  \\ 
&  &  \\ 
& C_{r,s}\left( 3,0\right) =r\left( r-1\right) \left( r-2\right) &  \\ 
& C_{r,s}\left( 3,1\right) =s\left( 3r^{2}+3r\left( s-2\right) +\left(
s-1\right) \left( s-2\right) \right) &  \\ 
& C_{r,s}\left( 3,2\right) =3s^{2}\left( r+s-1\right) &  \\ 
& C_{r,s}\left( 3,3\right) =s^{3} & 
\end{array}%
\end{equation*}

We have the explicit formula \cite[p. 335, ex. 35]{Ch2} 
\begin{equation}
C_{r,s}\left( n,k\right) =\frac{1}{k!}\sum_{j=0}^{k}\left( -1\right) ^{k+j}%
\binom{k}{j}\left( sj+r\right) ^{\underline{\,n\,}}.  \label{1.22}
\end{equation}

A connection of $C_{r,s}\left( n,k\right) $ with $S_{r}\left( n,k\right) $
is the following%
\begin{equation}
\lim_{s\rightarrow \infty }s^{-n}C_{r,s}\left( n,k\right) =S_{\rho }\left(
n,k\right) ,  \label{1.23}
\end{equation}%
where $\lim_{s\rightarrow \infty }\left( rs^{-1}\right) =\rho $ \cite[p.
335, ex. 37]{Ch2}. The particular case $s=-1$ of (\ref{1.21}) gives us the
coefficients $C_{r,-1}\left( n,k\right) $, called \underline{\textit{%
non-central Lah numbers}} and denoted by $L_{r}\left( n,k\right) $. We have 
\cite[p. 335, ex. 38]{Ch2}%
\begin{equation}
L_{r}\left( n,k\right) =\left( -1\right) ^{n}\frac{n!}{k!}\binom{n-r-1}{k-r-1%
}.  \label{1.24}
\end{equation}%
(Observe that we have also a relation between the Blasiak's generalized
Stirling numbers of the second kind and some generalized Lah numbers: see (%
\ref{1.9}).)

Formula (\ref{1.21}) can be written as%
\begin{equation}
\binom{sn+r}{p}=\frac{1}{p!}\sum_{k=0}^{p}k!C_{r,s}\left( p,k\right) \binom{n%
}{k},  \label{1.25}
\end{equation}%
which gives us the binomial coefficient $\binom{sn+r}{p}$ written as a
linear combination of the binomial coefficients $\binom{n}{k}$, $%
k=0,1,\ldots ,p$. (Again we see that (\ref{1.25}) has the same flavor of (%
\ref{1.021}).) Note that (\ref{1.02}) contains (\ref{1.021}) (with integer $%
p>1$) which is not contained in (\ref{1.25}). On the other hand, formula (%
\ref{1.25}) in the case $s\neq 1$, $r\neq 0$, is not contained in (\ref{1.02}%
). In fact, we have not any interesting intersection in both results, which
reminds us that, despite the shared flavor of both objects, they appear in
different directions. Some examples from (\ref{1.25}) are the following%
\begin{eqnarray*}
\binom{2n+1}{3} &=&\binom{n}{1}+8\binom{n}{2}+8\binom{n}{3}, \\
\binom{2n+1}{4} &=&5\binom{n}{2}+20\binom{n}{3}+16\binom{n}{4}, \\
\binom{3n+5}{4} &=&5+65\binom{n}{1}+195\binom{n}{2}+216\binom{n}{3}+81\binom{%
n}{4}.
\end{eqnarray*}
\end{remark}

\section{The main results\label{Sec.2}}

In the following proposition we consider a version of (\ref{1.3}) (and give
a simple induction argument to prove it). This result (\ref{2.1}), together
with the corresponding formula (\ref{2.5}) from proposition \ref{Prop.2.2},
are the starting point of the results we will obtain in the remaining
sections.

\begin{proposition}
\label{Prop.2.1}The following formula holds 
\begin{equation}
\prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}=\frac{1}{%
\left( s!\right) ^{p}}\sum_{k=s}^{ps}k!S_{r,s}\left( p,k\right) \dbinom{n}{k}%
.  \label{2.1}
\end{equation}
\end{proposition}

\begin{proof}
By induction on $p$. For $p=1$ formula (\ref{2.1}) is trivial. Suppose it is
true for a given $p\in \mathbb{N}$. Then, by using (\ref{1.7}) and the
induction hypothesis we have%
\begin{eqnarray*}
&&\frac{1}{\left( s!\right) ^{p+1}}\sum_{k=s}^{ps+s}k!S_{r,s}\left(
p+1,k\right) \dbinom{n}{k} \\
&=&\frac{1}{\left( s!\right) ^{p+1}}\sum_{k=s}^{ps+s}k!\sum_{i=0}^{s}\dbinom{%
p\left( r-s\right) +k+i-s}{i}\frac{s!}{\left( s-i\right) !}S_{r,s}\left(
p,k+i-s\right) \dbinom{n}{k} \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{k=s}^{ps+s}\sum_{i=0}^{s}\dbinom{%
p\left( r-s\right) +k-i}{s-i}\frac{k!}{i!}S_{r,s}\left( p,k-i\right) \dbinom{%
n}{k} \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{k=s}^{ps}\sum_{i=0}^{s}\dbinom{%
p\left( r-s\right) +k}{s-i}\frac{\left( i+k\right) !}{i!}S_{r,s}\left(
p,k\right) \dbinom{n}{i+k} \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{k=s}^{ps}k!S_{r,s}\left( p,k\right) 
\dbinom{n}{k}\sum_{i=0}^{s}\dbinom{p\left( r-s\right) +k}{s-i}\dbinom{n-k}{i}
\\
&=&\left( \prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}%
\right) \dbinom{n+p\left( r-s\right) }{s} \\
&=&\prod_{j=1}^{p+1}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s},
\end{eqnarray*}%
as expected.
\end{proof}

From expression (\ref{2.1}), we have at once the corresponding expression of
the partial sums of the sequence $\binom{n}{r}^{p}$ in terms of binomial
coefficients. In fact, by using that 
\begin{equation}
\binom{n}{r_{1}}\ast \binom{n}{r_{2}}=\binom{n+1}{r_{1}+r_{2}+1},
\label{2.35}
\end{equation}%
we have for $l\in \mathbb{N}$%
\begin{equation}
\dbinom{n}{r}^{p}\ast ^{l}1=\frac{1}{\left( r!\right) ^{p}}%
\sum_{j=r}^{pr}j!S_{r,r}\left( p,j\right) \dbinom{n+l}{j+l}.  \label{2.4}
\end{equation}

However, we are interested in expressions of the partial sums $\binom{n}{r}%
^{p}\ast ^{l}1$ as linear combinations of the binomials $\binom{n}{s}$,$%
\binom{n}{s+1}$,$\binom{n}{s+2}$,\ldots , since we want to study the number
arrays formed by the corresponding coefficients of these expressions.

\begin{proposition}
\label{Prop.2.2}For $k\geq pr-s+1$ we have%
\begin{equation}
\left( \prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}%
\right) \ast ^{k-pr+s-1}1=\sum_{j=s}^{k-p\left( r-s\right)
+s-1}a_{k,j}^{\left( r,s\right) }\dbinom{n}{j}.  \label{2.5}
\end{equation}%
where the coefficients $a_{k,j}^{\left( r,s\right) }$ (for $s\leq j\leq
k-p\left( r-s\right) +s-1$) are given by%
\begin{equation}
a_{k,j}^{\left( r,s\right) }=\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{j}%
\dbinom{k-pr+s-1}{j-i}i!S_{r,s}\left( p,i\right) .  \label{2.6}
\end{equation}
\end{proposition}

\begin{proof}
We proceed by induction on $k$. The case \thinspace $k=pr-s+1$ is (\ref{2.1}%
). Suppose the result is valid for a given $k\geq pr-s+1$. That is, suppose
that 
\begin{eqnarray}
&&\left( \prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}%
\right) \ast ^{k-pr+s-1}1  \label{2.7} \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{j=s}^{k-p\left( r-s\right)
+s-1}\sum_{i=s}^{j}\dbinom{k-pr+s-1}{j-i}i!S_{r,s}\left( p,i\right) \dbinom{n%
}{j}.  \notag
\end{eqnarray}%
By taking the convolution in both sides of (\ref{2.7}) with $1$ we obtain
that%
\begin{eqnarray}
&&\left( \prod_{j=1}^{p}\dbinom{n+\left( j-1\right) \left( r-s\right) }{s}%
\right) \ast ^{k+1-pr+s-1}1  \label{2.8} \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{j=s}^{k-p\left( r-s\right)
+s-1}\sum_{i=s}^{j}\dbinom{k-pr+s-1}{j-i}i!S_{r,s}\left( p,i\right) \dbinom{%
n+1}{j+1}.  \notag
\end{eqnarray}%
Thus, we need to prove that%
\begin{eqnarray}
&&\sum_{j=s}^{k-p\left( r-s\right) +s-1}\sum_{i=s}^{j}\dbinom{k-pr+s-1}{j-i}%
i!S_{r,s}\left( p,i\right) \dbinom{n+1}{j+1}  \label{2.9} \\
&=&\sum_{j=s}^{k+1-p\left( r-s\right) +s-1}\sum_{i=s}^{j}\dbinom{k+1-pr+s-1}{%
j-i}i!S_{r,s}\left( p,i\right) \dbinom{n}{j},  \notag
\end{eqnarray}%
and this is an easy exercise left to the reader.
\end{proof}

We form the $\left( r,s,p\right) $-arrays with the coefficients $%
a_{k,j}^{\left( r,s\right) }$ ($k$ for rows and $j$ for columns). The first $%
\left( pr-s\right) $ rows are cancelled, so the first non-cancelled row is
the $\left( pr-s+1\right) $-th one, which contains the coefficients of the
right-hand side of (\ref{2.1}) (corresponding to $k=pr-s+1$ of (\ref{2.5})).
The first $\left( s-1\right) $ columns are cancelled, and the corresponding
columns of the $k$-th row goes from the $s$-th column up to the $\left(
k-p\left( r-s\right) +s-1\right) $-th column (with zeros in the subsequent
places). Since the coefficients $a_{k,j}^{\left( r,s\right) }$ have the
Pascal's triangle property $a_{k,j}^{\left( r,s\right) }+a_{k,j+1}^{\left(
r,s\right) }=a_{k+1,j+1}^{\left( r,s\right) }$, we can also form the $\left(
r,s,p\right) $-arrays beginning with the first non-cancelled row, and
filling-out the array with the Pascal's triangle recurrence.

For example, in the case $r=s$ (formula (\ref{1.05})), if we set $k=s\left(
p-1\right) +2$, we can write%
\begin{equation}
\sum_{i=0}^{n}\dbinom{i}{s}^{p}=\frac{1}{\left( s!\right) ^{p}}%
\sum_{j=s}^{ps+1}\left( j-1\right) !\left( jS_{s,s}\left( p,j\right)
+S_{s,s}\left( p,j-1\right) \right) \dbinom{n}{j}.  \label{2.11}
\end{equation}

(The case $s=1$ of (\ref{2.11}) is the well-known formula $ \sum_{i=0}^{n}i^{p}=\sum_{j=1}^{p+1}\left( j-1\right) !S ( p+1,j ) 
\binom{n}{j}$ for sums of powers of integers.) In the case $r=s$ (\ref{1.05}%
), if we set $p=1$ we obtain that (for $k\geq 1$)%
\begin{equation}
\dbinom{n}{s}\ast ^{k-1}1=\sum_{j=0}^{k-1}\dbinom{k-1}{j}\dbinom{n}{j+s},
\label{2.14}
\end{equation}%
which gives us the partial sums of the sequence $\binom{n}{s}$ in terms of
the binomials $\binom{n}{j+s}$, $j=0,1,\ldots ,k-1$. For example, $%
\sum_{i=0}^{n}\binom{i}{s}=\binom{n}{s}+\binom{n}{s+1}$, $%
\sum_{j=0}^{n}\sum_{i=0}^{j}\binom{i}{s}=\binom{n}{s}+2\binom{n}{s+1}+\binom{%
n}{s+2}$, $\sum_{k=0}^{n}\sum_{j=0}^{k}\sum_{i=0}^{j}\binom{i}{s}=\binom{n}{s%
}+3\binom{n}{s+1}+3\binom{n}{s+2}+\binom{n}{s+3}$, and so on. That is, the $%
\left( s,s,1\right) $-array is a \textquotedblleft right-shifted Pascal's
triangle\textquotedblright , with its first column (the constant sequence $1$%
) in the $s$-th column of the array.

In the case $r=2s$ we can use (\ref{1.9}) to write (\ref{2.5}) as%
\begin{eqnarray*}
&&\left( \prod_{j=1}^{p}\dbinom{n+s\left( j-1\right) }{s}\right) \ast
^{k-2ps+s-1}1 \\
&=&\frac{\left( sp\right) !}{\left( s!\right) ^{p}}\sum_{j=s}^{k-s\left(
p-1\right) -1}\sum_{i=s}^{j}\dbinom{k-2ps+s-1}{j-i}\dbinom{s\left(
p-1\right) }{i-s}\dbinom{n}{j}.
\end{eqnarray*}%
or%
\begin{equation}
\left( \prod_{j=1}^{p}\dbinom{n+s\left( j-1\right) }{s}\right) \ast
^{k-2ps+s-1}1=\frac{\left( sp\right) !}{\left( s!\right) ^{p}}%
\sum_{j=s}^{k-s\left( p-1\right) -1}\dbinom{k-ps-1}{j-s}\dbinom{n}{j},
\label{2.15}
\end{equation}%
where $k\geq 2ps-s+1$. Note then that the $\left( 2s,s,p\right) $-array is a
\textquotedblleft piece of Pascal's triangle multiplied by $\frac{\left(
sp\right) !}{\left( s!\right) ^{p}}$\textquotedblright : the first
non-cancelled row of the array (the $\left( 2ps-s+1\right) $-th row)
contains the $s\left( p-1\right) $-th row of Pascal's triangle (multiplied
by the factor $\frac{\left( sp\right) !}{\left( s!\right) ^{p}}$); the first
non-cancelled column of the array (the $s$-th column) contains the first
column of Pascal's triangle (multiplied by the factor $\frac{\left(
sp\right) !}{\left( s!\right) ^{p}}$). That is, the $s$-th column of the
array is the constant sequence $\frac{\left( sp\right) !}{\left( s!\right)
^{p}}$. The subsequent rows and columns of the array are the corresponding
subsequent rows and columns of Pascal's triangle multiplied by $\frac{\left(
sp\right) !}{\left( s!\right) ^{p}}$. See for example the $\left(
2,1,3\right) $-array below, which is a Pascal's triangle beginning in row $%
s\left( p-1\right) =2$ and multiplied by $\frac{\left( sp\right) !}{\left(
s!\right) ^{p}}=6$, and the $\left( 4,2,3\right) $-array which is a Pascal's
triangle beginning in row $s\left( p-1\right) =4$ and multiplied by $\frac{%
\left( sp\right) !}{\left( s!\right) ^{p}}=90$.

Some $\left( r,s,p\right) $-arrays are the following:

%\begin{equation*}
\[
\begin{array}{ll}
\begin{tabular}{|c|}
\hline
$\left( 2,1,3\right) $-array, $k\geq 6,1\leq j\leq k-3$ \\ \hline\hline
$%
\begin{array}{cccccccc}
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
6 & 12 & 6 & 0 & 0 & 0 & 0 & \cdots \\ 
6 & 18 & 18 & 6 & 0 & 0 & 0 &  \\ 
6 & 24 & 36 & 24 & 6 & 0 & 0 &  \\ 
6 & 30 & 60 & 60 & 30 & 6 & 0 & \cdots \\ 
\vdots &  &  & \vdots &  &  & \vdots & 
\end{array}%
$ \\ \hline
\end{tabular}
%& 
\,\,
\begin{tabular}{|c|}
\hline
$\left( 2,2,3\right) $-array, $k\geq 5,2\leq j\leq k+1$ \\ \hline\hline
$%
\begin{array}{cccccccccc}
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & 1 & 24 & 114 & 180 & 90 & 0 & 0 & 0 & \cdots \\ 
\ast & 1 & 25 & 138 & 294 & 270 & 90 & 0 & 0 &  \\ 
\ast & 1 & 26 & 163 & 432 & 564 & 360 & 90 & 0 &  \\ 
\ast & 1 & 27 & 189 & 595 & 996 & 924 & 450 & 90 & \cdots \\ 
\ast & \vdots &  &  & \vdots &  &  & \vdots &  & 
\end{array}%
$ \\ \hline
\end{tabular}%
\end{array}%
%\end{equation*}%
\]
\begin{equation*}
\begin{array}{ll}
\begin{tabular}{|c|}
\hline
$\left( 3,2,3\right) $-array, $k\geq 8,2\leq j\leq k-2$ \\ \hline\hline
$%
\begin{array}{cccccccccc}
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{18} & _{126} & _{288} & _{270} & _{90} & _{0} & _{0} & _{0}
& _{\cdots } \\ 
_{_{\ast }} & _{18} & _{144} & _{414} & _{558} & _{360} & _{90} & _{0} & _{0}
&  \\ 
_{_{\ast }} & _{18} & _{162} & _{558} & _{972} & _{918} & _{450} & _{90} & 
_{0} &  \\ 
_{_{\ast }} & _{\vdots } &  &  & _{\vdots } &  &  &  &  & 
\end{array}%
$ \\ \hline
\end{tabular}
& 
\begin{tabular}{|c|}
\hline
$\left( 3,3,2\right) $-array, $k\geq 4,3\leq j\leq k+2$ \\ \hline\hline
$%
\begin{array}{cccccccccc}
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & 1 & 12 & 30 & 20 & 0 & 0 & 0 & \cdots \\ 
\ast & \ast & 1 & 13 & 42 & 50 & 20 & 0 & 0 &  \\ 
\ast & \ast & 1 & 14 & 55 & 92 & 70 & 20 & 0 &  \\ 
\ast & \ast & \vdots &  &  & \vdots &  &  & \vdots & 
\end{array}%
$ \\ \hline
\end{tabular}%
\end{array}%
\end{equation*}%
\begin{equation*}
\begin{tabular}{|c|}
\hline
$\left( 3,3,3\right) $-array, $k\geq 7,3\leq j\leq k+2$ \\ \hline\hline
$%
\begin{array}{cccccccccccccc}
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & 
\ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & 
\ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & 
\ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & 
\ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & 
\ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast & 
\ast & \ast & \ast \\ 
\ast & \ast & 1 & 60 & 690 & 2940 & 5670 & 5040 & 1680 & 0 & 0 & 0 & 0 & 
\cdots \\ 
\ast & \ast & 1 & 61 & 750 & 3630 & 8610 & 10710 & 6720 & 1680 & 0 & 0 & 0 & 
\\ 
\ast & \ast & 1 & 62 & 811 & 4380 & 12240 & 19320 & 17430 & 8400 & 1680 & 0
& 0 &  \\ 
\ast & \ast & 1 & 63 & 873 & 5191 & 16620 & 31560 & 36750 & 25830 & 10080 & 
1680 & 0 & \cdots \\ 
\ast & \ast & \vdots &  &  & \vdots &  &  & \vdots &  &  & \vdots &  & 
\end{array}%
$ \\ \hline
\end{tabular}%
\end{equation*}%
\begin{equation*}
\begin{array}{lll}
\begin{tabular}{|c|}
\hline
$\left( 4,1,3\right) $-array, $k\geq 12,1\leq j\leq k-9$ \\ \hline\hline
$%
\begin{array}{cccccccc}
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
\ast & \ast & \ast & \ast & \ast & \ast & \ast & \ast \\ 
28 & 24 & 6 & 0 & 0 & 0 & 0 & \cdots \\ 
28 & 52 & 30 & 6 & 0 & 0 & 0 &  \\ 
28 & 80 & 82 & 36 & 6 & 0 & 0 &  \\ 
28 & 108 & 162 & 118 & 42 & 6 & 0 & \cdots \\ 
\vdots &  &  & \vdots &  &  & \vdots & 
\end{array}%
$ \\ \hline
\end{tabular}
&  & 
\begin{tabular}{|c|}
\hline
$\left( 4,2,3\right) $-array, $k\geq 11,2\leq j\leq k-5$ \\ \hline\hline
$%
\begin{array}{cccccccccc}
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{90} & _{360} & _{540} & _{360} & _{90} & _{0} & _{0} & _{0} & 
_{\cdots } \\ 
_{\ast } & _{90} & _{450} & _{900} & _{900} & _{450} & _{90} & _{0} & _{0} & 
\\ 
_{\ast } & _{90} & _{540} & _{1350} & _{1800} & _{1350} & _{540} & _{90} & 
_{0} &  \\ 
_{\ast } & _{\vdots } &  &  & _{\vdots } &  &  & _{\vdots } &  & 
\end{array}%
$ \\ \hline
\end{tabular}%
\end{array}%
\end{equation*}%
\begin{equation*}
\begin{array}{ll}
\!\!%
\begin{tabular}{|c|}
\hline
$\left( 4,3,2\right) $-array, $k\geq 6,3\leq j\leq k$ \\ \hline\hline
$%
\begin{array}{cccccccccc}
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & 
_{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} & _{_{\ast }} \\ 
_{_{\ast }} & _{_{\ast }} & _{4} & _{24} & _{40} & _{20} & _{0} & _{0} & _{0}
& _{\cdots } \\ 
_{_{\ast }} & _{_{\ast }} & _{4} & _{28} & _{64} & _{60} & _{20} & _{0} & 
_{0} &  \\ 
_{_{\ast }} & _{_{\ast }} & _{4} & _{32} & _{92} & _{124} & _{80} & _{20} & 
_{0} &  \\ 
_{_{\ast }} & _{_{\ast }} & _{\vdots } &  &  & _{\vdots } &  &  &  & 
\end{array}%
$ \\ \hline
\end{tabular}
& \!\!%
\begin{tabular}{|c|}
\hline
$\left( 4,4,2\right) $-array, $k\geq 5,4\leq j\leq k+3$ \\ \hline\hline
$%
\begin{array}{cccccccccccc}
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } & 
_{\ast } & _{\ast } & _{\ast } & _{\ast } & _{\ast } \\ 
_{\ast } & _{\ast } & _{\ast } & _{1} & _{20} & _{90} & _{140} & _{70} & _{0}
& _{0} & _{0} & _{\cdots } \\ 
_{\ast } & _{\ast } & _{\ast } & _{1} & _{21} & _{110} & _{230} & _{210} & 
_{70} & _{0} & _{0} &  \\ 
_{\ast } & _{\ast } & _{\ast } & _{1} & _{22} & _{131} & _{340} & _{440} & 
_{280} & _{70} & _{0} &  \\ 
_{\ast } & _{\ast } & _{\ast } & _{\vdots } &  &  & _{\vdots } &  &  & 
_{\vdots } &  & 
\end{array}%
$ \\ \hline
\end{tabular}%
\end{array}%
\end{equation*}

Some examples from (\ref{2.7}) are shown in the following table.


\begin{table}[htpb] \centering%
$
\begin{tabular}{|c|c|c|}
\hline
Array & $k$ & Sequence (\ref{2.7}) \\ \hline\hline
$\left( 2,1,3\right) $ & $7$ & \multicolumn{1}{|l|}{$\sum\limits_{i=1}^{n}i%
\left( i+1\right) \left( i+2\right) =6\binom{n}{1}+18\binom{n}{2}+18\binom{n%
}{3}+6\binom{n}{4}$} \\ \hline
$\left( 2,2,3\right) $ & $6$ & \multicolumn{1}{|l|}{$\sum\limits_{i=1}^{n}%
\binom{i}{2}^{3}=\binom{n}{2}+25\binom{n}{3}+138\binom{n}{4}+294\binom{n}{5}%
+270\binom{n}{6}+90\binom{n}{7}$} \\ \hline
$\left( 3,2,3\right) $ & $8$ & \multicolumn{1}{|l|}{$\binom{n}{2}\binom{n+1}{%
2}\binom{n+2}{2}=18\binom{n}{2}+126\binom{n}{3}+288\binom{n}{4}+270\binom{n}{%
5}+90\binom{n}{6}$} \\ \hline
$\left( 3,3,2\right) $ & $5$ & \multicolumn{1}{|l|}{$\sum\limits_{i=1}^{n}%
\binom{i}{3}^{2}=\binom{n}{3}+13\binom{n}{4}+42\binom{n}{5}+50\binom{n}{6}+20%
\binom{n}{7}$} \\ \hline
$\left( 3,3,3\right) $ & $7$ & \multicolumn{1}{|l|}{$\binom{n}{3}^{3}=\binom{%
n}{3}+60\binom{n}{4}+690\binom{n}{5}+2940\binom{n}{6}+5670\binom{n}{7}+5040%
\binom{n}{8}+1680\binom{n}{9}$} \\ \hline
$\left( 4,1,3\right) $ & $12$ & \multicolumn{1}{|l|}{$n\left( n+3\right)
\left( n+6\right) =28\binom{n}{1}+24\binom{n}{2}+6\binom{n}{3}$} \\ \hline
$\left( 4,2,3\right) $ & $12$ & \multicolumn{1}{|l|}{$\sum\limits_{i=1}^{n}%
\binom{i+2}{2}\binom{i+4}{2}\binom{i+6}{2}=90\binom{n}{2}+450\binom{n}{3}+900%
\binom{n}{4}+900\binom{n}{5}+450\binom{n}{6}+90\binom{n}{7}$} \\ \hline
$\left( 4,3,2\right) $ & $6$ & \multicolumn{1}{|l|}{$\binom{n}{3}\binom{n+1}{%
3}=4\binom{n}{3}+24\binom{n}{4}+40\binom{n}{5}+20\binom{n}{6}$} \\ \hline
$\left( 4,4,2\right) $ & $6$ & \multicolumn{1}{|l|}{$\sum\limits_{i=1}^{n}%
\binom{i}{4}^{2}=\binom{n}{4}+21\binom{n}{5}+110\binom{n}{6}+230\binom{n}{7}%
+210\binom{n}{8}+70\binom{n}{9}$} \\ \hline
\end{tabular}%
$%
\caption{Examples of sequence (23). The coefficients come from k-th line of
the (r,s,p)-array. }\label{T.1}%
%TCIMACRO{\TeXButton{E}{\end{table}}}%
%BeginExpansion
\end{table}%
%EndExpansion

\section{Some properties (rows, columns, falling diagonals)\label{Sec.3}}

From the well-known properties of rows, columns and falling diagonals in
Pascal%
%TCIMACRO{\U{b4}}%
%BeginExpansion
\'{}%
%EndExpansion
s triangle, we state in this section the corresponding generalizations of
them for our $\left( r,s,p\right) $-arrays. Most of the proofs of these
facts are easy exercises left to the reader.

We begin by considering the rows of $\left( r,s,p\right) $-arrays.

\subsection{Rows}
%\medskip
%
%\begin{center}
%\underline{\textbf{Rows.}}
%\end{center}

We mention (leaving the proofs to the reader) the generalization of two
famous properties of rows of Pascal's triangle:

\begin{enumerate}
\item In a $\left( r,s,p\right) $-array, the sum of the elements of the $k$%
-th row ($k>pr-s+1$) is equal to $2^{k-pr+s-1}$ times the sum of the
elements of the $\left( pr-s+1\right) $-th row (the first non-cancelled
row). That is, we have%
\begin{equation}
\sum_{j=s}^{k-p\left( r-s\right) +s-1}a_{k,j}^{\left( r,s\right)
}=2^{k-pr+s-1}\sum_{j=s}^{ps}a_{pr-s+1,j}^{\left( r,s\right) },  \label{3.1}
\end{equation}%
or more explicitly%
\begin{equation}
\sum_{j=s}^{k-p\left( r-s\right) +s-1}\sum_{i=s}^{j}\dbinom{k-pr+s-1}{j-i}%
i!S_{r,s}\left( p,i\right) =2^{k-pr+s-1}\sum_{j=s}^{ps}j!S_{r,s}\left(
p,j\right) .  \label{3.101}
\end{equation}

\item Making the row $k\geq pr-s+1$ a single number $N_{k}$ (the elements of
the row being the digits, and carrying over when appear elements with more
than one digit), this is equal to $11^{k-pr+s-1}N_{pr-s+1}$, where $%
N_{pr-s+1}$ is the single number corresponding to row $k=pr-s+1$.
\end{enumerate}

In Table \ref{TableRows1} we have some examples taken from the first rows of
the $\left( 4,3,2\right) $-array.

%TCIMACRO{\TeXButton{B}{\begin{table}[htbp] \centering}}%
%BeginExpansion
\begin{table}[htbp] \centering%
%EndExpansion
$%
\begin{tabular}{|c|c|c|c|}
\hline
$%
\begin{array}{l}
\text{Beginning } \\ 
\text{in column} \\ 
j=3\text{ } \\ 
\text{and row }k%
\end{array}%
$ & The row is & $%
\begin{array}{l}
\text{Sum of the} \\ 
\text{elements}%
\end{array}%
$ & $%
\begin{array}{l}
\text{As a single} \\ 
\text{number}%
\end{array}%
$ \\ \hline\hline
$k=6$ & $\left( 4,24,40,20\right) $ & $88$ & $6820$ \\ \hline
$k=7$ & $\left( 4,28,64,60,20\right) $ & $176=2\left( 88\right) $ & $%
75020=11\left( 6820\right) $ \\ \hline
$k=8$ & $\left( 4,32,92,124,80,20\right) $ & $352=2^{2}\left( 88\right) $ & $%
825220=11^{2}\left( 6820\right) $ \\ \hline
$\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ \\ \hline
\end{tabular}%
$\caption{Properties of rows in the (4,3,2)-array.}\label{TableRows1}%
%TCIMACRO{\TeXButton{E}{\end{table}}}%
%BeginExpansion
\end{table}%
%EndExpansion

A more interesting question arises when one considers the sequence of
alternating sums of rows of a $\left( r,s,p\right) $-array. In the case $%
r=s=1$, we proved that this sequence is $\left( \left( -1\right)
^{p},0,0,0,\ldots \right) $ (see formula (26) in \cite{Pi}). In general, for
a $\left( r,s,p\right) $-array, we have only some partial answers, as we
show next.

First to all note that for $k>pr-s+1$, the alternating row sums of the $%
\left( r,s,p\right) $-array are equal to $0$ (easy proof left to the
reader). The alternating sum of the elements of row $k=pr-s+1$ is equal to%
\begin{equation}
\alpha _{r,s,p}=\frac{1}{\left( s!\right) ^{p}}\sum_{j=s}^{ps}\left(
-1\right) ^{j}j!S_{r,s}\left( p,j\right) .  \label{3.2}
\end{equation}

Thus, (ignoring the cancelled rows) the alternating sums of rows in a $%
\left( r,s,p\right) $-array, are sequences of the form $\left( \alpha
_{r,s,p},0,0,0,\ldots \right) $. Let us see some particular cases in which
the term $\alpha _{r,s,p}$ has a simple closed formula.

If $s=r$ we have $\alpha _{s,s,p}=\left( -1\right) ^{ps}$. That is, we have%
\begin{equation}
\frac{1}{\left( s!\right) ^{p}}\sum_{j=s}^{ps}\left( -1\right)
^{j}j!S_{s,s}\left( p,j\right) =\left( -1\right) ^{ps}.  \label{3.3}
\end{equation}

In fact, proceeding by induction on $p$, we see at once that this formula is
trivial if $p=1$. If it is true for a given $p\in \mathbb{N}$, then
(according to the recurrence (\ref{1.7}))%
\begin{eqnarray*}
&&\frac{1}{\left( s!\right) ^{p+1}}\sum_{j=s}^{\left( p+1\right) s}\left(
-1\right) ^{j}j!S_{s,s}\left( p+1,j\right) \\
&=&\frac{1}{\left( s!\right) ^{p+1}}\sum_{j=s}^{\left( p+1\right) s}\left(
-1\right) ^{j}j!\sum_{i=0}^{s}\dbinom{j+i-s}{i}\frac{s!}{\left( s-i\right) !}%
S_{s,s}\left( p,j+i-s\right) \\
&=&\frac{\left( -1\right) ^{s}}{\left( s!\right) ^{p}}\sum_{j=s}^{ps}%
\sum_{i=0}^{s}\frac{\left( -1\right) ^{j-i}\left( j-i+s\right) !j!}{\left(
s-i\right) !i!\left( j-i\right) !}S_{s,s}\!\left( p,j\right) \\
&=&\frac{\left( -1\right) ^{s}}{\left( s!\right) ^{p}}\sum_{j=s}^{ps}\!%
\left( -1\right) ^{j}j!\sum_{i=0}^{s}\!\left( -1\right) ^{i}\binom{s}{i}%
\binom{j-i+s}{s}S_{s,s}\!\left( p,j\right) .
\end{eqnarray*}

But we have $\sum_{i=0}^{s}\left( -1\right) ^{i}\binom{s}{i}\binom{j-i+s}{s}%
=1$ (see identity (\ref{3.17}) below, or identity (1.83) in \cite{Go2}).
Thus, by using the induction hypothesis we have%
\begin{eqnarray*}
\frac{1}{\left( s!\right) ^{p+1}}\sum_{j=s}^{\left( p+1\right) s}\left(
-1\right) ^{j}j!S_{s,s}\left( p+1,j\right)  & = & \frac{\left( -1\right) ^{s}}{ \left( s!\right) ^{p}}\sum_{j=s}^{ps}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \\ 
& = & \left( -1\right) ^{s}\left( -1\right) ^{ps}=\left( -1\right)
^{\left( p+1\right) s},
\end{eqnarray*}
as expected. That is, the alternating sum of rows in the $\left(
s,s,p\right) $-arrays is the sequence $\left( \left( -1\right)
^{ps},0,0,0,\ldots \right) $.

Other interesting particular case is when $s=1$. We claim that%
\begin{equation}
\alpha _{r,1,p}=\sum_{j=1}^{p}\left( -1\right) ^{j}j!S_{r,1}\left(
p,j\right) =\prod_{i=1}^{p}\left( \left( r-1\right) i-r\right) .
\label{3.4}
\end{equation}

The case $p=1$ is obvious. If this formula is true for a given $p\in \mathbb{%
N}$, we have (by using the recurrence (\ref{1.7}))%
\begin{eqnarray*}
&&\sum_{j=1}^{p+1}\left( -1\right) ^{j}j!S_{r,1}\left( p+1,j\right)  \\
&=&\sum_{j=1}^{p+1}\left( -1\right) ^{j}j!\left( S_{r,1}\left( p,j-1\right)
+\left( p\left( r-1\right) +j\right) S_{r,1}\left( p,j\right) \right)  \\
&=&\sum_{j=1}^{p}\!\left( -1\right) ^{j+1}j!\left( j+1\right) S_{r,1}\left(
p,j\right) +p\left( r-1\right) \!\sum_{j=1}^{p}\!\left( -1\right)
^{j}j!S_{r,1}\left( p,j\right)  \\
& & +\sum_{j=1}^{p}\!\left( -1\right)
^{j}j!jS_{r,1}\left( p,j\right)  \\
&=&\left( p\left( r-1\right) -1\right) \sum_{j=1}^{p}\left( -1\right)
^{j}j!S_{r,1}\left( p,j\right)  \\
&=&\left( p\left( r-1\right) -1\right) \prod_{i=1}^{p}\left( \left(
r-1\right) i-r\right) =\prod_{i=1}^{p+1}\left( \left( r-1\right) i-r\right)
.
\end{eqnarray*}%
as expected. Some examples (within the particular case $s=1$ we are
considering) are the following:

\begin{itemize}
\item If $r=1$ we have $\alpha _{1,1,p}=\left( -1\right) ^{p}$ (as we
obtained before).

\item If $r=2$ we have $\alpha _{2,1,p}=\prod_{i=1}^{p}\left( i-2\right) $,
so $\alpha _{2,1,p}=-1$ if $p=1$ and $\alpha _{2,1,p}=0$ if $p>1$. That is,
the alternating sums of rows in the $\left( 2,1,p\right) $-arrays is the
sequence $\left( -1,0,0,0,\ldots \right) $ when $p=1$, and the constant
sequence $0$ when $p>1$.

\item For $r>2$, we see from (\ref{3.4}) that $\alpha _{r,1,p}$ is the $%
\left( r-1\right) $-fold factorial:

\begin{itemize}
\item With $r=3$ we have the $p$-sequence 
\begin{equation*}
\left( \alpha _{3,1,p}\right) _{p\in \mathbb{N}}=\left(
\prod_{i=1}^{p}\left( 2i-3\right) \right) _{p\in \mathbb{N}}=\left(
-1,-1,-3,-15,\ldots \right) 
\end{equation*}
formed by the (negatives of \ the) double factorials $\left( 2p-3\right) !!$
(see \cite[A001147]{OEIS} for $\left( 2p-1\right) !!$). For example, the
alternating sums of rows in the $\left( 3,1,2\right) $-array is the sequence 
$\left( -1,0,0,0,\ldots \right) $, and the alternating sums of rows in the $%
\left( 3,1,3\right) $-array is the sequence $\left( -3,0,0,0,\ldots \right) $%
.

\item With $r=4$ we have the $p$-sequence 
\begin{equation*}
\left( \alpha _{4,1,p}\right) _{p\in \mathbb{N}}=\left(
\prod_{i=1}^{p}\left( 3i-4\right) \right) _{p\in \mathbb{N}}=\left(
-1,-2,-10,-80,\ldots \right) 
\end{equation*}
formed by the (negatives of \ the) triple factorials $\left( 3p-4\right) !!!$
(see \cite[A008544]{OEIS} for $\left( 3p-1\right) !!!$). For example, the
alternating sums of rows in the $\left( 4,1,2\right) $-array is the sequence 
$\left( -2,0,0,0,\ldots \right) $, and the alternating sums of rows in the $%
\left( 4,1,3\right) $-array is the sequence $\left( -10,0,0,0,\ldots \right) 
$.

\item With $r=5$ we have the $p$-sequence 
\begin{equation*}
\left( \alpha _{5,1,p}\right) _{p\in \mathbb{N}}=\left(
\prod_{i=1}^{p}\left( 4i-5\right) \right) _{p\in \mathbb{N}}=\left(
-1,-3,-21,-231,\ldots \right) 
\end{equation*}%
of the (negatives of \ the) quadruple factorials $\left( 4p-5\right) !!!!$
(see \cite[A008545]{OEIS} for $\left( 4p-1\right) !!!!$). For example, the
alternating sums of rows in the $\left( 5,1,2\right) $-array is the sequence 
$\left( -3,0,0,0,\ldots \right) $, and the alternating sums of rows in the $%
\left( 5,1,3\right) $-array is the sequence $\left( -21,0,0,0,\ldots \right) 
$.
\end{itemize}
\end{itemize}

\subsection{Columns}
%\medskip
%
%\begin{center}
%\underline{\textbf{Columns.}}
%\end{center}

According to (\ref{2.6}), the $s$-th column (first non-cancelled column) in
the $\left( r,s,p\right) $-array is the (constant) sequence%
\begin{equation}
a_{k,s}^{\left( r,s\right) }=\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{s}%
\dbinom{k-pr+s-1}{s-i}i!S_{r,s}\left( p,i\right) =\frac{S_{r,s}\left(
p,s\right) }{\left( s!\right) ^{p-1}}.  \label{3.5}
\end{equation}

For $j\geq s$, the $j$-th column is%
\begin{equation}
\left( \frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{j}\dbinom{l-1}{j-i}%
i!S_{r,s}\left( p,i\right) \right) _{l\in \mathbb{N}}.  \label{3.6}
\end{equation}

If $j\geq ps$, we can shift the corresponding sequence in order to show only
the non-zero terms:%
\begin{equation}
\left( \frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{j}\dbinom{l+j-ps-1}{j-i}%
i!S_{r,s}\left( p,i\right) \right) _{l\in \mathbb{N}}.  \label{3.7}
\end{equation}

For $j>ps$, we have the hockey-stick property (proof left to the reader)%
\begin{equation}
\sum_{l=1}^{m}\sum_{i=s}^{j}\dbinom{l+j-ps-1}{j-i}i!S_{r,s}\left( p,i\right)
=\sum_{i=s}^{j+1}\dbinom{m+j-ps}{j+1-i}i!S_{r,s}\left( p,i\right) .
\label{3.8}
\end{equation}

In general we have for $j>ps$ and $t,m\in \mathbb{N}$,%
\begin{equation}
\sum_{i_{t}=1}^{m}\cdots
\sum_{i_{2}=1}^{i_{3}}\sum_{i_{1}=1}^{i_{2}}\sum_{i=s}^{j}\dbinom{%
i_{1}+j-ps-1}{j-i}i!S_{r,s}\left( p,i\right) =\sum_{i=s}^{j+t}\dbinom{%
m+t-1+j-ps}{j+1-t}i!S_{r,s}\left( p,i\right) .  \label{3.9}
\end{equation}

\subsection{Falling Diagonals}
%\medskip
%
%\begin{center}
%\underline{\textbf{Falling Diagonals.}}
%\end{center}

We consider first falling diagonals of the $\left( r,s,p\right) $-array,
beginning in row $k\geq pr-s+1$ and column $j=s$. According to (\ref{2.6})
these diagonals are the sequences (where $n\in \mathbb{N}$)%
\begin{equation}
a_{k+n-1,s+n-1}^{\left( r,s\right) }=\frac{1}{\left( s!\right) ^{p}}%
\sum_{i=s}^{ps}\dbinom{k+n-pr+s-2}{s+n-1-i}i!S_{r,s}\left( p,i\right) .
\label{3.10}
\end{equation}

In particular, observe that the falling diagonal beginning in row $%
k=\allowbreak pr+1$ is%
\begin{equation}
a_{pr+n,s+n-1}^{\left( r,s\right) }=\frac{1}{\left( s!\right) ^{p}}%
\sum_{i=s}^{ps}\dbinom{n+s-1}{s+n-1-i}i!S_{r,s}\left( p,i\right) =\frac{1}{%
\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n+s-1}{i}i!S_{r,s}\left(
p,i\right) .  \label{3.11}
\end{equation}

That is, the falling diagonal beginning in row $k=\allowbreak pr+1$ and
column $j=s$ in the $\left( r,s,p\right) $-array is, according to (\ref{2.1}%
), the sequence%
\begin{equation}
\left( \prod_{j=1}^{p}\dbinom{n+s-1+\left( j-1\right) \left( r-s\right) }{s}%
\right) _{n\in \mathbb{N}},  \label{3.12}
\end{equation}%
(the original sequence involved in the array). For example, let us consider
the $\left( 3,2,3\right) $-array: the starting point of this array is the
sequence $\left( \binom{n+1}{2}\binom{n+2}{2}\binom{n+3}{2}\right) _{n\in 
\mathbb{N}}$. When writing this sequence in terms of the binomials $\binom{n%
}{2},\binom{n}{3},\ldots ,\binom{n}{6}$, we obtain the coefficients $%
18,126,288,270,90\,$\ (which form the \textquotedblleft
first\textquotedblright\ --non-cancelled-- row of the array). Then we form
the $\left( 3,2,3\right) $-array by using the Pascal's triangle recurrence.
In this array, the falling diagonal beginning in row $k=\allowbreak pr+1=10$
and column $j=s=2$ is precisely the original sequence $\left( \binom{n+1}{2}%
\binom{n+2}{2}\binom{n+3}{2}\right) _{n\in \mathbb{N}}=\left(
18,180,900,3150,\ldots \right) $.

From the falling diagonal beginning in row $k=\allowbreak pr+1$, the
subsequent diagonals can be obtained as the partial sums of the previous
diagonal. That is, we have the hockey-stick property (proof left to the
reader)%
\begin{equation}
\sum_{i_{l}=1}^{n}\cdots
\sum_{i_{2}=1}^{i_{3}}\sum_{i_{1}=1}^{i_{2}}\prod_{j=1}^{p}\dbinom{%
i_{1}+s-1+\left( j-1\right) \left( r-s\right) }{s}=\frac{1}{\left( s!\right)
^{p}}\sum_{i=s}^{ps}\dbinom{n+s+l-1}{i+l}i!S_{r,s}\left( p,i\right) ,
\label{3.13}
\end{equation}%
where $l\in \mathbb{N}$. In fact, observe that the falling diagonal
beginning in row $k=pr$ (and column $j=s$) is%
\begin{eqnarray*}
a_{pr+n-1,s+n-1}^{\left( r,s\right) } &=&\frac{1}{\left( s!\right) ^{p}}%
\sum_{i=s}^{ps}\dbinom{n+s-2}{s+n-1-i}i!S_{r,s}\left( p,i\right) \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n+s-2}{i-1}%
i!S_{r,s}\left( p,i\right) \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\left( \dbinom{n+s-1}{i}-%
\dbinom{n+s-2}{i}\right) i!S_{r,s}\left( p,i\right) \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n+s-1}{i}%
i!S_{r,s}\left( p,i\right) -\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}%
\dbinom{n+s-2}{i}i!S_{r,s}\left( p,i\right) \\
&=&\prod_{j=1}^{p}\dbinom{n+s-1+\left( j-1\right) \left( r-s\right) }{s}%
-\prod_{j=1}^{p}\dbinom{n+s-2+\left( j-1\right) \left( r-s\right) }{s},
\end{eqnarray*}%
and then the falling diagonal beginning in row $k=pr+1$ and column $j=s$,
that is, the sequence $\prod_{j=1}^{p}\binom{n+s-1+\left( j-1\right) \left(
r-s\right) }{s}$, can also be obtained as the partial sums of the falling
diagonal beginning in the previous row (the one beginning in row $k=pr$ and
column $j=s$).

The case $r=s$ of (\ref{3.13}) give us the \textquotedblleft hyper-sums of
powers of binomial coefficients\textquotedblright , namely%
\begin{equation}
\sum_{i_{l}=1}^{n}\cdots \sum_{i_{2}=1}^{i_{3}}\sum_{i_{1}=1}^{i_{2}}\dbinom{%
i_{1}+s-1}{s}^{p}=\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{%
n+s+l-1}{i+l}i!S_{s,s}\left( p,i\right) ,  \label{3.14}
\end{equation}

Some examples from (\ref{3.13}) are given in the following table:

%TCIMACRO{\TeXButton{B}{\begin{table}[htbp] \centering}}%
%BeginExpansion
\begin{table}[htbp] \centering%
%EndExpansion
$%
\begin{tabular}{|c|c|c|}
\hline
$\left( r,s,p\right) $ & $l$ & Sequence (\ref{3.13}) \\ \hline\hline
$\left( 3,2,2\right) $ & $1$ & \multicolumn{1}{|l|}{$\sum\limits_{i=1}^{n}%
\binom{i+1}{2}\binom{i+2}{2}=3\binom{n+2}{3}+9\binom{n+2}{4}+6\binom{n+2}{5}%
. $} \\ \hline
$\left( 4,3,2\right) $ & $2$ & \multicolumn{1}{|l|}{$\sum\limits_{j=1}^{n}%
\sum\limits_{i=1}^{j}\binom{i+2}{3}\binom{i+3}{3}=4\binom{n+4}{5}+24\binom{%
n+4}{6}+40\binom{n+4}{7}+20\binom{n+4}{8}.$} \\ \hline
$\left( 2,2,2\right) $ & $1$ & \multicolumn{1}{|l|}{$\sum\limits_{i=0}^{n}%
\binom{i+1}{2}^{2}=\binom{n+2}{3}+6\binom{n+2}{4}+6\binom{n+2}{5}.$} \\ 
\hline
$\left( 3,3,2\right) $ & $2$ & \multicolumn{1}{|l|}{$\sum\limits_{j=1}^{n}%
\sum\limits_{i=1}^{j}\binom{i+2}{3}^{2}=\binom{n+4}{5}+12\binom{n+4}{6}+30%
\binom{n+4}{7}+20\binom{n+4}{8}.$} \\ \hline
\end{tabular}%
$\caption{Hockey-stick property on falling diagonals of a (r,s,p)-array.}%
\label{TableFD}%
%TCIMACRO{\TeXButton{E}{\end{table}}}%
%BeginExpansion
\end{table}%
%EndExpansion

Now let us consider the falling diagonals beginning in row $k=pr-s+1$ and
column $j$, where $s<j\leq ps$. According to (\ref{2.6}) these are the
sequences (where $n\in \mathbb{N}$)%
\begin{equation}
\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n-1}{i-j}%
i!S_{r,s}\left( p,i\right) .  \label{3.15}
\end{equation}

From (\ref{3.15}) we see at once that the falling diagonal beginning in row $%
k=pr-s+1$ and column $j=sp$ is the constant sequence $\frac{\left( ps\right)
!}{\left( s!\right) ^{p}}$.

It turns out that the considered falling diagonals can be described in a
different way.

\begin{proposition}
\label{Prop.3.1}For $l=s,s+1,\ldots ,sp$, and $n\in \mathbb{N}$ we have the
identity%
\begin{equation}
\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n-1}{i-l}%
i!S_{r,s}\left( p,i\right) =\sum_{i=0}^{l}\left( -1\right) ^{i}\dbinom{l}{i}%
\prod_{j=1}^{p}\dbinom{n+l-1-i+\left( j-1\right) \left( r-s\right) }{s}.
\label{3.16}
\end{equation}
\end{proposition}

\begin{proof}
By induction on $n$. For $n=1$ we have to prove that (for $s\leq l\leq sp$)%
\begin{equation*}
\sum_{i=0}^{l}\left( -1\right) ^{i}\dbinom{l}{i}\prod_{j=1}^{p}\dbinom{%
l-i+\left( j-1\right) \left( r-s\right) }{s}=\frac{l!}{\left( s!\right) ^{p}}%
S_{r,s}\left( p,l\right) .
\end{equation*}

But this is an easy exercise left to the reader (by using the explicit
formula (\ref{1.2}) for $S_{r,s}\left( p,l\right) $). If (\ref{3.16}) is
true for all positive integers $\leq n$, where $n\in \mathbb{N}$ is given,
then%
\begin{eqnarray*}
&&\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n}{i-l}%
i!S_{r,s}\left( p,i\right) \\
&=&\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}\dbinom{n-1}{i-l}%
i!S_{r,s}\left( p,i\right) +\frac{1}{\left( s!\right) ^{p}}\sum_{i=s}^{ps}%
\dbinom{n-1}{i-l-1}i!S_{r,s}\left( p,i\right) \\
&=&\sum_{i=0}^{l}\!\left( -1\right) ^{i}\!\dbinom{l}{i}\!\prod_{j=1}^{p}%
\dbinom{n+l-1-i+\left( j-1\right) \left( r-s\right) }{s} \\
&&+\sum_{i=0}^{l+1}\!\left( -1\right) ^{i}\!\dbinom{l+1}{i}\!\prod_{j=1}^{p}%
\dbinom{n+l-i+\left( j-1\right) \left( r-s\right) }{s} \\
&=&\sum_{i=0}^{l+1}\left( -1\right) ^{i}\left( \dbinom{l+1}{i}-\dbinom{l}{i-1%
}\right) \prod_{j=1}^{p}\dbinom{n+l-i+\left( j-1\right) \left( r-s\right) }{%
s} \\
&=&\sum_{i=0}^{l}\left( -1\right) ^{i}\dbinom{l}{i}\prod_{j=1}^{p}\dbinom{%
n+l-i+\left( j-1\right) \left( r-s\right) }{s},
\end{eqnarray*}%
as wanted.
\end{proof}

The case $p=1$ of (\ref{3.16}) says that for any $n,s\in \mathbb{N}$ we have%
\begin{equation}
\sum_{i=0}^{s}\left( -1\right) ^{i}\dbinom{s}{i}\dbinom{n+s-1-i}{s}=1.
\label{3.17}
\end{equation}

Also, if we set $l=sp$ in (\ref{3.16}) (and use that the corresponding
falling diagonal is the constant sequence $\frac{\left( sp\right) !}{\left(
s!\right) ^{p}}$), we obtain the following identity valid for any $n\in 
\mathbb{N}$ 
\begin{equation}
\sum_{i=0}^{sp}\left( -1\right) ^{i}\dbinom{sp}{i}\prod_{j=1}^{p}\dbinom{%
n+sp-1-i+\left( j-1\right) \left( r-s\right) }{s}=\frac{\left( sp\right) !}{%
\left( s!\right) ^{p}}.  \label{3.18}
\end{equation}

The case $r=s$ of (\ref{3.18}) is%
\begin{equation}
\sum_{i=0}^{sp}\left( -1\right) ^{i}\dbinom{sp}{i}\dbinom{n-1+i}{s}%
^{p}=\left( -1\right) ^{sp}\frac{\left( sp\right) !}{\left( s!\right) ^{p}}.
\label{3.19}
\end{equation}

One more particular case of (\ref{3.16}) is when $r=2s$. By using (\ref{1.9}%
) we can write (\ref{3.16}) as%
\begin{equation}
\sum_{i=s}^{l+n-1}\dbinom{s\left( p-1\right) }{i-s}\dbinom{n-1}{i-l}%
=\sum_{i=0}^{l}\left( -1\right) ^{i}\dbinom{l}{i}\dbinom{n+l-1-i+s\left(
p-1\right) }{sp}.  \label{3.20}
\end{equation}

\section{Rising diagonals\label{Sec.4}}

In this section, related to rising diagonals of the $\left( r,s,p\right) $%
-arrays, Fibonacci and Lucas numbers will appear in a natural way. Recall
that the Fibonacci numbers sequence $F_{n}$ and Lucas numbers sequence $%
L_{n} $ satisfy the same second-order recurrence $a_{n+2}=a_{n+1}+a_{n}$,
with initial conditions $F_{0}=0$ and $F_{1}=1$ in the Fibonacci case, and $%
L_{0}=2$ and $L_{1}=1$ in the Lucas case. We will use without further
comments the Binet's formulas for $F_{n}$ and $L_{n}$, namely $F_{n}=\frac{1%
}{\sqrt{5}}\left( \alpha ^{n}-\beta ^{n}\right) $ and $L_{n}=\alpha
^{n}+\beta ^{n}$, where $\alpha =\frac{1}{2}\left( 1+\sqrt{5}\right) $ and $%
\beta =\frac{1}{2}\left( 1-\sqrt{5}\right) $ (observe that $\alpha +\beta =1$%
; we will use this in the proof of lemma \ref{Lemma.4.1}).

We consider the polynomials%
\begin{equation}
\sum_{k=r}^{rp}k!S_{r,r}\left( p,k\right) x^{k-r}.  \label{4.1}
\end{equation}

\begin{proposition}
\label{Prop.4.1}The following formula holds%
\begin{equation}
\sum_{k=r}^{rp}k!S_{r,r}\left( p,k\right) x^{k-r}=\sum_{k=r}^{rp}\left(
-1\right) ^{k-rp}k!S_{r,r}\left( p,k\right) \left( x+1\right) ^{k-r}.
\label{4.2}
\end{equation}
\end{proposition}

\begin{proof}
By induction on $p$. For $p=1$ it is easy to see that both sides of (\ref%
{4.2}) are equal to $r!$. Let us assume that (\ref{4.2}) is true for a given 
$p\in \mathbb{N}$. Then, by using (\ref{1.7}) we have that%
\begin{eqnarray*}
\sum_{k=r}^{r\left( p+1\right) }k!S_{r,r}\left( p+1,k\right) x^{k-r}
&=&\sum_{k=r}^{r\left( p+1\right) }k!\sum_{i=0}^{r}\dbinom{k+i-r}{i}r^{%
\underline{\,i\,}}S_{r,r}\left( p,k+i-r\right) x^{k-r} \\
&=&\sum_{i=0}^{r}\binom{r}{i}\sum_{k=r}^{rp}\frac{\left( k+r-i\right) !}{%
\left( k-i\right) !}k!S_{r,r}\left( p,k\right) x^{k-i} \\
&=&\sum_{i=0}^{r}\binom{r}{i}\sum_{k=r}^{rp}\left( \frac{d^{r}}{dx^{r}}%
x^{k+r-i}\right) k!S_{r,r}\left( p,k\right)  \\
&=&\frac{d^{r}}{dx^{r}}\left( x^{2r}\sum_{i=0}^{r}\binom{r}{i}%
x^{-i}\sum_{k=r}^{rp}k!S_{r,r}\left( p,k\right) x^{k-r}\right)  \\
&=&\frac{d^{r}}{dx^{r}}\left( x^{r}\left( x+1\right)
^{r}\sum_{k=r}^{rp}k!S_{r,r}\left( p,k\right) x^{k-r}\right)  \\
&=&\frac{d^{r}}{dx^{r}}\left( x^{r}\left( x+1\right)
^{r}\sum_{k=r}^{rp}\left( -1\right) ^{k-rp}k!S_{r,r}\left( p,k\right) \left(
x+1\right) ^{k-r}\right) .
\end{eqnarray*}%
Some further simplifications give us%
\begin{eqnarray*}
&&\sum_{k=r}^{r\left( p+1\right) }k!S_{r,r}\left( p+1,k\right) x^{k-r} \\
&=&\frac{d^{r}}{dx^{r}}\left( x^{r}\sum_{k=r}^{rp}\left( -1\right)
^{k-rp}k!S_{r,r}\left( p,k\right) \left( x+1\right) ^{k}\right)  \\
&=&\frac{d^{r}}{dx^{r}}\left( \sum_{i=0}^{r}\binom{r}{i}\left( -1\right)
^{i}\left( x+1\right) ^{r-i}\sum_{k=r}^{rp}\left( -1\right)
^{k-rp}k!S_{r,r}\left( p,k\right) \left( x+1\right) ^{k}\right)  \\
&=&\sum_{i=0}^{r}\binom{r}{i}\sum_{k=r}^{rp}\left( -1\right)
^{k-i-rp}k!S_{r,r}\left( p,k\right) \frac{d^{r}}{dx^{r}}\left( x+1\right)
^{k+r-i} \\
&=&\sum_{i=0}^{r}\binom{r}{i}\sum_{k=r}^{rp}\left( -1\right) ^{k+r-i-r\left(
p+1\right) }\left( k+r-i\right) !\frac{k!}{\left( k-i\right) !}S_{r,r}\left(
p,k\right) \left( x+1\right) ^{k-i} \\
&=&\sum_{k=r}^{r\left( p+1\right) }\left( -1\right) ^{k-r\left( p+1\right)
}k!\sum_{i=0}^{r}\binom{k+i-r}{i}\frac{r!}{\left( r-i\right) !}S_{r,r}\left(
p,k+i-r\right) \left( x+1\right) ^{k-r} \\
&=&\sum_{k=r}^{r\left( p+1\right) }\left( -1\right) ^{k-r\left( p+1\right)
}k!S_{r,r}\left( p+1,k\right) \left( x+1\right) ^{k-r},
\end{eqnarray*}%
as wanted (we used (\ref{1.7}) in the last step) .
\end{proof}

\begin{lemma}
\label{Lemma.4.1}(a) If $r$ is even, or $r$ and $p$ are odd, we have%
\begin{equation}
\sum_{k=r}^{rp}\left( -1\right) ^{k}k!S_{r,r}\left( p,k\right) F_{k-r}=0%
\text{.}  \label{4.3}
\end{equation}

(b) If $p$ is even and $r$ is odd, we have%
\begin{equation}
\sum_{k=r}^{rp}\left( -1\right) ^{k}k!S_{r,r}\left( p,k\right) L_{k-r}=0%
\text{.}  \label{4.4}
\end{equation}
\end{lemma}

\begin{proof}
In (\ref{4.2}) set $x=-\alpha $ to obtain%
\begin{equation}
\sum_{k=r}^{rp}\left( -1\right) ^{k-r}k!S_{r,r}\left( p,k\right) \alpha
^{k-r}=\sum_{k=r}^{rp}\left( -1\right) ^{k-rp}k!S_{r,r}\left( p,k\right)
\beta ^{k-r}.  \label{4.5}
\end{equation}

The desired formulas (\ref{4.3}) and (\ref{4.4}) are obtained from (\ref{4.5}%
) and the hypotheses made on $r$ and $p$ in each case.
\end{proof}

\begin{lemma}
\label{Lemma.4.2}(a) The following identity holds%
\begin{equation}
\sum\limits_{j=0}^{r\left( p-1\right) -1}\sum\limits_{i=r}^{r+j}\dbinom{%
r\left( p-1\right) -j-1}{r+j-i}i!S_{r,r}\left( p,i\right)
=\sum\limits_{i=r}^{rp-1}i!S_{r,r}\left( p,i\right) F_{rp-i}.  \label{4.8}
\end{equation}

(b) Let $r,l\in \mathbb{N}$ be given, such that $l\geq r$. Then%
\begin{equation}
\sum\limits_{i=0}^{l}\left( -1\right) ^{i+1}\binom{r}{l-i}F_{i}=F_{2r-l}.
\label{4.9}
\end{equation}
\end{lemma}

\begin{proof}
(a) Some simple manipulations on the left hand side of (\ref{4.8}) give us%
\begin{eqnarray*}
&&\sum\limits_{j=0}^{r\left( p-1\right) -1}\sum\limits_{i=r}^{r+j}\!%
\dbinom{r\left( p-1\right) -j-1}{r+j-i}i!S_{r,r}\!\left( p,i\right) \! \\
&=&\!\sum\limits_{j=0}^{r\left( p-1\right) -1}\sum\limits_{i=0}^{r\left(
p-1\right) -1}\!\dbinom{r\left( p-1\right) -j-1}{j-i}\left( i+r\right)
!S_{r,r}\!\left( p,i+r\right) \\
&=&\sum\limits_{i=0}^{r\left( p-1\right) -1}\left( i+r\right)
!S_{r,r}\left( p,i+r\right) \sum\limits_{j=0}^{r\left( p-1\right) -1-i}%
\dbinom{r\left( p-1\right) -j-i-1}{j} \\
&=&\sum\limits_{i=r}^{rp-1}i!S_{r,r}\left( p,i\right) F_{rp-i},
\end{eqnarray*}%
as wanted.

(b) We begin with the left-hand side of (\ref{4.9}) and Binet's formulas to
write%
\begin{eqnarray*}
\sum\limits_{i=0}^{l}\left( -1\right) ^{i+1}\binom{r}{l-i}F_{i} &=&\frac{%
\left( -1\right) ^{l+1}}{\sqrt{5}}\sum\limits_{i=0}^{r}\left( -1\right) ^{i}%
\binom{r}{i}\left( \alpha ^{l-i}-\beta ^{l-i}\right) \\
&=&\frac{\left( -1\right) ^{l+1}}{\sqrt{5}}\left( \alpha
^{l-r}\sum\limits_{i=0}^{r}\left( -1\right) ^{i}\binom{r}{i}\alpha
^{r-i}-\beta ^{l-r}\sum\limits_{i=0}^{r}\left( -1\right) ^{i}\binom{r}{i}%
\beta ^{r-i}\right) \\
&=&\frac{\left( -1\right) ^{l+1}}{\sqrt{5}}\left( \alpha ^{l-2r}-\beta
^{l-2r}\right) \\
&=&F_{2r-l},
\end{eqnarray*}%
as wanted.
\end{proof}

In the following proposition we will use the identities%
\begin{eqnarray}
F_{r\left( p-1\right) }F_{2r-l} &=&F_{r}F_{rp-l}+\left( -1\right)
^{r}F_{r\left( p-2\right) }F_{r-l},  \label{4.10} \\
L_{r\left( p-1\right) }F_{2r-l} &=&L_{r}F_{rp-l}-\left( -1\right)
^{r}L_{r-l}F_{r\left( p-2\right) },  \label{4.11}
\end{eqnarray}%
which are consequences of the index-reduction formula 
\begin{equation*}
G_{M}F_{N}-G_{M+K}F_{N-K}=\left( -1\right) ^{N-K}G_{M+K-N}F_{K},
\end{equation*}
where $G=F$ or $G=L$.

\begin{proposition}
\label{Prop.4.2}For\ $k\geq 2r\left( p-1\right) $ we have%
\begin{eqnarray}
&&\sum\limits_{j=0}^{\left\lfloor \frac{k-1}{2}\right\rfloor
}\sum\limits_{i=r}^{r+j}\dbinom{k-j-r\left( p-1\right) -1}{r+j-i}%
i!S_{r,r}\left( p,i\right)   \label{4.12} \\
&=&\left\{ 
\begin{array}{lll}
\left( \dfrac{1}{F_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left( p+1,i\right)
F_{i-r}\right) F_{k-r\left( p-1\right) } & \text{if} & 
\begin{array}{l}
r\text{ is even } \\ 
\text{or } \\ 
r\text{ and }p\text{ are odd}%
\end{array}
\\ 
&  &  \\ 
\left( \dfrac{1}{L_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i}\left( i-r\right) !S_{r,r}\left( p+1,i\right) F_{i-r}\right)
L_{k-r\left( p-1\right) } & \text{if} & 
\begin{array}{l}
r\text{ is odd } \\ 
\text{and } \\ 
p\text{ is even}%
\end{array}%
\end{array}%
\right.   \notag
\end{eqnarray}
\end{proposition}

\begin{proof}
Let us prove first that if $r$ is even or $r$ and $p$ are odd the formula%
\begin{eqnarray}
&&\sum\limits_{j=0}^{\left\lfloor \frac{k-1}{2}\right\rfloor
}\sum\limits_{i=r}^{r+j}\dbinom{k-j-r\left( p-1\right) -1}{r+j-i}%
i!S_{r,r}\left( p,i\right)  \label{4.13} \\
&=&\left( \dfrac{1}{F_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left( p+1,i\right)
F_{i-r}\right) F_{k-r\left( p-1\right) },  \notag
\end{eqnarray}%
holds for\ $k\geq 2r\left( p-1\right) $. We proceed by induction on $k$. For 
$k=2r\left( p-1\right) $ we have to prove that%
\begin{eqnarray}
&&\sum\limits_{j=0}^{r\left( p-1\right) -1}\sum\limits_{i=r}^{r+j}\dbinom{%
r\left( p-1\right) -j-1}{r+j-i}i!S_{r,r}\left( p,i\right)  \label{4.14} \\
&=&\left( \dfrac{1}{F_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left( p+1,i\right)
F_{i-r}\right) F_{r\left( p-1\right) }.  \notag
\end{eqnarray}

That is, according to (\ref{4.8}) we have to prove that 
\begin{equation}
\left( \dfrac{1}{F_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left( p+1,i\right)
F_{i-r}\right) F_{r\left( p-1\right)
}=\sum\limits_{i=r}^{rp-1}i!S_{r,r}\left( p,i\right) F_{rp-i}.  \label{4.15}
\end{equation}

We begin with the left-hand side of (\ref{4.15}) to write%
\begin{eqnarray*}
&&\left( \dfrac{1}{F_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left( p+1,i\right)
F_{i-r}\right) F_{r\left( p-1\right) } \\
&=&\dfrac{F_{r\left( p-1\right) }}{F_{r}}\sum\limits_{i=r}^{r\left(
p+1\right) }\left( -1\right) ^{i+r+1}\left( i-r\right) !\sum_{j=0}^{r}\binom{%
i+j-r}{j}\frac{r!}{\left( r-j\right) !}S_{r,r}\left( p,i+j-r\right) F_{i-r}
\\
&=&\dfrac{F_{r\left( p-1\right) }}{F_{r}}\sum\limits_{i=0}^{rp}\left(
-1\right) ^{i+1}i!\sum_{j=0}^{r}\binom{i+j}{j}\frac{r!}{\left( r-j\right) !}%
S_{r,r}\left( p,i+j\right) F_{i} \\
&=&\dfrac{F_{r\left( p-1\right) }}{F_{r}}\sum\limits_{i=0}^{rp}\left(
-1\right) ^{i+1}\sum_{j=0}^{r}\binom{r}{j}\left( i+j\right) !S_{r,r}\left(
p,i+j\right) F_{i} \\
&=&\dfrac{F_{r\left( p-1\right) }}{F_{r}}\sum\limits_{i=0}^{rp}\left(
-1\right) ^{i+1}\sum_{l=0}^{rp}\binom{r}{l-i}l!S_{r,r}\left( p,l\right) F_{i}
\\
&=&\dfrac{F_{r\left( p-1\right) }}{F_{r}}\sum_{l=0}^{rp}l!S_{r,r}\left(
p,l\right) F_{2r-l}
\end{eqnarray*}%
where we used (\ref{4.9}) in the last step. Now use (\ref{4.10}) and (\ref%
{4.3}) to obtain finally that if $r$ is even or $r$ and $p$ are odd we have%
\begin{eqnarray*}
&&\left( \!\dfrac{1}{F_{r}}\!\sum\limits_{i=r}^{r\left( p+1\right)
}\!\left( -1\right) ^{i+r+1}\!\left( i-r\right) !S_{r,r}\!\left(
p+1,i\right) \!F_{i-r}\!\right) \!F_{r\left( p-1\right) } \\
&=&\dfrac{1}{F_{r}}\sum_{l=0}^{rp}\!\,l!S_{r,r}\!\left( p,l\right) \!\left(
F_{r}F_{rp-l}+\left( -1\right) ^{r}F_{r\left( p-2\right) }F_{r-l}\right) \\
&=&\sum_{l=0}^{rp}l!S_{r,r}\left( p,l\right) F_{rp-l},
\end{eqnarray*}%
as expected. The rest of the induction argument is an easy exercise (left to
the reader) by using the recurrence of Fibonacci numbers (see Proposition 13
in \cite{Pi}).

Now we prove that if $r$ is odd and $p$ is even we have 
\begin{eqnarray}
&&\sum\limits_{j=0}^{\left\lfloor \frac{k-1}{2}\right\rfloor
}\sum\limits_{i=r}^{r+j}\dbinom{k-j-r\left( p-1\right) -1}{r+j-i}%
i!S_{r,r}\left( p,i\right)  \label{4.16} \\
&=&\left( \dfrac{1}{L_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i}\left( i-r\right) !S_{r,r}\left( p+1,i\right) F_{i-r}\right)
L_{k-r\left( p-1\right) }.  \notag
\end{eqnarray}

For $k=2r\left( p-1\right) $ we have to prove (according to (\ref{4.8}))
that if $r$ is odd and $p$ is even:%
\begin{equation}
\left( \dfrac{1}{L_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i}\left( i-r\right) !S_{r,r}\left( p+1,i\right) F_{i-r}\right)
L_{r\left( p-1\right) }=\sum\limits_{i=r}^{rp-1}i!S_{r,r}\left( p,i\right)
F_{rp-i}.  \label{4.17}
\end{equation}

We have%
\begin{eqnarray*}
&&\left( \dfrac{1}{L_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i}\left( i-r\right) !S_{r,r}\left( p+1,i\right) F_{i-r}\right)
L_{r\left( p-1\right) } \\
&=&\dfrac{L_{r\left( p-1\right) }}{L_{r}}\sum\limits_{i=0}^{rp}\left(
-1\right) ^{i+r}i!\sum_{j=0}^{r}\binom{i+j}{j}\frac{r!}{\left( r-j\right) !}%
S_{r,r}\left( p,i+j\right) F_{i} \\
&=&\dfrac{L_{r\left( p-1\right) }}{L_{r}}\sum\limits_{i=0}^{rp}\left(
-1\right) ^{i+r}\sum_{j=0}^{r}\binom{r}{j}\left( i+j\right) !S_{r,r}\left(
p,i+j\right) F_{i} \\
&=&\dfrac{L_{r\left( p-1\right) }}{L_{r}}\sum\limits_{i=0}^{rp}\left(
-1\right) ^{i+r}\sum_{l=r}^{rp}\binom{r}{l-i}l!S_{r,r}\left( p,l\right) F_{i}
\\
&=&\dfrac{L_{r\left( p-1\right) }}{L_{r}}\sum_{l=r}^{rp}l!S_{r,r}\left(
p,l\right) F_{2r-l},
\end{eqnarray*}%
where we used that $r$ is odd and (\ref{4.9}). Now use (\ref{4.11}) and\ (%
\ref{4.4}) to conclude finally that for $r$ odd and $p$ even we have 
\begin{eqnarray*}
&&\left( \dfrac{1}{L_{r}}\sum\limits_{i=r}^{r\left( p+1\right) }\left(
-1\right) ^{i}\left( i-r\right) !S_{r,r}\left( p+1,i\right) F_{i-r}\right)
L_{r\left( p-1\right) } \\
&=&\dfrac{1}{L_{r}}\sum_{l=r}^{rp}l!S_{r,r}\left( p,l\right) \left(
L_{r}F_{rp-l}+L_{r-l}F_{r\left( p-2\right) }\right) \\
&=&\sum_{l=r}^{rp}l!S_{r,r}\left( p,l\right) F_{rp-l},
\end{eqnarray*}%
as wanted. Again, the rest of the induction argument is an easy exercise
left to the reader.
\end{proof}

According to (\ref{2.6}), the sum of the elements of the rising diagonal
beginning in row $k\geq 2r\left( p-1\right) $ and column $j=s$ is%
\begin{equation}
\frac{1}{\left( r!\right) ^{p}}\sum\limits_{j=0}^{\left\lfloor \frac{k-1}{2}%
\right\rfloor }\sum\limits_{i=r}^{r+j}\dbinom{k-j-r\left( p-1\right) -1}{%
r+j-i}i!S_{r,r}\left( p,i\right) .  \label{4.18}
\end{equation}

Thus, in proposition \ref{Prop.4.2} we have proved that when $r$ is even or $%
r$ and $p$ are odd, the sum (\ref{4.18}) is equal to%
\begin{equation}
\left( \dfrac{1}{\left( r!\right) ^{p}F_{r}}\sum\limits_{i=r}^{r\left(
p+1\right) }\left( -1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left(
p+1,i\right) F_{i-r}\right) F_{k-r\left( p-1\right) },  \label{4.19}
\end{equation}%
and that when $r$ is odd and $p$ is even, this sum is equal to%
\begin{equation}
\left( \dfrac{1}{\left( r!\right) ^{p}L_{r}}\sum\limits_{i=r}^{r\left(
p+1\right) }\left( -1\right) ^{i}\left( i-r\right) !S_{r,r}\left(
p+1,i\right) F_{i-r}\right) L_{k-r\left( p-1\right) }.  \label{4.20}
\end{equation}

When $r=1$ the expressions in parenthesis in (\ref{4.19}) and (\ref{4.20})
are equal to the $p$-sequence%
\begin{equation}
\left( \sum\limits_{i=1}^{p+1}\left( -1\right) ^{i}\left( i-1\right)
!S\left( p+1,i\right) F_{i-1}\right) _{p\in \mathbb{N}}=\left(
1,1,7,13,151,421,\ldots \right) ,  \label{4.21}
\end{equation}%
called \textquotedblleft Stirling-Bernoulli transform of Fibonacci
numbers\textquotedblright\ (\cite[A050946]{OEIS}). This is the sequence that
appears in \cite{Pi} and the result is that (for $k\geq 2\left( p-1\right) $%
) the sum of rising diagonals is equal to the Stirling-Bernoulli transform (%
\ref{4.21}) times the Fibonacci number $F_{k-p+1}$ when $p$ is odd, and
times the Lucas number $L_{k-p+1}$ when $p$ is even. For $r>1$ we have the
following generalizations of the sequence (\ref{4.21}), which we call
\textquotedblleft $r$-Stirling-Bernoulli transform of Fibonacci
numbers\textquotedblright :

\begin{enumerate}
\item If $r$ is even, the $r$-Stirling-Bernoulli transform of Fibonacci
numbers is the $p$-sequence%
\begin{equation}
\left( \dfrac{1}{\left( r!\right) ^{p}F_{r}}\sum\limits_{i=r}^{r\left(
p+1\right) }\left( -1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left(
p+1,i\right) F_{i-r}\right) _{p\in \mathbb{N}}.  \label{4.22}
\end{equation}%
For example, if $r=2$ the sequence is $\left( 1,7,115,3499,170611,\ldots
\right) $ and if $r=4$ the sequence is $\left( 1,91,54091,116359591,\ldots
\right) $. In this case the result on rising diagonals is the following:

\textit{In a }$\left( r,r,p\right) $\textit{-array, where }$r$\textit{\ is
even, the sum of the elements of the rising diagonals (beginning in row }$%
k\geq 2r\left( p-1\right) $\textit{\ and column }$r$\textit{) is equal to
the }$r$\textit{-Stirling-Bernoulli transform of Fibonacci numbers (\ref%
{4.22}) times the Fibonacci number }$F_{k-r\left( p-1\right) }$\textit{.}

\item If $r$ is odd, we have two versions of $r$-Stirling-Bernoulli
transforms of Fibonacci numbers, depending on the parity of $p$:

\begin{enumerate}
\item For $p$ odd the sequence is 
\begin{equation}
\left( \dfrac{1}{\left( r!\right) ^{p}F_{r}}\sum\limits_{i=r}^{r\left(
p+1\right) }\left( -1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left(
p+1,i\right) F_{i-r}\right) _{p=1,3,5,\ldots },  \label{4.23}
\end{equation}%
and the result is the following:

\textit{In a }$\left( r,r,p\right) $\textit{-array, where }$r$\textit{\ and }%
$p$\textit{\ are odd, the sum of the elements of the rising diagonals
(beginning in row }$k\geq 2r\left( p-1\right) $\textit{\ and column }$r$%
\textit{) is equal to the }$r$\textit{-Stirling-Bernoulli transform of
Fibonacci numbers (\ref{4.23}) times the Fibonacci number }$F_{k-r\left(
p-1\right) }$\textit{.}

\item For $p$ even the sequence is%
\begin{equation}
\left( \dfrac{1}{\left( r!\right) ^{p}L_{r}}\sum\limits_{i=r}^{r\left(
p+1\right) }\left( -1\right) ^{i+r+1}\left( i-r\right) !S_{r,r}\left(
p+1,i\right) F_{i-r}\right) _{p=2,4,6,\ldots },  \label{4.24}
\end{equation}%
and the result is the following:

\textit{In a }$\left( r,r,p\right) $\textit{-array, where }$r$\textit{\ is
odd and }$p$\textit{\ is even, the sum of the elements of the rising
diagonals (beginning in row }$k\geq 2r\left( p-1\right) $\textit{\ and
column }$r$) \textit{is equal to the }$r$\textit{-Stirling-Bernoulli
transform of Fibonacci numbers (\ref{4.24}) times the Lucas number }$%
L_{k-r\left( p-1\right) }$\textit{.}
\end{enumerate}
\end{enumerate}

Note that in the case $r=1$, the two versions of $r$-Stirling-Bernoulli
transforms of Fibonacci numbers (\ref{4.23}) and (\ref{4.24}) are the
odd-indexed and even-indexed subsequences of the sequence (\ref{4.21}),
respectively. (This is the case of \cite{Pi}.) However, if $r>1$ these
sequences are different. For example, if $r=3$ the corresponding sequences (%
\ref{4.23}) and (\ref{4.24}) are $\left( 1,2371,303607891,\ldots \right) $
and $\left( 11,264431,124768775351,\ldots \right) $, respectively. Thus, in
the $\left( 3,3,p\right) $-array, the sum of the elements of the rising
diagonals (beginning in row $k\geq 6\left( p-1\right) $ and $3$rd column) is
equal to $F_{k}$ when $p=1$, to $11L_{k-3}$ when $p=2$, to $2371F_{k-6}$
when $p=3$, to $264431L_{k-9}$ when $p=4$, to $303607891F_{k-12}$ when $p=5$%
, and so on.

In tables \ref{TableR1} and \ref{TableR2} we have some more examples.

%TCIMACRO{\TeXButton{B}{\begin{table}[htbp] \centering}}%
%BeginExpansion
\begin{table}[htbp] \centering%
%EndExpansion
$%
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$r$ & $p$ & $%
\begin{array}{c}
\text{{\tiny Beginning} } \\ 
\text{{\tiny in row k}}%
\end{array}%
\!\!$ & {\small Rising diagonal} & $%
\begin{array}{c}
\text{{\tiny Sum} } \\ 
\text{{\tiny of}} \\ 
\text{{\tiny elements}}%
\end{array}%
$ & {\small Equal to} \\ \hline\hline
$2$ & $2$ & $k=4$ & $\left( 1,6\right) $ & $7$ & $7F_{2}$ \\ \hline
&  & $k=5$ & $\left( 1,7,6\right) $ & $14$ & $7F_{3}$ \\ \hline
&  & $k=6$ & $\left( 1,8,12\right) $ & $21$ & $7F_{4}$ \\ \hline
&  & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ \\ \hline
$2$ & $3$ & $k=8$ & $\left( 1,26,138,180\right) $ & $345$ & $115F_{4}$ \\ 
\hline
&  & $k=9$ & $\left( 1,27,163,294,90\right) $ & $575$ & $115F_{5}$ \\ \hline
&  & $k=10$ & $\left( 1,28,189,432,270\right) $ & $920$ & $115F_{6}$ \\ 
\hline
&  & $\vdots $ & $\vdots $ & $\vdots $ &  \\ \hline
$3$ & $3$ & $k=12$ & $\left( 1,64,873,4380,8610,5040\right) $ & $18968$ & $%
2371F_{6}$ \\ \hline
&  & $k=13$ & $\left( 1,65,936,5191,12240,10710,1680\right) $ & $30823$ & $%
2371F_{7}$ \\ \hline
&  & $k=14$ & $\left( 1,66,1000,6064,16620,19320,6720\right) $ & $49791$ & $%
2371F_{8}$ \\ \hline
&  & $\vdots $ & $\vdots $ & $\vdots $ &  \\ \hline
$4$ & $2$ & $k=8$ & $\left( 1,22,110,140\right) $ & $273$ & $91F_{4}$ \\ 
\hline
&  & $k=9$ & $\left( 1,23,131,230,70\right) $ & $455$ & $91F_{5}$ \\ \hline
&  & $k=10$ & $\left( 1,24,153,340,210\right) $ & $728$ & $91F_{6}$ \\ \hline
&  & $\vdots $ & $\vdots $ & $\vdots $ &  \\ \hline
\end{tabular}%
$%
\caption{Sums of rising diagonals in a (r,r,p)-array, when r is even, or r
and p are odd.}\label{TableR1}%
%TCIMACRO{\TeXButton{E}{\end{table}}}%
%BeginExpansion
\end{table}%
%EndExpansion

%TCIMACRO{\TeXButton{B}{\begin{table}[htbp] \centering}}%
%BeginExpansion
\begin{table}[htbp] \centering%
%EndExpansion
$%
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$r$ & $p$ & $\!\!\!\!%
\begin{array}{c}
\text{{\tiny Beginning} } \\ 
\text{{\tiny in row k}}%
\end{array}%
\!\!\!\!$ & {\small The rising diagonal is} & $%
\begin{array}{c}
\text{{\tiny Sum} } \\ 
\text{{\tiny of}} \\ 
\text{{\tiny elements}}%
\end{array}%
$ & \negthinspace \negthinspace {\small Equal to\negthinspace \negthinspace }
\\ \hline\hline
$3$ & $2$ & $k=6$ & $\left( 1,13,30\right) $ & $44$ & $11L_{3}$ \\ \hline
&  & $k=7$ & $\left( 1,14,42,20\right) $ & $77$ & $11L_{4}$ \\ \hline
&  & $k=8$ & $\left( 1,15,55,50\right) $ & $121$ & $11L_{5}$ \\ \hline
&  & $\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ \\ \hline
$3$ & $4$ & $k=18$ & $_{(1,259,10257,150000,\ldots,7203420,6186600,1663200)}$ & 
$_{20096756}$ & $_{264431L_{9}}$ \\ \hline
&  & $k=19$ & $_{(1,260,10515,160000,\ldots,11040960,4762800,369600)}$ & $%
_{32525013}$ & $_{264431L_{10}}$ \\ \hline
&  & $k=20$ & $\!_{(1,261,10774,170257,\ldots,18244380,10949400,2032800)}\!$ & $%
_{52621769}$ & $_{264431L_{11}}$ \\ \hline
&  & $\vdots $ & $\vdots $ & $\vdots $ &  \\ \hline
$5$ & $2$ & $k=10$ & $\left( 1,33,271,770,630\right) $ & $1705$ & $155L_{5}$
\\ \hline
&  & $k=11$ & $\left( 1,34,303,1010,1190,252\right) $ & $2790$ & $155L_{6}$
\\ \hline
&  & $k=12$ & $\left( 1,35,336,1281,1960,882\right) $ & $\allowbreak 4495$ & 
$155L_{7}$ \\ \hline
&  & $\vdots $ & $\vdots $ & $\vdots $ &  \\ \hline
\end{tabular}%
$%
\caption{Sums of rising diagonals in a (r,r,p)-array, when r is odd and p
is even.}\label{TableR2}%
%TCIMACRO{\TeXButton{E}{\end{table}}}%
%BeginExpansion
\end{table}%
%EndExpansion

\begin{remark}
\label{Remark4.1}We believe that for the general case $r>s$, the sums of
rising diagonals in a $\left( r,s,p\right) $-arrays are also related with
Fibonacci and/or Lucas numbers. Note that the sum of the elements of the
rising diagonal beginning in column $s$ and row $k\geq (s+r)p-2s$ is given by%
\begin{equation}
\frac{1}{\left( s!\right) ^{p}}\sum\limits_{j=0}^{\left\lfloor \frac{%
k-1-\left( r-s\right) p}{2}\right\rfloor }\sum_{i=s}^{s+j}\dbinom{k-j-pr+s-1%
}{s+j-i}i!S_{r,s}\left( p,i\right) .  \label{4.25}
\end{equation}

In the case $r=2s$ it is not difficult to see that the sum (\ref{4.25}) is
equal to $\frac{\left( sp\right) !}{\left( s!\right) ^{p}}F_{k-sp}$ (since,
as we noticed in section \ref{Sec.2}, the $\left( 2s,s,p\right) $-array is a
piece of Pascal's triangle multiplied by $\frac{\left( sp\right) !}{\left(
s!\right) ^{p}}$.) Then, we have for $k\geq 3sp-2s$ that%
\begin{equation}
\frac{1}{\left( s!\right) ^{p}}\sum\limits_{j=0}^{\left\lfloor \frac{k-1-sp%
}{2}\right\rfloor }\sum_{i=s}^{s+j}\dbinom{k-j-2sp+s-1}{s+j-i}%
i!S_{2s,s}\left( p,i\right) =\frac{\left( sp\right) !}{\left( s!\right) ^{p}}%
F_{k-sp}.  \label{4.26}
\end{equation}

However, formula (\ref{4.26}) is not an interesting result, since we can
write it (by using (\ref{1.9})) as%
\begin{equation}
\sum\limits_{j=0}^{\left\lfloor \frac{k-1-sp}{2}\right\rfloor
}\sum_{i=s}^{s+j}\dbinom{k-j-2sp+s-1}{s+j-i}\dbinom{s\left( p-1\right) }{i-s}%
=F_{k-sp},  \label{4.27}
\end{equation}%
and one can see at once that (\ref{4.27}) is consequence of the following
two easy to prove identities%
\begin{equation}
\sum_{j=0}^{m}\binom{m-j}{j}=F_{m+1},\text{ \ \ \ \ \ and \ \ \ \ \ \ }%
\sum_{i=0}^{m}\binom{m}{i}F_{r-i}=F_{m+r},  \label{4.28}
\end{equation}%
where $m$ is a non-negative integer and $r\in \mathbb{Z}$. We obtained
explicit values of the sum (\ref{4.25}) for some concrete values of $r,s,p,k$
in the remaining cases (when $1\leq s<r$, $r\neq 2s$) and observed that
Fibonacci and/or Lucas numbers appear in these sums. In table \ref{TableR3}
we have some examples.

{
\begin{table}[htbp] \centering%
\footnotesize
$%
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$r$ & $s$ & $p$ & $
\begin{minipage}[c]{2cm}
{\footnotesize 
\begin{center}
Beginning in column  $j$ and row $k$
\end{center}
}
\end{minipage}
%\begin{array}{c}
%\text{{\tiny Beginning in} } \\ 
%\text{{\tiny column j}}\!\text{{\tiny \ and row k }}%
%\end{array}%
$ &
\begin{minipage}{4cm} 
 {\footnotesize The rising diagonals are} 
 \end{minipage}
 & $%
\begin{minipage}[c]{1.8cm}
{\footnotesize Sum of elements}
\end{minipage}
%\begin{array}{c}
%\text{{\tiny Sum} } \\ 
%\text{{\tiny of}} \\ 
%\text{{\tiny elements}}%
%\end{array}%
$ & { \footnotesize Equals to} \\ \hline\hline
$3$ & $1$ & $2$ & $%
\begin{array}{l}
k=6, \\ 
k=7, \\ 
k=8,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
( 3) , \\ 
( 3,2) , \\ 
( 3,5) ,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
3, \\ 
5, \\ 
8,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
F_{4}, \\ 
F_{5}, \\ 
F_{6},\text{{\tiny \ldots }}%
\end{array}%
$ \\ \hline
&  & $3$ & $%
\begin{array}{l}
k=10, \\ 
k=11, \\ 
k=12,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
( 15,18) , \\ 
( 15,33,6) , \\ 
( 15,48,24) ,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
33, \\ 
54, \\ 
87,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
3L_{5}, \\ 
3L_{6}, \\ 
3L_{7},\text{{\tiny \ldots }}%
\end{array}%
$ \\ \hline
& $2$ & $3$ & $%
\begin{array}{l}
k=11, \\ 
k=12, \\ 
k=13,\text{{\tiny \ldots }}%
\end{array}%
$ & \multicolumn{1}{|c|}{$%
\begin{array}{l}
( 18,162,414,270) , \\ 
( 18,180,558,558,90) , \\ 
( 18,198,720,972,360) ,\text{{\tiny \ldots }}%
\end{array}%
$} & \multicolumn{1}{|c|}{$%
\begin{array}{l}
864, \\ 
1404, \\ 
2268,\text{{\tiny \ldots }}%
\end{array}%
$} & $%
\begin{array}{l}
108F_{6}, \\ 
108F_{7}, \\ 
108F_{8},\text{{\tiny \ldots }}%
\end{array}%
$ \\ \hline
$4$ & $1$ & $2$ & $%
\begin{array}{l}
k=8, \\ 
k=9, \\ 
k=10,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
( 4) , \\ 
( 4,2) , \\ 
( 4,6) ,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
4, \\ 
6, \\ 
10,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
2F_{3}, \\ 
2F_{4}, \\ 
2F_{5},\text{{\tiny \ldots }}%
\end{array}%
$ \\ \hline
&  & $3$ & $%
\begin{array}{l}
k=13, \\ 
k=14, \\ 
k=15,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
( 28,24) , \\ 
( 28,52,6) , \\ 
( 28,80,30) ,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
52, \\ 
86, \\ 
138,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
2( 7F_{4}+F_{5}) , \\ 
2( 7F_{5}+F_{6}) , \\ 
2( 7F_{6}+F_{7}) ,\text{{\tiny \ldots }}%
\end{array}%
$ \\ \hline
& $3$ & $3$ & $%
\begin{array}{l}
k=15, \\ 
k=16, \\ 
k=17,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
\text{{\tiny $( 40,800,5440,15840,19040,6720) $\textsc{,}}} \\ 
\text{{\tiny $( 40,840,6200,20560,30840,17360,1680) $\textsc{,}}}
\\ 
\text{{\tiny $( 40,880,7000,26000,46680,36400,8400) $\textsc{,}%
\ldots }}%
\end{array}%
$ & $\allowbreak 
\begin{array}{l}
47880, \\ 
77520, \\ 
125400,\text{{\tiny \ldots }}%
\end{array}%
$ & $%
\begin{array}{l}
2280F_{8}, \\ 
2280F_{9}, \\ 
2280F_{10},\text{{\tiny \ldots }}%
\end{array}%
$ \\ \hline
\end{tabular}%
$\caption{Sums of rising diagonals in (r,s,p)-arrays.}\label{TableR3}%
%TCIMACRO{\TeXButton{E}{\end{table}}}%
%BeginExpansion
\end{table}%
%EndExpansion
}

We mention that we were not able to propose and prove the results (of the
last column of Table \ref{TableR3}) corresponding to this general situation.
We believe that they are not as nice and simple as those we showed in this
section for the case $r=s$.
\end{remark}

\section{Some additional results\label{Sec.5}}

By equating coefficients of similar powers of $x$ in both sides of (\ref{4.2}%
) we see that%
\begin{equation}
S_{r,r}\left( p,j\right) =\frac{1}{j!}\sum_{i=j}^{rp}\left( -1\right) ^{i-rp}%
\binom{i-r}{j-r}i!S_{r,r}\left( p,i\right) .  \label{5.1}
\end{equation}

In this section we want to obtain some consequences of this identity. First
of all, observe that (\ref{1.021}) and (\ref{5.1}) give us that 
\begin{equation}
\dbinom{n}{s}^{p}=\frac{1}{\left( s!\right) ^{p}}\sum_{k=s}^{sp}%
\sum_{i=k}^{sp}\left( -1\right) ^{i-sp}i!S_{s,s}\left( p,i\right) \binom{i-s%
}{k-s}\dbinom{n}{k},  \label{5.2}
\end{equation}%
which can be written as%
\begin{equation}
\dbinom{n}{s}^{p}=\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}%
\sum_{k=s}^{sp}\left( -1\right) ^{k}k!S_{s,s}\left( p,k\right) \dbinom{n+k-s%
}{k}.  \label{5.3}
\end{equation}

After a simplification procedure (that we describe next), using Pascal's
triangle recurrence, the right-hand side of (\ref{5.3}) can be written in a
simpler form, in which we need only `the half' of the number of coefficients
involved in (\ref{5.3}). To prove this fact is one of the goals of this
section, and it is contained in corollary \ref{Cor.5.1}.

The mentioned simplification procedure is as follows.

\textsc{Procedure}: Begin with expression (\ref{5.3}) (this is our
\textquotedblleft starting point\textquotedblright\ expression) and perform
steps 1,2,3.

\begin{description}
\item[Step 1] Write the last (the $k\!=\!sp$-th) term of the right-hand side
of (\ref{5.3}), namely the term $\frac{\left( sp\right) !}{\left( s!\right)
^{p}}S_{s,s}\!\left( p,sp\right) \!\binom{n+sp-s}{sp}$, as 
\begin{equation}
\frac{\left( sp\right) !}{\left( s!\right) ^{p}}S_{s,s}\left( p,sp\right)
\left( \dbinom{n+sp-s-1}{sp-1}+\dbinom{n+sp-s-1}{sp}\right) .  \label{5.4}
\end{equation}

\item[Step 2] Write together the first summand of (\ref{5.4}), namely $\frac{%
\left( sp\right) !}{\left( s!\right) ^{p}}S_{s,s}\left( p,sp\right) \binom{%
n+sp-s-1}{sp-1}$, and the $k=\left( sp-1\right) $-th summand of (\ref{5.3}),
namely $-\frac{\left( sp-1\right) !}{\left( s!\right) ^{p}}S_{s,s}\left(
p,sp-1\right) \binom{n+sp-s-1}{sp-1}$, to get the new $k=\left( sp-1\right) $%
-th summand of (\ref{5.3}) as 
\begin{equation}
\left( \frac{\left( sp\right) !}{\left( s!\right) ^{p}}S_{s,s}\left(
p,sp\right) -\frac{\left( sp-1\right) !}{\left( s!\right) ^{p}}S_{s,s}\left(
p,sp-1\right) \right) \dbinom{n+sp-s-1}{sp-1}.  \label{5.5}
\end{equation}

\item[Step 3] Return to Step 1 with the new $k=\left( sp-1\right) $-th
summand (\ref{5.5}), to get (after the new Step 2) the new $k=\left(
sp-2\right) $-th summand as 
\begin{eqnarray}
&&\left( \frac{\left( sp\right) !}{\left( s!\right) ^{p}}S_{s,s}\left(
p,sp\right) -\frac{\left( sp-1\right) !}{\left( s!\right) ^{p}}S_{s,s}\left(
p,sp-1\right) +\frac{\left( sp-2\right) !}{\left( s!\right) ^{p}}%
S_{s,s}\left( p,sp-2\right) \right)   \label{5.7} \\
&&\times \dbinom{n+sp-s-2}{sp-2},  \notag
\end{eqnarray}%
and so on. That is, return to Step 1 with the new $\left( sp-2\right) $-th
summand (\ref{5.7}) to get a new $\left( sp-3\right) $-th summand.
Continue in this way until you perform Step 2 involving the first term (the $%
k=s$-th term) of the right-hand side of (\ref{5.3}).
\end{description}

At the end of \textsc{Procedure} we have the expression (\ref{5.3})
transformed in%
\begin{eqnarray}
\dbinom{n}{s}^{p} &=&\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}%
\sum_{t=0}^{1}\sum_{j=s+t}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left(
p,j\right) \dbinom{n}{s+t}  \label{5.8} \\
&&+\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}\sum_{t=2}^{s\left(
p-1\right) }\left( \sum_{j=s+t}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left(
p,j\right) \right) \dbinom{n+t-1}{s+t}.  \notag
\end{eqnarray}

Now we take (\ref{5.8}) as the starting point and repeat \textsc{Procedure},
beginning with the last term of the right-hand side of (\ref{5.8}) and going
backwards, \textit{up to the third term }(that is, we leave the first two
summands of the right-hand side of (\ref{5.8}) as they appear in this
expression, namely $\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}%
\sum_{t=0}^{1}\sum_{j=s+t}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left(
p,j\right) \binom{n}{s+t}$). (The reason why we stop now in the third term
will become clear shortly, and will be explained correctly in Proposition %
\ref{Prop.5.1}.) We get the expression%
\begin{eqnarray}
\dbinom{n}{s}^{p} &=&\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}%
\sum_{t=0}^{1}\sum_{j=s+t}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left(
p,j\right) \dbinom{n}{s+t}  \label{5.9} \\
&&+\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}\sum_{t=2}^{3}%
\sum_{j=s+t}^{sp}\left( -1\right) ^{j}j!\left( j-s-t+1\right) S_{s,s}\left(
p,j\right) \dbinom{n+1}{s+t}  \notag \\
&&+\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}\sum_{t=4}^{s\left(
p-1\right) }\left( \sum_{j=s+t}^{sp}\left( -1\right) ^{j}j!\left(
j-s-t+1\right) S_{s,s}\left( p,j\right) \right) \dbinom{n+t-2}{s+t}.  \notag
\end{eqnarray}

And again, we repeat \textsc{Procedure} with (\ref{5.9}) as starting point,
beginning with the last term and going backwards, \textit{up to the fifth
term}.

After $\left\lfloor \frac{1}{2}s\left( p-1\right) \right\rfloor $ times of
applying \textsc{Procedure} as described above, we obtain the expression%
\begin{eqnarray}
\dbinom{n}{s}^{p} &=&\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}%
\sum_{t=0}^{\left\lfloor \frac{1}{2}s\left( p-1\right) \right\rfloor
}\sum_{j=s+2t}^{sp}\left( -1\right) ^{j}j!\binom{j-s-t}{t}S_{s,s}\left(
p,j\right) \dbinom{n+t}{s+2t}  \label{5.10} \\
&&+\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}\sum_{t=0}^{\left%
\lfloor \frac{1}{2}s\left( p-1\right) \right\rfloor
}\sum_{j=s+2t+1}^{sp}\left( -1\right) ^{j}j!\binom{j-s-t-1}{t}S_{s,s}\left(
p,j\right) \dbinom{n+t}{s+2t+1}.  \notag
\end{eqnarray}

Let us see a concrete example. For $s=4$ and $p=3$, expression (\ref{5.3})
says that%
\begin{eqnarray}
\dbinom{n}{4}^{3} &=&\dbinom{n}{4}-120\dbinom{n+1}{5}+2640\dbinom{n+2}{6}%
-21840\dbinom{n+3}{7}+87570\dbinom{n+4}{8} \notag  \\
&&\!\!\!\!\!\!\!\!-189000\dbinom{n+5}{9}+224700\dbinom{n+6}{10}-138600\dbinom{n+7}{11}+34650%
\dbinom{n+8}{12}.  
\label{5.101}
\end{eqnarray}

We begin \textsc{Procedure} writing the term \ $34650\binom{n+8}{12}$ as $%
34650\left( \binom{n+7}{11}+\binom{n+7}{12}\right) $ and simplifying to%
\begin{eqnarray}
\dbinom{n}{4}^{3} &=&\dbinom{n}{4}-120\dbinom{n+1}{5}+2640\dbinom{n+2}{6}%
-21840\dbinom{n+3}{7}+87570\dbinom{n+4}{8}  \notag \\
&&\!\!\!\!\!\!\!\!-189000\dbinom{n+5}{9}+224700\dbinom{n+6}{10}-103950\dbinom{n+7}{11}+34650%
\dbinom{n+7}{12}.  
\label{5.1011}
\end{eqnarray}

Now we write the term $103950\binom{n+7}{11}$ of (\ref{5.1011}) as $%
103950\left( \binom{n+6}{10}+\binom{n+6}{11}\right) $ and simplify to%
\begin{eqnarray*}
\dbinom{n}{4}^{3} &=&\dbinom{n}{4}-120\dbinom{n+1}{5}+2640\dbinom{n+2}{6}%
-21840\dbinom{n+3}{7}+87570\dbinom{n+4}{8} \\
&&-189000\dbinom{n+5}{9}+120750\dbinom{n+6}{10}-103950\dbinom{n+6}{11}+34650%
\dbinom{n+7}{12}.
\end{eqnarray*}

Continuing in this way we obtain 
\begin{eqnarray}
\dbinom{n}{4}^{3} &=&\dbinom{n}{4}+120\dbinom{n+1}{6}-2520\dbinom{n+2}{7}%
+19320\dbinom{n+3}{8}-68250\dbinom{n+4}{9}  \notag \\
&&+120750\dbinom{n+5}{10}-103950\dbinom{n+6}{11}+34650\dbinom{n+7}{12}.
\label{5.102}
\end{eqnarray}

(This is (\ref{5.8}). Observe that the coefficient of the term $\binom{n+1}{5%
}$ is $0.$) Now we write the last term of the right-hand side of (\ref{5.102}%
) as $34650\left( \binom{n+6}{11}+\binom{n+6}{12}\right) $ and simplify.
Then, we repeat this step with the term $\left( -103950+34650\right) \binom{%
n+6}{11}$, and with all the subsequent terms obtained. We get 
\begin{eqnarray}
\dbinom{n}{4}^{3} &=&\dbinom{n}{4}+120\dbinom{n+1}{6}+2520\dbinom{n+2}{8}%
-16800\dbinom{n+3}{9}  \label{5.103} \\
&&+51450\dbinom{n+4}{10}-69300\dbinom{n+5}{11}+34650\dbinom{n+6}{12}.  \notag
\end{eqnarray}

(This is (\ref{5.9}). Observe that the coefficient of the term $\binom{n+2}{7%
}$ is $0$.) And again, we write the last term of the right-hand side of (\ref%
{5.103}) as $34650\left( \binom{n+5}{11}+\binom{n+6}{12}\right) $ and
simplify. And repeat with the subsequent terms to obtain%
\begin{eqnarray}
\dbinom{n}{4}^{3} &=&\dbinom{n}{4}+120\dbinom{n+1}{6}+2520\dbinom{n+2}{8}%
+16800\dbinom{n+3}{10}  \label{5.105} \\
&&-34650\dbinom{n+4}{11}+34650\dbinom{n+5}{12}.  \notag
\end{eqnarray}

One final procedure beginning with the last term of the right-hand side of (%
\ref{5.105}) take us to the formula%
\begin{equation}
\dbinom{n}{4}^{3}=\dbinom{n}{4}+120\dbinom{n+1}{6}+2520\dbinom{n+2}{8}+16800%
\dbinom{n+3}{10}+34650\dbinom{n+4}{12}.  \label{5.106}
\end{equation}

The fact that several coefficients are equal to $0$ makes (\ref{5.106}) a
simpler expression than the original one (\ref{5.101}). In general we have
the following result.

\begin{proposition}
\label{Prop.5.1}(a) If $s$ is even or $s$ and $p$ are odd, the following
formula holds for any non-negative integer $t$%
\begin{equation}
\sum_{j=s+2t+1}^{sp}\left( -1\right) ^{j}j!\dbinom{j-s-t-1}{t}S_{s,s}\left(
p,j\right) =0.  \label{5.11}
\end{equation}

(b) If $s$ is odd and $p$ is even, the following formula holds for any
non-negative integer $t$%
\begin{equation}
\sum_{j=s+2t+1}^{sp}\left( -1\right) ^{j}j!\dbinom{j-s-t-1}{t}S_{s,s}\left(
p,j\right) =2\sum_{j=s+2t}^{sp}\left( -1\right) ^{j}j!\dbinom{j-s-t}{t}%
S_{s,s}\left( p,j\right) .  \label{5.12}
\end{equation}
\end{proposition}

\begin{proof}
(a) Observe that (\ref{5.11}) is equivalent to%
\begin{align}\label{5.13}
\sum_{j=s}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right)
& \prod   \limits_{l=0}^{t-1}\left( j-s-t-1-l\right)  \\
& =\sum_{j=s}^{s+t}\left(
-1\right) ^{j}j!S_{s,s}\left( p,j\right) \prod\limits_{l=0}^{t-1}\left(
j-s-t-1-l\right) .  \notag
\end{align}

By using Gould's identity (3.50) in \cite{Go1}, we see that 
\begin{equation}
\binom{j-s-t-1}{t}=\sum_{k=0}^{t}\left( -1\right) ^{k+t}\binom{2t-k}{t}%
\binom{j-s}{k},  \label{5.14}
\end{equation}%
which can be written as%
\begin{equation}
\prod\limits_{l=0}^{t-1}\left( j-s-t-1-l\right) =\sum_{k=0}^{t}\frac{\left(
-1\right) ^{k}\left( t+k\right) !}{k!\left( t-k\right) !}\prod%
\limits_{l=0}^{t-1-k}\left( j-s-l\right) ,  \label{5.15}
\end{equation}

Beginning with the left-hand side of (\ref{5.13}) and using (\ref{5.15}) we
get%
\begin{eqnarray}
&&\sum_{j=s}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right)
\prod\limits_{l=0}^{t-1}\left( j-s-t-1-l\right)  \label{5.16} \\
&=&\sum_{k=0}^{t}\frac{\left( -1\right) ^{k}\left( t+k\right) !}{k!\left(
t-k\right) !}\sum_{j=s+t-k}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left(
p,j\right) \prod\limits_{l=0}^{t-1-k}\left( j-s-l\right) .  \notag
\end{eqnarray}

By setting $j=s+t-k$ in (\ref{5.1}) we see that 
\begin{equation}
S_{s,s}\left( p,s+t-k\right) =\frac{\left( -1\right) ^{sp}}{\left(
t-k\right) !\left( s+t-k\right) !}\sum_{j=s+t-k}^{sp}\left( -1\right)
^{j}j!S_{s,s}\left( p,j\right) \prod\limits_{l=0}^{t-1-k}\left(
j-s-l\right) .  \label{5.17}
\end{equation}

According to (\ref{5.17}) we can write (\ref{5.16}) as%
\begin{align*}
\sum_{j=s}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right)
& \prod\limits_{l=0}^{t-1}\left( j-s-t-1-l\right) \\
&=\left( -1\right) ^{sp}\sum_{k=0}^{t}\frac{\left( -1\right) ^{k}\left(
t+k\right) !}{k!}\left( s+t-k\right) !S_{s,s}\left( p,s+t-k\right) ,
\end{align*}%
or%
\begin{align}\label{5.18}
\sum_{j=s}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right)
& \prod\limits_{l=0}^{t-1}\left( j-s-t-1-l\right) \\
& =\left( -1\right) ^{s\left(
p+1\right) +t}\sum_{j=s}^{s+t}\frac{\left( -1\right) ^{j}\left(
2t+s-j\right) !}{\left( t+s-j\right) !}j!S_{s,s}\left( p,j\right) . \notag
\end{align}

The fact that 
\begin{equation*}
\prod\limits_{l=0}^{t-1}\left( j-s-t-1-l\right) =\left( -1\right) ^{t}\frac{%
\left( 2t+s-j\right) !}{\left( t+s-j\right) !},
\end{equation*}%
(easy proof, left to the reader), together with the hypothesis that $s$ is
even or $s$ and $p$ are odd, give us from (\ref{5.18}) the end of the proof
of (\ref{5.13}).

(b) The case $t=0$ of (\ref{5.12}) can be written as%
\begin{equation*}
\sum_{j=s}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right)
=s!S_{s,s}\left( p,s\right) .
\end{equation*}

This is a direct consequence of (\ref{4.2}) (with $x=0$) and the fact that $%
p $ is even. So let us consider the cases when $t\geq 1$. We can write (\ref%
{5.12}) as\textbf{\ }%
\begin{equation*}
\sum_{j=s+2t}^{sp}\left( -1\right) ^{j}j!\left( \dbinom{j-s-t-1}{t}+2\dbinom{%
j-s-t-1}{t-1}\right) S_{s,s}\left( p,j\right) =0,
\end{equation*}%
or%
\begin{equation}
\sum_{j=s+2t}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \left(
j-s\right) \prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right) =0.  \label{5.20}
\end{equation}

To prove (\ref{5.20}) is equivalent to prove that%
\begin{eqnarray}
&&\sum_{j=s+1}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \left(
j-s\right) \prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right)  \label{5.21} \\
&=&\sum_{j=s+1}^{s+t}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \left(
j-s\right) \prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right) .  \notag
\end{eqnarray}

From Gould's identity (3.50) in \cite{Go1} we see that 
\begin{equation*}
\binom{j-s-t-1}{t-1}=\sum_{k=0}^{t-1}\left( -1\right) ^{k+t+1}\binom{2t-2-k}{%
t-1}\binom{j-s-1}{k},
\end{equation*}%
which can be written as%
\begin{equation}
\prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right) =\sum_{k=0}^{t-1}\frac{%
\left( -1\right) ^{k}\left( t-1+k\right) !}{k!\left( t-1-k\right) !}%
\prod\limits_{l=0}^{t-2-k}\left( j-s-1-l\right) .  \label{5.22}
\end{equation}

Beginning with the left-hand side of (\ref{5.21}) we can write (by using (%
\ref{5.22}))%
\begin{eqnarray}
&&\sum_{j=s+1}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \left(
j-s\right) \prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right)  \label{5.23} \\
&=&\sum_{j=s+1}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \left(
j-s\right) \sum_{k=0}^{t-1}\frac{\left( -1\right) ^{k}\left( t-1+k\right) !}{%
k!\left( t-1-k\right) !}\prod\limits_{l=0}^{t-2-k}\left( j-s-1-l\right) 
\notag \\
&=&\sum_{k=0}^{t-1}\frac{\left( -1\right) ^{k}\left( t-1+k\right) !}{%
k!\left( t-1-k\right) !}\sum_{j=s+t-k}^{sp}\left( -1\right)
^{j}j!S_{s,s}\left( p,j\right) \prod\limits_{l=0}^{t-1-k}\left(
j-s-l\right) .  \notag
\end{eqnarray}

Now we use (\ref{5.17}) to write (\ref{5.23}) as%
\begin{eqnarray}
&&\sum_{j=s+1}^{sp}\left( -1\right) ^{j}j!S_{s,s}\left( p,j\right) \left(
j-s\right) \prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right)  \label{5.24} \\
&=&\sum_{k=0}^{t-1}\frac{\left( -1\right) ^{k+sp}\left( t-1+k\right) !}{k!}%
\left( t-k\right) \left( s+t-k\right) !S_{s,s}\left( p,s+t-k\right)  \notag
\\
&=&\left( -1\right) ^{t+s\left( p-1\right) }\sum_{j=s+1}^{s+t}\frac{\left(
-1\right) ^{j}\left( 2t-1-j+s\right) !}{\left( t-j+s\right) !}j!\left(
j-s\right) S_{s,s}\left( p,j\right) .  \notag
\end{eqnarray}

Finally, we use that 
\begin{equation*}
\prod\limits_{l=0}^{t-2}\left( j-s-t-1-l\right) =\left( -1\right) ^{t+1}%
\frac{\left( 2t-1-j+s\right) !}{\left( t-j+s\right) !},
\end{equation*}%
(proof left to the reader), together with the hypotheses on $s$ and $p$, to
obtain from (\ref{5.24}) the desired conclusion (\ref{5.21}).
\end{proof}

\begin{corollary}
\label{Cor.5.1}(a) If $s$ is even, or $s$ and $p$ are odd, expression (\ref%
{5.10}) can be written as%
\begin{equation}
\dbinom{n}{s}^{p}=\frac{\left( -1\right) ^{sp}}{\left( s!\right) ^{p}}%
\sum_{t=0}^{\frac{1}{2}s\left( p-1\right) }\sum_{j=s+2t}^{sp}\left(
-1\right) ^{j}j!\dbinom{j-s-t}{t}S_{s,s}\left( p,j\right) \dbinom{n+t}{s+2t}.
\label{5.25}
\end{equation}

(b) If $s$ is odd and $p$ is even, expression (\ref{5.10}) can be written as%
\begin{align}
& \dbinom{n}{s}^{p}  \label{5.26} \\ 
&=\frac{1}{\left( s!\right) ^{p}}\sum_{t=0}^{\left\lfloor 
\frac{1}{2}s\left( p-1\right) \right\rfloor }\sum_{j=s+2t}^{sp}\left(
-1\right) ^{j}j!\binom{j-s-t}{t}S_{s,s}\left( p,j\right) \left( \binom{n+t}{s+2t+1}+\binom{n+t+1}{s+2t+1}\right) .  \notag
\end{align}
\end{corollary}

\begin{proof}
Follows from (\ref{5.10}) and Proposition \ref{Prop.5.1}.
\end{proof}

Of course, formulas (\ref{5.25}) and (\ref{5.26}) have the following
versions for partial sums of $\binom{n}{s}^{p}$ (obtained by taking the
convolution $\ast ^{l}1$ in both sides of (\ref{5.25}) and (\ref{5.26}), and
using (\ref{2.35}))

(a) If $s$ is even, or $s$ and $p$ are odd, then for any integer $l\geq 0$
we have%
\begin{equation}
\dbinom{n}{s}^{p}\ast ^{l}1=\frac{\left( -1\right) ^{sp}}{\left( s!\right)
^{p}}\sum_{t=0}^{\frac{1}{2}s\left( p-1\right) }\sum_{j=s+2t}^{sp}\left(
-1\right) ^{j}j!\dbinom{j-s-t}{t}S_{s,s}\left( p,j\right) \dbinom{n+t+l}{%
s+2t+l}.  \label{5.27}
\end{equation}

(b) If $s$ is odd and $p$ is even, then for any integer $l\geq 0$ we have%
\begin{align}
&\dbinom{n}{s}^{p}\!\ast ^{l}\!1\!  \label{5.28} \\
&=\!\frac{1}{\left( s!\right) ^{p}}\!\!\sum_{t=0}^{\left\lfloor \frac{1}{2}%
s\left( p-1\right) \right\rfloor }\!\!\sum_{j=s+2t}^{sp}\!\!\!\left(
-1\right) ^{j}\!j!\dbinom{j-s-t}{t}S_{s,s}\left( p,j\right) \left( \!\!%
\dbinom{n+t+l}{s+2t+1+l}+\dbinom{n+t+1+l}{s+2t+1+l}\!\!\!\right) .  \notag
\end{align}

Some examples from (\ref{5.27}) and (\ref{5.28}) are the following (compare
them with similar expressions shown in Table \ref{T.1}):%
\begin{eqnarray*}
\sum\limits_{k=1}^{n}\sum\limits_{j=1}^{k}\sum\limits_{i=1}^{j}\dbinom{i}{2}%
^{2} &=&\dbinom{n+3}{5}+6\dbinom{n+4}{7}. \\
\sum\limits_{i=1}^{n}\binom{i}{2}^{3} &=&\dbinom{n+1}{3}+24\dbinom{n+2}{5}+90%
\dbinom{n+3}{7}. \\
\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{j}\binom{i}{3}^{3} &=&\dbinom{n+2}{5}%
\!+60\dbinom{n+3}{7}\!+630\dbinom{n+4}{9}\!+1680\dbinom{n+5}{11}. \\
\sum\limits_{i=1}^{n}\binom{i}{3}^{2} &=&\dbinom{n+1}{5}+\dbinom{n+2}{5}%
+10\left( \dbinom{n+2}{7}+\dbinom{n+3}{7}\right) . \\
\sum\limits_{i=1}^{n}\binom{i}{4}^{2} &=&\dbinom{n+1}{5}+20\dbinom{n+2}{7}+70%
\dbinom{n+3}{9}.
\end{eqnarray*}

\subsection*{Acknowledgments}

I thank the referee for calling my attention to some other important works
on generalizations of Stirling numbers.





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% then copy and past the contents of your .bbl file into your .tex file

\begin{thebibliography}{10}

\bibitem{Ar} H. Arod\'{z}. Comment on `Combinatorics of boson normal
ordering and some applications'. Concepts of Physics, Vol. I: 279--281,
2004. 
\url{http://merlin.fic.uni.lodz.pl/concepts/2004\_3\_4/2004\_3\_4\_279.pdf}

\bibitem{Bl1} P. Blasiak. Combinatorics of boson normal ordering and some
applications. PhD Thesis: University of Paris VI and Polish Academy of
Sciences, Krakow, Poland. arXiv:quant-ph/0507206v2

\bibitem{Bl2} P. Blasiak, A. Horzela, K. A. Penson, A. I. Solomon, and G. H.
E. Duchamp. Combinatorics and boson normal ordering: A gentle introduction. 
\textit{Am. J. Phys.} 75:639--646, 2007.

\bibitem{Bl3} P. Blasiak, K. A. Penson, and A. I. Solomon. The Boson Normal
Ordering Problem and Generalized Bell Numbers. \textit{Ann. Comb.}
7:127--139, 2003.

\bibitem{Ca} N. J. Calkin. Factors of sums of powers of binomial
coefficients. \textit{Acta Arith.} 86:17--26,\textbf{\ }1998.

\bibitem{Car} L. Carlitz. Weighted Stirling numbers of the first and second
kind -- I. \textit{Fibonacci Quart.} 18:147--162, 1980.

\bibitem{Ch1} Ch. A. Charalambides, Non-central generalized $q$-factorial
coefficients and $q$-Stirling numbers, \textit{Discrete Math.}, 275:67--85,
2004.

\bibitem{Ch2} Ch. A. Charalambides, \textit{Enumerative Combinatorics,}
Chapman \& Hall/CRC, Boca Raton, FL, 2002.

\bibitem{Ch3} Ch. A. Charalambides and M. Koutras, On the differences of the
generalized factorials at an arbitrary point and their combinatorial
applications, \textit{Discrete Math.}, 47:183--201, 1983.

\bibitem{Co} P. Codara, O. M. D'Antona, and P. Hell. A simple combinatorial
interpretation of certain generalized Bell and Stirling numbers.
arXiv:1308.1700v1

\bibitem{D} M. Dziemia\'{n}czuk. Generalizing Delannoy numbers via counting
weighted lattice paths. \textit{Integers} 13, \#A54, 2013.

\bibitem{Ga} N. Gauthier. Convolving the $m$-th Powers of the Consecutive
Integers with the General Fibonacci Sequence Using Carlitz's Weighted
Stirling Polynomials of the Second Kind. \textit{Fibonacci Quart.}
42:306--313, 2004.

\bibitem{Go1} H. W. Gould. \textit{Combinatorial Identities}. Morgantown, W.
Va. 1972.

\bibitem{Go2} H. W. Gould. \textit{Combinatorial Identities: Table I:
Intermediate Techniques for Summing Finite Series.} From the seven
unpublished manuscripts of H. W. Gould. Edited and Compiled by Jocelyn
Quaintance, 2010. Available at 
\url{http://www.math.wvu.edu/~gould/Vol.4.PDF}

\bibitem{G-K-P} R. L. Graham, D. E. Knuth, and O. Patashnik, \textit{%
Concrete Mathematics: A Foundation for Computer Science}. Second Edition,
Addison Wesley, 1993.

\bibitem{TXH} Tian-Xiao He, Generalized Stirling Numbers and Generalized
Stirling Functions. arXiv:1106.5251v1

\bibitem{Jo} Ch. Jordan, \textit{Calculus of Finite Differences,} Chelsea,
New York, 1965.

\bibitem{Ko} T. Komatsu, F. Luca, and C. Pita. Generalized poly-Cauchy
polynomials and their interpolating functions. Submitted.

\bibitem{Kou} M. Koutras, Non-central Stirling numbers and some
applications, \textit{Discrete Math.}, 42:73--89, 1982.

\bibitem{La1} W. Lang. Combinatorial Interpretation of Generalized Stirling
Numbers. \textit{J. Integer Seq.} 12, Article 09.3.3,\textbf{\ }2009.

\bibitem{La2} W. Lang. On Generalizations of the Stirling Number Triangles. 
\textit{J. Integer Seq.} 3, Article 00.2.4,\textbf{\ }2000.

\bibitem{Ma} T. Mansour, M. Schork, and M. Shattuck. On a new family of
generalized Stirling and Bell numbers. \textit{Electron. J. Combin.} 18,
\#P77, 2011.

\bibitem{Me} M. A. M\'{e}ndez, P. Blasiak, and K. A. Penson. Combinatorial
approach to generalized Bell and Stirling numbers and boson normal ordering
problem. \textit{J. Math. Phys.} 46, Article 083511, 2005.

\bibitem{Pi} C. Pita. Some Number Arrays Related to Pascal and Lucas
Triangles. \textit{J. Integer Seq. }16, Article 13.5.7,\textbf{\ }2013.

\bibitem{OEIS} N. J. A. Sloane. The On-line Encyclopedia of Integer
Sequences. \url{http://oeis.org}

\bibitem{So1} A. I. Solomon, P. Blasiak, G. Duchamp, A. Horzela, and K. A.
Penson. Combinatorial Physics, Normal Order and Model Feynman Graphs. 
\textit{Symmetries in Science XI, }pages 527--536. Kluwer Academic
Publishers, 2004.

\bibitem{So2} A. I. Solomon, G. Duchamp, P. Blasiak, A. Horzela, and K. A.
Penson. Normal Order: Combinatorial Graphs. arXiv:quant-ph/0402082v1


\end{thebibliography}

\end{document}
