text:
\section{Introduction and statement of results}

One of the more ubiquitous concepts in enumerative combinatorics is unimodality, as a large number of common combinatorial functions have this property (see Stanley's survey articles \cite{Stan80,Stan89} for many examples and applications).  In particular, a {\it unimodal sequence of size $n$} is a sequence of positive integers that sum to $n$ such that the terms are monotonically increasing until a peak is reached, after which the terms are monotonically decreasing.  This means that the sequence can be written in terms of parts
\begin{equation*}
a_1, a_2, \dots, a_r, c, b_s, b_{s-1}, \dots, b_1,
\end{equation*}
that satisfy the inequalities
\begin{align}
\label{E:ac}
& 1 \leq a_1 \leq a_2 \leq \dots \leq a_r \leq c, \qquad \text{and} \\
\label{E:cb}
& c \geq b_s \geq b_{s-1} \geq \dots \geq b_1 \geq 1.
\end{align}
When writing examples of unimodal sequences, we  typically write the sequences without commas in order to save space. For example, $1244322$ is a unimodal sequence of size $18$, and all of the unimodal sequences of size $4$ are concisely listed as follows:
\begin{equation*}
4; \; 31; \; 13; \; 22; \; 211; \; 121; \; 112; \; 1111.
\end{equation*}

This simple combinatorial definition and its variants have also appeared in the literature under many other names and contexts, including ``stacks'' \cite{Aul51, Wri68, Wri71, Wri72}, ``convex compositions'' \cite{And12}, integer partitions \cite{And98, And11}, and in the study of ``saw-toothed'' crystal surfaces  in statistical mechanics \cite{Tem52}.  Furthermore, it is also appropriate to mention the closely related concept of ``concave compositions'', which were studied in \cite{And84, And11, And12, ARZ12, Rho12}, although we will not further discuss these sequences.

The purposes of this paper are twofold.  First, we concisely review previous work on enumeration functions related to stacks, unimodal sequences, and compositions, including the corresponding generating functions, and their asymptotic behavior when known.  Second, we provide new asymptotic formulas that successfully fill in the notable missing cases from the literature.  Along the way we describe some of the analytic techniques that were used in proving the previously known asymptotic formulas, including the theory of modular forms, Euler-MacLaurin summation, and Tauberian theorems; broadly speaking, these methods allow us to use the analytic properties of generating series in order to determine the asymptotic behavior of their coefficients.
In order to prove the new asymptotic formulas, we also introduce additional techniques such as the Constant Term Method and Saddle Point Method, which have been less widely used in this combinatorial setting.

Before presenting the definitions of the many variants of unimodal sequences, we first highlight a potential ambiguity in the enumeration of sequences that satisfy \eqref{E:ac} and \eqref{E:cb}.  In particular, if the largest part is not unique, then there are multiple choices for the ``summit'' $c$.  There are therefore two natural enumeration functions, as these choices may or may not be counted distinctly, and the majority of the enumeration functions thus occur in closely related pairs, corresponding to whether or not the summit is distinguished.  For the first type of enumeration function, we  canonically choose the summit $c$ to be the last instance of the largest part in the sequence.

We now describe the various enumeration functions for unimodal sequences, adopting Wright's terminology as much as possible -- we refer to all examples as ``stacks'' of various types, and we represent unimodal sequences visually by diagrams in which a sequence of blocks are stacked in columns.  The first definition is that of a {\it stack}, which is simply a unimodal sequence with $c > b_s$; this means that each distinct sequence is counted once, without regard to the multiplicity of the largest part.
Following Wright's notation in \cite{Wri68}, we denote the number of stacks of size $n$ by $s(n)$ (this function has also been called $Q(n)$ in the enumeration of ``Type B partitions'' found in \cite{Aul51}, and $u(n)$ in Section 2.5 of \cite{Stan11}).

Following the alternative method of enumeration, a {\it stack with summit} is defined to be a stack in which one of the largest parts is marked (this represents the possible different choices of $c$ in \eqref{E:ac} and \eqref{E:cb}).  In other words, we distinguish between stacks in which a different instance of the largest part is marked as the summit.  We write the enumeration function for stacks with summits of size $n$ as $ss(n)$; this is also found as $\sigma\sigma(n)$ in Section 3 of \cite{And84}, as $v(n)$ in the study of ``V-partitions'' in Section 2.5 of \cite{Stan11}, and as $X(n-1)$ in \cite{And12}, where Andrews' definition of ``convex compositions'' (which implicitly have a unique largest part) is equivalent to stacks with summits, up to a minor normalization.

For example, the stacks of size $4$ were listed previously,
and if we represent the marked parts as overlined,  then the stacks with summits of size $4$ are
\begin{equation*}
\overline{4}; \; \overline{3}1; \; 1\overline{3}; \; \overline{2}2; \; 2\overline{2}; \; \overline{2}11; \; 1\overline{2}1;
\; 11\overline{2}; \; \overline{1}111; \; 1\overline{1}11; \; 11\overline{1}1; \; 111\overline{1},
\end{equation*}
so $s(4) = 8$ and $ss(4) = 12.$  See Figures \ref{F:Stack} and \ref{F:StackSummit} for examples of the diagrams of stacks and stacks with summits.
The relevant generating functions, which first appeared in \cite{Aul51} and \cite{And84}, respectively, are
\begin{align}
\label{E:sgen}
\mathscr{S}(q) := \sum_{n \geq 0} s(n) q^n & = 1+\sum_{m \geq 1} \frac{q^m}{(q)_{m-1}^2 (1 - q^m)}, \\
\label{E:ssgen}
\mathscr{S}_s(q) := \sum_{n \geq 0} ss(n) q^n & = \sum_{m \geq 0} \frac{q^m}{(q)_{m}^2}.
\end{align}
Here we have used the standard notation for $q$-factorials, namely
$$
(a)_n = (a;q)_n := \prod_{j=0}^{n-1} \left(1 - aq^j\right)
$$
 for $n \geq 0$.    In order to understand these generating functions, observe that the $m$th term generates all stacks (respectively, stacks with summits) with largest part exactly $m$.

\begin{figure}[here]
\begin{center}
\includegraphics[width = 175 pt]{StackExample}
\end{center}
\caption{\label{F:Stack}
Diagram for the stack $1244322$.}
\end{figure}

\begin{figure}[here]
\begin{center}
\includegraphics[width = 400 pt]{StackSummitExample}
\end{center}
\caption{\label{F:StackSummit}
Diagrams for the stacks with summits $12\overline{4}4322$ and $124\overline{4}322$, respectively, where the summits are distinguished by a mark at the top of the column.
}
\end{figure}

The next definition is that of a {\it receding stack}, which must satisfy \eqref{E:ac} and \eqref{E:cb} as well as the additional conditions
\begin{align}
\label{E:arec}
a_j & \geq a_{j+1} - 1 \quad \text{for } 1 \leq j \leq r-1, \\
\label{E:brec}
b_j & \leq b_{j+1} + 1 \quad \text{for } 1 \leq j \leq s-1.
\end{align}
In other words, the collections of $a_j$s and $b_j$s must both contain each part size from $1$ to $c-1$ at least once.  Also, as in the case of stacks above, we further require $c > b_s$ for this first type of enumeration function.  Again following Wright \cite{Wri71}, we denote the number of receding stacks of size $n$ by $g(n)$ (also written as $P(n)$ in the enumeration of ``Type A partitions'' in \cite{Aul51}).  Similarly, a {\it receding stack with summit} distinguishes one of the largest parts with a mark, and we denote the corresponding enumeration function by $gs(n)$.

For example, the receding stacks of size $4$ are
\begin{equation*}
121; \; 1111;
\end{equation*}
and the receding stacks with summits of size $4$ are
\begin{equation*}
1\overline{2}1;  \; \overline{1}111; \; 1\overline{1}11; \;
11\overline{1}1; \; 111\overline{1};
\end{equation*}
so $g(4) = 2$ and $gs(4) = 5.$
The generating functions for receding stacks (resp. with summits) are
\begin{align}
\label{E:ggen}
\mathscr{G}(q) := \sum_{n \geq 0} g(n) q^n & = 1 + \sum_{m \geq 1} \frac{q^{m^2}}{(q)_{m-1}^2 (1 - q^m)}, \\
\label{E:gsgen}
\mathscr{G}_s(q) := \sum_{n \geq 0} gs(n) q^n & = \sum_{m \geq 0} \frac{q^{m^2}}{(q)_{m}^2}.
\end{align}
The generating function for $g(n)$ was first given in \cite{Aul51}, but to our knowledge, receding stacks with summits have not previously been described.  This is likely due to the fact that \eqref{E:gsgen} is better known as the generating function for $p(n)$, Euler's integer partition function, which has been widely studied in its own right \cite{And98}.  We will describe the combinatorial relationship between partitions and receding stacks with summits in greater detail later in this paper.

Continuing with the definitions, a {\it shifted stack} is a sequence that satisfies \eqref{E:ac}, \eqref{E:cb}, and \eqref{E:arec}, with the additional condition $c > b_s$.  Although this definition might also naturally be called ``left-receding stacks'', the terminology is inspired by Auluck's  \cite{Aul51} and Wright's \cite{Wri71} observations that if each row is shifted left by half of a cell, then this is combinatorially equivalent to counting stacks of coins, or circular logs, with a single ``peak''.  See Figure \ref{F:Shifted} for an example of a shifted stack represented by both types of diagrams (cf. Figure 2 in Auluck's original work \cite{Aul51}).   The number of shifted stacks of size $n$ is denoted by $h(n)$ as in \cite{Wri71} (this function is also found as $R(n)$ in the enumeration of ``Type C partitions'' in \cite{Aul51}).  Similarly,  a {\it shifted stack with summit} is a shifted stack with a distinguished occurrence of the largest part, and the enumeration function is now denoted by $hs(n)$; these stacks previously appeared in Section 3 of \cite{And84}, where they were referred to as ``gradual stacks with summits'', with enumeration function $g\sigma(n)$.

\begin{figure}[here]
\begin{center}
\includegraphics[width = 300 pt]{Shifted}
\end{center}
\caption{\label{F:Shifted}
Two equivalent diagrams for the shifted stack $12231$.  The stack of coins diagram is obtained from the block diagram by successively shifting each row a half space to the left, and then replacing each block with a coin.}
\end{figure}

For example, the shifted stacks of size $4$ are
\begin{equation*}
121; \; 112; \; 1111;
\end{equation*}
and the shifted stacks with summits of size $4$ are
\begin{equation*}
1\overline{2}1; \; 11\overline{2}; \; \overline{1}111; \; 1\overline{1}11; \; 11\overline{1}1; \;
111\overline{1};
\end{equation*}
so $h(4) = 3$ and $hs(4) = 6.$
The generating functions for the two types of shifted stacks are first found in \cite{Aul51} and \cite{And84}, respectively, and are given by
\begin{align}
\label{E:hgen}
\mathscr{H}(q) := \sum_{n \geq 0} h(n) q^n & = 1 + \sum_{m \geq 1} \frac{q^{\frac{m(m+1)}{2}}}{(q)_{m-1}^2 (1 - q^m)}, \\
\label{E:hsgen}
\mathscr{H}_s(q) := \sum_{n \geq 0} hs(n) q^n & = \sum_{m \geq 0} \frac{q^{\frac{m(m+1)}{2}}}{(q)_{m}^2}.
\end{align}

The final variants that we discuss in this paper were introduced by Andrews in \cite{And12}, where he studied  ``strict convex compositions''.  We therefore define a {\it strict stack} to be a stack in which all of the inequalities in \eqref{E:ac} and \eqref{E:cb} are strict, and write the enumeration function as $d(n)$ (this appeared as $x_d(n)$ in \cite{And12}, and as the enumeration function $u^\ast(n)$ for ``strict unimodal sequences'' in \cite{Rho12}). Furthermore, a {\it semi-strict stack} (or ``left-strict'') requires strict inequalities only in \eqref{E:ac}, with the additional condition that the largest value is unique, so $c > b_s.$  We denote the enumeration function for semi-strict stacks by $dm(n)$ (this was denoted by $x_m(n)$ in \cite{And12}).   Note that the summits are already unique for strict and semi-strict stacks, so there are not enumeration functions of the second type.

For example, the strict stacks of size $4$ are
\begin{equation*}
4; \; 31; \; 13; \; 121;
\end{equation*}
and the semi-strict stacks of size $4$ are
\begin{equation*}
4; \; 31; \; 13; \; 211; \; 121;
\end{equation*}
so $d(4) = 4$ and $dm(4) = 5.$
The generating function for strict and semi-strict stacks are also found in \cite{And12}, and are given by
\begin{align}
\label{E:dgen}
\mathscr{D}(q) := \sum_{n \geq 0} d(n) q^n & = \sum_{m \geq 0} q^{m+1} \left(-q\right)_m^2, \\
\label{E:dmgen}
\mathscr{D}_m(q) := \sum_{n \geq 0} dm(n) q^n & = \sum_{m \geq 0} \frac{(-q)_m q^{m+1}}{(q)_m}.
\end{align}

All of the previously known asymptotic results for stacks are listed in Table \ref{Tab:stack}.  It is important to point out that in many cases the enumeration function is known to much greater precision than its main asymptotic term; this includes the full asymptotic expansion for $g(n)$ and $h(n)$ \cite{Wri71}, an exact formula for $gs(n)$ \cite{Rademacher}, and an asymptotic series with polynomial error for $d(n)$ \cite{Rho12}.

The asterisks in the second row indicate that in the case of $ss(n)$ the asymptotic formula is not explicitly stated in either of the references, but can be obtained by combining Stanley and Wright's results.  For completeness, the details and further context are discussed later in Section \ref{S:summit} of this paper.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l||c|c|c|c|c|} \hline
\quad Objects \quad & \quad Functions \quad & \quad $C$ \quad & \quad $\alpha$ \quad & \quad $\beta$ \quad & \quad References \quad \\ \hline \hline

Stacks & $s(n)$  & $2^{-3} 3^{-\frac{3}{4}}$ & $\frac{5}{4}$ & $\frac{4}{3}$ & Auluck \cite{Aul51} \\ \hline

\: \dots w/ summits & $ss(n)$ & $2^{-3} {3^{-\frac{3}{4}}}^\ast$ & $\frac{5}{4}^\ast$ & $\frac{4}{3}^\ast$ & $\text{Stanley}^\ast$ \cite{Stan11}, $\text{Wright}^\ast$ \cite{Wri71}  \\ \hline

Receding stacks & $g(n)$ & $2^{-3} 3^{-\frac{1}{2}}$ & $1$ & $\frac{2}{3}$ & Auluck \cite{Aul51} \\ \hline

\: \dots w/ summits & $gs(n)$ & $2^{-2} 3^{-\frac{1}{2}}$ & $1$ & $\frac{2}{3}$ & Hardy-Ramanujan \cite{HR} \\ \hline

Shifted stacks & $h(n)$ & ? & ? & $\frac{4}{5}$ & Wright \cite{Wri72} \\ \hline

\: \dots w/ summits & $hs(n)$ & ? & ? & ? & --- \\ \hline

Strict stacks & $d(n)$ & $2^{-\frac{13}{4}} 3^{-\frac{1}{4}}$ & $\frac{3}{4}$ & $\frac{2}{3}$ & Rhoades \cite{Rho12} \\ \hline

Semi-strict stacks & $dm(n)$ & ? & ? & ? & --- \\ \hline

\end{tabular}
\end{center}
\label{Tab:stack}
\caption{Previous asymptotic formulas for stack enumeration functions, where the main asymptotic term is written in the form $C n^{-\alpha} e^{\pi \sqrt{\beta n}}$ as $n \to \infty$.  See Section \ref{S:summit} for more details on how the formula for $ss(n)$ follows from \cite{Stan11} and \cite{Wri71}.}
\end{table}

Our new results fill in the missing values and incomplete rows of Table \ref{Tab:stack}.  These missing cases are of longstanding interest, as they represent some of the earliest examples of unimodal enumeration problems, and have inspired much of the subsequent work in the area.  Indeed, shifted stacks were first introduced in Temperley's original study of boundary cases for rectangular arrays in thermal equilibrium \cite{Tem52}.  Auluck then investigated $h(n)$ at Temperley's request \cite{Aul51}, and proved that the exponential order lay in the bounded range $\frac{2}{3} \leq \beta \leq \frac{5}{6}$. The study was subsequently taken up by Wright \cite{Wri72}, who improved the asymptotic analysis by finding the exact exponent, showing that
\begin{equation}
\label{E:logh}
\log \left( h(n) \right)\sim \pi \sqrt{\frac{4}{5} n} \qquad \text{as } n \to \infty.
\end{equation}
Our first result gives precise asymptotic formula for shifted stacks and shifted stacks with summits, which is a significant improvement over \eqref{E:logh}.  We use the standard notation $\phi := \frac{1 + \sqrt{5}}{2}$ for the Golden Ratio.
\begin{theorem}
\label{T:h}
As $n \to \infty$, we have the asymptotic formulae
\begin{align*}
h(n) & \sim \frac{\phi^{-1}}{2\sqrt{2} 5^{\frac{3}{4}} n} e^{2\pi \sqrt{\frac{n}{5}}}, \\
hs(n) & \sim \phi \cdot h(n).
\end{align*}
\end{theorem}
\begin{remark*}
As pointed out to us by R. Rhoades, Ramanujan's last letter to Hardy amazingly includes the generating function for shifted stacks with summits.  Although he did not discuss the coefficients, nor their combinatorial significance, Ramanujan did provide the asymptotic expansion for the generating function (as was often the case in Ramanujan's work, he did not write his results in a strictly rigorous manner).  See \cite{Wat36} and the further discussion in Section \ref{S:shifted:summit} of this paper for additional details.  Combined with the techniques discussed in Section \ref{S:Tauberian}, Ramanujan's result could also be used to prove the asymptotic formula for $hs(n)$.  However, we provide an additional proof of this case in Section \ref{S:shifted}, as our present techniques apply equally well to both formulas in Theorem \ref{T:h}.
\end{remark*}

Our next result does not directly appear in Table \ref{Tab:stack}, but is inspired by a closer investigation of the relationship between receding stacks with summits and integer partitions.  We will review Andrews' definition of ``Frobenius symbols'' in Section \ref{S:partns:Frob}, but for the purposes of stating our result we simply denote the number of Frobenius symbols of size $n$ with no zeros in the top row by $F_\phi(n).$   We also define the notation $F_0(n)$ for the number of Frobenius symbols of size $n$ that {\it do} contain a zero in the top row.  Since there are exactly $p(n)$ Frobenius symbols of size $n$ (again, see Section \ref{S:partns:Frob} for definitions), we have
\begin{equation}
\label{E:F+F=p}
F_\phi(n) + F_0(n) = p(n).
\end{equation}

The following theorem describes the asymptotic relationship between these two functions.
\begin{theorem}
\label{T:Frob}
As $n \to \infty$,
\begin{equation*}
F_\phi(n) \sim F_0(n) \sim \frac{1}{8 \sqrt{3} n} e^{\pi \sqrt{\frac{2}{3}n}}.
\end{equation*}
\end{theorem}
\begin{remark*}
In fact, the bijections in Section \ref{S:partns} further prove that $g(n) = F_0(n)$.  This means that Theorem \ref{T:Frob} equivalently states that roughly half of all partitions (according to the uniform measure on partitions up to a given size) correspond to a Frobenius symbol that does not contain a zero in the top row, just as the ``average'' receding stack has two copies of its largest part (since $gs(n) \sim 2 g(n)$).
\end{remark*}

Our final result fills in the last row of Table \ref{Tab:stack} by giving a formula for semi-strict stacks.  Among the known results in Table \ref{Tab:stack}, the case of strict stacks was perhaps the most difficult.  The analysis for most of the other rows relied on some combination of combinatorial arguments, the transformation theory of (modular) theta functions, and/or the use of Euler-MacLaurin summation for asymptotic expansions.  However, Andrews \cite{And12} showed that $\mathscr{D}(q)$ can be expressed in terms of modular forms and Ramanujan's  famous ``mock theta functions'', which were only recently placed into the modern framework of real-analytic automorphic forms thanks to Zwegers' seminal Ph.D. thesis \cite{Zw02} (see also \cite{Zag06}).  Rhoades \cite{Rho12} then proved the asymptotic series expansion for $d(n)$ by using an extended version of the Hardy-Ramanujan Circle Method that the present authors originally developed for the asymptotic analysis of such non-modular functions in \cite{BM11}.

Indeed, it is a very challenging problem to determine the automorphic properties of a generic hypergeometric $q$-series.  Zagier's work on Nahm's conjecture in \cite{Zag06} provides a classification of modular hypergeometric $q$-series written in so-called ``Eulerian'' form, and S. Zwegers pointed out to us that this criterion shows that the shifted stack generating functions, $\mathscr{H}(q)$ and $\mathscr{H}_s(q)$, are not modular forms either.  See Section \ref{S:shifted:shifted} for additional details.

In contrast, the case of semi-strict stacks is more similar to some of the earlier rows, as $\mathscr{D}_m(q)$ can be expressed in terms of modular forms.  We prove the final missing asymptotic formula using modular transformations and Ingham's Tauberian Theorem.
\begin{theorem}
\label{T:dm}
As $n \to \infty$, we have the asymptotic formula
\begin{equation*}
dm(n) \sim \frac{1}{16n} e^{\pi \sqrt{n}}.
\end{equation*}
\end{theorem}

The remainder of the paper is structured as follows.  In Section \ref{S:partns} we define the Frobenius symbol of a partition, give a combinatorial bijection between partitions and receding stacks with summits, and finally prove Theorem \ref{T:Frob} by relating Frobenius symbols with no zeros to receding stacks.  In Section \ref{S:Tauberian} we recall Ingham's Tauberian Theorem and then use it to prove Theorem \ref{T:dm}.  We next address the case of shifted stacks, using the Constant Term Method and Saddle Point Method to prove Theorem \ref{T:h} in Section \ref{S:shifted}.  Finally, we conclude in Section \ref{S:summit} by outlining the asymptotic analysis of stacks with summits and provide further discussion of Stanley's and Wright's works.

\section*{Acknowledgments}

The authors thank Rob Rhoades and Sander Zwegers for helpful comments regarding historical works relevant to shifted stacks.

\section{Receding stacks, partitions, and Frobenius symbols}
\label{S:partns}

This section describes the combinatorial properties and asymptotic behavior of receding stacks, partitions, and Frobenius symbols.
In particular, we present a simple bijection between partitions and receding stacks with summits, and then  describe the relationship between these stacks and Frobenius symbols.
Indeed, our map essentially gives a ``conjugated'' version of the Frobenius symbol (in the sense that certain row and columns are interchanged), and it therefore comes as no surprise that our study also includes asymptotic results relevant to Andrews' study of Frobenius symbols with no zeros in the top row from \cite{And11}.

\subsection{Frobenius symbols}
\label{S:partns:Frob}

In \cite{And84F}, Andrews defined a {\it Frobenius symbol} as two equal-length sequences of decreasing nonnegative integers, usually written in two rows as
\begin{equation*}
\begin{pmatrix}
\alpha_1 & \alpha_2 & \dots & \alpha_k \\
\beta_1 & \beta_2 & \dots & \beta_k
\end{pmatrix}.
\end{equation*}
The {\it size} of such a symbol is defined to be the sum $k + \alpha_1 + \dots + \alpha_k + \beta_1 + \dots + \beta_k$.

The Frobenius symbols of size $n$ are in bijective correspondence with the partitions of size $n$.  The {\it Ferrer's diagram} of a partition $\lambda_1 + \dots + \lambda_m$ begins with a row of $\lambda_1$ blocks, which is then followed below by (left-justified) rows of $\lambda_2, \dots, \lambda_m$ blocks.  Given such a diagram, the corresponding Frobenius symbol is constructed by first setting $k$ to be the length of the main (southeast) diagnoal.  The entry $\alpha_j$ is then the length of the $j$th row that lies to the right of the diagonal, while $\beta_j$ is the length of the $j$th column that is below the diagonal.  For example, the partition $4+4+3+3+1$ corresponds to the Frobenius symbol
$$
\begin{pmatrix}
3 & 2 & 0 \\
4 & 2 & 1
\end{pmatrix}.
$$

\begin{remark*}
\label{R:Frob0}
It is straightforward to see that a Frobenius symbol with a zero in the top row corresponds to a partition in which the $k$-th largest part is exactly $k$ for some $k \geq 1$.
\end{remark*}

\subsection{A bijection between partitions and receding stacks with summits}

We now describe a bijection between partitions and receding stacks with summits that provides a combinatorial proof of the following result.  See Figure \ref{F:PartitionBij} for an example of the bijection.
\begin{proposition}
\label{P:gs=p}
For all $n \geq 0$, the number of partitions of size $n$ is equal to the number of receding stacks with summits of size $n$, so that
\begin{equation*}
gs(n) = p(n).
\end{equation*}
\end{proposition}
\begin{remark*}
This equality can also be read directly from the generating function \eqref{E:gsgen}, but the combinatorial arguments that follow provide a more refined understanding of the relationship between partitions and receding stacks with summits.
\end{remark*}
\begin{figure}[here]
\begin{center}
\includegraphics[width = 425 pt]{PartitionBij}
\end{center}
\caption{\label{F:PartitionBij}
An illustration of the bijection that maps the partition $4+4+3+3+1$ to the receding stack with summit $1123\overline{3}221.$
Starting from the lower-left corner in the partition and proceeding clockwise along its boundary, the unimodal sequence is constructed by reading the blocks of the partition along diagonals.  The summit is marked on the main diagonal of the partition, which is also distinguished by the thickest dashed line in the figure.}
\end{figure}
\begin{proof}
As in the construction of the Frobenius symbol corresponding to a partition, we begin by considering separately the portion of the Ferrer's diagram to the right and below the main diagonal.  The summit $c$ of the receding stack is set to be the length of the main diagonal.  The $a_js$ are determined by reading the portion of the diagram below the main diagonal as diagonal ``rows'', beginning with the lower-left cell, which automatically gives $a_1 = 1$.  After reaching the summit $c$ along the main diagonal, the $b_js$ are then found by reading diagonal ``columns'' to the right of the main diagonal.  Proceeding along these columns, the lengths give the successive values of $b_s, b_{s-1}, \dots,$ ending with the upper-right cell, which gives $b_1 = 1$.

It is routine to verify that the receding stack conditions \eqref{E:ac}, \eqref{E:cb}, \eqref{E:arec}, and \eqref{E:brec} are always satisfied.  One needs only keep in mind that the cells in the Ferrer's diagram of a partition are fully left- and top-justified.  It is also clear that this map is a bijection, as the procedure is easily reversible, giving a well-defined inverse map from receding stacks with summits to partitions.
\end{proof}

\subsection{Frobenius symbols with no zeros}

In this section we prove Theorem \ref{T:Frob}.  As first defined in the introductory section, $F_\phi(n)$ denotes the number of Frobenius symbols of size $n$ with no zeros in the top row, and $F_0(n)$ enumerates the number of Frobenius symbols of size $n$ that do contain a zero in the top row (in particular, the final entry in the top row is the only one that can be zero).
\begin{proof}[Proof of Theorem \ref{T:Frob}]
Andrews showed in Lemma 12 (and the proof of Theorem 4) of \cite{And11} that the generating function for Frobenius symbols without zeros in the top row is
\begin{equation*}
\mathscr{F}_\phi(q) := \sum_{n \geq 0} F_\phi(n) q^n = \sum_{m \geq 0} \frac{q^{m^2 + m}}{(q)_m^2}.
\end{equation*}
The simple identity
\begin{equation*}
\sum_{m \geq 0} \frac{q^{m^2}}{(q)_m^2} = \sum_{m \geq 0} \frac{q^{m^2 + m}}{(q)_m^2} +
\left(1 + \sum_{m \geq 1} \frac{q^{m^2}}{(q)_{m-1} (q)_m}\right)
\end{equation*}
can be rewritten as
\begin{equation}
\label{E:G=F+G}
\mathscr{G}_s(q) = \mathscr{F}_\phi(q) + \mathscr{G}(q).
\end{equation}
Along with \eqref{E:gsgen} and \eqref{E:F+F=p}, this implies that
\begin{equation*}
\sum_{n \geq 0} F_0(n) q^n = \mathscr{G}(q).
\end{equation*}
We therefore have $F_0(n) = g(n)$ for all $n \geq 0$.  Furthermore, comparing coefficients in \eqref{E:G=F+G}, we also have
\begin{equation}
\label{E:gs=F+g}
gs(n) = F_\phi(n) + g(n).
\end{equation}
As alluded to in Table \ref{Tab:stack}, Auluck proved an asymptotic formula for $g(n)$; in particular, equation (15) of \cite{Aul51} states that
\begin{equation*}
g(n) \sim \frac{1}{8 \sqrt{3} n} e^{\pi \sqrt{\frac{2}{3}n}}.
\end{equation*}
Table \ref{Tab:stack} also contains Hardy and Ramanujan's famous asymptotic formula for the partition function,
\begin{equation*}
p(n)=
gs(n) \sim \frac{1}{4 \sqrt{3} n} e^{\pi \sqrt{\frac{2}{3}n}}.
\end{equation*}
Combined with \eqref{E:gs=F+g}, this completes the proof.
\end{proof}

\begin{remark*}
The discussion at the end of Section \ref{S:partns:Frob} and the proof of Proposition \ref{P:gs=p} also provide a combinatorial argument for the fact that $F_0(n) = g(n)$.  In particular, if the Frobenius symbol of a partition has a zero in its top row, then the main diagonal of its Ferrers diagram is longer than any diagonal to the right.  Using the bijection from Proposition \ref{P:gs=p}, the summit $c$ in the corresponding receding stack is therefore canonically larger than all of the $b_j$s.  The partition in Figure \ref{F:PartitionBij} is a relevant example, as the corresponding Frobenius symbol in this case was $\left(\begin{smallmatrix}
3 & 2 & 0 \\
4 & 2 & 1
\end{smallmatrix}\right).$
\end{remark*}

\section{Ingham's Tauberian Theorem and semi-strict stacks}
\label{S:Tauberian}

We begin this section by recalling a fundamental tool for analyzing the coefficients of combinatorial generating functions.  We then apply it to the case of semi-strict stacks, proving Theorem \ref{T:dm}.

\subsection{Ingham's Tauberian Theorem}
We require the following Tauberian Theorem from \cite{Ing41}, which allows us to describe the asymptotic behavior of the coefficients of a power series using the analytic nature of the series itself.
\begin{theorem}[Ingham]\label{T:Ingham}
Let $f(q)=\sum_{n\geq 0}a(n) q^n$ be a power series with weakly increasing nonnegative
coefficients and radius of convergence equal to $1$. If there are constants $A>0$,
$\lambda, \alpha\in\field{R}$ such that
\[
f \left(e^{-\varepsilon} \right)  \sim
\lambda   \varepsilon^\alpha
\exp\left(\frac{A}{\varepsilon}\right)
\]
as $\varepsilon \to 0^+$, then
\[
a(n) \sim\frac{\lambda}{2\sqrt{\pi}}\,
\frac{A^{\frac{\alpha}{2}+\frac14}}{n^{\frac{\alpha}{2}+\frac34}}\,
\exp\left(2\sqrt{An}\right)
\]
as $n\to\infty$.
\end{theorem}

In order to apply this theorem throughout the paper, we repeatedly use the following simple observation that guarantees increasing coefficients.
\begin{lemma}
\label{L:mono}
Suppose that $A(q) = \sum_{n \geq 0} a(n) q^n$ has non-negative coefficients $a(n).$  Then the series
\begin{equation*}
\displaystyle \sum_{n \geq 0} a'(n) q^n := \frac{1}{1-q} A(q)
\end{equation*}
has monotonically increasing non-negative coefficients.
\end{lemma}
\begin{remark*}
Lemma \ref{L:mono} (and therefore Theorem \ref{T:Ingham}) are easily adapted to any series $A(q)$ whose coefficients are strictly positive after some fixed index $n'$.
\end{remark*}

\subsection{Asymptotic behavior of semi-strict stacks}

In order to apply Ingham's Tauberian Theorem to the case of semi-strict stacks, we must first understand the asymptotic behavior of the generating function $\mathscr{D}_m(q)$ (cf. \eqref{E:dmgen}).  This is best accomplished not with the expression in \eqref{E:dmgen}, but by rather using an alternative expression given by Andrews in Theorem 1, equation (1.7) of \cite{And12} (see also (1.13)).  He proved that
\begin{equation}
\label{E:Dmeta}
\mathscr{D}_m(q) = \frac{q \left(q^2; q^2\right)_\infty}{(1+q) (q)_\infty^2},
\end{equation}
which is a modular form, up to a rational expression in $q$.
If we let $q = e^{-\varepsilon}$ as in Theorem \ref{T:Ingham}, then
the modular inversion formula for Dedekind's eta-function (page 121,
Proposition 14 of \cite{Ko84}) implies that as $\varepsilon \to 0^+$
\begin{equation}
\label{E:qinfty}
\left(e^{-\varepsilon}; e^{-\varepsilon} \right)_{\infty} \sim \sqrt{\frac{2\pi}{\varepsilon}}e^{-\frac{\pi^2}{6\varepsilon}}.
\end{equation}

\begin{proof}[Proof of Theorem \ref{T:dm}]
Applying \eqref{E:qinfty} to \eqref{E:Dmeta}, we find that
\begin{equation}
\mathscr{D}_m(q) \sim \frac{1}{4 \sqrt{\pi}} \varepsilon^{\frac{1}{2}} e^{\frac{\pi^2}{4 \varepsilon}}
\end{equation}
as $\varepsilon \to 0^+.$  Furthermore, Lemma \ref{L:mono} applied to \eqref{E:Dmeta} (or to the original generating function \eqref{E:dmgen})
 shows that the coefficients are monotonically increasing.  Theorem \ref{T:Ingham} now directly gives the theorem statement.
\end{proof}

\section{Asymptotic behavior of shifted stacks}
\label{S:shifted}

In this section we use a variety of tools from complex analysis and number theory to prove Theorem \ref{T:h}.  In particular, we begin with the Constant Term Method, inserting an additional variable $x$ in order to identify \eqref{E:hgen} as the constant term of the product of more recognizable number-theoretic functions.  We then use Cauchy's Theorem to recover this coefficient as a contour integral; the asymptotic behavior of the integral is obtained through the Saddle Point Method.  Finally, the asymptotic behavior of $h(n)$ is recovered from an application of Ingham's Tauberian Theorem.  The asymptotic analysis of shifted stacks with summits proceeds very similarly.  It should be noted that this approach is a standard (if not widely known) technique for determining the asymptotic behavior of hypergeometric $q$-series written in Eulerian form (cf. Proposition 5 in \cite{Zag07}).  This technique can also be applied in settings where there is not an Eulerian $q$-series, such as the double summation studied in \cite{BHMV12}.

It is interesting to note that the Constant Term Method already appears in Auluck's paper \cite{Aul51}, where he used this approach to derive alternative expressions for each of the generating functions \eqref{E:sgen}, \eqref{E:ggen}, and \eqref{E:hgen}.  However, in his asymptotic analysis of shifted stacks, he then relied on somewhat oblique comparisons to the (known) asymptotic behavior of integer partitions; the same is also true of Wright's work in \cite{Wri72}.  Our main contribution is that we immediately apply analytic approximation techniques (such as modular transformations and the Saddle Point Method) after using the Constant Term Method, rather than seeking to find exact expressions for generating functions.

\subsection{Asymptotic analysis of shifted stacks and the proof of Theorem \ref{T:h}}
\label{S:shifted:shifted}

In order to study shifted stacks and prove the first asymptotic formula in Theorem \ref{T:h}, we begin by writing
\begin{align}
\label{E:CConst}
\mathscr{H} (q) &= 1+ \operatorname{coeff} \left[x^0\right] \left( \sum_{r\geq 0} \frac{x^{-r} q^{\frac{r^2-r}{2}}}{(q)_r} \sum_{m\geq 1} \frac{x^m q^{m}}{(q)_{m-1}} \right)  \\
 &= 1 + \operatorname{coeff} \left[x^0\right] \left(qx \left( -x^{-1}\right)_{\infty}  \left(xq\right)_\infty^{-1} \right), \notag
\end{align}
where we have used the following two instances of the $q$-binomial theorem (cf. Theorem 2.1 in \cite{And98}):
\begin{align*}
(x)_\infty &= \sum_{m\geq 0}\frac{(-1)^m q^{\frac{m(m-1)}{2}}x^m}{(q)_m}, \\
\frac{1}{(x)_\infty} &= \sum_{m\geq 0}\frac{x^m}{(q)_m}.
\end{align*}

In order to further rewrite \eqref{E:CConst}, we recall Jacobi's Triple Product formula (Theorem 14.6 in \cite{Apo76}), which states that
\begin{equation*}
(-x^{-1})_\infty (-xq)_\infty (q)_\infty =
- q^{-\frac{1}{8}} x^{-\frac{1}{2}} \vartheta\left(u + \frac{1}{2}; \frac{i \varepsilon}{2 \pi} \right).
\end{equation*}
Here we write $x = e^{2 \pi i u}$, $q = e^{- \varepsilon}$ with $\varepsilon>0$, and the {\it Jacobi theta function} is given by
\begin{align}
\label{E:theta}
\vartheta(u; \tau) & := \sum_{n \in \frac12+\field{Z}}
e^{\pi i n^2\tau   + 2 \pi i n \left(u + \frac{1}{2}\right)}.
\end{align}
We  also use the {\it quantum dilogarithm} \cite{Zag07}, which is given by
\begin{align*}
\Li_2(x; q) & := -\log (x)_\infty = \sum_{m \ge 1} \frac{x^m}{m(1-q^m)}.
\end{align*}
We note that the quantum dilogarithm is a $q$-deformation of the ordinary \textit{dilogarithm}, defined for $|x|<1$ by
$$
\Li_2(x) := \sum_{m \geq 1}
\frac{x^m}{m^2}.
$$
To be more precise, for $|x|<1$ we have
$$
\lim_{\varepsilon \to 0^+} \varepsilon \Li_2\left(x; e^{-\varepsilon} \right)= \Li_2(x).
$$
Furthermore, if $|x|<1$ and $B\geq 0,$ then the Laurent expansion of the
quantum dilogarithm begins with the terms
\begin{align}
\label{TaylorLi}
\Li_2\left(e^{-B\varepsilon} x; e^{-\varepsilon}\right)
&=\sum_{m\geq 1}\frac{x^m e^{-Bm\varepsilon}}{m\left(1-e^{-\varepsilon m}\right)}
=\frac{1}{\varepsilon}\sum_{m\geq 1}\frac{x^m}{m^2}\left(1-m\varepsilon\left(B-\tfrac12\right)+
O\left(\varepsilon^2\right)\right)\\
&=\frac{1}{\varepsilon} \Li_2(x)+\left(B-\tfrac12\right)\log(1-x)+O(\varepsilon), \notag
\end{align}
where the series converges uniformly in $x$ as $\varepsilon \to 0^+$.

Using the above definitions, we can therefore rewrite \eqref{E:CConst} as
\begin{equation*}
\mathscr{H} (q)  = 1 - \frac{q^{\frac{7}{8}}}{ \left( q \right)_{\infty}} \mathscr{A} (q),
\end{equation*}
with
\begin{equation}
\label{E:AConst}
\mathscr{A} (q) := \operatorname{coeff} \left[x^0\right] \left( \vartheta \left( u+\frac12 ; \frac{i\varepsilon}{2\pi} \right) \exp \left( \pi i u + \text{Li}_2 \left(x^2 q^2 ; q^2 \right) \right) \right).
\end{equation}
The modular inversion formula for the Jacobi theta function (cf. Proposition 1.3 (8) in \cite{Zw02}) implies that
\begin{equation}
\label{E:thetaInv}
\vartheta \left( u+\frac12 ; \frac{i\varepsilon}{2\pi } \right)
=i \sqrt{ \frac{2 \pi}{\varepsilon}  } e^{- \frac{2\pi^2 }{\varepsilon}\left( u + \frac12 \right)^2}
\vartheta \left(\frac{2 \pi\left( u+\frac12  \right)}{i\varepsilon} ; \frac{2\pi i}{\varepsilon} \right).
\end{equation}
We use Cauchy's Theorem to recover the constant coefficient from \eqref{E:AConst} as a contour integral, and further simplify by plugging in \eqref{E:theta} and  \eqref{E:thetaInv}, obtaining
\begin{align}
\label{E:AInt}
\mathscr{A} (q) &= -\sqrt{\frac{2\pi}{\varepsilon}} \int_{[0,1] + iC}
\sum_{m\in\field{Z}} \exp\left(-\frac{2\pi^2}{\varepsilon}  (u+m)^2 + \pi i (u+m) + \text{Li}_2 \left(x^{2}q^2 ; q^2 \right) \right) du \\
&=-\sqrt{\frac{2\pi}{\varepsilon}} \int_{\field{R} + iC} \exp \left( -\frac{2\pi^2}{\varepsilon}  u^2 + \pi i u + \text{Li}_2 \left(x^{2}q^2 ; q^2 \right) \right) du. \notag
\end{align}
Here the contour is defined by a positive constant $C$ that will be specified later.  Note that $\text{Li}_2(x^{2}q^2; q^2)$ is (absolutely) convergent for all $u$ in the upper-half plane.

In order to continue the  analysis of the asymptotic behavior, we  require the Laurent expansion of the quantum dilogarithm. By (\ref{TaylorLi}), we have uniformly in $x$
\begin{equation}
\label{E:Li2Laurent}
\text{Li}_2 \left(x^{2}q^2; q^2\right) = \frac{1}{2\varepsilon} \text{Li}_2 \left( x^{2}\right) +\frac12 \log \left( 1- x^{2}\right) + O (\varepsilon).
\end{equation}

We are now prepared to apply the Saddle Point Method to \eqref{E:AInt}, and begin by defining the function
\begin{equation*}
f(u):= -2\pi^2 u^2 +\frac12 \text{Li}_2 \left( e^{4\pi i u} \right),
\end{equation*}
which is the leading Laurent coefficient ($\varepsilon^{-1}$-term) of the exponential argument in \eqref{E:AInt}.
Noting that
$$
\Li_2'(x) = - x^{-1} \log(1-x),
$$
we see that a critical point occurs when
\begin{equation*}
f'(u) =  2\pi i \left(2\pi i u - \log   \left( 1- e^{4\pi i u}\right) \right) = 0,
\end{equation*}
which has a unique solution in the upper half-plane given by $u = v := \frac{i}{2 \pi} \log\left(\phi\right)$, where $\phi := \frac{1 + \sqrt{5}}{2}$ is the golden ratio.  Equivalently, this point is also given by $e^{2 \pi i v} = \phi^{-1}.$  This also determines the most natural value for $C$ in \eqref{E:AInt}, namely $C = \frac{\log\left(\phi\right)}{2 \pi}$, so that the contour passes through the critical point.

At the critical point, we evaluate
\begin{equation}
\label{E:hCrit}
f(v)= \frac12 \log^2 (\phi)  +\frac12 \text{Li}_2 \left(\phi^{-2}  \right) = \frac{\pi^2}{30},
\end{equation}
using the special value (found in Section I.1 of \cite{Zag07})
\begin{equation*}
\text{Li}_2\left(\phi^{-2}\right)=\frac{\pi^2}{15}-\log^2(\phi).
\end{equation*}
We also need the value of the second derivative, so we calculate
\begin{equation*}
f''(u)
= -4\pi^2 \left( 1+\frac{2e^{4\pi i u}}{1-e^{4\pi i u}} \right).
\end{equation*}
Inserting the special value $v$,  we then find
$$
f''(v) = -4\pi^2 \left(\frac{1+\phi^{-2}}{1-\phi^{-2}} \right) = -4\pi^2 \sqrt{5}.
$$

Writing $u = v + \sqrt{\varepsilon}z$, then yields that the integrand in \eqref{E:AInt}  has the overall expansion
\begin{align*}
\exp & \left( \frac{f(u)}{\varepsilon} + \pi i u + \frac12 \log \left( 1-e^{4 \pi iu} \right)
+ O (\varepsilon) \right) \\
& = \phi^{-1} \exp \left( \frac{f(v)}{\varepsilon} + \frac{f''(v)}{2} z^2 +O \left( \varepsilon^{\frac12} \right)  \right)
= \phi^{-1} \exp\left(\frac{\pi^2}{30\varepsilon} + \frac{f''(v)z^2}{2} \right) \left( 1+O\left( \varepsilon^{\frac12} \right) \right).
\end{align*}
Inserting this into (\ref{E:AInt}) and changing variables gives  that
\begin{equation} \label{Aepsilon0}
\mathscr{A} \left(e^{-\varepsilon} \right)
=-\sqrt{2\pi} \phi^{-1} e^{\frac{\pi^2}{30\varepsilon}}
\left(1 + O\left(\varepsilon^{\frac{1}{2}}\right)\right)
 \int_{\field{R}} e^{\frac{f''(v) z^2}{2}} dz.
\end{equation}
The (Gaussian) integral evaluates to
$$
\sqrt{\frac{2 \pi}{-f''(v)}} =\sqrt{\frac{1}{2\pi\sqrt{5}}},
$$
which implies that
\begin{equation}
\label{E:AAsymp}
\mathscr{A} \left(e^{-\varepsilon} \right)
 =-\frac{\phi^{-1}}{5^{\frac14}} e^{\frac{\pi^2}{30\varepsilon}} \left(1 + O \left( \varepsilon^{\frac12} \right) \right).
\end{equation}

Recalling the asymptotic formula \eqref{E:qinfty} and combining with \eqref{E:AAsymp}, we finally have
\begin{equation}
\label{E:CAsymp}
\mathscr{H} \left(e^{-\varepsilon} \right) \sim
  \frac{ \phi^{-1}}{\sqrt{2\pi} \, 5^{\frac14}} \varepsilon^{\frac12} e^{\frac{\pi^2}{5\varepsilon}}.
\end{equation}

In order to finish the proof of the first asymptotic formula in Theorem \ref{T:h}, we first apply Lemma \ref{L:mono} to \eqref{E:hgen} in order to conclude that $\mathscr{H}(q)$ has monotonically increasing coefficients (this can also be shown directly from the definition of shifted stacks). We can then apply Theorem \ref{T:Ingham} to \eqref{E:CAsymp}, with the constants $\lambda= \frac{\phi^{-1}}{\sqrt{2\pi} 5^{\frac14}}, \alpha=\frac12,$ and $ A = \frac{\pi^2}{5}.$  This finally gives the desired formula
$$
h(n) \sim \frac{\phi^{-1}}{2\sqrt{2} \, 5^{\frac34} n} \, e^{2\pi \sqrt{\frac{n}{5}}}.
$$

We conclude this section by discussing Zagier's classification of modular $q$-series.  If the series in \eqref{E:AConst} is re-expanded using the $q$-binomial theorem and \eqref{E:theta}, we obtain the formula
\begin{equation*}
\mathscr{H}(q) = 1 + \frac{1}{(q)_\infty} \sum_{m \geq 0} \frac{q^{(2m+1)(m+1)}}{(q^2; q^2)_m}.
\end{equation*}
This is not one of the seven examples given in Table 1 of \cite{Zag06}, and is therefore not modular.

\subsection{Shifted stacks with summits}
\label{S:shifted:summit}
We proceed as in \eqref{E:CConst}, with the only difference being some small shifts in the second sum. In particular, we express the generating function as
\begin{align*}
\mathscr{H}_s(q) &= \operatorname{coeff} \left[x^0\right] \left(\sum_{r\geq 0} \frac{x^{-r} q^{\frac{r(r-1)}{2}}}{(q)_r}
\sum_{m\geq 0} \frac{x^{m}q^m}{(q)_m}\right)
=1 + \operatorname{coeff} \left[x^0\right] \left(\left( -x^{-1}\right)_{\infty}  \left(xq\right)_\infty^{-1} \right),
\end{align*}
and thereby conclude the asymptotic formula
\begin{align}
\label{E:GsimC}
\mathscr{H}_s  \left(e^{-\varepsilon} \right)   & = 1 + \frac{q^{-\frac18}}{(q)_{\infty}} \sqrt{\frac{2\pi}{\varepsilon}} \int_{\field{R} + iC} \exp \left(-\frac{2\pi^2 u^2}{\varepsilon} - \pi i u + \text{Li}_2 \left(x^{2}q^2;q^2 \right) \right) du \\
& \sim \phi \, \mathscr{H}  \left(e^{-\varepsilon} \right). \notag
\end{align}
The extra factor of $\phi$ arises from the fact that $e^{\pi i u}$ in the integrand of \eqref{E:AInt} has been replaced by $e^{- \pi i u}$, recalling further that the critical point is given by $e^{-2 \pi i v} = \phi$.
Arguing as before (using Lemma \ref{L:mono} on \eqref{E:hsgen} and applying Theorem \ref{T:Ingham}), we find a similar asymptotic relationship between the coefficients, concluding that
\begin{equation*}
h_s(n) \sim  \phi \cdot h (n).
\end{equation*}

\begin{remark*}
As described by Watson in \cite{Wat36} (see equation (C)), Ramanujan showed that
$$
\mathscr{H}_s(q) = (q)^{-1} G_{\frac{1}{2}}\left(q^2\right),
$$
where
\begin{equation*}
G_u(q) := \sum_{n \geq 0} \frac{q^{n(n+u)}}{(q)_n}.
\end{equation*}
The famous Rogers-Ramanujan identities \cite{RR19} state that if $u = 0$ or $1$, then
\begin{equation}
\label{E:RR}
G_u(q) = \frac{1}{\left(q^{1+r}; q^5\right)_\infty \left(q^{4-r}; q^5\right)_\infty}.
\end{equation}
The right-side of \eqref{E:RR} is a modular function (up to a rational power of $q$) , and as such its asymptotic expansion near $q=1$ is well-understood. Under the assumption that $G_u(q)$ has an asymptotic expansion of the same shape for both integral and non-integral $u$, Ramanujan then recovered an expression equivalent to \eqref{E:GsimC}, although he did not offer any justification for this assumption.
\end{remark*}

\section{Stacks with summits}
\label{S:summit}

In this final section we briefly outline how Stanley's  and Wright's work in \cite{Stan11} and \cite{Wri71}, respectively, together imply the asymptotic formula for stacks with summits from Table \ref{Tab:stack}, namely that as $n \to \infty,$
\begin{equation}
\label{E:stackssummits}
ss(n) \sim \frac{1}{2^3 3^{\frac{3}{4}} n^{\frac{5}{4}}} e^{2\pi \sqrt{\frac{n}{3}}}.
\end{equation}
Specifically, Stanley provided a formula for the generating function of stacks with summits, using an inclusion-exclusion argument to prove Proposition 2.5.1 in \cite{Stan11}, which gives that
\begin{equation}
\label{E:SsStan}
\mathscr{S}_s(q) = \frac{1}{(q)_\infty^{2}} \big( 1 - L(q)\big).
\end{equation}
Here we have adopted Wright's notation for the ``false theta function'' (see equation (1.4) in \cite{Wri71})
\begin{equation*}
L(q) := \sum_{n \geq 1} (-1)^{n-1} q^{\frac{n(n+1)}{2}}.
\end{equation*}
Wright's own analysis of stacks in \cite{Wri71} began with the closely related expression
\begin{equation}
\label{E:SAul}
\mathscr{S}(q) = \frac{1}{(q)_\infty^{2}} L(q);
\end{equation}
which had previously appeared as equation (20) of \cite{Aul51}.  Wright's proof can then essentially be followed verbatim with $L(q)$ replaced by $1 - L(q)$, which leads to the asymptotic formula for stacks with summits.

\begin{remark*}
Although Stanley and Wright's arguments provide additional combinatorial and analytic information, we point out that \eqref{E:SsStan} also follows immediately from a specialization of Heine's $_2 \phi_1$ transformation; in particular, set $a = b = 0$ and $c = t = q$ in Corollary 2.3 of \cite{And98}.
\end{remark*}

It is perhaps worthwhile (particularly for non-expert readers) to further discuss why Wright's proof can be modified in this manner.  Wright's general argument follows the philosophy of the Circle Method, where the coefficients of a generating series are recovered by applying Cauchy's theorem, with the analysis further amplified by using asymptotic properties of the series near roots of unity.  In the case of modular forms (such as the partition function, up to a power of $q$), Hardy and Ramanujan \cite{HR} (and later Rademacher \cite{Rademacher}) used cuspidal expansions in order to derive exact formulas for the coefficients.  However, the presence of the non-modular $L(q)$ in \eqref{E:SAul} prevents a similarly powerful analysis for stacks.

Wright's modified approach instead has just one ``Major Arc'', around the exponential singularity at the cusp $q = 1$, and is useful as it can be applied to any series that can be expressed as the product of a weakly holomorphic modular form with poles at the cusps and a series with a convergent asymptotic expansion (e.g., a meromorphic Laurent series such as $L(q)$).  In particular, the terms in the asymptotic expansion of $s(n)$ can then be found from the asymptotic expansion of $L(q)$ near $q=1$, which is proven by Euler-MacLaurin summation in Lemma 1 of \cite{Wri71}.
If $q = e^{-\varepsilon}$ and $\varepsilon \to 0^+$, then the first term of Wright's expansion is stated as
\begin{equation}
\label{E:Lasymp}
L(q) = \frac{1}{2} + O(\varepsilon);
\end{equation}
in other words, $L(q) \sim 1 - L(q)$, so the asymptotic main terms for $s(n)$ and $ss(n)$ are identical.

\begin{remark*}
Alternatively, \eqref{E:stackssummits} may also be proven by applying Ingham's Tauberian Theorem, which may be viewed as an even more broadly applicable version of the Circle Method, with a correspondingly weaker conclusion that only provides the main asymptotic term.  Starting with \eqref{E:SsStan} and \eqref{E:Lasymp}, one can use \eqref{E:qinfty}, Theorem \ref{T:Ingham} and Lemma \ref{L:mono} from the present paper in order to conclude the claimed asymptotic formula for $ss(n)$.
\end{remark*}

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The confirmation of John Russo, student body presidential candidate for the Vitality party, to the position of Board Member of the Office of Governmental Affairs (OGA) has sparked recent criticism from some student senators. OGA is a division of the executive branch that deals with the promotion of civic engagement on campus.
Last fall, Russo applied to the position of Board Member of OGA but was rejected by the Internal Affairs Committee before reaching the Student Senate for a final vote of confirmation. In January, Russo applied to the same position again but ultimately, Internal Affairs could not convene because they could not meet their own attendance requirement (quorum), and the interview with Internal Affairs was cancelled.
In cases where standing committees such as Internal Affairs cannot meet during their regularly scheduled time, a motion can be placed to move the Student Senate as a whole into a committee meeting. As a result, the Student Senate opted to adopt the role of Internal Affairs, taking on the responsibility of interviewing Russo before ultimately confirming him to the position of Board of OGA by a majority vote of 27 senators in favor, eight against with four abstaining.
Following the meeting, Senator Michael Moorhead published an opinion dissenting from the majority's decision to confirm Russo. Moorhead contends that the role of Internal Affairs is kept separate from the Student Senate as a whole for a reason and claims Russo may have leveraged his party's majority in the Senate to his advantage.
"I speculate that Mr. Russo did not attend the regularly scheduled Internal Affairs meeting because he thought he would be failed again, however, as his party is the controlling party of the Senate, he could pass Internal Affairs if it met in 'committee as a whole,'" Moorhead writes.
The application process of becoming a Board Member of OGA is regulated by a system of checks and balances. After a candidate submits his application with his resume attached, they are interviewed by a representative of the executive branch who selects certain candidates to then be interviewed by Internal Affairs, before being approved and forwarded to the Student Senate as a whole.
"I don't think Senate should be adjusting our schedule to vet a candidate because he's announcing he's running for something else," says Senator Hannah McLaughlin, who spoke out against Russo's confirmation during the meeting. McLaughlin added that it is not the Senate's responsibility to be acting as Internal Affairs, regardless of the circumstance.
"It is not the duty of all Senators to be well versed in the processes of other committees. For my committee specifically, I do not expect those Senators outside my committee to understand how the university budgeting process works," Moorhead, who also serves as the Budget Committee Chairman, writes. "Similarly, [Internal Affairs] interviewing process is not something all Senators are well versed in, nor are they expected to be."
Back in fall, during the vetting process for Russo's initial application, Internal Affairs cited Russo's inexperience as the reason he was not forwarded to the Senate for final approval.
"[Russo] ultimately wasn't forwarded by the Internal Affairs committee because they felt he did not know enough about the Office of Governmental Affairs," Senate President Pro Tempore Justin Ravelo said. He added that Russo, at the time, did not know specifically about the official lobbyist that works in cooperation with SGA to get university lobbying efforts accomplished at the Capital, a vital component of OGA.
According to Austin Engelbrecht, Director of OGA, Russo had later reached to him personally and discussed with him in specifics about OGA's duties and the position.
During the interview process, which took place on the floor of the Senate, senators also pointed out two violations in statutes regarding Russo's reapplication process, the first being that the forwarding letter did not contain a resume and the second being that the position was not advertised properly.
Per SGA statue 307.2 (A)(1) and 304.2 (C) respectively, all open/vacant positions within SGA must be published by the SGA Webmaster, a non-student employed by the university whose job it is to maintain the SGA website, and "all positions shall be advertised for at least one week."
According to Engelbrecht, it was determined on the floor that SGA President Jean Tabares had emailed the Webmaster to post the vacancies in fall and while it had still not been posted by spring, it was out of the student's hands.
Currently there are four vacancies in the Office of Governmental Affairs: Deputy Director, Internal Assistant Director and two open board positions – none of which are currently advertised on the SGA website as of the publishing of this article.
"SGA is entrusted by students to be open and transparent, and we break that trust when we don't care about the importance of advertising positions," says Senator Isaac Urichko, who also disagreed with the Senate's majority decision to ignore statutes.
As Director of OGA, Engelbrecht did not agree with the senator's reservations. He noted that there are other steps that can be taken within the system of checks and balances.
"John's forwarding and confirmation to the Board of Directors was completely legitimate," Engelbrecht said. "If there were truly any statutes broken, I don't know why it wasn't taken to the Supreme Court."
"We, as a result of this forwarding have begun down this slippery-slope, and must aim to get a grip before our feet are completely out from under us," Moorhead, who concluded his dissent by respectfully disagreeing with Russo's confirmation, writes.
SGA elections will take place Wednesday, Feb. 24 at 8 a.m. at sga.fsu.edu or Union Green.
David Walker contributed to this report

redpajama_set_name: RedPajamaCommonCrawl


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Be Afraid, Be Very Afraid: Roger Federer Isn't Planning To Go Anywhere
By Marianne Bevis
Rumors of Roger Federer's demise in tennis have peppered the media since the early days of 2008. For that was when he failed to reach his first Grand Slam final in almost three years.
It was the Australian Open, and he fell short—in the semifinals. He would go on to win just four games in the French Open final, would lose his Wimbledon crown, and would not win a Masters title throughout the entire year.
It mattered not that he was recovering from the debilitating effects of glandular fever during the early months of that year, nor that he struggled with a back injury in the latter stages. Nor, indeed, that he had made the finals of three out of the year's four Slams.
There was, for a brief moment, a little back-tracking when Federer won the title at Flushing Meadows for a record fifth consecutive time, before the muttering started up at his failure to pass the Round Robin phase of the World Tour Finals.
Just a month into 2009, and the clouds settled over the Federer story again. It was not so much his failure to win the first Slam of the year but that he lost it, in a long, tough final, to Rafael Nadal.
If the clay-court supremo could take the Australian title, on the surface always considered to favor the Swiss, this really did mark the handing over of the baton to the younger man, a transition to a new order.
Federer left the tour to rehabilitate his niggling back while Nadal soared away from the field in the rankings, taking title after title.
With the Australian Open in the bag, and the French on the horizon, it was even possible that Nadal could do what Federer had failed to: win the Grand Slam.
As if to add grist to the media mill, Federer announced that he was to become a father, and then married his childhood sweetheart. If ever there was sign that he had entered a different phase of his life, this was it.
How many fathers had won Grand Slams? Of the eig...


Saxophonist Fred Madison feels weird uneasy feeling of paranoia, when it detects that his house was under surveillance. In addition, he suspects his wife of infidelity. At a party he has a strange encounter with a mysterious man will only intensify the fear musician. The next morning, Fred discovers a corpse in his bed his wife, and he was arrested for murder. In death row awaiting execution Fred his sentence. But when jailers come for him on the spot Madison they discover some of Pete Dayton, who does not understand how I got here. With help of this site you can download Lost Highway in HD free full Movie.


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Сравнить время
Атомные часы
Время в г. Тунис
Текущее местное время
Querying time server...
--:------
-- ---- ----
Подробности о Тунис
Часовой пояс ---
Лето (летнее время) ---
Часовые пояса 1
Длинное имя al-Jumhūriyyah at-Tūnisiyyah
Столица Tunis
Валюта TND
Сокращения TN, TUN
Телефонный код 216
Язык Arabic
Города в Тунис
Время сейчас
Поиск страны / города, чтобы проверить местное время
Погода в Тунис
Тунис Фотографии
Больше информации о Тунис
Tunisia is a country on the Mediterranean coast of North Africa. It is the smallest and easternmost of the states that extend along the Atlas Mountains and is bordered by Algeria to the west and Libya to the southeast. 40% of the country is occupied by the Sahara desert and the rest comprising fertile soils and accessible shores. The geography of the country played an important role in ancient times, first with Carthage, a city of the Phoenicians, and later as an African province, as was the granary of the Roman Empire.Tunisia is a constitutional presidential republic with the President head of state and the Prime Minister as a leader. The new Constitution adopted on January 26, 2014 and guaranteeing women's rights, and stipulates that the religion of the President "should be Islam." Executive power is exercised by the government, while the legislative from the government and from the unicellular House of Representatives. The mandate of the president is five years. In October 2014 took place in the country, the first elections under the new constitution after the Arab Spring. Voting in elections are many and those who are aged 18 and over.
Life expectancy for the entire population were consistent with 2015 estimates of 75.89 years (73.79 years as the men and women 78.14)
Official currency: Tunisian Dinar (TND)
Тунис Карта
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3 stars Archives - Page 2 of 7 - thefatrumpirate.com the!
Flor de Cana Extra Seco 4. Flor de Cana are a rum producer from Nicaragua. They are one of the most popular and recognised brands worldwide.
Bounty Fiji Rum Premium Overproof Rum. It might well be the case that we never see this rum for sale in Europe (or the US). It seems to say firmly in the Oceania area of the Southern Hemisphere.
Rathlee Distilling Co Golden Barrel Aged Rum Aged 3 Years. Rathlee Distilling Co are the brainchild of husb and and wife team Stuart and Paolo Leather. Formed in 2012 they originally released their rum under the Revolver rum brand.


Is sensitive skincare really that sensitive?
Now Avène have come with something new, the Tolérance EXTRÊME collection that has been created for those of us with sensitive skin.
There are loads of 'sensitive' skin products out there with many of them still containing parabens, alcohol and even fragrance. If your skin is prone to sensitivity, you'll already know to avoid these ingredients like the plague! So Tolerance is not only a collection with kind ingredients, but the most smarty-pants-clever packaging you've seen yet.
When using products, particularly for sensitive skin, that are either in a pot or a tube, they are prone to contamination. With a pot it's when you stick your fingers inside, and when it's a tube, it's the air that's forced back in.
Normally this is fine, because there are often ingredients that kill bacteria and contaminants, but these are the ingredients that can easily irritate sensitive skin. With me so far?
So, Avène Tolérance EXTRÊME comes in tubes that once you squeeze, the air is blocked from going back inside… Cool, right?
If you haven't seen Avène before, you'd be forgiven. They're enormous on the continent as you can see below in this relatively tiny French pharmacy – I sadly didn't buy anything as only had hand luggage which makes it a pain to bring back. In the UK, Avène can now happily be found in most Boots stores – in the past it used to just be flagships, but now they're everywhere!
Skincare is sold very different within Europe – although you'll have your regular beauty hall brands, the pharmacy skincare brands are sold by Pharmacists who have a deep medical knowledge and understanding of acute skin concerns.
Avène are actually owned and produced by a pharmaceutical company, and this means that we are governed by certain rules and everything that comes with it. Pharmaceutical companies operate a practice called 'pharmacovigilance' which means that any reaction (itchiness, redness, rash, etc) must be reported to their head office.
Other than mentioning their kind ingredients and packaging, I'm purposely not talking about the ingredients, how it feels or what it does as I have been challenged to try the skincare out for 3 straight weeks and to report back. It's also worth mentioning that the Cleansing Lotion with 7 ingredients is available in 2017 (Jan) but the one right now has 8 ingredients and is available.
In the meantime, you can check out the full Avène range here at Escentual.
This article first appeared on MANFACE written by Thom Watson. It was published on Wednesday 31st August 2016 and was last updated on Monday 3rd April 2017 at 12:46pm. It's categorised under Avene, Dry / Dehydrated Skin, Sensitive Skin, Sponsored where you'll find more related content.


It has been stated that in Heaven there is only one law, Love. "Love God Above All Else" is a tribute to the First Commandment. A gentle, bold and celestial reminder to adore that which gives adoration. If we could just remember this – we probably would not even need the other 9 commandments. Let Love rule!


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What Should the Rangers Do With Smith?
By Jordan Goldberg January 29th, 2019
The New York Rangers will have a very important decision to make regarding defenseman Brendan Smith. Smith's defensive troubles are persistent and are often a primary cause of the team's defensive woes. He was acquired by the Rangers before the 2017 Trade Deadline in exchange for two draft picks. The team went on to make the playoffs that season, falling to the Ottawa Senators in the second round. Smith's play during the postseason was widely regarded as positive.
But as the 2017-18 season drew on, Smith's play slowly deteriorated until Rangers' management could take it no longer, assigning him to the Hartford Wolf Pack of the AHL. Several weeks following his demotion, it was reported that Smith got into an altercation and punched Wolf Pack teammate Vinni Lettieri.
Brendan Smith, New York Rangers, September 27, 2018 (Amy Irvin / The Hockey Writers)
As bad as Smith was last season, he ended up with a positive plus/minus rating and eight points with the Rangers. He managed to lose weight this past offseason and make the team at the start of the season.
Smith's Struggles Continue
Smith has been undoubtedly the Rangers' worst defenseman this season. He has the second-lowest plus/minus rating, ahead of only Kevin Shattenkirk. He also has among the lowest number of takeaways on the Rangers' blue line.
Another consistent issue Smith has faced is staying out of the penalty box. He has the second-most penalty minutes while playing significantly fewer games than the majority of the team. Delicately put, this has been a season for him to forget.
Related: Rangers' Power Play Surges, Defense Collapses
So what do the Rangers do from here? The ideal solution would be to trade him before he loses his entire trade value. But even at that, he is not an attractive asset at the moment. A trade for Smith would return to the Rangers a late-round pick at best.
I question if a team like the Senators would consider trading for Smith due to the NHL experience he would bring to a mostly mediocre blue line. Or perhaps a team with an already solid defense pushing towards the playoffs could look to add Smith as a depth defenseman.
Rangers' Brendan Smith. (Eric Bolte-USA TODAY Sports)
A trade sending Smith out of New York is certainly not impossible, but it is still far fetched. From his poor stats to his undisciplined play, he is not a player any team would consider their first choice in a trade.
Surely there should be at least one team desperate enough to change their team and take on Smith's big contract in what would be a new reclamation project. What complicates this is his modified no-trade clause through the conclusion of his contract.
Smith's NTC allows him to submit a list of 15 teams to which he'd be willing to accept a trade through the 2019 season and a 10-team list after that.
Trade Alternatives
There is also the buyout option, which seems more likely than a trade. Smith has a cap hit of $4.35 million over the next two years. The Rangers, who will be looking to re-sign some of their veterans like Chris Kreider, may need that money available.
A four-year buyout of Smith would cost the Rangers approximately $1.145 million per year. This transaction would save several million dollars in the long-run and allow the Rangers to allocate more money to re-signing their free agents.
John Gilmour, New York Rangers, Mar. 22, 2018 (Amy Irvin / The Hockey Writers)
Another option the Rangers have, which is one they've already enacted, is to leave Smith in Hartford. He has made the trip between Connecticut and New York far too many times for a player with his salary. But when it comes down to it, only the players that give a team the best chance to win should be in the lineup.
At this point in time, there are several players who should be playing instead of Smith. I would even go as far as to say Wolf Pack defensemen John Gilmour and Libor Hajek should be given opportunities to take Smith's spot. Gilmour was Hartford's All-Star representative and has notched 34 points in 45 games.
Related from the Archives: Gilmour & Pionk are Fitting Quite Nicely with the Rangers
For now, it will be interesting to see if a team on Smith's acceptable trade list actually does bite, which would send a late-round pick to New York. Until then, the Rangers are back in action tonight against the Philadelphia Flyers with Neal Pionk expected to return to the lineup.
Brendan Smith


The Garden Island newspaper had a front page story yesterday about the state Department of Land and Natural Resources (DLNR) warning Inter-Island Helicopter to stop illegal landings. In another example of helicopter companies flaunting the laws, they are advertising the tour on their website, and DLNR staff found a clearing and "temporary" structures to provide shade in the Moloa'a Forest Reserve, both presumably illegal. The DLNR adds that permits are needed for landings in state managed lands, though it's likely they wouldn't be granted for quasi-wilderness areas such as the remote forest reserves.
Inter-Island Helicopters are the bad boys of the Kauai helicopter industry. They don't have the sleek modern helicopter fleet that most other operators have, but they were the first to fly with the doors off, which many people found thrilling. They fly out of Port Allen airport, where they have a bunch of makeshift offices and structures with questionable permits (scroll down on the linked page). However, they are not just a tourist ride, their utilitarian choppers and skilled pilots are contracted by the county for mountain rescue and fire fighting. The owner's son perished in an accident last Christmas while refilling a fire-fighting bucket at a reservoir near Lihue.
But the competition in the industry is driving operators to seek new thrills to sell, and remote waterfalls and forests are easy targets. One of the Robinson Family members recently applied to the planning commission for a landing permit, saying that their Niihau helicopters would stop at a botanical reserve that they own in the hills above Hanapepe. At the hearing, it turned out that another company with far more flights wanted to share the landing permit and that the location happens to be in the Jurassic Park movie.
Helicopters used to fly tourists to the Kalalau valley for the day, and I'm not sure when that stopped or why. But between the noise impact and the danger of spreading invasive species, it's easy to see why landings are undesireable. What's clear is that the operators are looking for new products, and allowing landings for one will make them all feel entitled, resulting in more flights and more nuisance for residents, hikers, and wildlife.
This entry was posted on Thursday, July 13th, 2006 at 11:11 pm and is filed under Helicopters . You can follow any responses to this entry through the RSS 2.0 feed. You can skip to the end and leave a response. Pinging is currently not allowed.
i'm glad i found your blog here. i commented on another article i found via google, and then stumbled on to this topic as well. are there still issues with helicopters, and in particular, the kalalau? we were out there two weeks ago, and an inter-island helicopter landed on the beach twice, no more than 75 yards from where we were camping.
those bastards should lose their permit or something. seems totally ridiculous.
The only helicopter issues I'm aware of in Kalalau is early morning (8am) flyovers that are loud enough to wake up campers–quite annoying. I hadn't heard of any landings before you reported it here.
I do know that Inter-Island does rescue work for the county, and perhaps also maintenance flights for the state park service. So perhaps they were assisting someone, training for rescues, or doing some sort of cleanup.
Definitely weren't assisting anyone. The guys landed, let off a passenger who appeared to go use the restroom. They then flew around in the valley for perhaps 20 minutes, and then landed a second time, to pick the passenger up.
It appeared they were doing a flight-seeing tour, because we saw them flying extremely low in the valley earlier in the day. Possibly they were seeing if there was any damage from the flood?
At any rate, the low flying helicopters were ridiculous. There were two other instances where they were within 75-100 feet of the trail or beach. Certainly there must be a ceiling they should fly at?
likely "buds". Alot of money involved too.
on Earth and that is some serious real estate, can even see the missle silos. Awesome beaches! What's up with the airstrip, what flies outta there?
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Home Must Reads
Anxious Parents Speak Out Over Worry For Son Who Faces Three Years In UAE Jail For Accidentally Touching Man's Hip
The parents of a Brit arrested in Dubai for accidentally touching another man's hip to avoid bumping into him have warned others to stay away from the country. Jamie Harron is facing three years in a UAE jail for "public indecency" after he placed his hand on a man so they did not bump and […]
by Jess Young
in Must Reads, World News
Jamie Harron parents Graham and Patricia are distraught with fear for Jamie's future.
The parents of a Brit arrested in Dubai for accidentally touching another man's hip to avoid bumping into him have warned others to stay away from the country.
Jamie Harron is facing three years in a UAE jail for "public indecency" after he placed his hand on a man so they did not bump and spill their drinks.
And as his trial approaches Jamie's mother Patricia says she cannot even sleep out of worry for her son – and his father is warning Brits against visiting Dubai.
"I'm up all night worrying about what is going to happen to Jamie in court. There's no chance of sleep," she said.
And his father Graham, a caterer, added: "Patricia and I have not been able to relax for a moment. We never dreamed we would have to face something like this.
"We are a very close family and it is killing us to think of him spending even three nights in jail, let alone three years.
"People have to stop visiting that country. I think it is unacceptable that the FCO actually promotes the UAE to British tourists."
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27-year-old Jamie has been held in Dubai for the last three months following the incident in June.
He has spent over £32,000 in expenses and legal fees trying to resolve the cultural misunderstanding and has been stripped of his passport and his job in Afghanistan.
He even spent five days sharing a cell and a single mattress with eight men in a notorious prison.
He said: "The whole thing is like a horrible dream and I just don't know when it is going to end. I thought it would be over by now but it feels like it will never be.
"This has broken me, financially and emotionally."
And his anxious parents have said that the whole family's finances are now in "tatters", after they stretched their own finances to breaking point to help their son.
Jamie's legal battle began in June, when he was enjoying a night out with friends at the popular Rock Bottom bar in Dubai.
Jamie and his friend had bought their first drink in the crowded bar when he placed his hand on the hip of man to ensure they didn't bump and spill their drinks.
The Brit denies public indecency, saying he was simply crossing the dance floor to avoid confrontation.
But now his family, and his representative Radha Stirling at NGO Detained in Dubai, are all waiting "anxiously" to hear whether he will be jailed for missing a court hearing – that neither he nor his lawyer were made aware of.
Distraught Jamie said: "Of course if I had known there was a change in the court date, I would have been there.
"I have been waiting for months to appear and now it looks as though I will be
arrested when I go to the hearing.
"It is completely unfair and I can't understand how I can be sentenced for failing to appear when I was not even informed of the hearing."
Jamie's dad Graham added: "We can't believe this nightmare has gone on for three months. Jamie is a good boy, he is never a problem and never in trouble.
"The UK embassy has not done anything to help and seem to think it is absolutely fine for British nationals to be abused in this way.
"Since Jamie was arrested I have researched and found that this is more common than any of us think.
"Where are the embassy staff who should be offering emotional and practical support to both Jamie in the UAE and his family back home in Stirling?"
In a statement, Radha Stirling of Detained In Dubai said: "This case has been brought to the attention of the UAE government, who we hope will intervene and stop the injustice.
"I have also been contacted by Jamie's local MP, Stephen Kerr, who asked whether he should intervene, and we have requested he do so.
"More and more British nationals have written Dubai off as a possible holiday destination after several recent high profile arrests and we are inundated with more stories of legal abuse.
"We are hoping to work with the UAE government to make improvements to make the country safer for tourists, expats and investors."
She added: "We are all anxiously awaiting to see whether Jamie is jailed for previously not turning up to a court hearing that neither he, nor his lawyer, were advised of."
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Video : Altenew – Stamping a Sentiment Background with Inlaid Die Cutting!
Today I have 'A Lot' of stamping for the Altenew Blog and it's all from the same stamp set too. Basically I've created my own sentiment background and die cut a hidden message within. I created a video of the process to show you how.
I chose some fun bright Crisp Dye Ink Colours and stamped the Label Love sentiments in lines over nearly the whole card front panel.
Then I inlay die cut my hidden sentiment using the Bold Alphabet Die Set.
And as a final touch I added some Altenew enamel dots (which I am happy to say can now be bought from the online store).
I thought it would be fun to add some depth to the letters by using black fun foam as a base.
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Next > Video : Altenew July Inspiration Challenge!
COOL card!! What a great idea! Thanks for sharing.
Wow you always amaze me with what you come up with!!! I always love!!
Love the colour combo and oh so much patience in creating the background :). Looks fab.
Love your card. Thanks so much for the video.
That looks so cool! I love those label sentiments and your die cutting technique! Have a great day!
How imaginative! I love what you do with words and die cutting. Thanks.
Absolutely fabulous Therese. Love how you stamped and stamped and stamped your background. Stunning result with the popped sentiment.
The colours you chose for this card are awesome! I love these alpha dies and can't wait to get them myself. I have a hard time making large cards (A2 is my go to size) but I usually scrapbook in 12×12! Go figure!
Oh my gosh!! Therese, this card is amazing!!! what patience you have to line up all of those sentiments so beautifully and ever so perfectly without the aid of any positioning tool!! my hand would be shaking so much I would ruin the panel.
Love it when you do sentiments like this, specially if they are bold and big, just fabulous.
I do want to ask you if you could possibly tell me where did you get that glue dispenser, love how precisely applies the glue in just the right amount.
Thank you for another fantastic and beautiful Masculine card, like you mentioned, this types of cards are hard to make, this one is just amazing.
I did remove the top from the bottle and added it directly to the matte medium bottle.
Oh my this looked like serious work. Gonna have to check out that video.
Fantastic idea… gorgeous card…love it..!!
Oh, how I love these giant alphas with this stamp set, Therese! Another fabulous card!!!
Great card, Therese! it's so clever how by using the black fun foam, you get a shadow that highlights the letters. ☺ I love the stamped background as well.


Editor's Note: This article appeared in the Spring 2012 edition of Wisconsin People & Ideas. The mining bill we reference in this piece, AB 426, had passed through the Wisconsin Assembly and moved into the Senate in February 2012.
Mining in the Penokee Hills — Midwest Environmental Advocates is part of the legal team representing the Bad River Band of Lake Superior Chippewa in their effort to protect the Penokee Range from an open-pit iron mine.
Iron county was named for the -red gold - iron ore that was mined in the rugged Penokee Iron Range from the 1880s until the 1960s. The most famous mine in the range was the Montreal Iron Mine - the deepest iron ore mine in the world.
Penokee Hills Mining Summit - River Alliance of WI. Last week, I followed two of my coworkers, Allison Werner and Helen Sarakinos, up north to Iron County for a mining summit that Allison was facilitating.
Iron mining deposit report for "Penokee Deposit" (#USGS10204529) in Ashland County, Wisconsin.
Gogebic Taconite and the Penokee Hills: The Battle Rages On Mary Annette Pember • January 28, 2014 The battle over building a huge open pit iron ore mine in the Penokee Hills in Northern Wisconsin is heating up on a number of fronts.
In a state where the landscape tends towards low-lying swamps, flat fields and rolling hills, northern Wisconsin's Penokee Range of mountains are a dramatic outlier.
The first indication that GTac might pull the plug on the Penokee mine project was in January ... Secretary of the Wisconsin Resources Protection Council.
The biggest group against the mine are Tribal leaders from the Penokee range area, these tribes feel they would be directly affected by the mine.
The Penokee Photo Album; ... Where are the Penokee Hills located? Location of the proposed Penokee Hills Mining Project.
A group of people are planning a snowshoe hike through the Penokee Range this weekend that crosses land closed to the public because of mining-related activity.. The hike meets Sunday at the site of the Harvest Camp, or HELP Village on Moorepark Road in the town of Anderson in Iron County.


how to change keyboard shortcuts: Mac Wireless Keyboard Trouble! HELP!?
I recently purchased a wireless keyboard for my iMac computer. and my old wired keyboard had the keyboard shortcuts-meaning F1-the birghtness would work when pressing on the key. or F7, F8, F9, F10, F11, F12 would work for my iTunes when I have music playing. Now that i have this new wireless keyboard when i press the brightness key, nothing happens. when i press F7 the replay key for iTunes nothing happens, when i press F8 it takes me to spaces. i tried to go to system preferences keyboard %26amp; go to keyboard shortcuts %26amp; try and change stuff but its all screwed up. how do i change it so that all the F1's-F12s or any other key is back to the way it was on my wired keyboard? i tried to look it up on google and forums but i havent found anything that made it work!
PLEASE HELP!Mac Wireless Keyboard Trouble! HELP!?
If the behavior is random, the battery is low.
http://support.apple.com/kb/HT3059Mac Wireless Keyboard Trouble! HELP!?
If it is a wireless Mac keyboard, then everything should be working. Just go to the keyboard section and set everything to the default settings. If it is not a Mac keyboard, check the website to see if there are new drivers for it.


What other people say about Roxana – testimonials from clients.
Your every movement, expression and action resonates in those around you. The gloriously infamous butterfly effect means your actions can change both the fates and the future. I help you establish this vibrant personal resonance.
Roxana Kia is a human dynamo with a rare ability to "see" people – rapidly, intensely and positively.
She helps open the floodgates for what you can achieve in your organisation, your profession, your team and your personal relations.
Using simple yet powerful appreciative techniques that focus on people's strengths, Roxana trains clients in identifying and expressing personal conviction, excellence and impact.
Her approach is direct, hands-on, challenging and experiential. More energy, with clearer focus. Better leadership – clearer decisions – more vibrant communication – authentic behaviour that inspires and encourages.
Roxana's subtle encouragement helps you unleash new realms of dynamic interaction and purpose-driven leadership and impact that power out of your personal qualities and your own unique capabilities, so you're able to engage, energise and inspire those around you.


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yes its getting exciting, my project is really coming on now, have started putting the lights in, they all worked, but when i came to take photos,would you believe it only one lit up, to tired to sort it today, so its on my list to do tomorrow.
I made a scissor lift (well thats what my son called it) to me its s thingy that you put the motor bikes on so you can work on them, I used balsa wood and bits from my stash, I then painted it in blue paint, I am very pleased with how it came out.
This is really fun! I hope your lights are easy to fix. The scissor lift is fantastic!


There are some, and this is my attempt at pulling them together in one easy to find place. This list will by no means be complete, but it's hoped that in time it will grow to become something worth while.
I take no credit for the publications, many are written in conjunction with local motorcyclists their representative organisations or local councils. I have credited the authors where known.
If you know of an Australian town, region or guide, and can provide a website or PDF file for same, drop me a note by clicking the "Got Sumik Ta Say?" link at the top right of this page.
Ballina Ballina Base Motorcycle Touring Guide (7.6 Mb PDF) – Peter 'The Bear' Thoeming & Ballina Shire Council.
Lithgow Motorcycle Touring Guide (15.6 Mb PDF) – Motorcycle Council of NSW & Lithgow City Council.
Motorcycling The Snowy Mountains (1 Mb PDF) – Snowy Region Motorcycle Safety Group.


The popular radio host has confirmed that she is expecting her second baby.
Taking to Instagram last night to share the happy news, Fifi Box has announced that she is expecting her second baby! The baby will be a sibling for Fifi's five year old daughter Trixie Belle from her relationship with ex-partner Grant Kenny.
Fifi wrote a lengthy caption alongside her Instagram post saying that she feels incredibly lucky to be expecting a baby at 41, with the help of IVF and an anonymous donor. "A few years ago, I made an appointment to see an IVF specialist…and I began an IVF journey to extend our little family. I am still pinching myself that I have been this blessed and that this little miracle is growing within me, a much wanted and longed for sibling for a very excited big sister," she said. Fifi was quick to acknowledge just how fortunate she is, with many couples facing the pain and struggle of multiple IVF attempts. "It is not lost on me that I am one of the lucky ones," she wrote.
Congratulations to Fifi and Trixie on this exciting news!
Did you have your children after 40? Let us know in the comments below!
Yes, I had kids after 40, and I too was acutely aware how lucky I was.
Amazing. A true blessing indeed.
I think the headline should read fifi box pregnant, rather than bring her age into it.
So nice for her. Congratulations!
More and more women become pregnant at 40, sometimes with the first child.
Congratulations Fifi,wishing you all the best for the future!


After scanning Tim's blog I now have the phases of the moon on my sidebar. I may swap it later for the one that shows you a new 'wow look at this' astronomical sight each day, if I can locate it again.
The moon really is beautiful though. Sure, it might just be a hunk of rock not even spinning in space, but some things shouldn't be analysed too carefully.


Paul Wasserman, a publicist for the Rolling Stones, Bob Dylan, U2, the Who and other top musical acts whose career collapsed when he was caught using some of his famous clients' names in a financial scheme, died on Sunday in Los Angeles. He was 73.
The cause was respiratory failure, said Joan Myers, a friend and fellow publicist who had once worked for Mr. Wasserman.
Known throughout the music industry as Wasso, Mr. Wasserman was one of the most sought-after publicists from the mid-1960s into the '90s. He was a sophisticated operator who began his career at a time when public relations for pop music often meant little more than setting up airport news conferences and photo shoots.
Born in the Bronx, Mr. Wasserman moved with his family to Los Angeles at the age of 3. A Stanford University law-school dropout, he was working as a journalist when he entered public relations. After writing what he thought was a tough article on Bob Hope, Mr. Wasserman was offered a job by Hope's publicist.
He quickly joined with another Hollywood public relations man, Jim Mahoney, and was assigned to work with Frank Sinatra, the Beach Boys and the Beatles' company, Apple Records. By the late '60s Mr. Mahoney had made Mr. Wasserman a partner, and the firm was renamed Mahoney/Wasserman Public Relations.
Mr. Wasserman was a tenacious advocate for his clients who could also go to great lengths for journalists. He once waited outside Mick Jagger's hotel room until 4 a.m. to make sure that a reporter got his interview.
His client list over the years included Neil Diamond, Tom Petty, Paul Simon, the Mamas and the Papas, Linda Ronstadt, James Taylor, Jack Nicholson and Dennis Hopper.
But after three decades as one of the most powerful publicists in Hollywood, Mr. Wasserman was arrested in 2000 on charges of grand theft.
He had been using his clients' names — including those of Yahoo and Mr. Nicholson — to swindle tens of thousands of dollars from friends and business associates by selling nonexistent stock options and other phony financial instruments.
He pleaded guilty and was sentenced to six months in jail and ordered to pay $87,000 in restitution.


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52 Insights
20th/Jun/2019
Junya Ishigami
"I want to create architecture as the new nature."
Japanese born architect Junya Ishigami wants to reinvent architecture altogether, intent on bringing back a new type of inspiration for the city, that is the power of nature. Here Junya talks about his philosophy behind his work, why there is power in forgetting rather than remembering and what does an architect do living in one of the most earthquake-prone countries in the world.
Intelligence. Inspiration. Conversations
Exploring the divide between popular culture and science once a week.
52 Insights publish one interview with one truly unique individual or collective each week. 52 Insights will introduce you to the many authors, creators, researchers, leaders and various other innovators out there who are changing the way we see the world. Read More
31st/Jan/2019
White Jesus is one of the most destructive forces for black people.
He's one-half of the explosive hip-hop group Run The Jewels, Killer Mike has been using his own words to articulate his people's disempowerment for decades. Now it all comes to a head in his latest hit TV show on Netflix, Trigger Warning.
Hip-hop,
Killer Mike,
Netflix,
Politics,
Run The Jewels,
22nd/Jan/2019
Willo Perron
"Difficult people make beautiful things."
From humble beginnings, Canadian multi-disciplinary designer Willo Perron has risen up to become a creative conscience to the stars, imagining worlds for Jay-Z, Kanye West and St. Vincent. In this exclusive interview, he grants us a rare insight into his world and what it's like to be the designer for Kanye West's mind.
Es Devlin,
Jay-Z,
Kanye West,
Set design,
Op-Ed: Are China and America Headed For a 21st Century Cold War?
If you were to open the newspapers as of late, you might be mistaken for thinking it was the 1950s all over again. Headlines talking of a…
World's Biggest Guaranteed Income Experiment Announced In Tiny Indian State
High in the hilltops of the Himalayas, one of the most ambitious experiments in human history is about to begin. Sikkim, one of the smallest states…
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Interviewied one of the greatest scientists of our time @geochurch about extending life, eliminating disease and th… https://t.co/v0iNnXV9DW, 2nd/May/2018
Our review of the anticipated @McQueen documentary has been selected for their new trailer take a look below. https://t.co/2B0hVjzDHw, 2nd/May/2018
Alex Honnold
"There's no real secret."
Politics, Science
Why Our World Is Getting Better
"We are destroying the world."
27th/Dec/2018
Coco Capitán
'Our entire life is built on remembering'
Art & Culture, Fashion
'Mayhem is the only that's going to change it.'
Politics, Uncategorised
7th/Dec/2018
'I'm an introvert who got corrupted.'
Art & Culture, Uncategorised
Trump,
Hollywood,
Acting,
Biology,
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David W. Lawless
September 17, 1956 - April 2, 2020
David Wesley Lawless, 64 of Princeton, WV went to be with the Lord on Thursday, April 2, 2020 following an illness. David was born on September 17, 1956 in McDowell County, WV to the late Phillip Lawless and Pearl Dove Lawless. David attended Princeton High School and received a scholarship for tennis to Bluefield State College where he received a degree in Business and was inducted to the Tennis Hall of Fame. David went to further his education at Bluefield College obtaining a master's degree in Business Administration. David began a career at AEP and has been a loyal employee for 40+ years. David was a lover of sports and his favorite teams included the WVU Mountaineers, Notre Dame Fighting Irish, Wake Forest Demon Deacons, and the Cincinnati Reds. He loved to play his guitar and wrote some of his own music. Those left to cherish his memory include two brothers, Jimmy Lawless and wife Violet, Michael Lawless and wife Jane, two sisters, Margaret McCormick, and Deborah Lawless, two sister in laws, Lois Lawless and Helen Lawless, two special cousins, Barbara Scott and Rusty Kitts. David is also survived by a host of precious nieces' nephews and cousins. Due to the ongoing circumstances surrounding COVID-19 the family will have a private celebration of David's life at Roselawn Funeral Home and Memorial Gardens. An online obituary and guest book may be signed by visiting roselawnmontevista.com Roselawn Funeral Home 450 Courthouse Road on Princeton is serving the Lawless Family
David Wesley Lawless, 64 of Princeton, WV went to be with the Lord on Thursday, April 2, 2020 following an illness. David was born on September 17, 1956 in McDowell County, WV to the late Phillip Lawless and Pearl Dove Lawless. David attended... View Obituary & Service Information
The family of David W. Lawless created this Life Tributes page to make it easy to share your memories.
David Wesley Lawless, 64 of Princeton, WV went to be with the...
Send flowers to the Lawless family.


Support Naya Jeevan in this exciting journey through perilous waters. You'll have to be agile and quick to tap your way clear of the stinging tentacles of Bernie the Jellyfish and the suffocating grasp of Scary Octy the Octopus. Be extra careful as you weave through caves of razor-sharp coral while avoiding the sharp poisonous barbs of Spike the Urchin and Puffy the Blowfish who could blow you away in an instant! Collect gleaming pearls to unlock the other colourful characters to swim as in this grand underwater world, and make the best use of your diver's armor. These, and many more challenges await, so answer the siren call of the sea and tap your way to subaquatic fame and support Naya Jeevan while doing it!
The vision of Naya Jeevan is to provide low income families with affordable access to quality healthcare through a model of health insurance combined with value added medical services.
Naya Jeevan looks to create a pool of rising stars in the health care profession who could become a part of this innovative journey and to create in them, a sense of need for social entrepreneurship and innovation as one of the tools for rebuilding Pakistan.
Read more on the Naya Jeevan website.
When you play & interact with Karmafy-enabled games, you are part of a global community for doing good. Karmafy-enabled games all direct a portion of the Game Developer's revenue to good causes - causes you help select.
The more you play and the more Karmafy points you collect, the more support goes to your favorite cause! It's that simple. Find out more at karmafy.com or join in on the discussion on facebook or twitter.


This root FS is for ARM architecture, not for a general PC. So for cross compile purpose, I am using Suse host machine.
But it is throwing me an error while doing cp, " No space left on device". So I have come to one colclusion that the 10MB is not the actual size of the image.
Please give me advice ..