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text:
\section*{Abstract}
\noindent A barred preferential arrangement is a preferential arrangement, onto which in-between the blocks of the preferential arrangement a number of identical bars are inserted. We offer a generalisation of barred preferential arrangements by making use of the generalised Stirling numbers proposed by Hsu and Shiue (1998). We discuss how these generalised barred preferential arrangements offer a unified combinatorial interpretation of geometric polynomials. We also discuss asymptotic properties of these numbers.

\noindent\fontsize{10}{1}\selectfont
Mathematics Subject Classifications : 05A15, 05A16, 05A18, 05A19,   11B73, 11B83
\\\textbf{Keyword(s)}:preferential arrangement, barred preferential arrangement, geometric polynomial.
\normalsize
\vspace{2mm}

\begin{section}{Introduction}\end{section}

A \textit{barred preferential arrangement} of an $n$-element set $X_n$ is a preferential arrangement(ordered set partition) on which a number of bars are inserted in-between the formed blocks of $X_n$.

The following are two examples of barred preferential arrangements of $X_5$ having two and three bars respectively,

\RNum{1}).\:\:  $24\:\:  | 1 \quad53  | $

\RNum{2}). $|\:\: 2\:\: | 5\quad 4|\: 3\quad 1.$

The barred preferential arrangement in \RNum{1}, has two bars hence three sections. The first section from(left to right) has a single block i.e the block formed by the elements \{2,4\}. The second section (the one between the two bars) has two blocks which are, \{1\} and,  \{3,5\}. The third section(to the right of the second bar) is empty. The barred preferential arrangement in \RNum{2} has three bars hence four sections. The first section is empty. The second section has a single block. The third section has two blocks, and the fourth section also has two blocks. Barred preferential arrangements(BPA) having multiple bars seem to first appear in \cite{barred:2013}.

The generalised Stirling numbers $S(n,k,\alpha,\beta,\gamma)$ are defined in \cite{A unified approach to generalized Stirling numbers} in the following way, $(t|\alpha)_n=\sum\limits_{k=0}^{n}S(n,k,\alpha,\beta,\gamma)(t-\gamma|\beta)_n$, such that $(t|\alpha)_n$ is the generalised factorial polynomial $(t|\alpha)_n=\prod\limits_{k=0}^{n-1}(t-k\alpha)$, where $n\geq1$ and $\alpha,\beta,\gamma$ real or complex not all equal to zero.
\vspace{3mm}

In \cite{Higher order geometric polynomials} the authors proposed as a way of generalising geometric polynomials proposed the following generating function;
\begin{equation}\label{equation:1}\sum\limits_{n=0}^{\infty} T_n^{\lambda,x}(\alpha,\beta,\gamma)\frac{t^n}{n!}=
       \frac{(1+\alpha t)^{\gamma/\alpha}}{(1-x((1+\alpha t)^{\beta/\alpha}-1))^{\lambda}}
       .\end{equation}

Using  non-combinatorial methods the authors  recognised $T_n^{\lambda,x}(\alpha,\beta,\gamma)$ as generalisation of the counting sequence of barred preferential arrangements. The authors ask for a study of combinatorial properties of these numbers, which we provide in the current paper.

 Nelsen and Schmidt proposed the family of generating functions (see~\cite{Chains in power set Nelsen 91}),\begin{equation}\label{equation:15} \frac{e^{\gamma x}}{2-e^x}.\end{equation} In the manuscript for $\gamma=2$  Nelsen and Schmidt interpreted the generating function as being that of the number of chains in the power set of $X_n$. The generating function for $\gamma=0$ is known to be that of number of preferential arrangements/number of outcomes in races with ties (see~\cite{gross:1962,mendelson:1982}). In the manuscript Nelsen and Schmidt then asked, ``could there be combinatorial structures associated with either $X_n$ or the power set of $X_n$ whose integer sequences are generated by  members of the family in \eqref{equation:15}, for other values of $\gamma$?" We will now refer to this question as the Nelsen-Schmidt question. In answering the  Nelsen-Schmidt question, the authors in \cite{Nkonkobe & Murali Nelesn-Schmidt generating function,Our paper on Generalised barred preferential arrangements}, offered combinatorial interpretations of integer sequences arising from the more general generating functions $\frac{e^{\gamma t}}{(2-e^t)^{\lambda}}$ and $\frac{e^{\gamma t}}{(2-e^{\beta t})^{\lambda}}$, where $\lambda,\beta,\gamma$ are in $\mathbb{N}_0$ (non-negative integers). In this study we offer a further generalisation of their results in answering the Nelsen-Schmidt question by interpreting combinatorially integer sequences arising from the generating function $\frac{(1+\alpha t)^{\gamma/\alpha}}{(1-x((1+\alpha t)^{\beta/\alpha}-1))^{\lambda}}$. Both generating functions $\frac{e^{\gamma t}}{(2-e^t)^{\lambda}}$, $\frac{e^{\gamma t}}{(2-e^{\beta t})^{\lambda}}$ arise as special cases of the generating function $\frac{(1+\alpha t)^{\gamma/\alpha}}{(1-x((1+\alpha t)^{\beta/\alpha}-1))^{\lambda}}$.

Geometric polynomials go far back as Euler's work on the year 1755 (see page 389 on Part \RNum{2} of \cite{Euler's work on geometric polynomials}). These polynomials are  well known in the literature, and it is also well known that these polynomials arise from variations of the  generating function $\frac{e^{r t}}{(1-x(e^{t}-1))^r}$, for instance in \cite{On generalised Bell polynomials,A series transformation formula,Apostol-Bernoulli functions,close encounters with stirling numbers of the second kind,Investing Geometric and exponential polynomials,Geometric Polynomials: Properties and Applications to Series with Zeta Values,Polynomials Related to Harmonic Numbers and Evaluation of Harmonic Number Series II,Fourier series of sums of products of ordered Bell and poly-Bernoulli functions,An explicit formula for Bernoulli polynomials in terms of $ r $-Stirling numbers of the second kind,Generalization of Mellin derivative and its applications}. In this study our combinatorial interpretation of the integer sequences arising from the generating function $\frac{(1+\alpha t)^{\gamma/\alpha}}{(1-x((1+\alpha t)^{\beta/\alpha}-1))^{\lambda}}$ offers a unified combinatorial interpretation of these geometric polynomials.

\begin{section}{When one bar is used.}

In this section we study the numbers $T_n^{1,x}(\alpha,\beta,\gamma)$.

      \begin{lemma}\cite{Corcino Rorberto 2001}\label{lemma:3} For real/complex $\alpha,\beta,\gamma$, \fontsize{10}{1}

      $S(n,k,\alpha,\beta,\gamma)=\frac{1}{\beta^kk!}\Delta^k(\beta k+\gamma|\alpha)_n\big|_{s=0}=\frac{1}{\beta^kk!}\sum\limits_{s}^{}(-1)^{k-s}\binom{i}{s}(\beta s+\gamma|\alpha)_n$.
      \end{lemma}

      \begin{lemma}\label{lemma:10}\cite{Higher order geometric polynomials}For $\alpha,\beta,\gamma,x\in\mathbb{N}_0$ such that $(\alpha,\beta,\gamma,x)\not=(0,0,0,0)$,
      \begin{equation}\label{equation:9}
      G_n^x(\alpha,\beta,\gamma)=\sum\limits_{k=0}^{n}\beta^kk!x^kS(n,k,\alpha,\beta,\gamma).
      \end{equation}
      \end{lemma}
      \begin{lemma}\label{lemma:11}\cite{Higher order geometric polynomials,On generalised Bell polynomials}For real/complex $\alpha,\beta,\gamma,x$ such that $(\alpha,\beta,\gamma,x)\not=(0,0,0,0)$,
      \begin{equation}\label{equation:10}
      \sum\limits_{n=0}^{\infty}T_n^{1,x}(\alpha,\beta,\gamma)=\frac{(1+\alpha t)^{\gamma/\alpha}}{1-x((1+\alpha t)^{\beta/\alpha}-1)}.
      \end{equation}
      \end{lemma}
      \begin{theorem} \label{Theorem:15
       \label{theorem:10}\cite{Paper with interpretation of G-stirling numbers}
     Given $\alpha,\beta,\gamma$ are non-negative integers such that $\alpha$ divides both $\beta$, and $\gamma$. Given $k+1$ distinct cells such that the first $k$ cells each contains $\beta$ labelled compartments, and the $(k+1){th}$ cell contains $\gamma$ labelled compartments. In each cell the compartments are having cyclic ordered numbering. The capacity of each compartment is limited to one ball. The number $\beta^k k!S(n,k,\alpha,\beta,\gamma)$, is the number of ways of distributing $n$ distinct elements into the $k+1$ cells one ball at a time such that only the cell having $\gamma$ compartments may be empty. compartments.
      \end{theorem}

We extend a special case of Theorem~\ref{Theorem:15} to the following property (see \cite{On generalised Bell polynomials}).

 \begin{property}\label{property:1} The number of ways of distributing $n$ balls into $k$ distinct cells one ball at a time (where $k$ runs from o to $n$), such that all the cells are non-empty, and each cell  has $\beta$ labelled compartments having cyclic ordered numbering such that for each consecutive available $\alpha$ compartments only the first compartment gets a ball, where each of the $k$ cells is colored with one of $x$ available colors is,
  $\sum\limits_{k=0}^{n}\beta^kk!x^kS(n,k,\alpha,\beta,0)=G_n^x(\alpha,\beta,0)$.
               \end{property}

        \begin{property}\cite{Paper with interpretation of G-stirling numbers}\label{property:2}
                          Given a single cell with $\gamma$ compartments such that $\alpha|\gamma$, where the cells are given cyclic ordered numbering such that on each consecutive $\alpha$  compartments only the first compartment gets a ball. For any given $n$ balls the number of ways of placing the balls into the cell, one ball at a time is
                         $(\gamma|\alpha)_n$. \end{property}

                  \begin{remark}
                  It is clear that whether you consider barred preferential arrangements as a result of first forming blocks of elements and then inserting bars, or you first place the bars and then distribute the elements into the resultant sections, the results are the same. In our arguments in Theorems~\ref{theorem:2} to Theorem~\ref{theorem:12}, the latter way of viewing barred preferential arrangements is used, this technique is also used in \cite{Nkonkobe & Murali Nelesn-Schmidt generating function,Our paper on Generalised barred preferential arrangements}. In Theorem \ref{theorem:8} and \ref{theorem:11}, the former way of looking at barred preferential arrangements is used, this is the same technique used by the authors in \cite{barred:2013}.
                  \end{remark}
                    \begin{remark}
                    In our arguments in the following theorems we assume that the section having property~\ref*{property:2} i.e the one having $\gamma$ compartments is the first section from(left to right). We will sometimes refer to this section as the special section.
                    \end{remark}
               \begin{theorem}\label{theorem:2}
               The generating function $\frac{(1+\alpha t)^{\gamma/\alpha}}{1-x((1+\alpha t)^{\beta/\alpha}-1)}$ for $\alpha,\beta,\gamma$ in non-negative integers such that $\alpha|\beta$ and $\alpha|\gamma$, is that of the number of barred preferential arrangements with one bar, such that one section has property~\ref{property:2} and the other section has property~\ref{property:1}.
               \end{theorem}                   \begin{proof}
               By equation~\ref{equation:10} we have,

               \begin{equation}\label{equation:12}
               T_n^{1,x}(\alpha,\beta,\gamma)=\sum\limits_{k=0}^{n}\binom{n}{k}G_k^x(\alpha,\beta,0)(\gamma|\alpha)_{n-k}.
               \end{equation}
               \end{proof}

                 \begin{theorem}For $\alpha,\beta,\gamma,x\in\mathbb{N}_0$ such that $(\alpha,\beta,\gamma,x)\not=(0,0,0,0)$,\label{theorem:3}
                 \fontsize{10}{14}
                 \begin{equation}
                 T_{n+1}^{1,x}(\alpha,\beta,\gamma)=\gamma T_n^{1,x}(\alpha,\beta,\gamma-\alpha)+x\beta\sum\limits_{k=0}^{n}\binom{n}{k}T_{n-k}^{1,x}(\alpha,\beta,\gamma)T_{k}^{1,x}(\alpha,\beta,\beta-\alpha).\end{equation}
                 \end{theorem}
                 \begin{proof}
                 The argument is based on the position of the $(n+1)th$ element.

   If the $(n+1)th$ element is on the special section (with $\gamma$ compartments), then we may choose in $\gamma$ ways a compartment for the element. There are $\gamma-\alpha$ free compartment left for the other elements to occupy within this section. The remainder $n$ elements can be arranged among the two sections in $T_n^{1,x}(\alpha, \beta, \gamma- \alpha)$ ways.

    Next, we consider the case when $(n+1)$th goes into the other section with Property~\ref{property:1} and let $B$ be the block that includes it. Further, let $k$ be the number of elements that are included in the block $B$ and blocks arranged right to $B$ within the section. This part of the preferential arrangement can be viewed as one on $k$ elements having a special block $B$ with $\beta-\alpha$ compartments. The other part of the preferential arrangement, the blocks that go to the left of $B$ including those in the special section with $\gamma$ compartments, can be viewed simply as a preferential arrangement on the remainder $n-k$ elements. Hence, we choose the $k$ elements in $\binom{n}{k}$ ways, select a compartment for the $(n+1)$th element in $\beta$ ways, a color for the block $B$ in $x$ ways, and construct the preferential arrangements on the $k$ element in $T_k^{1,x}(\alpha,\beta,\gamma+\beta-\alpha)$ ways, and on the remainder $n-k$ elements in $T_{n-k}^{1,x}(\alpha,\beta,\gamma)$ ways.

                 \end{proof}

                  Theorem~\ref{theorem:3} is a generalisation of Theorem 8 of \cite{Our paper on Generalised barred preferential arrangements}.

                 \begin{theorem}For $\alpha,\beta,\gamma,x\in\mathbb{N}_0$ such that $(\alpha,\beta,\gamma,x)\not=(0,0,0,0)$,\fontsize{10}{14}
                 \begin{equation}
                 T_{n+1}^{1,x}(\alpha,\beta,\gamma)=\gamma T_n^{1,x}(\alpha,\beta,\gamma-\alpha)+x\beta\sum\limits_{k=0}^{n}\binom{n}{k}T_k^{1,x}(\alpha,\beta,\gamma+\beta-\alpha)T_{n-k}^{1,x}(\alpha,\beta,0).
                 \end{equation}
                 \end{theorem}

                \begin{proof}The proof is similar to the previous theorem. For the case the $(n+1)th$ element is on the special section having $\gamma$ compartments is as in the previous theorem.

                We now consider the case where the $(n+1)th$ element is in the section having property~\ref{property:1}, say the element is part of block $B$. Now consider the block $B$, and the  special section having $\gamma$ compartments as a single unit. Let $k$ be the number of elements on the left to the block $B$, within the section. This unit having $\beta+\gamma-\alpha$ available compartments.

                Given $k$ elements, there are $T_k^{1,x}(\alpha,\beta,\gamma+\beta-\alpha)$ possibilities that the elements are on either the unit or on the left of $B$ within the same section  that $B$ is on. The remainder $n-k$ elements may be preferential arrangements to the right of $B$ in  $T_{n-k}^{1,x}(\alpha,\beta,0)$ ways.
                \end{proof}

\end{section}

\section{When multiple bars are considered}

In this section we examine the numbers $T_n^{\lambda,x}(\alpha,\beta,\gamma)$, for an arbitrary $\lambda$.

\begin{theorem}
\label{theorem:5}
 The number $T_n^{\lambda,x}(\alpha,\beta,\gamma)$  for $\alpha,\beta,\gamma,\lambda$ in non-negative integers where $\alpha|\beta$ and $\alpha|\gamma$, is the number of barred preferential arrangements having $\lambda$ bars, such that one section has property~\ref{property:2} and  $\lambda$ other sections have property~\ref{property:1}.

\end{theorem}

\begin{proof}
 By \eqref{equation:1} we have
 \fontsize{10}{14}
 \begin{equation}\label{equation:14}T_n^{\lambda,x}(\alpha,\beta,\gamma)=\sum\limits_{r_1+\cdots+r_{\lambda+1}}\binom{n}{r_1,\ldots,r_{\lambda+1}}(\gamma|\alpha)_{r_1}\prod\limits_{i=2}^{\lambda+1}G^x_{r_i}(\alpha,\beta,0)\end{equation}
\end{proof}

For the case $\lambda=1$ the statement of Theorem~\ref{theorem:5} can be derive from the one given for the numbers $B_n(\alpha,\beta,\gamma)$ in \cite{On generalised Bell polynomials}.

 \begin{corollary}\label{corolloary:1}
                                           For $\lambda,\beta,\gamma, x\in\mathbb{N}_0$,
                                           $$\frac{e^{\gamma t}}{[1-x(e^{\beta t}-1)]^\lambda},$$
 is the generating function for the number of barred preferential arrangements, having $\lambda$ bars. Where elements on each block on fixed $\lambda$ sections are colored with $\beta$ available colors, and elements on one section are colored with $\gamma$ available colors. Also the blocks themselves on the first $\lambda$ sections being colored with one of $x$ available colors. This is a generalisation of the work done in \cite{Our paper on Generalised barred preferential arrangements}.                                       \end{corollary}

                           \begin{remark}
                           [Geometric Polynomials]
    Geometric polynomials go far back as Euler's work on the year 1755 (see page 389 on Part \RNum{2} of \cite{Euler's work on geometric polynomials}). These polynomials are
    well known in the literature, and it is also well known that these polynomials arise from variations of the  generating function $\frac{e^{r t}}{(1-x(e^{t}-1))^r}$, for instance in \cite{On generalised Bell polynomials,A series transformation formula,Apostol-Bernoulli functions,close encounters with stirling numbers of the second kind,Investing Geometric and exponential polynomials,Geometric Polynomials: Properties and Applications to Series with Zeta Values,Polynomials Related to Harmonic Numbers and Evaluation of Harmonic Number Series II,Fourier series of sums of products of ordered Bell and poly-Bernoulli functions,An explicit formula for Bernoulli polynomials in terms of $ r $-Stirling numbers of the second kind,Generalization of Mellin derivative and its applications}.
                         Hence, our results in Theorem~\ref{theorem:5} offers a generalised combinatorial interpretation of these geometric polynomials.
                         .
                           \end{remark}

          \begin{remark}[Nelsen-Schmidt question] Both generation functions, $\frac{e^{\gamma t}}{(2-e^t)^{\lambda}}$ and $\frac{e^{\gamma t}}{(2-e^{\beta t})^{\lambda}}$ previously studied in answering the Nelsen-Schmidt question considered by the authors in  \cite{Nkonkobe & Murali Nelesn-Schmidt generating function,Our paper on Generalised barred preferential arrangements} arising as special cases of the generating function we study here given in \eqref{equation:1}.
                                                                       \end{remark}

\begin{theorem}\label{theorem:7}For $\alpha,\lambda,\beta,x\in\mathbb{N}_0$ such that $(\alpha,\beta,\gamma,x)\not=(0,0,0,0)$,\fontsize{10}{14}
\begin{equation}
T_{n+1}^{\lambda,x}(\alpha,\beta,\gamma)=\gamma T_n^{\lambda,x}(\alpha,\beta,\gamma-\alpha)+x\lambda\beta T_n^{\lambda+1,x}(\alpha,\beta,\gamma+\beta-\alpha).
\end{equation}
\end{theorem}
\begin{proof}

Consider the $(n+1)$th element. If it is in the special block, we have to select a compartment in $\gamma$ ways and to arrange the remainder $n$ elements in $ T_n^{\lambda,x}(\alpha,\beta,\gamma-\alpha)$ ways.

In the other case the $(n+1)th$ element is in one of the sections with property~\ref{property:1}.
Let $B$ be the block with $\beta$ compartments which the $(n+1)$th element forms part-off. Consider the first special block and $B$ as united special block. However, to be able to reconstruct the original situation, mark the place of $B$, with an extra bar. Viewing our object this way, we have a preferential arrangement on $n$ elements having a special block with $\gamma+\beta-\alpha$ (after dropping the $(n+1)$th element, $\alpha$ compartments are closed from $\beta$ compartments of $B$) with $\lambda +1$ sections having property~\ref{property:1}.
\end{proof}

Theorem~\ref{theorem:7} is a generalisation of Theorem 9 of \cite{Our paper on Generalised barred preferential arrangements}.
\begin{theorem}For $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that
$(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0)$,\fontsize{10}{14}
\begin{equation}
T_{n+1}^{\lambda,x}(\alpha,\beta,\gamma)=\gamma T_n^{\lambda,x}(\alpha,\beta,\gamma-\alpha)+x\lambda\beta\sum\limits_{s=0}^{n}\binom{n}{s}(\gamma|\alpha)_sT_{n-s}^{\lambda+1,x}(\alpha,\beta,\beta-\alpha).
\end{equation}

\end{theorem}
\begin{proof}
We consider again the position of the $(n+1)$th element. If it is in the first special block, we have $\gamma T_n^{\lambda,x}(\alpha,\beta,\gamma-\alpha)$ possibilities. Now, assume it is in one of the sections where each block has $\beta$ compartments.
We let $B$ denote the block of which the $(n+1)$th element is part-off. Clearly, we need to select the color for the block $B$ in $x$ ways, the compartment for $(n+1)$th element in $\beta$ ways and the section of the block in $\lambda$ ways. We build the special block from $s$ chosen element (in $\binom{n}{s}(\gamma|\alpha)_s$ ways). Now treating $B$ as a special block with $\beta-\alpha$ available compartments, the remainder $n-s$ elements can be arranged $T^{\lambda+1,x}_{n-s}(\alpha,\beta,\beta-\alpha)$ ways, this is from the fact that within the section in which the $(n+1)th$ element is in, each of the left side and the right side of the block $B$ excluding $B$ gives rise to a single section having property~\ref{property:1}.
\end{proof}

\begin{theorem}For $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that $(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0)$,
\begin{equation}
(1+x)T_n^{\lambda+1,x}(\alpha,\beta,0)= T_n^{\lambda,x}(\alpha,\beta,0)+ xT_n^{\lambda+1,x}(\alpha,\beta,\beta).
\end{equation}
\end{theorem}
\begin{proof}
We reorder the equation in the following way,
\begin{equation}\label{equation:7}
xT_n^{\lambda+1,x}(\alpha,\beta,\beta)-xT_n^{\lambda+1,x}(\alpha,\beta,0)=T_n^{\lambda+1,x}(\alpha,\beta,0)-T_n^{\lambda,x}(\alpha,\beta,0).
\end{equation}

The left hand side of \eqref{equation:7} counts the preferential arrangements having a non-empty special block with $\beta$ compartments, which is colored by one of the $x$ colors. Since the color is also chosen in the case when the first block is empty, one can think of that as a marked block color.
On the other hand, from a preferential arrangement with $\lambda+1$ sections and empty first section, we obtain a preferential arrangement with a non-empty first section, if we shift the first bar to the right, directly after the first block. This block has a color (out of $x$) originally. However, if there are no blocks directly to the righ of the first bar, this can not be done, we need to reduce by the number of these preferential arrangements, which is exactly the number of the same preferential arrangements with one less bar.
\end{proof}
\begin{theorem}For $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that
 $(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0)$,\label{theorem:9}
\fontsize{10}{14}
\begin{equation}
T_{n+1}^{\lambda,x}(\alpha,\beta,\gamma)=\gamma T_n^{\lambda,x}(\alpha,\beta,\gamma-\alpha)+x\beta \lambda\sum\limits_{k=0}^{n}\binom{n}{k}T_k^{\lambda,x}(\alpha,\beta,\gamma)T_{n-k}^{1,x}(\alpha,\beta,\beta-\alpha).\end{equation}
\end{theorem}

\begin{proof}
Consider the $(n+1)$th element. The proof goes similarly to the previous ones. Let $B$ denote the block containing the $(n+1)$th element, and let this block fall between the bars $|^*$(to the left of $B$) and $|^{**}$(to the right of $B$). The block $B$ can be interpreted as a special block for those elements that are either part of the block $B$ or arranged to the right of $B$ before the bar $|^{**}$, $k$ elements can be arranged here in $T_{k}^{1,x}(\alpha,\beta,\beta-\alpha)$ ways. The remaining $n-k$ elements can be arranged on the other sections in $T^{\lambda,x}_{n-k}$ ways, where the extra section is the section to the left of $B$ excluding $B$ before the bar $|^*$. Clearly, we need to choose the $k$ elements out of the $n$ in $\binom{n}{k}$ ways, the section in $\lambda$ ways, the compartment in $\beta $ ways and the color of the block $B$ in $x$ ways.
\end{proof}

Theorem~\ref{theorem:9} is a generalisation of Theorem 8 of \cite{Our paper on Generalised barred preferential arrangements}.
\begin{theorem}For $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that
 $(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0)$,
\fontsize{10}{14}
\begin{equation}
T_{n+1}^{\lambda,x}(\alpha,\beta,\gamma)=\gamma\sum\limits_{k=0}^{n}\binom{n}{k}(\gamma-\alpha|\alpha)_kT_{n-k}^{\lambda,x}(\alpha,\beta,0)+x\lambda \beta\sum\limits_{k=0}^{n}\binom{n}{k}(\gamma+\beta-\alpha|\alpha)_k T_{n-k}^{\lambda+1,x}(\alpha,\beta,0).
\end{equation}
\end{theorem}
\begin{proof}
We locate the position of the $(n+1){th}$ element.

When the $(n+1){th}$ element is on the special section having $\gamma$ compartments, there are $\gamma$ ways of choosing a compartment for the element. Off the $n$ other elements, $k$ of them can also form part of the special section in $(\gamma-\alpha|\alpha)_k$ ways. The remaining $n-k$ elements can be arranged on the other sections with property~\ref{property:1} in $T^{\lambda,x}_{n-k}(\alpha,\beta,0)$ ways.

In the second case the $(n+1)th$ element is in one of the sections with property~\ref{property:1}.  We denote the block of which the $(n+1){th}$ element is part of by $B$. Clearly, a section, compartment and color for the block $B$ can be chosen in $x\lambda\beta$ ways. Now treating the block $B$ and the special block having $\gamma$ compartments as a single unit, $k$ elements can be arranged in this unit in $(\gamma+\beta-\alpha)_k$ ways. The remaining $n-k$ elements can be arranged on the other places in $T^{\lambda+1,x}_{n-k}(\alpha,\beta,0)$ ways.
\end{proof}

\begin{theorem}\label{theorem:12}For $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that
$(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0),$
\fontsize{10}{14}
\begin{equation}
T_{n+1}^{\lambda,x}(\alpha,\beta,\gamma)=\gamma T_n^{\lambda,x}(\alpha,\beta,\gamma-\alpha)+\sum\limits_{k=0}^{n}\binom{n}{k}(\gamma|\alpha)_kT_{n-k+1}^{\lambda,x}(\alpha,\beta,0).
\end{equation}
\end{theorem}
\begin{proof}

Similar to the previous theorems, we locate the position of the $(n+1)^{th}$ element.

The $(n+1){th}$ element is either on the special block or in one of the sections with property~\ref{property:1}. When the $(n+1){th}$ element is not on the special section, $k$ elements can be chosen and arranged on the special section in $(\gamma|\alpha)_k$ ways. The $n-k$ other elements plus the $(n+1)th$ element, can be arranged on the other sections having property~\ref{property:1} in $T^{\lambda,x}_{n-k+1}(\alpha,\beta,0)$ ways.
\end{proof}

           In \cite{Higher order geometric polynomials} the authors proposed the following identity, \\             $T_n^{\lambda+1,x}(\alpha,\beta,\gamma)=\sum\limits_{k=0}^{n}\binom{k+\lambda}{k}k!S(n,k,\alpha,\beta,\gamma)
                       \beta^kx^k$. In the following theorem we give a combinatorial interpretation of the result.

           \begin{theorem}\label{theorem:8}For $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that
           $(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0),$\fontsize{10}{1}
           \begin{equation}\label{equation:18}
           T_n^{\lambda,x}(\alpha,\beta,\gamma)=\sum\limits_{k=0}^{n}\binom{k+\lambda-1}{k}k!S(n,k,\alpha,\beta,\gamma)
           \beta^kx^k.
           \end{equation}
           \end{theorem}
           \begin{proof}
            In this theorem we use the notion of forming a barred preferential arrangement by first forming an ordered set partition and then insert bars. The $n$ elements can be distributed into $k+1$ cells where the first $k$ cells have property~\ref{property:1} and the $(k+1)th$ cell has property~\ref{property:2}, such that the first $k$ cells are non-empty in $\beta^kk!S(n,k,\alpha,\beta,\gamma)$ ways (see Theorem~\ref{Theorem:15}). The first $k$ cells can be coloured in $x^k$ ways. Then, $\lambda-1$ bars can be inserted in-between the first $k$ cells having property~\ref{property:1} in $\binom{k+\lambda-1}{k}$.
           \end{proof}

           Theorem~\ref{theorem:8} is a generalisation of theorem 3 of \cite{barred:2013}.

              \vspace{4mm}
 \noindent Combining \eqref{equation:18} and the following recurrence relation from \cite{A unified approach to generalized Stirling numbers},
               \fontsize{10}{14}
          \begin{equation}\label{equation:17}
   S(n+1,k,\alpha,\beta,\gamma)=S(n,k-1,\alpha,\beta,\gamma)+(k\beta-n\alpha+\gamma)S(n,k,\alpha,\beta,\gamma),
\end{equation}\normalsize
we obtain \eqref{equation:19} below. In Theorem~\ref{theorem:11} we give a combinatorial interpretation of the result.

  \begin{theorem}\label{theorem:11} For $\lambda\geq 1$, and $\alpha,\beta,\gamma,\lambda,x\in\mathbb{N}_0$ such that
  $(\alpha,\beta,\gamma,\lambda,x)\not=(0,0,0,0,0),$
   \fontsize{10}{1}
          \begin{equation}\label{equation:19}
          T_{n+1}^{\lambda+1,x}(\alpha,\beta,\gamma)=(\gamma|\alpha)_{n+1}+\sum\limits_{k=1}^{n}\binom{\lambda+k}{k}[S(n,k-1,\alpha,\beta,\gamma)+(k\beta-n\alpha+\gamma)S(n,k,\alpha,\beta,\gamma)]\beta^kk!x^k.
          \end{equation}
          \end{theorem}
         \begin{proof}
         We divide the proof into three cases.

         Case 1: all the elements are on the special section, there are $(\gamma|\alpha)_{n+1}$ ways of arranging the elements.

         Case 2: the $(n+1)th$ element is on its own block on one of the sections having property 1.
         Of the $n$ elements $k$ blocks can be formed in $\beta^{k-1}(k-1)!S(n,k-1,\alpha,\beta,\gamma)x^{k-1}$ ways where only the $(k)th$  block is allowed to be empty, satisfying the conditions of Theorem~\ref{theorem:10}. The $(n+1)th$ element can be put to form its own block on the $k$ spaces in-between the $k-1$ blocks that are not allowed to be empty and a color of this block in $k\beta x$ ways. Now $\lambda$ bars can be inserted in-between the spaces of the $k$ non-empty blocks in $\binom{k+\lambda}{k}$. Hence, the total number of possibilities in this case is $\sum\limits_{k=1}^n\binom{k+\lambda}{k}k!\beta^kS(n,k-1,\alpha,\beta,\gamma)x^k$.

         Case 3: the $(n+1)th$ element is either on the special block or on a block with other elements.
        By Theorem~\ref{theorem:10} the $n$ other elements can form $k+1$ blocks in $\beta^kk!S(n,k,\alpha,\beta,\gamma)$ ways, where only the special block with $\gamma$ compartments is allowed to be empty. The non-empty blocks can be colored in $x^k$ ways. As the last element to be placed the $(n+1)th$ element can be placed on one of these $k+1$ blocks in $k\beta+\gamma-\alpha n$ ways. Then, $\lambda$ bars can be placed in between the non-empty blocks in $\binom{k+\lambda}{k}$ ways. Hence, the total number of possibilities in this case is $\sum\limits_{k=1}^n\binom{k+\lambda}{k}\beta^kk!(k\beta+\gamma-\alpha n)S(n,k,\alpha,\beta,\gamma)x^k$.

         \end{proof}

       \begin{section}{Asymptotic Analysis}

        In this section, an asymptotic expansion for the higher order geometric polynomials will be derived using the known result of Hsu \cite{Power-type} on asymptotic expansion formula for the coefficients of power-type generating functions involving large parameters.

       \smallskip
       Let $\mathbb{N}$ be the set of positive integers and $\sigma(n)$ be the set of partitions of integer $n\in\mathbb{N}$, represented by $1^{k_1}2^{k_2}\ldots n^{k_n}$ such that
       $$1k_1+2k_2+\ldots +nk_n=n, k_i\geq0 \;\;(i=1,2,\ldots, n)\;\; \& \;\;k=k_1+k_2+\ldots+k_n.$$
       The parameter $k$ denotes the numbers of parts of the partition. Let $\sigma(n,k)$ be the subset of $\sigma(n)$ consisting of partitions of $n$ with $k$ parts.

       \smallskip
       Consider a formal power series $\phi(t)=\sum\limits_{n=0}^{\infty}a_nt^n$ over the complex field $\mathbb{C}$ with $a_0=\phi(0)=1$. For every $j$ $(0\leq j<n)$, let $W(n,j)$ be equal to the following sum,
       \begin{equation}\label{eq1}
       W(n,j)=\sum_{\sigma(n,n-j)}\frac{a_1^{k_1}a_2^{k_2}\ldots a_n^{k_n}}{k_1!k_2!\ldots k_n!}.
       \end{equation}
       Using the notation $\left[t^n\right]\left(\phi(t)\right)^{\lambda}$ for the coefficient of $t^n$ in the power series expansion of $\left(\phi(t)\right)^{\lambda}$,
       \begin{equation}\label{eq2}
       \frac{1}{(\lambda)_n}\left[t^n\right]\left(\phi(t)\right)^{\lambda}=\sum_{j=0}^s\frac{W(n,j)}{(\lambda-n+j)_j}\;\;+o\left(\frac{W(n,s)}{(\lambda-n+s)_s}\right).
       \end{equation}
       To derive the said asymptotic formula, we need to consider first the generating function in Lemma~\ref{lemma:11}. We use $\phi(t)$ to denote this generating function. That is,
       \begin{equation}\label{eq3}
       \phi(t) = \sum_{n=0}^{\infty} T_{n}^{1,x} (\alpha,\beta,\gamma) \frac{t^n}{n!} = \frac{(1+\alpha t)^{\frac{\gamma}{\alpha}}}{\left(1-x\left((1+\alpha t)^{\frac{\beta}{\alpha}}-1\right)\right)}.
       \end{equation}
       Then,
       \begin{equation*}
       (\phi(t))^{\lambda} = \frac{(1+\alpha t)^{\frac{\lambda\gamma}{\alpha}}}{\left(1-x\left((1+\alpha t)^{\frac{\beta}{\alpha}}-1\right)\right)^{\lambda}} = \sum_{n=0}^{\infty} T_{n}^{\lambda,x} (\alpha,\beta,\gamma) \frac{t^n}{n!}.
       \end{equation*}
       By making use of \eqref{eq2}, we have
       \begin{equation*}
       \frac{T_{n}^{\lambda,x} (\alpha,\beta,\gamma)}{(\lambda)_nn!} = \sum_{j=0}^{s} \frac{W(n,j)}{(\lambda-n+j)_j} + o\left(\frac{W(n,s)}{(\lambda-n+s)_s}\right),
       \end{equation*}
       where $n=o\left(\lambda^{1/2}\right)(\lambda\to\infty)$ and the numbers $W(n,j)$ are given in \eqref{eq1} with $a_j$ being determined by \eqref{eq3}, namely,
       \begin{align*}
       a_j &= [t^j]\phi(t) = \frac{T_{j}^{1,x} (\alpha, \beta, \gamma)}{j!} \\
           &= [t^j]\frac{(1+\alpha t)^{\frac{\gamma}{\alpha}}}{\left(1-x\left((1+\alpha t)^{\frac{\beta}{\alpha}}-1\right)\right)}.
       \end{align*}
       Note that
       \begin{align*}
       \sum_{n=0}^{\infty} T_{n}^{1,x} (\alpha,\beta,\gamma) \frac{t^n}{n!}&=
       \frac{(1+\alpha t)^{\frac{\gamma}{\alpha}}}{\left(1-x\left((1+\alpha t)^{\frac{\beta}{\alpha}}-1\right)\right)} \\
       		&= (1+\alpha t)^{\frac{\gamma}{\alpha}} \left(1-x\left((1+\alpha t)^{\frac{\beta}{\alpha}}-1\right)\right)^{-1} \\
       		&= (1+\alpha t)^{\frac{\gamma}{\alpha}} \sum_{k=0}^{\infty} x^k \left[(1+\alpha t)^{\frac{\beta}{\alpha}}-1\right]^k \\
       		&= (1+\alpha t)^{\frac{\gamma}{\alpha}} \sum_{k=0}^{\infty} x^k \left[\left((1+\alpha t)^{\frac{\beta}{\alpha}}-1\right)^k\right] \\
       		&= (1+\alpha t)^{\frac{\gamma}{\alpha}} \sum_{k=0}^{\infty} x^k \sum_{j=0}^{k} \paren{k}{j} (1+\alpha t)^{(\frac{\beta}{\alpha})j} (-1)^{k-j} \\
       		&= \sum_{k=0}^{\infty} x^k \sum_{j=0}^{k} \paren{k}{j} (1+\alpha t)^{(\frac{\beta}{\alpha})j + \frac{\gamma}{\alpha}} (-1)^{k-j} \\
       		&= \sum_{k=0}^{\infty} \sum_{j=0}^{k} x^k \paren{k}{j} (-1)^{k-j} (1+\alpha t)^{(\frac{\beta}{\alpha})j + \frac{\gamma}{\alpha}} \\
       		&= \sum_{k=0}^{\infty} \sum_{j=0}^{k} x^k \paren{k}{j} (-1)^{k-j} \sum_{i=0}^{\infty} \paren{\frac{\beta j+\gamma}{\alpha}}{i} (\alpha t)^i \\
       		\end{align*}

       So, \begin{align*}
       \sum_{n=0}^{\infty} T_{n}^{1,x} (\alpha,\beta,\gamma) \frac{t^n}{n!}&= \sum_{i=0}^{\infty}\sum_{k=0}^{\infty}\sum_{j=0}^{k} x^k \paren{k}{j} (-1)^{k-j} \paren{\frac{\beta j+\gamma}{\alpha}}{i} (\alpha t)^i \\
              		&= \sum_{i=0}^{\infty}\sum_{k=0}^{\infty}\sum_{j=0}^{k} x^k \paren{k}{j} (-1)^{k-j} \paren{\frac{\beta j+\gamma}{\alpha}}{i} \alpha^i t^i.
       \end{align*}
       Comparing coefficients,
       \begin{equation*}
       \frac{T_{i}^{1,x} (\alpha,\beta,\gamma)}{i!} = \sum_{k=0}^{\infty}\sum_{j=0}^{k} x^k \paren{k}{j} (-1)^{k-j} \paren{\frac{\beta j+\gamma}{\alpha}}{i} \alpha^i
       \end{equation*}
       or
       \begin{align*}
       T_{n}^{1,x} (\alpha,\beta,\gamma) &= \sum_{k=0}^{\infty}\sum_{j=0}^{k} n! x^k \paren{k}{j} (-1)^{k-j} \paren{\frac{\beta j + \gamma}{\alpha}}{n} \alpha^n \\
       &= n!\alpha^n \sum_{k=0}^{\infty}\sum_{j=0}^{k} x^k \paren{k}{j} (-1)^{k-j} \paren{\frac{\beta j + \gamma}{\alpha}}{n} \\
       &= n!\alpha^n \sum_{k=0}^{\infty} \left\lbrace\sum_{j=0}^{k} \paren{k}{j} (-1)^{k-j} \paren{\frac{j\beta + \gamma}{\alpha}}{n}\right\rbrace x^k \\
       &= \frac{n!}{n!}\alpha^n \sum_{k=0}^{\infty} \left\lbrace \sum_{j=0}^{k} \paren{k}{j} (-1)^{k-j} \left(\frac{j\beta + \gamma}{\alpha}\right)_n \right\rbrace x^k \\
       &= \alpha^n \sum_{k=0}^{\infty} \left\lbrace \sum_{j=0}^{k} \paren{k}{j} (-1)^{k-j} \frac{1}{\alpha^n} (j\beta + \gamma | \alpha)_n \right\rbrace x^k.
       \end{align*}
       Using the explicit formula for the unified generalization of Stirling numbers \cite{Corcino Rorberto 2001}, we have
       \begin{equation*}
       T_{n}^{1,x} (\alpha,\beta,\gamma)= \sum_{k=0}^{\infty} \beta^k k! S(n,k;\alpha,\beta,\gamma)x^k.
       \end{equation*}
       By Lemma~\ref{lemma:10}, we obtain
       \begin{equation*}
       T_{n}^{1,x} (\alpha,\beta,\gamma) =  G_n^x(\alpha, \beta, \gamma).
       \end{equation*}
       When $x=1$, these polynomials yield the numbers in \cite{On generalised Bell polynomials}, denoted by $B(n;\alpha, \beta, \gamma)$. That is,
       \begin{equation*}
       T_{n}^{1,1} (\alpha,\beta,\gamma) =  B_n(\alpha, \beta, \gamma).
       \end{equation*}
       The following theorem formally states the above asymptotic formula.

       \begin{theorem}
       There holds the asymptotic formula
       \begin{equation*}
       \frac{T_{n}^{\lambda,x} (\alpha,\beta,\gamma)}{(\lambda)_nn!} = \sum_{j=0}^{s} \frac{W(n,j)}{(\lambda-n+j)_j} + o\left(\frac{W(n,s)}{(\lambda-n+s)_s}\right),
       \end{equation*}
       for $\lambda\to\infty$ with $n=o\left(\lambda^{1/2}\right)$, where the numbers $W(n,j)$ are defined in \eqref{eq1} with $a_j$ being given by
       $$a_j=\frac{1}{j!}G_j^x(\alpha, \beta, \gamma).$$
       \end{theorem}

       \smallskip
       Assume $\gamma=0$ and take $s=2$. Notice that the computation of $W(n,j)$ is based on the number of partitions $\sigma(n,n-j)$ of $n$ with $k=n-j$ parts (cf. Hsu and Shiue \cite{A unified approach to generalized Stirling numbers}).  So, when $j=0$, we need to compute $\sigma(n,n)$, the number of partion of $n$ with $n$ parts. That is, finding $k_i$'s satisfying
       $$k=n = k_1 + k_2 + \cdots + k_n=k_1\ \mbox{and}\ n=k_1 + 2k_2 + \cdots + nk_n = k_1.$$
       Hence, we have
       \begin{equation*}
       W(n,0) = \sum_{\sigma(n,n)} \frac{a_{1}^{k_1}a_{2}^{k_2} \cdots a_{n}^{k_n}}{k_{1}!k_{2}! \cdots k_{n}!} = \frac{1}{n!} a_{1}^n.
       \end{equation*}
       For $j=1$, we need to compute $\sigma(n,n-1)$. That is, finding $k_i$'s satisfying
       \begin{align*}
       k&=n-1 = k_1 + k_2 + \cdots + k_n=k_1+k_2=(n-2)+1\\
       n& = k_1 + 2k_2 + ... + nk_n=(n-2)+2(1).
       \end{align*}
       That is, $k_1=n-2$ and $k_2=1$. Hence, we have
       \begin{equation*}
       W(n,1) = \sum_{\sigma(n,n-1)} \frac{a_{1}^{k_1}a_{2}^{k_2} \cdots a_{n}^{k_n}}{k_{1}!k_{2}! \cdots k_{n}!} = \frac{1}{(n-2)!} a_{1}^{n-2} a_2
       \end{equation*}
       Now, for $j=2$,
       \begin{align*}
       k&=n-2 = k_1 + k_2 + \cdots + k_n=k_1+k_2=(n-3)+1\\
       n&= k_1 + 2k_2 + 3k_3+\ldots+ nk_n=(n-3)+3(1).
       \end{align*}
       and
       \begin{align*}
       k&=n-2 = k_1 + k_2 + \cdots + k_n=k_1+k_2=(n-4)+2\\
       n&= k_1 + 2k_2 + 3k_3+\ldots + nk_n=(n-4)+2(1).
       \end{align*}
       Hence, the sum is composed of terms: one term is using the index
       $$(k_1, k_2, k_3, k_4, \cdots , k_n)=(n-3, 0, 1, 0, \cdots , 0)$$
       and the other term is using the index
       $$(k_1, k_2, k_3, k_4, \cdots , k_n)=(n-4, 2, 0, 0, \cdots , 0).$$
       Hence, we have
       \begin{align*}
       W(n,2) &= \sum_{\sigma(n,n-2)} \frac{a_{1}^{k_1}a_{2}^{k_2} \cdots a_{n}^{k_n}}{k_{1}!k_{2}! \cdots k_{n}!}\\
       &= \frac{1}{(n-3)!} a_{1}^{n-3} a_3 + \frac{1}{2!(n-4)!} a_{1}^{n-4} a_{2}^{2} .
       \end{align*}
       Now, we can compute the approximate value of $T_{n}^{\lambda,x} (\alpha,\beta,0)$ as follows:
       \begin{equation*}
       \frac{T_{n}^{\lambda,x} (\alpha,\beta,0)}{n!} \sim (\lambda)_n W(n,0) + (\lambda)_{(n-1)} W(n,1) + (\lambda)_{(n-2)} W(n,2)
       \end{equation*}
       where $n = o(\lambda^{1/2})$ as $\lambda \rightarrow \infty$. With
       \begin{align*}
       W(n,0) &= \frac{1}{n!}a_{1}^{n} = \frac{1}{n!} \left( \sum_{k=0}^1 \beta^k k! S(1,k;\alpha,\beta,0)x^k \right)^n \\
       &= \frac{1}{n!} \left\lbrace S(1,0;\alpha,\beta,0) + \beta S(1,1;\alpha,\beta,0)x \right\rbrace^n \\
       &= \frac{1}{n!} \lbrace 0 + \beta x \rbrace^n = \frac{\beta^n x^n}{n!} \\
       W(n,1) &= \frac{1}{(n-2)!} a_{1}^{n-2} a_2,
       \end{align*}
       where $a_1 = \beta x$ and
       \begin{align*}
       a_2 &= T_{2}^{1,x} (\alpha,\beta,0) = \sum_{k=0}^2 \beta^k k! S(2,k;\alpha,\beta,0)x^k \\
       		&= 0 + \beta S(2,1;\alpha,\beta,0)x + \beta^2(2!) S(2,2;\alpha,\beta,0)x^2 \\
       		&= \beta S(2,1;\alpha,\beta,0)x + 2\beta^2(1)x^2 \\
       		&= \beta(\beta - \alpha)x + 2\beta^2 x^2
       \end{align*}
       \begin{equation*}
       W(n,1) = \frac{1}{(n-2)!} \lbrace (\beta x)^{n-2} \left( \beta(\beta - \alpha)x + 2\beta^2 x^2 \right) \rbrace
       \end{equation*}
       \begin{equation*}
       W(n,2) = \frac{1}{(n-3)!} a_{1}^{n-3} a_3 + \frac{1}{2!(n-4)!} a_{1}^{n-4} a_{2}^{2},
       \end{equation*}
       where $a_1 = \beta x$, $a_2 = \beta(\beta - \alpha)x + 2\beta^2 x^2$, and
       \begin{align*}
       a_3 &= T_{3}^{1,x} (\alpha,\beta,0) \\
       		&= \sum_{k=0}^{3} \beta^k k! S(3,k;\alpha,\beta,0)x^k \\
       		&= S(3,0;\alpha,\beta,0) + \beta S(3,1;\alpha,\beta,0)x\\
       		&\;\;\;+ 2\beta^2 S(3,2;\alpha,\beta,0)x^2 +3!\beta^3 S(3,3;\alpha,\beta,0)x^3,
       \end{align*}
       where
       \begin{align*}
       S(3,0;\alpha,\beta,0) &= 0 \\
       S(3,1;\alpha,\beta,0) &= S(2,0;\alpha,\beta,0) + (\beta - 2\alpha)S(2,1;\alpha,\beta,0)\\
       &= (\beta - 2\alpha)(\beta - \alpha) \\
       S(3,2;\alpha,\beta,0) &= S(2,1;\alpha,\beta,0) + (2\beta - 2\alpha)S(2,2;\alpha,\beta,0) \\
       &= (\beta - \alpha) + 2\beta - 2\alpha = 3\beta - 3\alpha \\
       S(3,3;\alpha,\beta,0) &= 1.
       \end{align*}
       Hence,
       \begin{equation*}
       a_3 = \beta(\beta -  \alpha)(\beta - 2\alpha)x + 2\beta^2(3\beta - 3\alpha)x^2 + 6\beta^3 x^3
       \end{equation*}
       \begin{multline*}
       W(n,2) = \frac{1}{(n-3)!} (\beta x)^{n-3} \left\lbrace (\beta | \alpha)_3 x + 6\beta^2(\beta - \alpha)x^2 + 6\beta^3 x^3 \right\rbrace \\
       					+ \frac{1}{2!(n-4)!} (\beta x)^{n-4} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace^2
       \end{multline*}

       Consequently,
       \begin{equation*}
       \begin{split}
       \frac{T_{n}^{\lambda,x} (\alpha,\beta,0)}{n!} &\sim (\lambda)_n \frac{\beta^n x^n}{n!} + (\lambda)_{n-1} \frac{1}{(n-2)!} (\beta x)^{n-2} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace \\
       &+ (\lambda)_{n-2} \frac{1}{(n-3)!} (\beta x)^{n-3} \left\lbrace (\beta | \alpha)_3 x + 6\beta^2(\beta - \alpha)x^2 + 6\beta^3 x^3 \right\rbrace \\
       &+ (\lambda)_{n-2} \frac{1}{2!(n-4)!} (\beta x)^{n-4} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace
       \end{split}																		\end{equation*}
       \begin{equation*}
       \begin{split}
       T_{n}^{\lambda,x} (\alpha,\beta,0) &\sim (\lambda)_n \beta^n x^n + (\lambda)_{n-1} n(n-1) (\beta x)^{n-2} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace \\
       &+ (\lambda)_{n-2} n(n-1)(n-2) (\beta x)^{n-3} \left\lbrace (\beta | \alpha)_3 x + 6\beta^2(\beta - \alpha)x^2 + 6\beta^3 x^3 \right\rbrace \\
       &+ (\lambda)_{n-2} \frac{n(n-1)(n-2)(n-3)}{2} (\beta x)^{n-4} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace
       \end{split}
       \end{equation*}
       \begin{equation*}
       \begin{split}
       T_{n}^{\lambda,x} (\alpha,\beta,0) &\sim (\lambda)_n \beta^n x^n + (\lambda)_{n-1} (n)_2 (\beta x)^{n-2} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace \\
       &+ (\lambda)_{n-2} (n)_3 (\beta x)^{n-3} \left\lbrace (\beta | \alpha)_3 x + 6\beta^2(\beta - \alpha)x^2 + 6\beta^3 x^3 \right\rbrace \\
       &+ (\lambda)_{n-2} \frac{(n)_4}{2} (\beta x)^{n-4} \left\lbrace \beta(\beta - \alpha)x + 2\beta^2 x^2 \right\rbrace
       \end{split}
       \end{equation*}
       where $n=o(\lambda^{1/2})$, $\lambda \rightarrow \infty$. Now, we look at the general case of $T_{n}^{\lambda,x} (\alpha,\beta,\gamma)$. Obtain few terms of the asymptotic expansion.
       \begin{equation*}
       \frac{T_{n}^{\lambda,x} (\alpha,\beta,\gamma)}{(\lambda)_n (n!)} = \frac{W(n,0)}{(\lambda - n)_0} + \frac{W(n,1)}{(\lambda - n + 1)_1}
       + \frac{W(n,2)}{(\lambda - n + 2)_2} + o\left( \frac{W(n,2)}{(\lambda - n + 2)_2} \right)
       \end{equation*}
       \begin{align*}
       a_1 &= \sum_{k=0}^{1} \beta^k k! S(1,k;\alpha,\beta,\gamma)x^k \\
       		&= S(1,0;\alpha,\beta,\gamma) + \beta S(1,1;\alpha,\beta,\gamma)x \\
       		&= \gamma + \beta x
       \end{align*}
       \begin{align*}
       a_2 &= \sum_{k=0}^{2} \beta^k k! S(2,k;\alpha,\beta,\gamma)x^k \\
       		&= S(2,0;\alpha,\beta,\gamma) + \beta S(2,1;\alpha,\beta,\gamma)x + \beta^2 2! S(2,2;\alpha,\beta,\gamma)x^2 \\
       		&= (\gamma | \alpha)_2 + \beta S(2,1;\alpha,\beta,\gamma)x + \beta^2 2! x^2 \\
       		&= (\gamma | \alpha)_2 + \beta(\beta - \alpha + 2\alpha)x + 2\beta^2 x^2\\
       a_3 &= \sum_{k=0}^{3} \beta^k k! S(3,k;\alpha,\beta,\gamma)x^k \\
       		&= S(3,0;\alpha,\beta,\gamma) + \beta S(3,1;\alpha,\beta,\gamma)x + 2!\beta^2 S(3,2;\alpha,\beta,\gamma)x^2\\
       		& + 3!\beta^3 S(3,0;\alpha,\beta,\gamma)x^3 \\
       		&= (\gamma|\alpha)_3 + \left[(\gamma|\alpha)_2 + (\beta - 2\alpha + \gamma)(\beta - \alpha + 2\gamma)\right]\beta x\\
       		&+ 6(\beta - \alpha + \gamma)\beta^2 x^2 + 6\beta^3 x^3
       \end{align*}
       \begin{align*}
       W(n,0) &= \frac{(\gamma + \beta x)^n}{n!} \\
       W(n,1) &= \frac{1}{(n-2)!} (\gamma + \beta x)^{n-2} \left\lbrace (\gamma|\alpha)_2 + \beta(\beta - \alpha + 2\gamma)x + 2\beta^2 x^2 \right\rbrace \\
       W(n,2) &= \frac{1}{(n-3)!} (\gamma + \beta x)^{n-3} \left\lbrace (\gamma|\alpha)_3 + \left[ (\gamma|\alpha)_2 + (\beta - 2\alpha + \gamma)(\beta - \alpha + 2\gamma) \right]\beta x \right\rbrace \\
       &+ \frac{1}{(n-3)!} (\gamma + \beta x)^{n-3} \left[ 6(\beta - \alpha + \gamma)\beta^2 x^2 + 6\beta^3 x^3 \right] \\
       &+ \frac{1}{2(n-4)!} (\gamma + \beta x)^{n-4} \left\lbrace (\gamma|\alpha)_2 + \beta(\beta - \alpha + 2\gamma)x + 2\beta^2 x^2  \right\rbrace^2
       \end{align*}
       Thus,
       \begin{equation*}
       \begin{split}
       &T_{n}^{\lambda,x} (\alpha,\beta,\gamma) \sim n! \left[(\lambda)_n W(n,0) + (\lambda)_{n-1} W(n,1) + (\lambda)_{n-2} W(n,2) \right] \\
       													&= (\lambda)_n (\gamma + \beta x)^n + (\lambda)_{n-1} (n)_2 (\gamma + \beta x)^{n-2} \left\lbrace (\gamma|\alpha)_2 + \beta(\beta - \alpha + 2\gamma)x + 2\beta^2 x^2 \right\rbrace \\
       													&+ (\lambda)_{n-2} (n)_3 (\gamma + \beta x)^{n-3} \left\lbrace (\gamma|\alpha)_3 + \left[ (\gamma|\alpha)_2 + (\beta - 2\alpha + \gamma)(\beta - \alpha + 2\gamma) \right]\beta x \right\rbrace\\
       													&+ (\lambda)_{n-2} (n)_3 (\gamma + \beta x)^{n-3} \left\lbrace 6(\beta - \alpha + \gamma)\beta^2 x^2 + 6\beta^3 x^3\right\rbrace \\
       													&+ (\lambda)_{n-2} (n)_4 (\gamma + \beta x)^{n-4} \left\lbrace (\gamma|\alpha)_2 + \beta(\beta - \alpha + 2\gamma)x + 2\beta^2 x^2  \right\rbrace^2.
       \end{split}
       \end{equation*}

      \end{section}

      \iffalse
\subsection{Generalised Barred Preferential Arrangements \RNum{2}}

 $$ \sum\limits_{n=0}^{\infty} T^{\lambda}_n(1,\alpha,\beta,\gamma)\frac{t^n}{n!}=(1+\alpha t)^{\gamma/\alpha}\begin{pmatrix}
         \frac{1}{1-((1+\alpha t)^{\beta/\alpha}-1)}
         \end{pmatrix}^{\lambda}$$

         For natural number $\lambda$.

         \subsection{Study Generalised BPA's where compartments split into $r$ compartments as done in \cite{Paper with interpretation of G-stirling numbers}. }
  \subsection{Special generalised barred preferential arrangements}

      \vspace*{10mm}
     Study $w^{\lambda}_n(1,\alpha,\beta,\gamma)$ where all the sections are all non-empty, as Pippenger et al has done in \cite{barred:2013}. \fi

redpajama_set_name: RedPajamaArXiv


Green, OH –The City of Green's annual community event, FreedomFest is Friday, June 24 from 6 p.m. to 9 p.m. and Saturday, June 25 from 12 p.m. to 11 p.m. at Boettler Park, 5300 Massillon Road. Fireworks are scheduled for approximately 10:15 p.m. on Saturday, June 25. Admission is FREE. The City is collecting donations of school supplies to benefit Green Good Neighbors and its back to school drive.
The festival begins Friday evening at 6 p.m. with food vendors and live music performed by Cleghorn, an America's Celtic rock band at 6:30 p.m. along with the MacConmara Irish Dancers.
On Saturday, family-fun activities begin at 12 p.m. and will close at 8 p.m. Activities include bounce houses and other inflatable activities, zip line, climbing wall, laser maze, farm yard playland, NASCAR simulators and more. In addition, local businesses will be exhibiting their goods and services, and food will be available for purchase throughout the day.
At 5 p.m., the City is pleased to host MAPS Air Museum's annual MAPS Legacy Awards, which honor museum volunteers who have served distinguished military careers. MAPS will also be displaying several vehicles at FreedomFest on Saturday.
On Saturday, live musical performances will begin at 6:30 p.m. with Cleghorn, along with the MacConmara Irish Dancers, followed by One, a U2 tribute band, at 8:30 p.m. Prior to the 8:15 p.m. performance, skydivers* will descend on the park – a crowd favorite. Fireworks* will begin at 10:15 p.m.
Free parking for FreedomFest will be available at Boettler Park and at various parking lots along International Parkway. Handicap parking will be available at Boettler Park. Local businesses and not-for-profits interested in exhibiting at FreedomFest may contact the Parks and Recreation office at 330-896-6621 or email kgoodhart@cityofgreen.org. Fee to exhibit is $125.
*In the event of inclement weather, the city reserves the right to launch fireworks prior to the 10:15 scheduled start. If weather determines the fireworks show should be canceled, the fireworks show will be held Sunday, June 26. Please follow the City of Green on Facebook (www.facebook.com/cityofgreenohio) or Twitter (@cityofgreen) or sign up for text message alerts by texting GREENOH to 888-777 for day of event cancellation or rescheduling information in the event of rain or severe weather.


text:
Formlabs Expands Large Format 3D Printing Line, Making Intuitive, Cost-Effective Large-Scale Printing a Reality
Formlabs launches Form 3BL, for biocompatible printing, Wash L + Cure L, and ships Form 3L to enable engineers, designers, dental labs, and more to create large prints in-house, slashing lead times and costs.
Europe, Sep. 15, 2020 – Formlabs, a leading 3D printing company, today announces the Form 3BL, a dental and medical 3D printer optimised for biocompatible materials. Today, Formlabs will also start shipping its flagship large format printer, the Form 3L. Both printers signify the expansion of the company's line of large format stereolithography (SLA) 3D printers to enable manufacturers and creators to print high-detail large-scale parts.
Formlabs also announces Wash L + Cure L, a fully automated post-processing system created for large-scale prints, which will ship in 2021. Already the best-selling professional desktop printer company in the world with over 70,000 printers sold, Formlabs is now bringing its renowned ease-of-use, value, and professional-grade quality to large format printers.
Max Lobovsky, CEO and Co-Founder of Formlabs, said: "Our large format printers are poised to have the same disruptive effect on the 3D printing industry as our Form 1 launch did. The companies that have dominated large format printing for the past 20+ years have lacked the innovation and agility needed to move the 3D printing industry forward. At Formlabs, we are changing that. With the Form 3L and Form 3BL, we are opening this powerful technology up to a whole new group of users at an accessible price while maintaining a professional quality. We are looking forward to seeing how this new technology will be used in industries like automotive and aerospace that need the large build platform to best utilise 3D printing."
The Form 3L, Formlabs' flagship large format printer announced in spring 2019, delivers accurate, repeatable prints across the printers' larger build platform without compromising cost or quality. The advanced Low Force Stereolithography (LFS) print process and larger build platform enabled Formlabs user Black Diamond, an outdoor equipment company known for its products' durability and performance, to rethink its R+D process. Given early access to Form 3L, Black Diamond determined it would be able to cut prototype development lead times from seven to three days and slashed costs by about 85%.
"The Form 3L makes the product design process more cohesive," said Matt Tetzl, R+D technician at Black Diamond. "The larger printer enables our designers to make tweaks in CAD and work off a full-scale model to close the gap between feedback and design. We have reduced our costs on full-size prototypes from $425 a print to $70 a print - with that savings, Form 3L would pay for itself in only 3 months."
The Form 3L and Form 3BL's features and benefits include:
3D print large-scale parts or many smaller parts in a single build, with stunning surface finish and detail that resemble final products
A build volume of 33.5 × 20 × 30 cm, five times larger build volume than the Formlabs Form 3 desktop SLA 3D printer
An intuitive workflow, with automatic resin dispensing, simple print preparation software, and an interface anyone can use
A robust end-to-end system designed for nonstop throughput with minimal intervention
Access to nearly all of Formlabs' extensive materials library, ensuring a stable supply chain and access to the resins needed for any project — including biocompatible materials on the Form 3BL
The Form 3BL, similar to the Form 3B, is optimised for biocompatible materials for use in the dental and healthcare industries. The Form 3BL balances throughput, quality, and reliability, and is thoroughly validated by Formlabs' dental team to meet the accuracy requirements for dental aligner production, with consistent part quality and accuracy across the entire build plane.
In medical settings, the Form 3BL is capable of printing human-scale anatomical models, in-house, on a single build platform. With the Form 3BL, healthcare providers can meet the tight deliverable windows needed to produce the customised surgical guides that are used to improve care and reduce the time spent in the operating room.
The Form 3L is available and all pre-ordered 3L printers are now being shipped, the Form 3BL is available for pre-order, and Wash L + Cure L will ship in 2021.
Click here for media assets.
Formlabs is expanding access to digital fabrication, so anyone can make anything. Headquartered in Somerville, Massachusetts with offices in Germany, Japan, China, Singapore, Hungary, and North Carolina, Formlabs is the professional 3D printer of choice for engineers, designers, manufacturers, and decision makers around the globe. Formlabs products include the Form 3, Form 3B, Form 3L, and Form 3BL powered by an advanced form of stereolithography (SLA) called Low Force Stereolithography (LFS)™ 3D printing, Form Wash and Form Cure post-processing solutions, Fuse 1 SLS 3D printer, and Form Cell manufacturing solution. Through its Factory Solutions offering for industrial users, Formlabs provides the factories of tomorrow with the flexibility and versatility needed for demanding, evolving industrial applications. Formlabs also develops its own suite of high-performance materials that continue to push the boundaries for 3D printing, as well as best-in-class 3D printing software.
Media Contact (UK)
+49 1573 5993322 (EU)
+34 955 00 33 40 (ES)
+44 330 027 0040 (IE)

redpajama_set_name: RedPajamaCommonCrawl


text:
module Oden.Output.Unification where

import           Text.PrettyPrint.Leijen

import           Oden.Infer.Unification
import           Oden.Output
import           Oden.Pretty             ()

instance OdenOutput UnificationError where
  outputType _ = Error

  name UnificationFail{}         = "Infer.UnificationFail"
  name RowFieldUnificationFail{} = "Infer.UnificationFail"
  name InfiniteType{}            = "Infer.InfiniteType"
  name UnificationMismatch{}     = "Infer.UnificationMismatch"

  header (UnificationFail _ t1 t2) s = text "Cannot unify types"
    <+> code s (pretty t1) <+> text "and" <+> code s (pretty t2)
  header (RowFieldUnificationFail _ (l1, _) (l2, _)) s = text "Cannot unify fields"
    <+> code s (pretty l1) <+> text "and" <+> code s (pretty l2)
  header InfiniteType{} _ = text "Cannot construct an infinite type"
  header UnificationMismatch{} _ = text "Types do not match"

  details UnificationFail{} _ = empty
  details (RowFieldUnificationFail _ (l1, t1) (l2, t2)) s =
    text "The field" <+> code s (pretty l1) <+> text "has type" <+> code s (pretty t1)
    <+> text "and the field" <+> code s (pretty l2) <+> text "has type" <+> code s (pretty t2)
  details (InfiniteType _ v t) s = code s (pretty v) <+> equals <+> code s (pretty t)
  details (UnificationMismatch _ left right) s = formatTypes left right
    where formatTypes (t1:ts1) (t2:ts2)
            | t1 == t2  = vcat [
                code s (pretty t1) <+> text "==" <+> code s (pretty t2),
                formatTypes ts1 ts2
              ]
            | otherwise = vcat [
                code s (pretty t1) <+> text "!=" <+> code s (pretty t2),
                formatTypes ts1 ts2
              ]
          formatTypes (t1:ts1) [] = vcat [
              code s (pretty t1) <+> text "!= <none>",
              formatTypes ts1 []
            ]
          formatTypes [] (t2:ts2) = vcat [
              text "<none> !=" <+> code s (pretty t2),
              formatTypes [] ts2
            ]
          formatTypes [] [] = empty

  sourceInfo (UnificationFail si _ _)            = Just si
  sourceInfo (RowFieldUnificationFail si _ _)    = Just si
  sourceInfo (InfiniteType si _ _)               = Just si
  sourceInfo (UnificationMismatch si _ _)        = Just si

redpajama_set_name: RedPajamaGithub


Vitamin C is often regarded as a wrinkle fighter or an anti-aging agent. The main objective of 'Vitamin C skin care', in scientific terms, is to increase the synthesis of collagen (a structural protein that is found in skin). The additional benefit is related to its capability of countering free radicals which cause damage to the skin.
Vitamin C skin care, however, faces a major challenge today. This is related to the oxidation tendency of skin care products. On coming in contact with any oxidising agent (e.g. air), the Vitamin C in the skin care products, gets oxidised; thus making the Vitamin C skin care product useless (in fact counter-effective). The oxidised Vitamin C imparts a yellowish-brown colour to the skin care product. This is something that you need to check before buying a Vitamin C skin care product. Even after you buy a skin care product, you need to store it properly and keep checking that it's still good to use (i.e. it hasn't attained a yellowish-brown texture).
The manufacturers of Vitamin C skin care products have tried to deal with this (oxidation) problem in multiple ways (and research on Vitamin C skin care products is on the top of their list). One such method of retaining effectiveness of skin care products for a long term is to keep a high concentration (say 10%) of Vitamin C. However, this makes the skin care products even more expensive. The Vitamin C skin care products are already pretty inexpensive and making them even more expensive will throw the product manufacturers out of business.
The other way is to use Vitamin C derivatives (like ascorbyl palmitate and magnesium ascorbyl phosphate). These are not only more stable but also inexpensive. Even though the derivatives based products are not as effective as the Vitamin C skin care products, their stability against oxidation is a very desirable feature that makes them very attractive. Moreover, these are known to be less irritating too.
Talking of effectiveness of Vitamin C skin care products, it's important to mention that not everyone responds to Vitamin C treatments. So it's not a magic potion in any way. If you don't see a noticeable difference in your skin, it might be because of your skin not responding to the treatment (and the skin care products might not be at fault, at all).
As more research goes on, one can only keep our fingers crossed and wait for a complete solution to the challenges faced by 'Vitamin C skin care'today.
According to a recent article by BeautyWorldNews.com, sleep is not only important, it is also most effective if you pair it with a skin care regime to focus on replenishing, repairing and renewing your skin.
To Repair your skin while you sleep, the article author Dr. Vish Banthia, (founder of ZendyBeauty) recommends you boost collagen. If you didn't know, collagen is actually a protein that makes up the connective tissues of the human body. As we age there is a loss of collagen, which can lead to wrinkles and sagging skin. Dr. Banthia suggests to "apply vitamin C to help promote build-up of collagen." This powerful antioxidant has been shown to minimize fine lines and, in some cases, even out pigmentation.
The article then went on to recommend The Aloe Source Vitamin C Antioxidant Crème ($40). This luxuriously rich crème hydrates parched skin, while a stabilized vitamin C complex and Aloe vera provide antioxidant benefits for a glowing, youthful-looking complexion.
To take the full advantage of all the regenerative properties that vitamin c and Aloe vera offer, we recommend you use it as part of our Anti Oxidant Kit. The kit allows you to really repair fine lines and wrinkles while evening out skin tone. Plus, you'll love the fresh scent. For best results, apply twice daily, after cleansing and toning by massaging 1-2 pumps of the Vitamin C Antioxidant Complex. Allow the complex to dry completely. Then, follow with the Vitamin C Antioxidant Crème.


text:
Le Sénégali de Salvadori (Cryptospiza salvadorii) est une espèce de passereau appartenant à la famille des Estrildidae.

Liens externes

Estrildidae

redpajama_set_name: RedPajamaWikipedia


Mold Exterminators is a remediation services company with over 25 years of industry experience.Founded in 2007,we are committed to providing our Residental and Commerical clients with expert service at a value to restore their properties,businesses,and homes.Our services include mold remediation,water extraction,dehumidification,fire and smoke cleanup and reconstruction.Headquartered in Atlanta,Georgia our experienced staff and emergency equipment are avaliable 24 hours a day,7 days a week,and 365 days a year to respond to all of your restoration services.
We here at Mold Exterminators Atlanta pledge to treat you, our customer, with respect and individual attention.
At Mold Exterminators Atlanta, our mission is simple: Professionalism,Quality,and care while restoring your valued property.


So today I thought I'd let you have a nosey around my bedroom. I personally love watching these videos a) because I'm nosey and b) they give me ideas and inspiration on storage and decor etc. I did my first ever room tour video back in 2012 which I've also linked down below in case you're interested. I've since moved but you may notice things from my first tour in the second video.
Do you love your bedroom?


text:
Home Comment Press Releases 202111
RE: 'NaSIA Struggling Due To Square Peg In Round Hole'
/ Press Releases, Comment
As an education policy analyst, I keenly follow the ongoing policy reforms at the Ministry of Education, especially those relating to curriculum implementation, teacher education and professional learning and school standards and regulations.
Just last week, I read an article which essentially asserted some grievances regarding the leadership and operations of the National Schools Inspectorate Authority (NaSIA). The said article contained a litany of accusations and concerns by an association referred to as the Private Education Coalition (PEC). It had been published (and for that matter shared) in a number of online platforms.
It is important to address these concerns in order to carry all stakeholders along the path of building an accountable education policy reform space. While the decision of an official response lies at NaSIA's discretion, I am duty-bound as a citizen to speak to the issue based on my accumulated knowledge and experience in the industry. It is instructive to say that the emergence of NaSIA, formerly the National Inspectorate Board (NIB) as a statutory body created by the Education Regulatory Bodies Act 2020 (Act 1023) has redefined the supervision and inspection architecture of public and private pre-tertiary schools in Ghana.
NaSIA has unambiguously espoused mandate, part of which is to set and enforce standards to be observed at the public and private pre-tertiary education level in Ghana. NaSIA has in recent times, in a Covid-19 era, evolved the use of innovative and data-driven measures despite the odds to implement its mandate. Before I proceed further, let me indicate government's commitment, since 2017, to completely overhaul the pre-tertiary education system with the view to improving learning outcomes in schools across all grade levels. Right from the National Teaching Council (NTC) which targets teacher professional development and quality through rigorous processes of teacher registration and licensing, to NaCCA which focuses on standardizing the school curriculum and then to NaSIA which aims at enforcing quality education standards and architecture, our educational system has been put on the right pedestal for improved learning and teaching standards.
Agitations
As often the case, the implementation of these reforms has been met with marginal strains of difficulties and more importantly confrontations from other parties. It is thus not surprising that the Private Education Coalition (PEC), has amplified their challenges with these reforms. However, it is appropriate and fair for such concerns to be evaluated through the lenses of objectivism in order to ascertain their weight of veracity. The main concern of the PEC group is the allegation of "exorbitant and unapproved fees under the pretext of licensing and registering [schools] in order to be able to operate". Having read the ACT that established NaSIA, I have been constrained to say that the allegation by PEC is far from the truth. The reasons behind my position are espoused as follows. The functions of NaSIA as captured in clause 89 (2) of the Act 1023 states that;
The Authority shall, in the performance of functions under subsection (1),
publish the school effectiveness standards and guidelines developed for pre-tertiary education institutions under paragraph (a) and periodically review the standards and guidelines;
prescribe disciplinary measures for a pre-tertiary education institution that fails to comply with the standards and guidelines issued by the Authority; publish reports and findings on the performance of pre-tertiary education institutions; temporarily or permanently in consultation with the Minister and the Regional Education Directorate or District Assembly concerned close down a pre-tertiary institution that fails to meet the standards set by the Authority; close down a school in emergency situations after consultations with the Minister and the Regional Education Directorate or District Assembly concerned; recommend that pre-tertiary institutions be managed through Public Private Partnerships; and
register private pre-tertiary institutions.
From the above, one may question which of the clauses may have been violated by NaSIA as alleged by PEC. If NaSIA's functions include the registration through licensing of private pre-tertiary institutions as well as reprimanding pre-tertiary institutions that fail to meet the requisite standards, then it is quite retrogressive and unfair for NaSIA to be castigated for carrying out its mandate. Licensing by regulatory bodies demand some form of financial commitment in order to be efficient in their delivery.
Thus, it appears fuzzy for anyone to claim that NaSIA is charging unapproved fees. In any case, NaSIA's board has representation from the Teacher Unions and the Ghana National Association of Private schools (GNAPS) and such decisions are sanctioned by the Board for implementation. Again, what charges as licensing fees would be reckoned appropriate for private schools? I believe the private schools and Teacher Unions that have a representation on the Board are in position to argue fairly on what best is reasonable. We did not include them for nothing when we crafted the Act governing NaSIA's operations.
Another concern of the group is the unfortunate remark of a "square peg in a round hole". I would ordinarily have ignored this claim because it appears to have been borne out of mischief and an attempt directed at smearing certain respected individuals at the institution. NaSIA's board is chaired by Professor Eugene K. Marfo Darteh of the University of Cape Coast, an accomplished academic with extensive background in driving change.
Membership of the Board which was recently appointed by President Akufo-Addo and inaugurated by the Hon Minister, Dr Yaw Adutwum in accordance with the expectation of Act 1023 also includes Dr. Christian Addai-Poku (Registrar, NTC), Dr. Damasus Tuurosong, President of the Ghana National Association of Private Schools, Mr. Anthony Boateng, Deputy Director-General of the Ghana Education Service(GES), Mr. Victor de Graft-Etsien, representative from the Ghana National Association of Teachers (GNAT).
There are also representatives from the West African Examinations Council (WAEC) and Technical, Vocational Education and Training Service (TVETS). Additionally, Dr. Haggar Hilda Ampadu, the Inspector-General of Schools who is also a member of the Board, is an accomplished individual with in-depth knowledge on project management, data management, monitoring and evaluation, policy and regulation, etc. Her expertise was put to good use in her previous employment in the U.S pharmaceutical sector as a Data Scientist for 13 years working with the likes of Pfizer and Jonhson & Johnson. She also worked as a Regulatory Policy Advisor to 34 countries in Africa in her seven (7) years of working with the World Health Organisation (WHO). This is just to validate her credentials. It cannot be said that these eminent personalities at NaSIA are square pegs in round holes.
In a nutshell, the points raised in the release by the Private Education Coalition (PEC) are littered with inaccuracies. Rather than being combative and accusatory, it is best they reconsider the merit of the reforms spearheaded by NaSIA and open up to progressive policies. NaSIA is a budding institution and would need the encouragement and support from its actors.
The sanitization of the school system is a load of work and it's important for all stakeholders to accommodate change and be constructive in their criticisms. NaSIA's success would ensure learners in pre-tertiary schools in the country meet learning outcomes in safe and serene learning environment and all progress-minded Ghanaians should welcome that.
Source: Prince Hamid Armah, PhD
Ade Coker: Response To False Publication
Obey Court Orders, You Have No Application Pending - Lawyer For Daniel Ofori Tells Ecobank
Stop The Stage Craft Antics And Undemocratic Intimidation Of Party Members
National Liberation Congress Appoints Mohammed Bashir Tijani As Interim National Communication Director
DRIFAN, A Digital Road Safety Platform Launched In Ghana
Impact Africa Elevator Pitch Competition Shortlisted Businesses Announced
President Akufo-Addo to host Presidential Business Summit 2021
Chamber Of Pharmacy Calls For Collaboration In Vaccine Search
Other Press Releases Stories
Health Minister Must Resign Or Be Sacked Immediately – London South Branch of NDC


text:
Rima Alsammarae
Freelance journalist // Available for work, just check out some of my articles here and get in touch if you like what you read // Based between Dubai, Beirut and Chicago // Ralsammarae3@gmail.com
A Bit of a Pickle
Turshi, takhlil, mouneh. There are many Arabic terms used to describe the region's different preserving techniques, a process that in the past has provided much needed sustenance during long, harsh winters and sweltering, humid summers. Nowadays, it's so common to supplement a meal with pickled vegetables in Middle Eastern cuisine that restaurants dress up empty tables with complementary plates of turshi. Whether made in-house or sourced from elsewhere, the pickled invitation is a commonly used one – often enticing even the most stubborn of customers.
Walid and Maysoun Nasserdin, the king and queen of organic mouneh in Mount Lebanon's Chouf District, have learned to rely on their crafty culinary techniques, and opened their restaurant Coara in 2013. While the couple's joint offers a vast menu of vegetarian goods, preserved vegetables from tomatoes to cabbage are also on sale. However, the bright pink pickled turnip – falafel's partner in crime – is more of a family-only production.
Maysoun notes, 'We like to pickle turnips, and we always include a chopped beetroot in the jar to make the vegetable turn pink. We don't do large quantities, but if there's a big order, we can do that. We mostly pickle for ourselves, while in Coara, we offer other forms of mouneh.'
Organic food production is a modern-day trend, and the Nasserdins are local pioneers. In line with their culinary ideology, Maysoun explains that the mixture in her batches of pickled turnips only consists of turnip, beetroot, salt and water – no vinegar, a common component of pickling recipes. 'The taste doesn't change at all,' she insists. 'And I think that it tastes even better this way.'
She adds, 'But we pickle everything, and everything is picklable. Some people even make vegetable cocktails. Our own specialty is that we only use salt water. It takes 10 to 15 days this way, and when you taste it, you don't even notice that there's no vinegar.'
However, for the average chef, pickling turnips is a more reliable process when using vinegar, and deciding on white or grape vinegar can be left up to preference. But salt remains the key to a perfect batch of pickled turnips. A fellow Lebanese pickler, Nada Saber, explains that the 'amount of salt used in the water mixture needs to be perfect so the turnip doesn't become soft. You have to prepare the ingredients in specific proportions, and each part needs to be exact.'
Pickled turnips are a favourite falafel topping across the Middle East, and the neon pink finger of turnip is a common staple found in most, if not all, falafel joints. While Saber loves the pickled vegetable ('turnips are electrifying'), her family, like the Nasserdins, tend to pickle only for themselves, or for neighbouring families.
Maysoun and Walid are quite strict about their organic, vegetarian diet, and ensure that everything made and served at Coara follows their healthy lifestyles. From baked falafel to organic dried fruits (like strawberries and peaches), the Nasserdins are unyielding in their pursuit of better and better vegetarian solutions.
'We don't spray the growing vegetables or use any chemicals,' says Maysoun. 'The vegetables come with certificates given by the organisations that monitor organic farmers. We began preserving foods in the early 2000s as a personal initiative between my husband and I. But then it grew from there. The workload is spread evenly between us – we don't have specific tasks that we take care of, but he leans more towards the garden.'
The Nasserdins source many of their vegetables and fruits from a patch of land that's located near their restaurant in Chouf. And while this humble garden doesn't meet all the demands of the restaurant, Maysoun is happy reaching out to other Lebanese farmers – as long as they follow the Nasserdins' dietary guidelines.
At the moment, the Nasserdins are on a mission to promote vegetarianism, and do so through the use of Coara's kitchen. By tackling common Lebanese foods like shawarma sandwiches with meat substitutes, they're looking to start a culinary revolution. 'People are moving towards vegetarianism, but they're afraid because they don't know what to eat. But when they start trying different vegetarian diets, it has a huge impact, and people really like it. They're open to new ideas.'
The older generation of picklers in Lebanon is concerned that the culinary tradition will soon be lost. According to Saber, younger generations are not pickling at all, and are instead falling into a 'buy it, don't raise it' mentality. To fight this, many picklers are looking to pass on the practice to their children, by teaching recipes and techniques that help them stay in touch with local culture, wherever in the world they may be.
Maysoun says, 'My kids all have scholarships and they went to study far away from us. They're studying and working in unrelated fields, but I think that once they're a bit older, they'll come back. That's what I hope for anyway. At the end of the day, it's their lives and they're free to live how they want, but I think they'll join us.'
This article was originally published in the July/August 2015 issue of Brownbook magazine. Photo credit: Roland Ragi.
Case Study: Danish Embassy, Riyadh
Mohamed Elshahed


text:
Music | NW Life | local | Arts | Entertainment | Sessions
Frankie Cosmos Live From Type Foundry
by Jerad Walker opbmusic Nov. 14, 2019 5 a.m. | Updated: Jan. 8, 2020 1:44 p.m. | Portland, Ore.
As a teenager growing up in New York, Greta Kline did what most young musicians do: She began writing songs, honing her craft by experimenting with recording techniques, lyrics and sounds. What started as a trickle of very short, lo-fi demos, quickly turned into a torrent of creativity as the budding songwriter recorded hundreds of songs in just a few short years.
Her productivity was notable, but what made Kline's evolution as a songwriter truly unique was the very public way in which it played out. Unlike most musicians who guard their early work, Kline put it all out there. Along the way, she became one of the first prominent bedroom pop artists to emerge on Bandcamp, the now much-loved digital marketplace for musicians.
Almost a decade later, Kline fronts the critically-acclaimed band Frankie Cosmos. The group is signed to storied indie-rock label Sub Pop and released their fourth full-length album earlier this year, an impressively polished record called "Close It Quietly" that's easily one of the year's best.
While Kline may no longer be recording exclusively in her bedroom, some things clearly haven't changed. You can still hear her music on Bandcamp. And like her earliest solo demos, Frankie Cosmos songs are short by modern standards, tethered to an almost obsessive love of efficiency. Although that might sound like a stifling philosophy for creative output, the result is surprisingly refreshing. The band makes a point to not recycle lyrics and choruses, no matter how insanely catchy they may be. The song arrangements are never boring because Frankie Cosmos makes absolutely every second count.
Frankie Cosmos (Greta Kline, Alex Bailey, Lauren Martin and Luke Pyenson) performed songs from "Close It Quietly" at Type Foundry Studio in Portland, Oregon, on Halloween. Kline also chatted with us about her love of short songs, coping with stage fright, and the struggle to overcome her wunderkind reputation as she enters adulthood.
Jerad Walker: You're an incredibly prolific songwriter. The first line in the song "41st," which you played for us, also directly references that. How many songs did you write during that album cycle?
Greta Kline: I think it really was 40.
Walker: Was that the very last song?
Kline: No. It was one of the more recent ones [but] I do think "41st" was one of the later [songs] written in that cycle …
I definitely, since [my previous full-length album] "Vessel," had written probably 40 songs and scrapped 20 of them.
Walker: Well, I'm really glad it made it onto this album. It's one of my favorite songs on the record. You pumped so much into that song, but it's only a brisk 2 minutes and 6 seconds long. By modern standards, you write really short songs — 45 seconds or a minute and a half isn't out of the ordinary on your albums. Why do you prefer shorter compositions?
Kline: I think, partially, it's just that I have a short attention span and that's how much of a song I want to listen to or play. But also, I guess I just am kind of averse to the normal sort of pop song structure of repeating choruses. I feel like unless there's a reason that I want to repeat a chorus, like if I think it would have a different meaning, I don't really see a point to it. I just like to keep them moving.
Walker: As a listener, it certainly makes me repeat your songs more often than other bands that I listen to. And you said it — you don't recycle lyrics. Choruses may only happen once in a song. Bridges are truly special moments where you seem to give a lot of attention to the writing. I find that approach really refreshing. In particular, there's one song on the album called "So Blue" where I just don't want it to stop. Basically, it's this close to becoming an anthem when the bottom just falls out. Do you ever find yourself fighting the urge to extend tracks?
Kline: Oh, you know it's funny [you say that] because actually "So Blue" was even shorter than that and we extended it when I brought it to the full band. Originally, it was just like the parts with the lyrics and there weren't any instrumental measures at all.
So, to me, "So Blue" was kind of extended. It was kind of stretched thin, I would say. I mean not really, because we added so much and Alex on the guitar made it just a totally different song. But I mean, lyrically, I like keeping it concise.
With "So Blue," especially, the lyrics only work because it's concise. I think it would get boring to kind of take the same structure and stretch it thinner. So yeah, sometimes it's hard to edit stuff down, but with that one, it's definitely the longest lyrically that it could have been. We could probably stretch out the instrumental [portion] longer if we wanted to, but I like them as just little nuggets, you know?
Close It Quietly by Frankie Cosmos
Walker: You and drummer Luke Pyenson seem to have a particularly strong bond in the music that you make in Frankie Cosmos. I think I first noticed it on the song "Jesse" from your 2018 record "Vessel." And that interplay is really striking throughout the new record. It stands out the most on "Windows" and "Wannago," which you also played for us. How collaborative is your writing and arranging on a song like "Wannago"?
Kline: I write the basic sort of bones of the songs for the guitar and the vocals. I usually have it to a point where I feel like it's done in terms of the format of it. And then I bring it in [to the band] and we'll tweak it a little bit.
With "Wannago" there was actually more tweaking than usual. Luke had the idea to cut a verse from it, and I was totally down because I felt like it was too long and that there was a verse that was unnecessary. Generally, that stuff doesn't really happen, but when it comes to arranging it's just really collaborative and everybody either writes their own parts or sometimes helps write each other's parts.
We spend a lot of time talking about what we want it to be in terms of the arrangement and I think that's when it starts to feel really collaborative — when we start adding in other instruments. With the drums, two voices, and the guitar ["Wannago"] feels like a totally different song.
Walker: In the past you've said that your favorite thing about music is arranging songs, but not necessarily playing them or touring. And you've even said that you get stage fright in the past. You come from a family of performers [Kline is the daughter of noted actors Phoebe Cates and Kevin Kline], so I was surprised to hear that. Is that still true for you? Do you still get nervous when you're on stage?
Kline: Yeah. I mean, it depends on the night and like where I'm at in my life. But I think as I've gotten older I've definitely gotten more relaxed about telling myself what I think a show means in the context of my life.
When I was 19, if I played a show and I thought I did a bad job I'd be feeling really worthless and crying. And now I'm just like, 'Well, yeah, it's just one show' and now it doesn't say anything about my songs or who I am if I play a bad show. So I feel like I feel a little bit more grounded when it comes to my reactions to performing.
But I definitely get really nervous and especially when we go on tour with artists that we really respect … So I think I've just gotten more reasonable as I get older in terms of using logic to sort of talk myself down from stage fright.
Walker: You mentioned maturing as a performer, and I guess, not necessarily getting over your stage fright, but learning to cope with it. From a songwriting standpoint, I feel like this record is a much more mature album as well. One theme that I've noticed throughout your work is a mention of place and age. The first Frankie Cosmos record "Zentropy" starts with a lyric about high school. And you have a song called "I'm 20" on another one of your earlier records. There's virtually none of that on "Close It Quietly." Was that by design? Are you just kind of eschewing the importance of age, lyrically?
Kline: Huh. I think when I was younger, it definitely felt like that was a gimmick. Not personally, but I felt like that was people's reaction to me making music- — like, 'Oh, it's this young girl.'
And I think when you're 19, especially if you're a woman, it's like if you can write any song, people are applauding you like, 'Wow, look at this little girl that wrote a song.' And then when you get older, especially the more records you put out, I think people aren't as impressed because they're used to it. So it's less exciting or impressive for a 25-year-old to write 20 good songs than it is for a 19-year-old or something like that. I don't know.
I think people love gimmicks like that. And I think from a press perspective — even this time around, like I'm 25. All my bandmates are in their late 20s. We've all been touring for seven years. It doesn't seem like it would be a very interesting gimmick anymore, but the Pitchfork review of this album mentioned how old I was when I started making music and like how I used to write about being young and it's like 'Yeah, yeah, I'm aging.' It's not that interesting. [Laughter]
I just don't think it's that good of a gimmick anymore. I'm just an adult.
Audio mixes: Steven Kray
Recording engineer: Nalin Silva
Cameras: Jarratt Taylor, Elias Williamson, Nick McClurg, Jennifer Sowell
Video Editor: Elias Williamson
Interview: Jerad Walker
Executive Producer: David Christensen
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Report: Dolphins QB Rosen drawing trade interest
FILE PHOTO: Nov 17, 2019; Miami Gardens, FL, USA; Miami Dolphins quarterback Josh Rosen (3) warms up before a game against the Buffalo Bills at Hard Rock Stadium. Mandatory Credit: Steve Mitchell-USA TODAY Sports
Dolphins quarterback Josh Rosen is attracting interest from teams who expect the former first-round pick to be the odd man out after Miami drafted Tua Tagovailoa, NFL Network reported.
Tagovailoa could be limited as a rookie to fully recover from recent injuries, including hip and ankle surgery in 2019, and the Alabama product repeatedly has made reference to entering the league ready to learn as a backup. The Dolphins also have Jake Rudock, who spent 2019 on the practice squad, on the roster. Ryan Fitzpatrick is the incumbent starter.
NFL Network reported the Dolphins are in "no hurry" to move Rosen. Drafted No. 10 overall by the Arizona Cardinals in 2018, Rosen was 3-10 as a starter as a rookie. Arizona landed the No. 1 pick in 2019, changed coaches and drafted Heisman Trophy winner Kyler Murray, moving Rosen in a deal with the Dolphins.
Rosen, 23, was unable to unseat the veteran Fitzpatrick for the starting role. He went 0-3 as a starter last season.


The Klinikum rechts der Isar university hospital of the Technical University of Munich is a major European center of medical care and education and teaching with 1,161 beds and around 5,500 employees. Every year around 60,000 patients benefit from inpatient and around 250,000 patients from out patient care. The hospital is a supra-maximal care home that covers the entire spectrum of modern medicine. Since 2003, the Klinikum rechts der Isar is an institute of public law of the Free State of Bavaria.
at the Center for Translational Cancer Research (TranslaTUM) in Munich.
We are deploying genetic models and bioinformatics to study molecular, immunological and translational aspects of pancreatic cancer. One focus in the past few years was the development of novel preclinical models and organoid-based orthotopic trans plantation strategies to mechanistically analyze sub types of pancreatic cancer. The focus of the advertised projects will be on the identification and functional validation of novel therapeutic targets with a focus on tumor cell intrinsic as well as inflammatory path ways that are required for tumor initiation and progression in vivo (e.g. Nature 2018, 554:62-68; Nat. Rev. Cancer 2017 17:239-253; Nat. Commun. 2016, 7:10770; Nat Med. 2014, 20:1340-7; Cancer Cell. 2013,24:15-29, Cancer Cell. 2013,23:406-20, Nat Commun. 2013;4:1630; PNAS 2011, 108:9945-50).
We are a young and enthusias tic team and offer intensive training and mentoring.
The successful candidates (f/m/d) will have the opportunity to learn a wide spectrum of genetic, molecular and cell biology, biochemical as well as immunological and bioinformatics techniques.
They will be exposed to cutting edge methods to study translational aspects of cancer (e.g. Cre / loxP and Flp / FRT based dual- and triple recombinase systems, organoid culture, orthotopic trans plantations of organoids, CRISPR / Cas9, molecular in vivo imaging, in vivo transposon mutagenesis, RNA and exome sequencing, single cell sequencing, histocytometry).
They will also be able to gain insights into clinically relevant aspects of cancer development through proximity to the clinical facilities and through the inter disciplinary composition of our team and closely collaborating groups.
We are looking for self-motivated bright individuals with a Diploma or Masters Degree in bioinformatics, bio chemistry, biology, molecular / translational medicine or related subjects.
Previous exposure to and experience with basic biochemical and molecular biologytechniques, proteomics or bioinformatics as well interests in molecular aspects of cancer research is essential.
Payment is according to tarif (TV-L). We give priority to severely disabled applicants (f/m/d) with essentially equal qualifications.
We are looking for ward for your application!
Please send your application including a letter of motivation, CV, certificates, recommendation letters in one PDF file to Prof. Dieter Saur (e-mail: dieter.saur@tum.de).


I went to Devon for a short break and one of the activities on offer was making picture frames, its actually quite difficult and quite a skill but the results are very rewarding. Something I definitely want to do more of!
This entry was posted in Arts and Crafts and tagged Crafts.


Congratulations to the Commack girls soccer team on their win against North Babylon in the first playoff game.
For this second half of this playoff game I moved to the east end of the field, which is something I don't usually do when it's sunny out. This puts the sun at the players' backs and puts their faces in the shade. Backlighting your subject can make for a nice photo but requires some additional post-processing work. But, since most of the action in this game was at the North Babylon goal, it was either move to that goal or stand around waiting at the Commack goal.
When I have players with the sun at their backs I'm basically using the same exposure setting as with the sun in their faces but I open up the shadows in post. This way, I'm not blowing out the backs of the player that are in the sunlight. This is especially important since the Commack players have white uniforms; if both teams have darker uniforms you might consider increasing your exposure. Lightening shadows can make photos look a bit flat but there's not much of a choice. You can only increase the highlights so much without blowing them out and you don't want to darken the shadows because that would include their faces. Sometimes you might be able to get some definition back by increasing the clarity slightly.
One more thing you may want to do with late afternoon games in sunlight is to reduce the orange saturation. Most players' skin tones will be very orange due to the warmth of the sunlight. You may or may not like this effect. I tend to dial it back a bit by lowering the orange saturation.
That's it for today. Another playoff game this afternoon.