TITLE: Will moving observer see time dilation?
QUESTION [1 upvotes]: Please look at this animation.

The green dots and red dots in the animation represent spaceships. The ships of the green fleet have no velocity relative to each other, so for the clocks onboard of the individual ships, the same amount of time elapses relative to each other, and they can set up a procedure to maintain a synchronized standard fleet time. The ships of the "red fleet" are moving with a velocity of 0.866 of the speed of light with respect to the green fleet.
  The blue dots represent pulses of light. One cycle of light-pulses between two green ships takes two seconds of "green time", one second for each leg.
As seen from the perspective of the reds, the transit time of the light pulses they exchange among each other is one second of "red time" for each leg. As seen from the perspective of the greens, the red ships' cycle of exchanging light pulses travels a diagonal path that is two light-seconds long. (As seen from the green perspective the reds travel 1.73 $( {\displaystyle {\sqrt {3}}} {\displaystyle {\sqrt {3}}})$ light-seconds of distance for every two seconds of green time.)
The animation cycles between the green perspective and the red perspective, to emphasize the symmetry.

Image and Extract Source Time Dilation Wikipedia
I have calculated, how many ticks moving clock makes during time of travel and how many ticks any clock at rest.
While moving makes 3, any clock at rest makes 6. Well, moving clock dilates. But observer at moving RED clock will think like that: when I started, my clock showed 12 and GREEN clock at rest showed 12. Then, some later, my RED clock shows 3 and GREEN clock at rest shows 6. That means, clocks at rest (time in reference frame) runs faster.
Then RED clock forgets, that it was in motion and becomes one at rest. The GREEN clock now moves in RED frame and dilates. But, I am afraid to say, now the GREEN clock measures that RED clocks run faster.
Well, let’s imagine a sand clock with two flasks A and B. The sand is in the bottom in flask B, flask A is empty. Then we turn the clock upside down. Sand falls into flask A. Now flask B is empty.
Flask A is empty and flask B is empty. There is less sand in the flask B from the point of view of A and vice versa. Paradox. Reciprocity of observations. Symmetry.
If two observers move relatively to each other with velocity 0.9 c, how each of them can be “at rest”? Who moves with velocity 0.9 c? They both? Nobody? GREEN first, RED then?
Isn’t it the same sort of symmetry as in the sand clock?
What is physical reality? One mutual space or different frames that describe it?
Is it necessary to replace once chosen reference frame which describes mutual for the both observers space with a new one?
Finally, can observer think that he is in a state of motion, but not at rest?

REPLY [3 votes]: If two observers move relatively to each other with velocity 0.9 c, how each of them can be “at rest”? Who moves with velocity 0.9 c? They both? Nobody? GREEN first, RED then?

Let me call the Red observer "Alice" and the Green observer "Bob." Alice sees herself at rest, and relative to her, Bob moves with velocity $+0.9c.$ Bob sees himself at rest, and relative to him, Alice moves with velocity $-0.9c.$ They both agree on the relative speed between them.

What is physical reality? One mutual space or different frames that describe it?

The best way to describe the underlying reality is as a 4-dimensional space $(w, x, y, z)$ where we interpret $w$ as $ct,$ a time coordinate multiplied by the speed of light. Let me write the symbol $\Delta X$ as a shorthand for $X_1 - X_0,$ the difference in the symbol $X$ from some initial state $X_0$ to some final state $X_1.$
In normal 3D space, the distance between two points is related to these coordinates by the 3D Pythagorean theorem, $s^2 = \Delta x^2 + \Delta y^2 + \Delta z^2,$ and the transforms which preserve this are translations and rotations. Every coordinate system which is a translation or rotation of another agrees on these distances. The reason why we say it's all one space, and do not focus on this coordinate or that coordinate, is because these translations and rotations are allowed.
In this 4D space, the distance between two events (points in spacetime, positions plus a moment in time) is best understood as the "spacetime interval" $I = \Delta w^2 - \Delta x^2 - \Delta y^2 - \Delta z^2.$ The transforms which preserve this are the rotations, translations, time-translations, and the so-called Lorentz boosts, which mix a space coordinate with the time coordinate. Preserving this interval means, among other things, that everyone agrees that a bubble expanding with speed $c$ maps to another bubble expanding at speed $c$, in other words, everyone agrees on the speed of things moving at the speed of light. Similar to the 3D space example, the fact that these transforms are possible means that this is one unified space, not just separate coordinates.
In terms of the hyperbolic sine and cosine these boosts are (mixing $z$ with $w$ for example)  $$w' = w \cosh\phi - z\sinh\phi,\\z'=z\cosh\phi - w \sinh\phi.$$
For very small $\phi$, $\cosh\phi \approx 1$ while $\sinh\phi\approx\phi.$ The low-velocity limit where Alice moves past Bob with velocity $v$ in the $z$-direction has $z' = z - v t$ and so for small velocities this corresponds to $$t' = t - \frac{v}{c^2} z,\\z' = z - v~ t.$$ This means that there is a feature of our world that you have never noticed before, which is that when you start moving forward with speed $c$, clocks that you thought were in sync are no longer in-sync. If one of them is "in front" of the other (in terms of the direction that you accelerated), then the one "in front" now shows an earlier time than the one "in back". This "de-synchronization" is not just important, it is at the core of relativity.
This is important so I will say it outright: all of the rest of relativity, all of the time dilation and length contraction stuff, is just built out of a lot of little de-synchronizations one on top of another on top of another, as the $z - v t$ terms shift through the $t - vz/c^2$ terms and vice versa. You need to know how to do matrix diagonalization and exponentiation to prove this, but even if you can't prove this, you can appreciate it. The only mystery in special relativity is that clocks de-synchronize.
In turn this is probably the core of your misunderstanding of this graphic. You are probably thinking implicitly that the clocks of one side see the clocks of the other side as synchronized, but they don't.


TITLE: Interval of monotonicity of $ f(x) = \frac{|x-1|}{x^{2}}$
QUESTION [0 upvotes]: $ f(x) = \frac{|x-1|}{x^{2}}\ \ $ Opening this function we have
$$
f(x)=
\begin{cases}
 \frac{1-x}{x^{2}} &, x< 1\\
 \frac{x-1}{x^{2}}&, x>1\
\end{cases}
$$
Afterwards we take the derivative we get
$$
f'(x)=
\begin{cases}
 \frac{x-2}{x^{3}} &, x< 1\\
 \frac{2-x}{x^{3}}&, x>1\
\end{cases}
$$
the question asked for the interval where the function is increasing and decreasing. We already know that If the $f(x)$ is increasing then $f'(x) > 0$ and if decreasing then $f'(x) <0$. I don't know how to figure out the interval from here though.

REPLY [1 votes]: Hint:
$\space f'(x)=\cases{\frac{x-2}{x^3}, \space x<1\\
\frac{2-x}{x^3}, \space x>1}$
$f'(x) > 0?$
$f'(x) > 0 \iff \cases{\frac{x-2}{x^3}>0 \space \land \space x<1\\
\frac{2-x}{x^3}>0 \space \land \space x>1} \iff \cases{x>2 \space \land \space x<0 \space \land \space x<1\\
0<x<2 \space \land \space x>1}\\ \iff\cases{x<0\\ 1<x<2}\iff\boxed{x\in(-\infty, 0) \cup (1, 2)}.$


TITLE: The 2011 Nobel Prize in Chemistry
QUESTION [4 upvotes]: As far as I understand a new pattern of crystal growth has been found experimentally. How does it relate to the known 2D and 3D nucleation and growth of crystals? The dominating theory of crystal growth nowadays is that initiated by Cabrera and Frank. How is the new finding related to what we know so far or is it at odds with what we know and how?

REPLY [5 votes]: The Cabrera-Frank theory is strictly speaking valid for crystals only. Quasicrystals – which were awarded by the Nobel prize – are strictly speaking not crystals. Crystals, by definition, have to be periodic while quasicrystals are not. So details of the theory for crystals of course don't have to hold for quasicrystals. But there are still many similarities in their growth and structure.
Indeed, Shechtman's 1982 discovery falsified some assumptions that people used to believe, e.g. that non-periodic quasicrystals with five-fold symmetries (otherwise behaving just like crystals for which five-fold symmetries are forbidden) were impossible. He was being humiliated by many people before he was admitted to be clearly right.


TITLE: Second order Cone and Brownian motion
QUESTION [2 upvotes]: Denote $D = \{ x\in \mathbb{R}^{d+1}: \sqrt{x_1^2+\dots+x_d^2 }\leq x_{d+1}\}$ which is the second order cone ( ice-cream cone) in $\mathbb{R}^{d+1}$. My question is that if a standard brownian motion $W_t=(X^1_t,\dots, X^{d+1}_t)$ in $\mathbb{R}^{d+1}$ is started at $x_0=(1,0,\dots,0)$, will it finally hit $D$, i.e.,
$$P^{x_0}( \exists t>0, W_t \in D)=1.$$
In the case $d=1$, it is true as $\{W_t:t>0\}$ is dense in $\mathbb{R}^2$. For $d>1$, the Brownian motion is transient, i.e., $\|W_t\|_2$ goes to infinity. How can I deal with this situation?

REPLY [2 votes]: A sketch of a proof, based on the comment of Christian Remling
at a now deleted MathOverflow question.

Let $x\in\mathbb{R}^d$ be any starting point, and define
$S_1=\{y\in\mathbb{R}^d: \|y\|=2\|x\|\}$.
Let $\tau_1$ be the hitting time of $S_1$ and $C$
be your cone. The function
$z\mapsto\mathbb{P}_z(B_{\tau_1}\in C)$ is continuous
and strictly positive for $z$ with $\|z\|=\|x\|$ and so, for such $z$,
$$ \mathbb{P}_z(B_{\tau_1}\in C)\geq p>0.$$
Now define $S_n=\{y\in\mathbb{R}^d: \|y\|=2^n\|x\|\}$
so that, by scaling, $\mathbb{P}_z(B_{\tau_n}\in C)\geq p>0$
for $z$ with $\|z\|=2^{n-1}\|x\|$.
Using the strong Markov property gives
$$\mathbb{P}_x(B_{\tau_n}\notin C\mbox{ for }n=1,\dots, N)\leq (1-p)^N,$$
and letting $N\to\infty$ we get $\mathbb{P}_x(\tau_C<\infty)=1.$


text:
\begin{document}

\title{A Fourfold Refined Enumeration of Alternating Sign Trapezoids}

\author{Hans H{\"o}ngesberg\thanks{The author acknowledges support from the Austrian Science Foundation FWF, SFB grant F50.}
	}

\date{}

\maketitle

\begin{abstract}
	Alternating sign trapezoids have recently been introduced as a generalisation of alternating sign triangles. Fischer established a threefold refined enumeration of alternating sign trapezoids and provided three statistics on column strict shifted plane partitions with the same joint distribuition. In this paper, we are able to add a new pair of statistics to these results. More precisely, we consider the number of $-1$s on alternating sign trapezoids and introduce a corresponding statistic on column strict shifted plane partitions that has the same distribution. More generally, we show that the joint distributions of the two quadruples of statistics on alternating sign trapezoids and column strict shifted plane partitions, respectively, coincide.
\end{abstract}

\section{Introduction}

Since their introduction in the early 1980s, alternating sign matrices have aroused great interest among combinatorialists. Mills, Robbins, and Rumsey \cite{MRR83} conjectured them to be equinumerous with descending plane partitions, which had been enumerated by Andrews \cite{And79} a few years earlier; it was finally proved over a decade later first by Zeilberger \cite{Zei96a} and shortly thereafter by Kuperberg \cite{Kup96}. Since then, the spellbinding research of alternating sign matrices has revealed new equinumerous classes of combinatorial objects but finding bijections remains one of the most challenging problems. Equally distributed statistics on these objects might finally lead to those eagerly awaited bijections. Embracing this idea, we provide a fourfold refined enumeration of alternating sign trapezoids, a recently defined  generalisation of alternating sign triangles. Moreover, we establish four statistics on certain column strict shifted plane partitions with the same joint distribution. Thus, we generalise the recent refined enumerations of alternating sign trapezoids and of column strict shifted plane partitions by Fischer \cite{Fis}.

We start by introducing alternating sign trapezoids and column strict shifted plane partitions together with four statistics on each of these classes of objects.

\begin{definition}
	For given integers~$n \geq 1$ and $l \geq 2$, an \emph{$(n,l)$-alternating sign trapezoid} is an array of $-1$s, $0$s, and $+1$s in a trapezoidal shape with $n$ rows

	of the following form
	\begin{equation*}
	\begin{array}[t]{ccccccccc}
	a_{1,1}&a_{1,2}&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&a_{1,2n+l-2}\\
	&a_{2,2}&\cdots&\cdots&\cdots&\cdots&\cdots&a_{2,2n+l-3}&\\
	&&\ddots&&&&\reflectbox{$\ddots$}&&\\
	&&&a_{n,n}&\cdots&a_{n,n+l-1}&&&
	\end{array}
	\end{equation*}
	such that the following four conditions hold: the nonzero entries alternate in sign in each row and each column; the topmost nonzero entry in each column is $1$ (if existent); the entries in each row sum up to $1$; and the entries in the central $l-2$ columns sum up to $0$.

	An \emph{$(n,1)$-alternating sign trapezoid} is defined as above with the exception that the bottom row, which consists of a single entry in this case, does not have to add up to $1$, but where this entry can either be $1$ or $0$.
\end{definition}

Note that the notion of $(n,1)$-alternating sign trapezoids coincide with the notion of \emph{quasi alternating sign triangles of order $n$}. Furthermore, $(n,3)$-alternating sign trapezoids are in bijective correspondence with \emph{alternating sign triangles of order $n+1$}. Both alternating sign triangles and quasi alternating sign triangles were introduced by Ayyer, Behrend, and Fischer \cite{ABF}.

From the definition, it follows that the entries in each column of an alternating sign trapezoid sum up to $0$ or $1$. A column whose entries sum up to $1$ is called a \emph{$1$-column}. If, in addition, the bottom entry of a $1$-column is $0$, we call the column a \emph{$10$-column}.

Note that the number of $1$-columns in any $(n,l)$-alternating sign trapezoid is exactly $n$ if $l \neq 1$ because the sum of all entries in an $(n,l)$-AST is $n$ as all rows sum up to $1$; otherwise, it is $n$ or $n-1$.

\begin{figure}[ht]
	\centering
	\begin{equation*}
	\begin{array}[t]{cccccccccc}
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
	& 1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 &  \\
	&   & 0 & 0 & 0 & 0 & 1 & 0 &   &  \\
	&   &   & 1 & 0 & 0 & 0 &   &   &  \\
	\end{array}
	\end{equation*}
	\caption{$(4,4)$-alternating sign trapezoid $A$ with $q(A)=1$, $r(A)=2$, $s(A)=0$, and $t(A)=2$}
	\label{fig:AST}
\end{figure}

Let $\AST_{n,l}$ denote the set of  $(n,l)$-alternating sign trapezoids and $A \in \AST_{n,l}$. We introduce four different statistics on alternating sign trapezoids. First, we define
\begin{align*}
q(A) &\coloneqq \text{\# $-1$s in A,}\\
r(A) &\coloneqq \text{\# $1$-columns within the $n$ leftmost columns of A.}
\end{align*}

Then, we distinguish two cases: For $l \ge 2$, we define
\begin{align*}
s(A) &\coloneqq \text{\# $10$-columns within the $n$ leftmost columns of A,}\\
t(A) &\coloneqq \text{\# $10$-columns within the $n$ rightmost columns of A.}
\end{align*}

An example of a $(4,4)$-alternating sign trapezoid is given in Figure~\ref{fig:AST}. We set the generating function of $(n,l)$-alternating sign trapezoids associated to the statistics above to be:

\begin{equation*}
\mathcal{Z}_{\AST}(n,l;Q,R,S,T) \coloneqq \sum_{A \in \AST_{n,l}}^{} Q^{q(A)} R^{r(A)} S^{s(A)} T^{t(A)}.
\end{equation*}

Table~\ref{tab:24ASTs} shows all $(2,4)$-alternating sign trapezoids and their respective weights. The sum of these weights yield $\mathcal{Z}_{\AST}(2,4;Q,R,S,T) = 1 + 2QR + 2R + R^2 + RS + RT$.

\begin{table}[ht]
	\centering
	\caption{$(2,4)$-alternating sign trapezoids}
	\begin{math}
		\begin{array}{cccc}
		\toprule
		\begin{array}[t]{cccccc}
			1  &  0  &  0  &  0  &  0  &  0 \\
			&  1  &  0  &  0  &  0  &
		\end{array}

		&

		\begin{array}[t]{cccccc}
			0  &  0  &  0  &  0  &  1  &  0 \\
			&  1  &  0  &  0  &  0  &
		\end{array}

		&

		\begin{array}[t]{cccccc}
			0  &  0  &  0  &  0  &  0  &  1 \\
			&  1  &  0  &  0  &  0  &
		\end{array}

		&

		\begin{array}[t]{cccccc}
			1  &  0  &  0  &  0  &  0  &  0 \\
			&  0  &  0  &  0  &  1  &
		\end{array}

		\\\midrule

		R^2 & R T & R & R

		\\\midrule\midrule

		\begin{array}[t]{cccccc}
			0  &  1  &  0  &  0  &  0  &  0 \\
			&  0  &  0  &  0  &  1  &
		\end{array}

		&

		\begin{array}[t]{cccccc}
			0  &  0  &  0  &  0  &  0  &  1 \\
			&  0  &  0  &  0  &  1  &
		\end{array}

		&

		\begin{array}[t]{cccccc}
			0  &  0  &  1  &  0  &  0  &  0 \\
			&  1  &  -1  &  0  &  1  &
		\end{array}

		&

		\begin{array}[t]{cccccc}
			0  &  0  &  0  &  1  &  0  &  0 \\
			&  1  &  0  &  -1  &  1  &
		\end{array}

		\\\midrule

		R S & 1 & Q R & Q R

		\\\bottomrule
		\end{array}
	\end{math}
	\label{tab:24ASTs}
\end{table}

For $l=1$, however, we adapt the statistics in the following way
\begin{align*}
\tilde{s}(A) &\coloneqq \text{\# $10$-columns within the $n-1$ leftmost columns of A,}\\
\tilde{t}(A) &\coloneqq \text{\# $10$-columns within the $n-1$ rightmost columns of A.}
\end{align*}

and define
\begin{equation*}
\mathcal{Z}_{\AST}(n,1;Q,R,S,T) \coloneqq \sum_{A \in \AST_{n,1}}^{} Q^{q(A)} R^{r(A)} S^{\tilde{s}(A)} T^{\tilde{t}(A)} (S+T-Q)^{\left[\text{central column is a $10$-column}\right]},
\end{equation*}

where we use the \emph{Iverson bracket}: For a logical proposition $P$, $\left[P\right]=1$ if $P$ holds true and $\left[P\right]=0$ otherwise.

Table~\ref{tab:31ASTs} provides a complete list of $(3,1)$-alternating sign trapezoids and their respective weights. By adding up all these weights, we see that the generating function $\mathcal{Z}_{\AST}(3,1;Q,R,S,T)$ equals
\begin{equation}
1 - QR - QR^2 - QR^2S - QRT + 3R + 3R^2 + R^3 + 3RS + RST + 3R^2S + R^2ST + R^2S^2 + 3RT + RT^2 + 3R^2T.
\end{equation}

\begin{table}[ht]
	\centering
	\caption{$(3,1)$-alternating sign trapezoids}
	\begin{math}
	\begin{array}{ccccc}
		\toprule
		\begin{array}[t]{ccccc}
			1  &  0  &  0  &  0  &  0 \\
			&  1  &  0  &  0  & \\
			&  &  1  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  1  &  0 \\
			&  1  &  0  &  0  & \\
			&  &  1  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  0 &  1 \\
			&  1  &  0  &  0  & \\
			&  &  1  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			1  &  0  &  0  &  0  &  0 \\
			&  0  &  0  &  1  & \\
			&  &  1  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  1  &  0  &  0  &  0 \\
			&  0  &  0  &  1  & \\
			&  &  1  &  & \\
		\end{array}

		\\\midrule

		R^3 & R^2 T & R^2 & R^2 & R^2 S

		\\\midrule\midrule

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  0  &  1 \\
			&  0  &  0  &  1  & \\
			&  &  1  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  1  &  0  &  0 \\
			&  1  &  -1  &  1  & \\
			&  &  1  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			1  &  0  &  0  &  0  &  0 \\
			&  1  &  0  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  1  &  0  &  0 \\
			&  1  &  0  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  1  &  0 \\
			&  1  &  0  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		\\\midrule

		R & Q R & R^2 & R^2 (S+T-Q) & R T

		\\\midrule\midrule

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  0  &  1 \\
			&  1  &  0  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			1  & 0  &  0  &  0  &  0 \\
			&  0  &  1  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  1  &  0  &  0  &  0 \\
			&  0  &  1  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  1  &  0 \\
			&  0  &  1  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  0  &  1 \\
			&  0  &  1  &  0  & \\
			&  &  0  &  & \\
		\end{array}

		\\\midrule

		R & R^2 (S+T-Q) & R^2 S (S+T-Q) & R T (S+T-Q) & R (S+T-Q)

		\\\midrule\midrule

		\begin{array}[t]{ccccc}
			1  &  0  &  0  &  0  &  0 \\
			&  0  &  0  &  1  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  1  &  0  &  0  &  0 \\
			&  0  &  0  &  1  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  1  &  0  &  0 \\
			&  0  &  0  &  1  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  0  &  0  &  1 \\
			&  0  &  0  &  1  & \\
			&  &  0  &  & \\
		\end{array}

		&

		\begin{array}[t]{ccccc}
			0  &  0  &  1  &  0  &  0 \\
			&  1  &  -1  &  1  & \\
			&  &  0  &  & \\
		\end{array}

		\\\midrule

		R & R S & R (S+T-Q) & 1 & Q R

		\\\bottomrule
	\end{array}
	\end{math}
	\label{tab:31ASTs}
\end{table}

Clearly, $s(A) \le r(A)$ if $l \ge 2$ and $\tilde{s}(A) \le r(A)$ if $l = 1$ for any $(n,l)$-alternating sign trapezoid $A$. Moreover, by reflecting along the vertical symmetric axis, we see that
\[\mathcal{Z}_{\AST}(n,l;Q,R,S,T) = R^n \mathcal{Z}_{\AST}(n,l;Q,R^{-1},T,S).\]

Ayyer, Behrend, and Fischer \cite{ABF} showed that $n \times n$-alternating sign matrices are equinumerous with $(n-1,3)$-alternating sign trapezoids. As a corollary of \cite[Theorem 1.2]{ABF}, the statistic~$Q$ generalises -- in the most natural way -- a statistic on alternating sign triangles which has the same distribution as the number of $-1$s on alternating sign matrices.

\begin{definition}
	For a \emph{strict partition} $\lambda = \left( \lambda_1,\dots,\lambda_m \right)$, that is, a sequence $\lambda_1 > \dots > \lambda_m > 0$ of strictly decreasing positive integers, a \emph{shifted Young diagram} of \emph{shape} $\lambda$ is a finite collection of cells arranged in $m$ rows such that row~$i$ has length $\lambda_i$ and each row is indented by one cell compared to the row above.

	\begin{figure}[ht]
		\begin{center}
			\hfill\begin{minipage}{.4\textwidth}
				\ydiagram{5,1+3,2+2}
			\end{minipage}
			\begin{minipage}{.4\textwidth}
				\begin{ytableau}
					9 & 8 & 8 & 7 & 3 \\
					\none & 7 & 7 & 5 \\
					\none & \none & 6 & 1 \\
				\end{ytableau}
			\end{minipage}
			\caption{A shifted Young diagram of shape $(5,3,2)$ and a column strict shifted plane partition of the same shape and of class $4$ with $q_3(P)=2$, $r(P)=3$, $s_3(P)=1$, and $t(P)=1$}
			\label{fig:CSSPP}
		\end{center}
	\end{figure}

	A filling of a shifted Young diagram with positive integers such that the entries, termed \emph{parts}, weakly decrease along each row and strictly decrease down each column is called a \emph{column strict shifted plane partition}. It is of \emph{class~k} if the first part of each row~$i$ is exactly $k+\lambda_i$, that is, exactly $k$ plus its corresponding row length.

\begin{comment}
	a \emph{column strict shifted plane partition} is an array of positive integers in $m$ rows of row lengths given by $\lambda$ and arranged in the following form
	\begin{equation*}
	\begin{array}[t]{ccccccccc}
	p_{1,1}&p_{1,2}&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&p_{1,\lambda_{1}}\\
	&p_{2,2}&\cdots&\cdots&\cdots&\cdots&\cdots&p_{2,\lambda_{2}+1}&\\
	&&\ddots&&&&\reflectbox{$\ddots$}&&\\
	&&&p_{m,m}&\cdots&p_{m,\lambda_{m}+m-1}&&&
	\end{array}
	\end{equation*}
	such that the entries, termed \emph{parts}, weakly decrease along each row and strictly decrease down each column. It is of \emph{class k} if the first part of each row $i$ is exactly $k+\lambda_i$, that is, exactly $k$ plus its corresponding row length.

		\begin{figure}[ht]
	\centering
	\begin{equation*}
	\begin{array}[t]{ccccc}
	9&8&8&7&3\\
	&7&7&5&\\
	&&6&1&
	\end{array}
	\end{equation*}
	\caption{Column strict shifted plane partition $P$ of class $4$ with $q_3(P)=2$, $r(P)=3$, $s_3(P)=1$, and $t(P)=1$}
	\label{fig:CSSPP}
	\end{figure}
\end{comment}
\end{definition}

For any $k$, we consider the collection of zero cells to be a column strict shifted plane partition with zero rows and of class $k$. Note that we usually omit the drawings of the cells. Furthermore, we cannot always associate a class to a given column strict shifted plane partition. Column strict shifted plane partitions of class $2$ correspond to \emph{descending plane partitions} as defined by Andrews \cite{And79}.

Let $\CSSPP_{n,k}$ denote the set of column strict shifted plane partitions of class $k$ with at most $n$ parts in the first row. Following the standard labelling of entries of matrices, we refer to the part in row~$i$ and column~$j$ as $p_{i,j}$. We introduce four different statistics on $P \in \CSSPP_{n,k}$ of which two depend on a fixed parameter~$d\in \{1,\dots,k\}$:

\begin{align*}
	q_d(P) &\coloneqq \text{\# parts $p_{i,j}\in\{2,3,\dots,j-i+k\} \setminus \{j-i+d\}$,}\\
	r(P) &\coloneqq \text{\# rows of $P$,}\\
	s_d(P) &\coloneqq \text{\# parts $p_{i,j}=j-i+d$, and}\\
	t(P) &\coloneqq \text{\# parts of $p_{i,j}=1$.}
\end{align*}

We define the generating function associated to these statistics:
\begin{equation*}
\mathcal{Z}_{\CSSPP}(n,k,d;Q,R,S,T) \coloneqq \sum_{P \in \CSSPP_{n,k}}^{} Q^{q_d(A)} R^{r(A)} S^{s_d(A)} T^{t(A)}.
\end{equation*}

For $d=0$, we add another family of statistics by defining

\begin{align*}
	q_0(P) &\coloneqq \text{\# parts $p_{i,j}\in\{2,3,\dots,j-i+k\} \setminus \{j-i\}$}\\
	s_0(P) &\coloneqq \text{\# parts $p_{i,j}=j-i>1$, and}\\
	t_0(P) &\coloneqq \text{\# parts $p_{i,j}=1$ such that $j-i >1$.}
\end{align*}

In this case, the generating function is:
\begin{multline*}
\mathcal{Z}_{\CSSPP}(n,k,0;Q,R,S,T) \\ \coloneqq \sum_{P \in \CSSPP_{n,k}}^{} Q^{q_0(A)} R^{r(A)} S^{s_0(A)} T^{t_0(A)} (S+T-Q)^{\left[\text{$p_{i,j}=1$ such that $j-i=1$}\right]}.
\end{multline*}

Note that $j-i=1$ means that the part $p_{i,j}$ is in the second position of a row. A $1$ at the second position of a row can only occur in the bottom row. Also note that there is never a $1$ at the first position of a row if the class of the column strict shifted plane partition is greater than $0$.

An example of a column strict shifted plane partition is presented in Figure~\ref{fig:CSSPP}. Note that the parts counted by the statistic~$q$ generalise the \emph{special parts} in descending plane partitions that are defined by Mills, Robbins, and Rumsey \cite{MRR83} and enumerated by Behrend, Di Francesco, and Zinn-Justin \cite{BFZ12}.

\begin{table}[ht]
	\centering
	\caption{Column strict shifted plane partitions of class 3 with at most two parts in the first row}
	\begin{math}
	\begin{array}{lcccccccc}
	\toprule

	&

	\begin{array}{c}
	\emptyset
	\end{array}

	&

	\begin{array}{c}
	4
	\end{array}

	&

	\begin{array}{cc}
	5 & 5
	\end{array}

	&

	\begin{array}{cc}
	5 & 4
	\end{array}

	&

	\begin{array}{cc}
	5 & 3
	\end{array}

	&

	\begin{array}{cc}
	5 & 2
	\end{array}

	&

	\begin{array}{cc}
	5 & 1
	\end{array}

	&

	\begin{array}{cc}
	5 & 5 \\
	& 4
	\end{array}

	\\\midrule

	d=0: & 1 & R & R & Q R & Q R & Q R & R (S+T-Q) & R^2

	\\

	d=1: & 1 & R & R & Q R & Q R & R S & R T & R^2

	\\

	d=2: & 1 & R & R & Q R & R S & Q R & R T & R^2

	\\

	d=3: & 1 & R & R & R S & Q R & Q R & R T & R^2

	\\\bottomrule
	\end{array}
	\end{math}
	\label{tab:23CSSPPs}
\end{table}

Table~\ref{tab:23CSSPPs} presents all column strict shifted plane partitions of class 3 with at most two parts in the first row. The weights with respect to all possible choices of the parameter $d\in\{0,1,2,3\}$ are listed below. By summing the weights for each $d$, we obtain $\mathcal{Z}_{\CSSPP}(2,3,d;Q,R,S,T) = 1 + 2QR + 2R + R^2 + RS + RT$. Similarly, the column strict shifted plane partitions of class 0 with at most three parts in the first row are shown in Table~\ref{tab:30CSSPPs}. We see in both cases that the generating function does not depend on $d$.

\begin{table}[ht]
	\centering
	\caption{Column strict shifted plane partitions of class 3 with at most two parts in the first row}
	\begin{math}
	\begin{array}{lccccc}
	\toprule

	&

	\begin{array}{c}
	\emptyset
	\end{array}

	&

	\begin{array}{c}
	1
	\end{array}

	&

	\begin{array}{cc}
	2 & 2
	\end{array}

	&

	\begin{array}{cc}
	2 & 1
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 3
	\end{array}

	\\\midrule

	d=0: & 1 & R & R & R (S+T-Q) & R

	\\\midrule\midrule

	&

	\begin{array}{ccc}
	3 & 3 & 2
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 1
	\end{array}

	&

	\begin{array}{ccc}
	3 & 2 & 2
	\end{array}

	&

	\begin{array}{ccc}
	3 & 2 & 1
	\end{array}

	&

	\begin{array}{ccc}
	3 & 1 & 1
	\end{array}

	\\\midrule

	d=0: & R S & R T & R S & R T & R T (S+T-Q)

	\\\midrule\midrule

	&

	\begin{array}{cc}
	2 & 2\\
	& 1
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 3\\
	& 1 &
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 3\\
	& 2 & 2
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 3\\
	& 2 & 1
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 2\\
	& 1 &
	\end{array}

	\\\midrule

	d=0: & R^2 & R^2 & R^2 & R^2 (S+T-Q) & R^2 S

	\\\midrule\midrule

	&

	\begin{array}{ccc}
	3 & 3 & 2\\
	& 2 & 1
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 1\\
	& 1 &
	\end{array}

	&

	\begin{array}{ccc}
	3 & 2 & 2\\
	& 1 &
	\end{array}

	&

	\begin{array}{ccc}
	3 & 2 & 1\\
	& 1 &
	\end{array}

	&

	\begin{array}{ccc}
	3 & 3 & 3\\
	& 2 & 2\\
	&   & 1
	\end{array}

	\\\midrule

	d=0: & R^2 S (S+T-Q) & R^2 T & R^2 S & R^2 T & R^3

	\\\bottomrule
	\end{array}
	\end{math}
	\label{tab:30CSSPPs}
\end{table}

It turns out that the generating functions $\mathcal{Z}_{\AST}(2,4;Q,R,S,T)$ and $\mathcal{Z}_{\CSSPP}(2,3,d;Q,R,S,T)$ for any $d \in \{0,1,\dots,3\}$ as well as $\mathcal{Z}_{\AST}(3,1;Q,R,S,T)$ and $\mathcal{Z}_{\CSSPP}(3,0,0;Q,R,S,T)$ coincide. Fischer \cite{Fis} established refined enumerations of alternating sign trapezoids and column strict shifted plane partitions by showing that

\begin{equation*}
\mathcal{Z}_{\AST}(n,l;1,R,S,T) = \mathcal{Z}_{\CSSPP}(n,l-1,d;1,R,S,T)
\end{equation*}

for any $d \in \{0,1,\dots,l-1\}$. We extend her proof by adding the fourth statistic and show that the joint distribution of the corresponding statistics on alternating sign trapezoids and on column strict shifted plane partitions coincide:

\begin{theorem}
	\label{thm:maintheorem}
	Let $n,l \ge 1$ and $0 \le d \le l-1$. Then
	\begin{equation*}
	\mathcal{Z}_{\AST}(n,l;Q,R,S,T) = \mathcal{Z}_{\CSSPP}(n,l-1,d;Q,R,S,T).
	\end{equation*}
\end{theorem}

\section{Weighted Enumeration of Alternating Sign Trapezoids}
\label{sec:EnumAST}

First, we provide a formula for the generating function of alternating sign trapezoids. For this purpose, we heavily exploit the correspondence between alternating sign trapezoids and certain truncated \emph{monotone triangles}, where the latter is defined below.

\subsection{Correspondence between trees and alternating sign trapezoids}

\begin{definition}
	For a given integer~$n \geq 1$, a \emph{monotone triangle of order $n$} is an array of integers in a triangular shape with $n$ rows of the following form

\begin{center}
		\begin{tikzpicture}[xscale=2]
		\mt{{$a_{n,1}$,$a_{n,2}$,$a_{n,3}$,\dots,$a_{n,n}$},
			{$a_{n-1,1}$,$a_{n-1,2}$,\dots,$a_{n-1,n-1}$},
			{\dots,\dots,\dots},
			{$a_{2,1}$,$a_{2,2}$},
			{$a_{1,1}$}}
		\end{tikzpicture}
\end{center}

	such that the entries strictly increase along rows and weakly increase both along $\nearrow$-diagonals and $\searrow$-diagonals.

\end{definition}

\begin{definition}
	For given integers $p,q \geq 0$ and $n \geq 1$ such that $p+q \le n$ as well as a weakly decreasing sequence $\mathbf{s}=(s_1,s_2,\dots,s_p)$ and a weakly increasing sequence $\mathbf{t}=(t_{n-q+1},t_{n-q+2},\allowbreak \dots,\allowbreak t_n)$ of nonnegative integers, we define an \emph{$(\mathbf{s}$,$\mathbf{t})$-tree} as an array of integers which arises from a  monotone triangle of order~$n$ by truncating the diagonals as follows: for each $1 \leq i \leq p$, we delete the $s_i$ bottom entries of the $i^{\text{th}}$ $\nearrow$-diagonal; for each $n-q+1 \leq i \leq n$, we delete the $t_i$ bottom entries of the $i^{\text{th}}$ $\searrow$-diagonal. All diagonals are counted from left to right.

	We say that an $(\mathbf{s}$,$\mathbf{t})$-tree has \emph{bottom row $\mathbf{k}=(k_1, \dots, k_{n})$} if the following holds true: for all $i$ such that $1 \le i \le n-q$ or $n-q+1 \le i \le n$, the integer $k_i$ is the bottom entry of the $i^{\text{th}}$ $\nearrow$-diagonal or the $i^{\text{th}}$ $\searrow$-diagonal, respectively.
\end{definition}

Figure~\ref{fig:TreeExample} gives an examples of a $((6,3,1,1),(1,4,4))$-tree. By way of illustration, we indicate the deleted parts of the truncated diagonals by $\color{gray}\bullet$. In addition, we add an edging to the shape of the tree in order to visualise the naming.

\begin{figure}[ht]
	\centering
	\begin{tikzpicture}
	\mt{{\color{gray}$\bullet$,\color{gray}$\bullet$,\color{gray}$\bullet$,\color{gray}$\bullet$,$6$,\color{gray}$\bullet$,\color{gray}$\bullet$,\color{gray}$\bullet$},
		{\color{gray}$\bullet$,\color{gray}$\bullet$,$4$,$5$,$7$,\color{gray}$\bullet$,\color{gray}$\bullet$},
		{\color{gray}$\bullet$,\color{gray}$\bullet$,$4$,$7$,\color{gray}$\bullet$,\color{gray}$\bullet$},
		{\color{gray}$\bullet$,$3$,$6$,\color{gray}$\bullet$,\color{gray}$\bullet$},
		{\color{gray}$\bullet$,$3$,$8$,$9$},
		{\color{gray}$\bullet$,$5$,$8$},
		{$1$,$7$},
		{$6$}}

	\draw (4.25,0.25) --++ (-2,0) --++ (1,1) --++ (-1,0) --++ (1.5,1.5) --++ (-1,0) --++ (1.25,1.25) --++ (2.25,-2.25) --++ (-2,0) --++ (1.5,-1.5) --++ (-1,0) --++ (0.5,-0.5) --++ (-1.5,0) --++ (0.5,0.5) ;
	\end{tikzpicture}
	\caption{$((6,3,1,1),(1,4,4))$-tree with eight rows and bottom row $(1,3,4,5,6,7,8,9)$}
	\label{fig:TreeExample}
\end{figure}

We can transform an alternating sign trapezoid into a tree: Given an $(n,l)$-alternating sign trapezoid such that $l \ge 2$, we pad the array with additional zeroes to obtain a rectangular shape of size $n \times (2n+l-2)$. To each entry, we add the entries in the same column above it. This yields an array  consisting merely of $0$s and $1$s with exactly $i$ many $1$s in the $i^{\text{th}}$ row. We want to record the positions of these $1$s. Therefore, we number the columns from $-n$ to $n+l-3$ from left to right and list the corresponding numbers row by row in a triangular shape. By removing the entries that correspond to the initially added ones, we finally obtain a truncated monotone triangle. In Figure~\ref{fig:Tree}, we exemplify the construction with the $(4,4)$-alternating sign trapezoid in Figure~\ref{fig:AST}.

\begin{figure}[ht]
	\centering
	\begin{gather*}
	\begin{array}{cccccccccc}
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
	 & 1 & 0 & 0 & 0 & -1 & 0 & 1 & 0 &  \\
	 &  & 0 & 0 & 0 & 0 & 1 & 0 &  &  \\
	 &  &  & 1 & 0 & 0 & 0 &  &  &  \\
	\end{array}
	\qquad\longleftrightarrow\qquad
	\begin{array}{cccccccccc}
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
	{\color{gray}0} & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & {\color{gray}0} \\
	{\color{gray}0} & {\color{gray}1} & 0 & 0 & 0 & 0 & 1 & 1 & {\color{gray}0}  & {\color{gray}0} \\
	{\color{gray}0} & {\color{gray}1} & {\color{gray}0} & 1 & 0 & 0 & 1 & {\color{gray}1} & {\color{gray}0} & {\color{gray}0} \\
	\end{array}
	\\
	\longleftrightarrow
	\begin{tikzpicture}[baseline=(current bounding box),outer sep=0pt,inner sep=0pt]
		\mt{{,-1,2,},{,2,3},{-3,3},{1}}
		\end{tikzpicture}
	\end{gather*}
	\caption{$(4,4)$-alternating sign trapezoid with $1$-column vector $(-3,-1,1,2)$ and corresponding $\left((2),(1)\right)$-tree with bottom row $(-3,-1,2,3)$}
	\label{fig:Tree}
\end{figure}

To illustrate the main features of this construction, we number (differently than before) the $n$ leftmost columns of an $(n,l)$-alternating sign trapezoid from $-n$ to $-1$ and the $n$ rightmost columns from $1$ to $n$. The \emph{$1$-column vector} $\mathbf{c}=\left(c_1,\dots,c_n\right)$ records the positions of the $1$-columns of the alternating sign trapezoid; hence, $-n \le c_1 < \dots < c_m < 0 < c_{m+1} < \dots < c_n\le n$ for some $0 \le m \le n$.  The construction above yields an $(\mathbf{s}$,$\mathbf{t})$-tree with bottom row $(c_1,\dots,c_m,c_{m+1}+l-3,\dots,c_{n}+l-3)$ such that $\mathbf{s} = ( -c_1-1,\dots,-c_m-1 )$ and $\mathbf{t} = ( c_{m+1}-1,\dots,c_n-1 )$.

Regarding the statistics of alternating sign trapezoids, we make the following observations:
\begin{compactitem}
	\item A $-1$ in the alternating sign trapezoid corresponds to an entry $a_{i,j}$ in the tree which has two neighbouring entries $a_{i+1,j}$ and $a_{i+1,j+1}$ in the row below such that $a_{i+1,j} < a_{i,j} < a_{i+1,j+1}$.
	\item The positions of  $1$-columns are reflected in the bottom row of the tree.
	\item $10$-columns cause the corresponding diagonals in the tree to have twice the same bottom entries.
\end{compactitem}

\subsection{Enumeration of trees}

To enumerate monotone triangles and trees, we use operator formulae and constant term expressions. To this end, we need to introduce several operators and notations. First, we define the \emph{symmetriser}~$\sym$ and the \emph{antisymmetriser}~$\asym$ of a function $f(X_1,\dots,X_n)$. Let $\mathfrak{S}_n$ be the symmetric group of degree $n$. Then

\begin{align*}
\sym_{X_1,\dots,X_n} f(X_1,\dots,X_n) &\coloneqq \sum_{\sigma\in\mathfrak{S}_n} f(X_{\sigma(1)},\dots,X_{\sigma(n)})\,\text{and}\\
\asym_{X_1,\dots,X_n} f(X_1,\dots,X_n) &\coloneqq \sum_{\sigma\in\mathfrak{S}_n} \sgn(\sigma) f(X_{\sigma(1)},\dots,X_{\sigma(n)}).
\end{align*}

We use $\sym_{\mathbf{X}}$ and $\asym_{\mathbf{X}}$ as an abbreviation if $\mathbf{X}=(X_1,\dots,X_n)$ is clear from the context. Furthermore, $\ct_{\mathbf{X}} f(\mathbf{X})=\ct_{X_1,\dots,X_n} f(X_1,\dots,X_n)$ denotes the constant term of the function $f$ with respect to the variables $X_1,\dots,X_n$. Finally, we define the \emph{shift operator}~$\E_X$, the \emph{forward difference operator}~$\fd_X$, and the \emph{backward difference operator}~$\delta_X$:

\begin{align*}
\E_X f(X) &\coloneqq f(X+1),\\
\fd_X &\coloneqq \E_X - \id,\\
\delta_X &\coloneqq \id - \E_X^{-1},
\end{align*}

where $\id$ denotes the standard identity operator. The \emph{$Q$-forward difference operator}~$\Qfd_X$ and the \emph{$Q$-backward difference operator}~$\Qbd_X$ are defined as follows:

\begin{align*}
\Qfd_X &\coloneqq (Q - (1-Q)\fd_X)^{-1}\fd_X,\\
\Qbd_X &\coloneqq (Q - (Q-1)\delta_X)^{-1}\delta_X.
\end{align*}

They generalise the previous difference operators as they reduce to $\Qfd_X$ and $\Qbd_X$, respectively, if we set $Q=1$. We use the notation $\E_x f(x) \coloneqq \left. \E_X f(X)\right|_{X=x}$ for a given a variable~$X$ and an integer~$x$. This abbreviatory notation is correspondingly used for other operator expressions. Note that all the operators commute independently of the variables they refer to.

Fischer and Riegler \cite{FR13} provided a weighted enumeration of monotone triangles:

\begin{theorem}
	\label{thm:Q-M}
	The generating function of monotone triangles $M$ of order $n$ with bottom row $\mathbf{k}=(k_1, \dots,\allowbreak k_{n})$ with respect to the weight~$Q^{q(M)}$ is given by $M_n(\mathbf{k})$ with $M_n \left(\mathbf{x}\right)$ defined as

	\begin{equation}
	\label{eq:Q-M}
	\ct_{\mathbf{Y}} \left( \asym_{\mathbf{Y}} \left( \prod_{i=1}^n \left(1+Y_i\right)^{x_i} \prod_{1 \le i < j \le n} \left( Q - (1-Q)Y_i + Y_j + Y_i Y_j \right) \right) \prod_{1 \le i < j \le n} \left( Y_j - Y_i \right)^{-1} \right).
	\end{equation}

\end{theorem}

To obtain a enumeration formula for trees, we need to apply generalised difference operators to $M_n(\mathbf{x})$. The crucial observation is that if we repeatedly apply $-\fd_{x_i}$ and $\delta_{x_i}$ to $\left. M_n(\mathbf{x}) \right|_{Q=1}$, we enumerate monotone triangles with truncated diagonals. By using the aforementioned generalisations of the difference operators, \cite[Theorem 5]{Fis18} implies the following result:

\begin{theorem}
	\label{thm:genfunktrees}
	The generating function of $(\mathbf{s}$,$\mathbf{t})$-trees $T$ with $\mathbf{s}=(s_1,s_2,\dots,s_p)$, $\mathbf{t}=(t_{n-q+1},t_{n-q+2},\allowbreak \dots,\allowbreak t_n)$ and bottom row $\mathbf{k}=(k_1, \dots, k_{n})$ with respect to the weight~$Q^{q(T)}$ is given by

	\begin{equation}
	\label{eq:genfunktrees}
	\prod_{i=1}^{p} \left( -\Qfd_{k_i} \right)^{s_i} \prod_{i=n-q+1}^{n} \leftidx{^\textit{Q}}{\delta}{}_{k_i}^{t_i} M_n \left(\mathbf{k}\right).
	\end{equation}
\end{theorem}

\subsection{Enumeration of alternating sign trapezoids with prescribed \texorpdfstring{\boldmath$1$}{1}-column vector}

We use the correspondence between alternating sign trapezoids and trees to obtain enumeration formulae. First, we consider alternating sign trapezoids with prescribed $1$-column vectors. For this purpose, we make use of Theorems~\ref{thm:Q-M} and \ref{thm:genfunktrees}.

\begin{lemma}
	The generating function of $(n,l)$-alternating sign trapezoids $A$ with $1$-column vector $\mathbf{c}$ with respect to the weight~$Q^{q(A)} S^{s(A)} T^{t(A)}$ is given by

	\begin{equation}
	\label{eq:QST-AST1columns}
	\prod_{i=1}^{m} \left( \id - \frac{S}{Q} \delta_{c_i} \right) \left( \id + \Qfd_{c_i} \right) \left( -\Qfd_{c_i} \right)^{-c_i-1} \hspace*{-1ex} \prod_{i=m+1}^{n} \left( \id + \frac{T}{Q} \fd_{c_i} \right) \left( \id - \leftidx{^\textit{Q}}{\delta}{}_{c_i} \right) \leftidx{^\textit{Q}}{\delta}{}_{c_i}^{c_i-1} M_n \left( \mathbf{\tilde c} \right),
	\end{equation}

	where $\mathbf{\tilde c}=\left(c_1,\dots,c_m,c_{m+1}+l-3,\dots,c_{n}+l-3\right)$.
\end{lemma}

\begin{proof}
	In order to enumerate $(n,l)$-alternating sign trapezoids with $1$-column vector $\mathbf{c}$ such that $-n \le c_1 < \dots < c_m < 0 < c_{m+1} < \dots < c_n\le n$ for some $0 \le m \le n$, it suffices to enumerate $(( -c_1-1,\dots,-c_m-1 ),( c_{m+1}-1,\dots,c_n-1 ))$-tree with bottom row $\mathbf{\tilde c}=(c_1,\dots,c_m,c_{m+1}+l-3,\dots,c_{n}+l-3)$.

	Let $L_{10} \subseteq \{1,\dots,m\}$ and $R_{10} \subseteq \{m+1,\dots,n\}$. Consider an $(n,l)$-alternating sign trapezoid with $1$-column vector $\mathbf{c}$ such that $c_i$ is a  $10$-column if and only if $i \in L_{10} \cup R_{10}$. By the correspondence between alternating sign triangles and trees, we have to apply a generalised difference operator for every $10$-column, namely $-\Qfd_{c_i}$ if $i \in L_{10}$ and $\Qbd_{c_i}$ if $i \in R_{10}$. For every every other column, we apply $\id+\Qfd_{c_i}$ or $\id-\Qbd_{c_i}$, respectively. In total, we obtain the following generating function by Theorem~\ref{thm:genfunktrees}:

	\begin{equation}
	 \label{eq:genfunkASTcvecor}
		\prod_{i \in L_{10}} -\Qfd_{c_i} \prod_{\substack{1 \le i \le m,\\i \notin L_{10}}} \id+\Qfd_{c_i} \prod_{i=1}^{m} \left( -\Qfd \right)^{-c_i-1}
		\prod_{i \in R_{10}} \Qbd_{c_i} \prod_{\substack{m+1 \le i \le n,\\i \notin R_{10}}} \id-\Qbd_{c_i} \prod_{i=m+1}^{n} \Qbd^{c_i-1} M_n \left( \mathbf{\tilde c} \right).
	\end{equation}

	Since $-\Qfd_X = -Q^{-1} \bd_X (\id+\Qfd_X)$ and $\Qbd_X = Q^{-1} \fd_X (\id-\Qbd_X)$, \eqref{eq:genfunkASTcvecor} is equal to

	\begin{equation*}
		\prod_{i \in L_{10}} -Q^{-1} \bd_{c_i} \prod_{i=1}^{m} \left( \id+\Qfd_{c_i} \right) \left( -\Qfd \right)^{-c_i-1}
		\prod_{i \in R_{10}} Q^{-1} \fd_{c_i} \prod_{i=m+1}^{n} \left( \id-\Qbd_{c_i} \right) \Qbd^{c_i-1} M_n \left( \mathbf{\tilde c} \right).
	\end{equation*}

	As an intermediate step, we enumerate all alternating sign trapezoids with $s$ many $10$-columns on the left side and $t$ many on the right side, that is, $\left|L_{10}\right| = s$ and $\left|R_{10}\right| = t$. The correpsonding generating function is given by

	\begin{multline*}
	e_s \left( -Q^{-1}\bd_{c_1},\dots,-Q^{-1}\bd_{c_m} \right)  \prod_{i=1}^{m} \left( \id+\Qfd_{c_i} \right) \left( -\Qfd \right)^{-c_i-1}\\
	\times e_t \left( Q^{-1}\fd_{c_{m+1}},\dots,Q^{-1}\fd_{c_n} \right) \prod_{i=m+1}^{n} \left( \id-\Qbd_{c_i} \right) \Qbd^{c_i-1} M_n \left( \mathbf{\tilde c} \right),
	\end{multline*}

	where $e_p$ denotes the $p^{\text{th}}$ elementary symmetric function. To sum over all possible positions of $10$-columns, we notice that $e_p \left( X_1,\dots,X_n \right)$ is the coefficient of $P^p$ in $\prod_{i=1}^n \left( 1+ P X_i \right)$. Hence, we finally obtain \eqref{eq:QST-AST1columns}.
\end{proof}

\begin{figure}[ht]
	\centering
	\begin{equation*}
	\begin{array}{ccccccccc}
	0 & 0 & 0 &  0 & 0 & 0 & 0 & 1 & 0 \\
	& 0 & 0 &  1 & 0 & 0 & 0 & 0 &   \\
	&   & 1 & -1 & 1 & 0 & 0 &   &   \\
	&   &   &  0 & 0 & 1 &   &   &   \\
	&   &   &    & 0 &   &   &   &   \\
	\end{array}
	\longleftrightarrow

	\begin{tikzpicture}[baseline=(current bounding box),outer sep=0pt,inner sep=0pt]
	\mt{{,$-1$,$-1$,,},{,$-1$,$0$,},{$-3$,$-1$,},{$-2$,$2$},{$2$}}
	\end{tikzpicture}
	\end{equation*}
	\caption{$(5,1)$-alternating sign trapezoid and corresponding ((2),(1,3))-tree}
	\label{fig:Examplel1}
\end{figure}

Let us have a look at the case $l=1$. We distinguish three sets of quasi alternating triangles: those with bottom entry $1$, those with bottom entry $0$ and central column sum $0$, and those with bottom entry $0$ and central column sum $1$. It was shown in the proof of \cite[Lemma 2.2]{Fis19} that these sets correspond to the following sets of trees: the first one corresponds to the case $c_m = -1$ and $c_{m+1} > 1$, the second one corresponds to the case $c_m < -1$ and $c_{m+1} = 1$, and the third one corresponds to the case $c_m = -1$ and $c_{m+1} = 1$. Figure~\ref{fig:Examplel1} illustrates an example of the last case. If $c_m = -1$ and $c_{m+1} = 1$, then all but two entries in the bottom row of the monotone triangle are truncated. The two remaining entries are identically equal to $-1$ and force an entry in the upper row to be $-1$, too. In this case, Equation~\eqref{eq:genfunktrees} produces a factor $2-Q$, whereas \eqref{eq:QST-AST1columns} gives a factor $S+T-Q$. This justifies the modification of the weights in the generating function of $(n,l)$-alternating sign trapezoids for $l=1$.

Instead of evaluating the Polynomial~\eqref{eq:QST-AST1columns} at $\mathbf{\tilde c}$, we can shift the argument by suitable operators and evaluate it at $\mathbf{x}=\mathbf{0}$. Therefore, using also the fact that

\begin{equation*}
\id + \Qfd_x = \frac{Q \left(\id + \fd_x\right)}{Q-(1-Q)\fd_x} = Q \frac{\E_x}{Q-(1-Q)\fd_x}
\end{equation*}
and
\begin{equation*}
\id - \Qbd_x = \frac{Q \left(\id-\bd_x\right)}{Q+(1-Q)\delta_x} = Q \frac{\E_x^{-1}}{Q+(1-Q)\delta_x},
\end{equation*}

\eqref{eq:QST-AST1columns} is equal to

\begin{multline}
\label{eq:QST-AST1columns-shifted}
\prod_{i=1}^{m} \E_{x_i}^{c_i} \left( \id - \frac{S}{Q} \delta_{c_i} \right) \left( \id + \Qfd_{c_i} \right) \left( -\Qfd_{c_i} \right)^{-c_i-1}\\

\times \left. \prod_{i=m+1}^{n} \E_{x_i}^{c_i+l-3} \left( \id + \frac{T}{Q} \fd_{c_i} \right) \left( \id - \leftidx{^\textit{Q}}{\delta}{}_{c_i} \right) \leftidx{^\textit{Q}}{\delta}{}_{c_i}^{c_i-1} M_n \left( \mathbf{x} \right) \right|_{\mathbf{x}=\mathbf{0}}\\

\shoveleft = \prod_{i=1}^{m} \E_{x_i}^{-1} \frac{Q-(S-Q)\fd_{x_i}}{Q-(1-Q)\fd_{x_i}} \left( \frac{-\delta_{x_i}}{Q-(1-Q)\fd_{x_i}} \right)^{-c_i-1}\\

\times \left. \prod_{i=m+1}^{n} \E_{x_i}^{l-2} \frac{Q+(T-Q)\delta_{x_i}}{Q+(1-Q)\delta_{x_i}} \left( \frac{\fd_{x_i}}{Q+(1-Q)\delta_{x_i}} \right)^{c_i-1} M_n \left( \mathbf{x} \right) \right|_{\mathbf{x}=\mathbf{0}}.
\end{multline}

We analyse how the operators in \eqref{eq:QST-AST1columns-shifted} interact with the argument of the antisymmetriser in \eqref{eq:Q-M}: The effect of the shift operator $\E_{x_i}$ is the multiplication by $1 + Y_i$. Therefore, the application of the forward difference operator $\fd_{x_i}$ or of the backward difference operator $\delta_{x_i}$ is equivalent to the multiplication by $Y_i$ or by $Y_i(1+Y_i)^{-1}$, respectively.

\begin{comment}
As a consequence, the application of
		$\frac{-\delta_{x_i}}{Q-(1-Q)\fd_{x_i}}$
		 and
		$\frac{Q-(S-Q)\fd_{x_i}}{Q-(1-Q)\fd_{x_i}}$
results in the multiplication by
		$\frac{-Y_i}{(1+Y_i)(Q-(1-Q)Y_i)}$
		and $\frac{Q-(S-Q)Y_i}{Q-(1-Q)Y_i}$,
respectively.

Similarly, the effect of
		$\frac{-\fd_{x_i}}{Q+(1-Q)\delta_{x_i}}$
		and
		$\frac{Q+(T-Q)\fd_{x_i}}{Q+(1-Q)\fd_{x_i}}$
	is the multiplication by
		$\frac{Y_i (1+Y_i)}{Q+Y_i}$
		and
		$\frac{Q+T Y_i}{Q+Y_i}$,
respectively.
\end{comment}

\begin{comment}
As a consequence, the application of
\begin{center}
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{-\delta_{x_i}}{Q-(1-Q)\fd_{x_i}}
		\end{equation*}
	\end{minipage}
	\quad and \quad
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{Q-(S-Q)\fd_{x_i}}{Q-(1-Q)\fd_{x_i}}
		\end{equation*}
	\end{minipage}\hfill
\end{center}
results in the multiplication by
\begin{center}
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{-Y_i}{(1+Y_i)(Q-(1-Q)Y_i)}
		\end{equation*}
	\end{minipage}
	\quad and \quad
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{Q-(S-Q)Y_i}{Q-(1-Q)Y_i},
		\end{equation*}
	\end{minipage}\hfill
\end{center}
respectively.

Similarly, the effect of
\begin{center}
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{-\fd_{x_i}}{Q+(1-Q)\delta_{x_i}}
		\end{equation*}
	\end{minipage}
	\quad and \quad
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{Q+(T-Q)\fd_{x_i}}{Q+(1-Q)\fd_{x_i}}
		\end{equation*}
	\end{minipage}\hfill
\end{center}
is the multiplication by
\begin{center}
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{Y_i (1+Y_i)}{Q+Y_i}
		\end{equation*}
	\end{minipage}
	\quad and \quad
	\begin{minipage}{.3\textwidth}
		\begin{equation*}
		\frac{Q+T Y_i}{Q+Y_i},
		\end{equation*}
	\end{minipage}\hfill
\end{center}
respectively.
\end{comment}

This observation implies that \eqref{eq:QST-AST1columns-shifted} is equal to

\begin{multline}
\label{eq:QST-AST1columns-CT}
\ct_{\mathbf{Y}} \left( \asym_{\mathbf{Y}} \left( \prod_{i=1}^m \left(-Y_i\right)^{-c_i-1} \left(1+Y_i\right)^{c_i} \left(Q-(1-Q)Y_i\right)^{c_i+1} \frac{Q-(S-Q)Y_i}{Q-(1-Q)Y_i} \right. \right.\\

\times \prod_{i=m+1}^n Y_i^{c_i-1} \left(1+Y_i\right)^{c_i+l-3} \left(Q+Y_i\right)^{-c_i+1} \frac{Q+T Y_i}{Q+Y_i}\\

\left. \left. \times \prod_{1 \le i < j \le n} \left( Q - (1-Q)Y_i + Y_j + Y_i Y_j \right) \right) \prod_{1 \le i < j \le n} \left( Y_j - Y_i \right)^{-1} \right).
\end{multline}

\subsection{Sum over all \texorpdfstring{\boldmath$1$}{1}-column vectors}

Thus far, we have considered $(n,l)$-alternating sign trapezoids with prescribed $1$-column vector $\mathbf{c}$.

\begin{comment}
To obtain an expression for the total amount of  $(n,l)$-alternating sign trapezoids, we sum over all $0 \le m \le n$ after summing over all $c_i$.
\end{comment}

To sum over all $c_i$ such that $-n \le c_1 < \dots < c_m < 0 < c_{m+1} < \dots < c_n \le n$,

\begin{comment}
with $-n \le c_1 < \dots < c_m < 0 < c_{m+1} < \dots < c_n \le n$
\end{comment}

we ignore the upper and lower bound in the summation since the polynomial in \eqref{eq:QST-AST1columns-CT} has no constant term if $c_1 < n$ or $c_n > n$. We use the following identity which we obtain by repeated geometric series evaluations:

\begin{equation}
\label{eq:GeometricSeries}
\sum_{0 \le x_1 < \dots < x_k} X_1^{x_1} \dots X_k^{x_k} = \prod_{i=1}^{k} X_i^{i-1} \left( 1 - \prod_{j=i}^{k} X_j \right)^{-1}.
\end{equation}

Hence, by applying \eqref{eq:GeometricSeries}, we obtain that

\begin{multline*}
\sum_{c_1 < \dots < c_m < 0} \left( \prod_{i=1}^m \left(-Y_i\right)^{-c_i-1} \left(1+Y_i\right)^{c_i} \left(Q-(1-Q)Y_i\right)^{c_i+1} \left( \frac{Q-(S-Q)Y_i}{Q-(1-Q)Y_i} \right) \right)\\

\times \sum_{0 < c_{m+1} < \dots < c_n} \left( \prod_{i=m+1}^n Y_i^{c_i-1} \left(1+Y_i\right)^{c_i+l-3} \left(Q+Y_i\right)^{-c_i+1} \left( \frac{Q+T Y_i}{Q+Y_i} \right) \right)\\

\times \prod_{1 \le i < j \le n} \left( Q - (1-Q)Y_i + Y_j + Y_i Y_j \right) \left( Y_j - Y_i \right)^{-1}.
\end{multline*}

is equal to

\begin{multline}
	\label{eq:QST-ASTsumOverC}
\prod_{i=1}^m \frac{1}{1+Y_i} \left( \frac{-Y_i}{\left( 1+Y_i \right) \left( Q-(1-Q)Y_i \right)} \right)^{m-i} \\

\times \left( 1 - \prod_{j=1}^i \left( \frac{-Y_j}{\left( 1+Y_j \right) \left( Q-(1-Q)Y_j \right)} \right) \right)^{-1} \frac{Q-(S-Q)Y_i}{Q-(1-Q)Y_i}\\

\times \prod_{i=m+1}^n \left( 1+Y_i \right)^{l-2} \left( \frac{Y_i \left( 1+Y_i \right)}{Q+Y_i} \right)^{i-m-1} \left( 1 - \prod_{j=i}^n \left( \frac{Y_j \left( 1+Y_j \right)}{Q+Y_j} \right) \right)^{-1} \frac{Q+T Y_i}{Q+Y_i}\\

\times \prod_{1 \le i < j \le n} \left( Q - (1-Q)Y_i + Y_j + Y_i Y_j \right) \left( Y_j - Y_i \right)^{-1}.
\end{multline}

Before summing over all $m$ such that $0 \le m \le n$, we apply the symmetriser to the expression~\eqref{eq:QST-ASTsumOverC}. To this end, we use the following trick by Fischer \cite{Fis}: We set

\begin{equation*}
\mathfrak{S}_n^m \coloneqq \{ \sigma \in \mathfrak{S}_n \mid \sigma (i) < \sigma (j)\, \forall\, 1 \le i < j \le m\, \vee\, m+1 \le i < j \le n \}
\end{equation*}

and define

\begin{equation*}
\subsets_{X_1,\dots,X_m}^{X_{m+1},\dots,X_n} f(X_1,\dots,X_n) \coloneqq \sum_{\sigma \in \mathfrak{S}_n^m} f\left( X_{\sigma(1)},\dots,X_{\sigma(n)} \right).
\end{equation*}

It follows that

\begin{equation*}
\sym_{X_1,\dots,X_n} f(X_1,\dots,X_n) = \subsets_{X_1,\dots,X_m}^{X_{m+1},\dots,X_n} \sym_{X_1,\dots,X_m} \sym_{X_{m+1},\dots,X_n} f(X_1,\dots,X_n).
\end{equation*}

That is, we first apply $\sym_{Y_1,\dots,Y_m}$ and $\sym_{Y_{m+1},\dots,Y_n}$ to \eqref{eq:QST-ASTsumOverC} by means of the following antisymmetriser lemma \cite[Lemma~5.9]{Hon} and its variation \cite[(5.19)]{Hon} by setting $X_i \mapsto -(X_{m+1-i}) (1+X_{m+1-i})^{-1}$:

\begin{lemma}
	Let $n \ge 1$. Then

	\begin{multline*}
	\asym_{\mathbf{X}} \left( \prod_{i=1}^{n} \frac{\left( \frac{X_i(1+X_i)}{Q+X_i} \right)^{i-1}}{1-\prod_{j=i}^{n} \frac{X_j(1+X_j)}{Q+X_j} } \prod_{1\leq i<j \leq n} ( Q - (1-Q) X_i + X_j + X_i X_j ) \right)\\

	= \prod_{i=1}^{n} \frac{Q+X_i}{Q - X_i^2} \prod_{1\leq i<j \leq n} \frac{(Q(1+X_i)(1+X_j)-X_i X_j)(X_j-X_i)}{Q-X_i X_j}.
	\end{multline*}

\begin{comment}
	\begin{multline*}
	\asym_{\mathbf{X}} \left( \prod_{i=1}^{n} \frac{\left( \frac{-X_i}{(Q-(1-Q)X_i)(1+X_i)} \right)^{n-i}}{1-\prod_{j=1}^{i} \left( \frac{-X_j}{(Q-(1-Q)X_j)(1+X_j)} \right)} \prod_{1\leq i<j \leq n} \left( Q - (1-Q) X_i + X_j + X_i X_j \right) \right)\\

	= \prod_{i=1}^{n} \frac{(1+X_i) (Q-(1-Q)X_i)}{Q(1+X_i)^2 - X_i^2} \prod_{1\leq i<j \leq n} \frac{\left(Q-X_i X_j\right) \left(X_j - X_i\right)}{Q(1+X_i)(1+X_j)-X_i X_j}.
	\end{multline*}
\end{comment}
\end{lemma}

Eventually, we obtain

\begin{multline}
\label{eq:QST-ASTsym}
\prod_{i=1}^m \frac{Q-(S-Q)Y_i}{Q(1+Y_i)^2-Y_i^2} \prod_{1 \le i < j \le m} \frac{Q-Y_i Y_j}{Q(1+Y_i)(1+Y_j)-Y_i Y_j} \prod_{i=m+1}^n \left( 1+Y_i \right)^{l-2} \frac{Q+T Y_i}{Q-Y_i^2}\\

\times \prod_{m+1 \le i < j \le n} \frac{Q(1+Y_i)(1+Y_j)-Y_i Y_j}{Q-Y_i Y_j} \prod_{i=1}^m \prod_{j=m+1}^n \frac{Q-(1-Q)Y_i+Y_j+Y_i Y_j}{Y_j-Y_i}.
\end{multline}

Next, we need to apply the operator $\subsets_{Y_1,\dots,Y_m}^{Y_{m+1},\dots,Y_n}$ to \eqref{eq:QST-ASTsym} and evaluate at $\mathbf{Y} = 0$. To simplify the computation, we divide \eqref{eq:QST-ASTsym} by the polynomial $\prod_{1 \le i < j \le n} ( Q(1+Y_i)(1+Y_j)-Y_i Y_j ) ( Q-Y_i Y_j )$, which is symmetric and, thus, invariant under the application of $\subsets_{Y_1,\dots,Y_m}^{Y_{m+1},\dots,Y_n}$. However, we need to incorporate its constant term $Q^{2 \binom{n}{2}}$. We obtain

\begin{multline}
\label{eq:QST-ASTsymSimplified}
Q^{2 \binom{n}{2}} \prod_{i=1}^m \left( Q-(S-Q)Y_i \right) \prod_{i,j=1}^{m} \frac{1}{Q(1+Y_i)(1+Y_j)-Y_i Y_j} \prod_{i=m+1}^n \left( 1+Y_i \right)^{l-2} \left( Q+T Y_i \right)\\

\times \prod_{i,j=m+1}^{n} \frac{1}{Q-Y_i Y_j} \prod_{i=1}^m \prod_{j=m+1}^n \frac{Q-(1-Q)Y_i+Y_j+Y_i Y_j}{\left( Y_j-Y_i \right) \left( Q(1+Y_i)(1+Y_j)-Y_i Y_j \right) \left( Q-Y_i Y_j \right)}.
\end{multline}

This expression can be written in determinantal form. For this purpose, we consider the \emph{Cauchy determinant}

\begin{equation*}
\det_{1 \le i,j \le n} \left( \frac{1}{X_i + Y_j} \right) = \frac{\prod_{1 \le i < j \le n} \left( X_j - X_i \right) \left( Y_j - Y_i \right)}{\prod_{i,j=1}^{n} \left( X_i + Y_j \right)}
\end{equation*}

and set $X_i = \frac{Q(1+Y_i)}{Q-(1-Q)Y_i}$ for all $1 \le i \le m$ and $X_i = -\frac{Q}{Y_i}$ for all $m+1 \le i \le n$. This yields that

\begin{equation*}
\det_{1 \le i,j \le n} \left(\begin{cases}
\frac{Q-(1-Q)Y_i}{Q(1+Y_i)(1+Y_j) - Yi Y_j}, & 1 \le i \le m\\
\frac{-Y_i}{Q - Y_i Y_j}, & m+1 \le i \le n
\end{cases}\right)
\end{equation*}

is equal to

\begin{multline*}
(-1)^{n-m} Q^{\binom{n}{2}} \prod_{i=1}^m \left( Q-(1-Q)Y_i \right) \prod_{i,j=1}^{m} \frac{1}{Q(1+Y_i)(1+Y_j)-Y_i Y_j} \prod_{1 \le i < j \le m} \left( Y_j - Y_i \right)^2\\

\times \prod_{i=m+1}^n Y_i \prod_{i,j=m+1}^{n} \frac{1}{Q-Y_i Y_j} \prod_{m+1 \le i < j \le n} \left( Y_j - Y_i \right)^2\\

\times \prod_{i=1}^m \prod_{j=m+1}^n \frac{\left( Y_j-Y_i \right) \left( Q-(1-Q)Y_i+Y_j+Y_i Y_j \right)}{\left( Q(1+Y_i)(1+Y_j)-Y_i Y_j \right) \left( Q-Y_i Y_j \right)}.
\end{multline*}

Row and column transformations of the determinant's underlying matrix show that \eqref{eq:QST-ASTsymSimplified} is equal to

\begin{equation*}
\frac{Q^{\binom{n}{2}}}{\prod_{1 \le i < j \le n} \left( Y_j - Y_i \right)^2} \det_{1 \le i,j \le n} \left(\begin{cases}
\frac{Q-(S-Q)Y_i}{Q(1+Y_i)(1+Y_j) - Yi Y_j}, & 1 \le i \le m\\
\left( 1+Y_i \right)^{l-2} \frac{Q + T Y_i}{Q - Y_i Y_j}, & m+1 \le i \le n
\end{cases}\right).
\end{equation*}

To apply the subset operator, we observe that the antisymmetry of the determinant implies that it follows for any $\sigma \in \mathfrak{S}_n^m$ that

\begin{equation*}
\frac{Q^{\binom{n}{2}}}{\prod_{1 \le i < j \le n} \left( Y_j - Y_i \right)^2} \det_{1 \le i,j \le n} \left(\begin{cases}
\frac{Q-(S-Q)Y_i}{Q(1+Y_i)(1+Y_j) - Yi Y_j}, & i \in \sigma \left(\{1,\dots,m\}\right)\\
\left( 1+Y_i \right)^{l-2} \frac{Q + T Y_i}{Q - Y_i Y_j}, & i \in \sigma \left(\{m+1,\dots,n\}\right)
\end{cases}\right)\\
\end{equation*}

is equivalent to applying $\sigma$ to the variables~$Y_1,\dots,Y_n$ of \eqref{eq:QST-ASTsymSimplified}. Hence, the application of $\subsets_{Y_1,\dots,Y_m}^{Y_{m+1},\dots,Y_n}$ and the summation over all $1 \le m \le n$ yield...

\begin{equation}
\label{eq:QRST-ASTdet}
\frac{Q^{\binom{n}{2}}}{\prod_{1 \le i < j \le n} \left( Y_j - Y_i \right)^2} \det_{1 \le i,j \le n} \left(
R \frac{Q-(S-Q)Y_i}{Q(1+Y_i)(1+Y_j) - Y_i Y_j} + \left( 1+Y_i \right)^{l-2} \frac{Q + T Y_i}{Q - Y_i Y_j}
\right),
\end{equation}

where the exponent of $R$ takes account of $m$.

\subsection{Transformation of the determinantal formula}

The determinantal formula~\eqref{eq:QRST-ASTdet} is our first expression for the fourfold refined enumeration of $(n,l)$-alternating sign trapezoids. We transform it into a determinant involving binomial coefficients. Our key tool is the following formula by Behrend, Di Francesco, and Zinn-Justin \cite[(43)-(47)]{BFZ12}:

\begin{lemma}
		\label{lem:BFZtrick}
For a given power series $f$ in variables~$X$ and $Y$, it holds that

\begin{equation*}
	\label{eq:BFZtrick}
\left. \frac{\det_{1 \le i,j \le n} \left( f(X_i,Y_j) \right)}{\prod_{1 \le i < \le n} \left( X_j - X_i \right) \left( X_j - X_i \right)} \right|_{\mathbf{X} = \mathbf{Y} = \mathbf{0}} = \det_{0 \le i,j \le n-1} \left( \left[ X^i Y^j \right] f(X,Y) \right);
\end{equation*}

here, $\left[ X^i Y^j \right] f(X,Y)$ denotes the coefficient of $X^i Y^j$ in the series expansion of $f$.
\end{lemma}

We set

\begin{equation*}
f(X,Y) = R \frac{Q-(S-Q)X}{Q(1+X)(1+Y) - X Y} + \left( 1+X \right)^{l-2} \frac{Q + T X}{Q - X Y}
\end{equation*}

and extract the coefficient $\left[ X^i Y^j \right] f(X,Y)$:

\begin{equation*}
R (-1)^{i+j} \sum_{k \ge 0}  \binom{j}{k} Q^{-k} \left( \binom{i-1}{k-1} + \binom{i-1}{k} S Q^{-1} \right) + \binom{l-2}{i-j} Q^{-j} + \binom{l-2}{i-j-1} T Q^{-j-1};
\end{equation*}

note that we set the binomial coefficient $\binom{n}{k} \coloneqq 0$ for $k<0$.

The determinant of the matrix $\left( \left[ X^i Y^j \right] f(X_i,Y_j) \right)_{1 \le i,j \le n}$ remains invariant under the left multiplication by the triangular matrix $\left( \binom{2-l}{i-j} \right)_{0 \le i,j \le n-1}$ with determinant $1$. Thus, we obtain

\begin{equation*}
R \sum_{k \ge 0}  (-1)^{j-k} \binom{j}{k} Q^{-k} \left( \binom{2-l-k}{i-k} - \binom{1-l-k}{i-k-1} S Q^{-1} \right)  + Q^{-j} \delta_{i,j} + T Q^{-j-1} \delta_{i,j+1}.
\end{equation*}

In addition, we multiply the $i^{\text{th}}$ row by $(-1)^i$ and the $j^{\text{th}}$ row by $(-1)^j$ for all $0 \le i,j \le n-1$ and use $\binom{n}{k} = (-1)^{k} \binom{k-n-1}{k}$ to obtain

\begin{equation*}
R \sum_{k \ge 0} \binom{j}{k} Q^{-k} \left( \binom{i+l-3}{i-k} + \binom{i+l-3}{i-k-1} S Q^{-1} \right) + Q^{-j} \delta_{i,j} - T Q^{-j-1} \delta_{i,j+1}.
\end{equation*}

Next, define the matrix

\begin{equation*}
K(n) \coloneqq \left( Q^{-j} \delta_{i,j} - Q^{-j-1} T \delta_{i,j+1} \right)_{0 \le i,j \le n-1},
\end{equation*}

whose determinant evaluation yields $Q^{-\binom{n}{2}}$, and whose inverse is

\begin{equation*}
K(n)^{-1} =
\begin{cases}
T^{i-j} Q^{j}, & i \ge j,\\
0, & i < j.
\end{cases}
\end{equation*}

We left multiply by $K(n)^{-1}$, and, thus, we finally obtain by Lemma~\ref{lem:BFZtrick} that
$\mathcal{Z}_{\AST}(n,l;Q,R,S,T)$ is equal to

\begin{equation}
\label{eq:QRST-ASTbinom}
\det_{0 \le i,j \le n-1} \left(R \sum_{k = 0}^{i} T^{i-k} \sum_{m=0}^{j} \binom{j}{m} Q^{k-m} \left( \binom{k+l-3}{k-m} + \binom{k+l-3}{k-m-1} S Q^{-1} \right) + \delta_{i,j} \right).
\end{equation}

\section{Weighted Enumeration of Column Strict Shifted Plane Partitions}

In order to enumerate column strict shifted plane partitions, we interpret them as a family of nonintersecting lattice paths which can be enumerated by the Lindstr{\"o}m-Gessel-Viennot lemma.

\subsection{Interpretation of column strict shifted plane partitions as families of nonintersecting lattice paths}

We transform column strict shifted plane partitions into a family of nonintersecting lattice paths as follows: Each row corresponds to a path that consists of vertical and horizontal unit steps. If $p$ is the first entry of the row, then the path starts at $(-1,p-1)$ and ends on the $x$-axis; the heights of the vertical steps are the parts of the corresponding row diminished by $1$. Figure~\ref{fig:LatticePaths} displays the family of nonintersecting lattice paths corresponding to the column strict shifted plane partition in Figure~\ref{fig:CSSPP}.

\begin{figure}[ht]
	\centering
		\begin{tikzpicture}[scale=.5]
		\draw [help lines,step=1cm, dashed] (-1.75,-.75) grid (4.75,8.75);

		\draw[->,thick] (-1.75,0)--(4.75,0) node[right]{$x$};
		\draw[->,thick] (0,-.75)--(0,8.75) node[above]{$y$};

		\fill (4,0) circle (5pt);
		\fill (2,0) circle (5pt);
		\fill (1,0) circle (5pt);

		\fill (0,8) circle (5pt);
		\fill (0,6) circle (5pt);
		\fill (0,5) circle (5pt);

		\draw[ultra thick] (-1,8) -- (0,8) -- (0,7) -- (2,7) -- (2,6) -- (3,6) -- (3,2) -- (4,2) -- (4,0);

		\node at (-0.5,8.5) {$9$};
		\node at (0.5,7.5) {$8$};
		\node at (1.5,7.5) {$8$};
		\node at (2.5,6.5) {$7$};
		\node at (3.5,2.5) {$3$};

		\draw[ultra  thick] (-1,6) -- (1,6) -- (1,4) -- (2,4) -- (2,0);

		\node at (-0.5,6.5) {$7$};
		\node at (0.5,6.5) {$7$};
		\node at (1.5,4.5) {$5$};

		\draw[ultra thick] (-1,5) -- (0,5) -- (0,0) -- (1,0);

		\node at (-0.5,5.5) {$6$};
		\node at (0.5,0.5) {$1$};

		\end{tikzpicture}
		\caption{Family of lattice paths corresponding to the corresponding to the column strict shifted plane partition in Figure~\ref{fig:CSSPP}}
		\label{fig:LatticePaths}
\end{figure}

This construction yields a bijective correspondence between column strict shifted plane partitions of class $l-1$ with at most $n$ parts in the first row and the family of nonintersecting lattice paths consisting of horizontal and vertical unit steps with start points $S \subseteq \{ S_i \coloneqq (i,0) \mid  0 \le i \le n-1\}$ and end points $E \subseteq \{ E_i \coloneqq (0,i+l-1) \mid  0 \le i \le n-1\}$ such that $S_i \in S$ if and only if $E_i \in E$.

By this interpretation of column strict shifted plane partitions as a family of nonintersecting lattice paths and by the Lindstr{\"o}m-Gessel-Viennot lemma, it can be shown that column strict shifted plane partitions of class $l-1$ with at most $n$ parts in the first row are enumerated by

\begin{equation}
\label{eq:AndrewsDet}
\det_{0 \le i,j \le n-1} \left( \binom{i+j+l-1}{i} + \delta_{i,j} \right).
\end{equation}

This was first proved by Andrews \cite{And79}. In fact, the determinant~\eqref{eq:AndrewsDet} can be obtained from \eqref{eq:QRST-ASTbinom} by setting $Q=R=S=T=1$.

\subsection{Weighted enumeration for {\boldmath$d \ge 1$}}

We generalise Andrews' result. To this end, consider an element of $\CSSPP_{n,l-1}$ and $L$ be the corresponding family of nonintersecting lattice paths. For the time being, fix $d \in \{1,\dots,l-1\}$. We investigate how to translate the statistics on column strict shifted plane partitions into statistics on lattice paths: The weight of $L$ is

\begin{equation*}
Q^{q_d(L)} R^{r(L)} S^{s_d(L)} T^{t(L)},
\end{equation*}

where

\begin{align*}
q_d(L) &\coloneqq \text{\# horizontal steps below the line $y=x+l-1$ neither touching the line $y=x+d$}\\
&\phantom{\coloneqq\qquad} \text{from the right with a horizontal step nor lying on the $x$-axis,}\\
r(L) &\coloneqq \text{\# paths of $L$,}\\
s_d(L) &\coloneqq \text{\# horizontal steps touching the line $y=x+d$ from the right with a horizontal step, and}\\
t(L) &\coloneqq \text{\# horizontal steps of $L$ on the $x$-axis.}
\end{align*}

\begin{figure}[ht]
	\centering
	\begin{tikzpicture}[scale=.5]
	\draw [help lines,step=1cm, dashed] (-.75,-.75) grid (10.75,12.75);

	\draw[->,thick] (-.75,0)--(10.75,0) node[right]{$x$};
	\draw[->,thick] (0,-.75)--(0,12.75) node[above]{$y$};

	\fill (10,0) circle (5pt);
	\node at (10,-.5) {$(i,0)$};

	\fill (8,1) circle (5pt);
	\node at (9,1) {$(k,1)$};

	\fill (6,6) circle (5pt);
	\node at (8.25,6) {$(v,v+d-1)$};

	\fill (6,7) circle (5pt);
	\node at (7.75,7) {$(v,v+d)$};

	\fill (4,10) circle (5pt);
	\node at (6.5,10) {$(m,m+l-1)$};

	\fill (0,11) circle (5pt);
	\node at (-2.25,11) {$(0,j+l-1)$};

	\draw[ultra thick] (0,11) --++ (1,0) --++ (1,0) --++ (1,0) --++ (0,-1) --++ (1,0) --++ (0,-1) --++ (1,0) --++ (1,0) --++ (0,-1) --++ (0,-1) --++ (0,-1) --++ (0,-1) --++ (1,0) --++ (0,-1) --++ (1,0) --++ (0,-1) --++ (0,-1) --++ (0,-1) --++ (0,-1) --++ (1,0) --++ (1,0);

	\draw[thick] (-.5,5.5) -- (6,12);
	\node at (-2.75,5.5) {$y=x+l-1$};

	\draw[thick] (-.5,.5) -- (10,11);
	\node at (11.25,10.5) {$y=x+d$};

	\end{tikzpicture}
	\caption{Lattice path from $(i,0)$ to $(0,j+l-1)$ with $i-k$ horizontal steps at height $0$ that intersects the line~$y=x+d$ from the right side with a vertical step}
	\label{fig:LatticePathsStatistics}
\end{figure}

Figure~\ref{fig:LatticePathsStatistics} shows a single lattice path from $(i,0)$ to $(0,j+l-1)$ that has $i-k$ horizontal steps at height $0$. In addition, the line $y=x+d$ intersects the path with a vertical step from the right side. Let $\mathcal{P}( (x_1,y_1) \rightarrow (x_2,y_2) )$ denote the number of paths from $(x_1,y_1)$ to $(x_2,y_2)$ with horizontal steps $(-1,0)$ and vertical steps $(0,1)$; that is, $\mathcal{P}( (x_1,y_1) \rightarrow (x_2,y_2) ) = \binom{(x_1-x_2)+(y_2-y_1)}{x_1-x_2}$. The generating function of the path is given by

\begin{multline*}
R \sum_{k=0}^{i} T^{i-k} \sum_{v=0}^{k} \mathcal{P} \left((k,1) \rightarrow (v,v+d-1)\right) \\ \times\sum_{m=0}^{j} Q^{k-m} \mathcal{P} \left((v,v+d) \rightarrow (m,m+l-1)\right) \mathcal{P} \left((m,m+l-1) \rightarrow (0,j+l-1)\right)\\
=R \sum_{k=0}^{i} T^{i-k} \sum_{v=0}^{k} \binom{k+d-2}{k-v} \sum_{m=0}^{j} Q^{k-m} \binom{l-1-d}{v-m} \binom{j}{m},
\end{multline*}

which we simplify to

\begin{equation}
	\label{eq:GenFuncCSSPPwithoutS}
R \sum_{k=0}^{i} T^{i-k} \sum_{m=0}^{j} Q^{k-m} \binom{j}{m} \binom{k+l-3}{k-m}
\end{equation}

by using the Chu-Vandermonde identity.
If the line $y=x+d$ intersects the path with a horizontal step from the right side, the generating function is given by

\begin{multline*}
R S \sum_{k=0}^{i} T^{i-k} \sum_{v=0}^{k} \mathcal{P} \left((k,1) \rightarrow (v+1,v+d)\right) \\ \times\sum_{m=0}^{j} Q^{k-m-1} \mathcal{P} \left((v,v+d) \rightarrow (m,m+l-1)\right) \mathcal{P} \left((m,m+l-1) \rightarrow (0,j+l-1)\right),
\end{multline*}

which is equal to
\begin{equation}
	\label{eq:GenFuncCSSPPwithS}
R S\sum_{k=0}^{i} T^{i-k} \sum_{m=0}^{j} Q^{k-m-1} \binom{j}{m} \binom{k+l-3}{k-m-1}.
\end{equation}

The sum of \eqref{eq:GenFuncCSSPPwithoutS} and \eqref{eq:GenFuncCSSPPwithS} yields the generating function of a single lattice path from $(i,0)$ to $(0,j+l-1)$:

\begin{equation}
	\label{eq:GenFuncPath}
R \sum_{k = 0}^{i} T^{i-k} \sum_{m=0}^{j} \binom{j}{m} Q^{k-m} \left( \binom{k+l-3}{k-m} + \binom{k+l-3}{k-m-1} S Q^{-1} \right).
\end{equation}

By the well-known Lindstr{\"o}m-Gessel-Viennot lemma \cite{GV85,Lin73}, $\mathcal{Z}_{\CSSPP}(n,l-1,d;Q,R,S,T)$ is equal to

\begin{equation*}
\sum_{\substack{0 \le u_1 < \dots < u_r \le n-1,\\0 \le r \le n-1}}^{} \det_{1 \le i,j \le r} \left( R \sum_{k = 0}^{u_i} T^{u_i-k} \sum_{m=0}^{u_j} \binom{u_j}{m} Q^{k-m} \left( \binom{k+l-3}{k-m} + \binom{k+l-3}{k-m-1} S Q^{-1} \right) \right),
\end{equation*}
which simplifies to \eqref{eq:QRST-ASTbinom}.

\subsection{The case {\boldmath$d = 0$}}

In the case $d=0$, a family $L$ of nonintersecting lattics paths is assigned the weight

\begin{equation*}
Q^{q_0(P)} R^{r(P)} S^{s_0(P)} T^{t_0(P)} (S+T-Q)^{\left[ \text{$P$ contains $(1,0) \rightarrow (0,0)$} \right]},
\end{equation*}

where

\begin{align*}
q_0(L) &\coloneqq \text{\# horizontal steps below the line $y=x+l-1$ neither touching the line $y=x$}\\
&\phantom{\coloneqq\qquad} \text{from the right with a horizontal step nor lying on the $x$-axis,}\\
s_0(L) &\coloneqq \text{\# horizontal steps touching the line $y=x$ from the right with a horizontal step}\\
&\phantom{\coloneqq\qquad} \text{but not on the $x$-axis, and}\\
t_0(L) &\coloneqq \text{\# horizontal steps of $L$ on the $x$-axis except $(1,0) \rightarrow (0,0)$.}
\end{align*}

To derive the generating function of a singe lattice path from $(i,0)$ to $(0,j+l-1)$, we distinguish several cases: If the path contains the horizontal step $(1,0) \rightarrow (0,0)$, it has the weight $R T^{i-1} (S+T-Q)$. If the path does not contain the horizontal step $(1,0) \rightarrow (0,0)$, it
can intersect the line $y=x$ with a vertical step from below. In this case, the generating function is given by

\begin{multline*}
R \sum_{k=1}^{i} T^{i-k} \sum_{v=1}^{k} \mathcal{P} \left((k,1) \rightarrow (v,v-1)\right) \\ \times\sum_{m=0}^{j} Q^{k-m} \mathcal{P} \left((v,v) \rightarrow (m,m+l-1)\right) \mathcal{P} \left((m,m+l-1) \rightarrow (0,j+l-1)\right),
\end{multline*}

which is equal to
\begin{equation}
	\label{eq:GenFuncCSSPPwithoutSd}
R T^{i-1}Q + R \sum_{k=1}^{i} T^{i-k} \sum_{m=0}^{j} Q^{k-m} \binom{j}{m} \binom{k+l-3}{k-m}.
\end{equation}

If the path intersects the line $y=x$ with a horizontal step from the right side, the generating function is given by

\begin{multline*}
R S \sum_{k=2}^{i} T^{i-k} \sum_{v=1}^{k} \mathcal{P} \left((k,1) \rightarrow (v+1,v)\right) \\ \times\sum_{m=0}^{j} Q^{k-m-1} \mathcal{P} \left((v,v) \rightarrow (m,m+l-1)\right) \mathcal{P} \left((m,m+l-1) \rightarrow (0,j+l-1)\right),
\end{multline*}

which is equal to
\begin{equation}
	\label{eq:GenFuncCSSPPwithSd}
R S\sum_{k=2}^{i} T^{i-k} \sum_{m=0}^{j} Q^{k-m-1} \binom{j}{m} \binom{k+l-3}{k-m-1}.
\end{equation}

As before, the sum of \eqref{eq:GenFuncCSSPPwithoutSd}, \eqref{eq:GenFuncCSSPPwithSd}, and $R T^{i-1} (S+T-Q)$ is equal to \eqref{eq:GenFuncPath}.

As a result, \eqref{eq:QRST-ASTbinom} is the generating function of column strict shifted plane partitions $P$ of class $l-1$ with at most $n$ entries in the first row with respect to the weight~$Q^{q(P)} R^{r(P)} S^{s(P)} T^{t(P)}$. This completes the proof of Theorem~\ref{thm:maintheorem}.

\section{The \texorpdfstring{\boldmath$x$}{x}-Enumeration}

For an integer $x\in\mathbb{Z}$, we define the \emph{$x$-enumeration} of $(n,l)$-alternating sign trapezoids as

\begin{equation*}
\sum_{A\in\AST_{n,l}} x^{q(A)}.
\end{equation*}

Ayyer, Behrend, and Fischer \cite{ABF} proved that the analogously defined $x$-enumeration of $n \times n$-alternating sign matrices is equal to the $x$-enumeration of $(n-1,3)$-alternating sign trapezoids. For specific values of $x$, the $x$-enumeration of alternating sign matrices surprisingly yields round numbers: Mills, Robbins, and Rumsey \cite{MRR83} presented a closed expression of the $2$-enumeration and Kuperberg \cite{Kup96} for the $3$-enumeration.

We conjecture that the $2$-enumeration of $(n,l)$-alternating sign trapezoids yields round numbers in general, too.

\begin{theorem}
	\label{thm:2Enum}
	For $l \ge 2$, the $2$-enumeration of $(2n+1,l)$-alternating sign trapezoids is given by
	\begin{equation*}
		2^{n+1} \left( \prod_{1 \le i \le j \le n} \frac{2(i+j)+(l-3)}{2i-1} \right)^2
	\end{equation*}
	and the $2$-enumeration of $(2n+2,l)$-alternating sign trapezoids is given by
	\begin{equation*}
	2^{n+1} \prod_{1 \le i \le j \le n} \frac{2(i+j)+(l-3)}{2i-1} \prod_{1 \le i \le j \le n+1} \frac{2(i+j)+(l-3)}{2i-1} .
	\end{equation*}
\end{theorem}

\begin{proof}
Let $l \ge 1$. To prove the conjecture, we provide a product formula for $\mathcal{Z}_{\AST}(n,l;2,1,1,1)$. To this end, we set

\begin{equation*}
\label{eq:funkdet}
f(l;i,j) \coloneqq \sum_{k = 0}^{i} \sum_{m=0}^{j} 2^{k-m} \binom{j}{m} \left( \binom{k+l-3}{k-m} + \frac{1}{2} \binom{k+l-3}{k-m-1} \right).
\end{equation*}

Then

\begin{equation*}
\mathcal{Z}_{\AST}(n,l;2,1,1,1) = \det_{0 \le i,j \le n-1} \left( f(l;i,j) + \delta_{i,j} \right).
\end{equation*}

First, we simplify the expression for $f(l;i,j)$. We split the sum up, shift the index in the second summand, and make use of $ \binom{j}{m} + \binom{j}{m-1} = \binom{j+1}{m}$ in order to see that

\begin{equation*}
\sum_{m=0}^{j} 2^{k-m} \binom{j}{m} \left( \binom{k+l-3}{k-m} + \frac{1}{2} \binom{k+l-3}{k-m-1} \right)

= \sum_{m=0}^{j+1}  2^{k-m} \binom{j+1}{m} \binom{k+l-3}{k-m},
\end{equation*}

which can be restated as an ordinary hypergeometric function\footnote{The \emph{ordinary hypergeometric function} is defined as $\pFq{2}{1}{a,b}{c}{z} \coloneqq \sum_{n=0}^{\infty} \frac{\left(a\right)_n \left(b\right)_n}{\left(c\right)_n} \frac{z^n}{n!}$, where $\left(a\right)_n \coloneqq a \left(a+1\right) \cdots \left(a+n-1\right)$ if $n \ge 1$ and $\left(a\right)_0 \coloneqq 1$ denote the \emph{Pochhammer symbol}.} as follows by means of \cite[p. 11]{Bai35}:

\begin{equation*}
2^k \binom{k+l-3}{k} \pFq{2}{1}{-k,-j-1}{l-2}{\frac{1}{2}} = \binom{k+l-3}{k} \pFq{2}{1}{-k,j+l-1}{l-2}{-1}.
\end{equation*}

Finally, by resorting the order of summation, we further deduce that $f(l;i,j)$ can be expressed as
\begin{multline*}
\sum_{k = 0}^{i} \binom{k+l-3}{k} \pFq{2}{1}{-k,j+l-1}{l-2}{-1}\\
= \sum_{k = 0}^{i} \binom{k+l-3}{k} \sum_{m=0}^{k} \binom{k}{m} \frac{\left(j+l-1\right)_m}{\left(l-2\right)_m} = \sum_{m=0}^{i}  \frac{\left(j+l-1\right)_m}{\left(l-2\right)_m} \sum_{k = m}^{i} \binom{k+l-3}{k} \binom{k}{m}\\
= \sum_{m=0}^{i}  \frac{\left(j+l-1\right)_m}{\left(l-2\right)_m} \binom{m+l-3}{m} \sum_{k = 0}^{i-m} \binom{k+m+l-3}{k} = \sum_{m=0}^{i}  \frac{\left(j+l-1\right)_m}{m!} \binom{i+l-2}{m+l-2},
\end{multline*}

which is equal to
\begin{equation}
	\label{eq:HypDet}
\binom{i+l-2}{l-2} \pFq{2}{1}{-i,j+l-1}{l-1}{-1}.
\end{equation}

However, if we set $a=l-2$, then
\begin{equation*}
g(a;i,j) \coloneqq \sum_{m = 0}^{i} \binom{m+j+a}{m}\binom{i+a}{m+a}
\end{equation*}

equals \eqref{eq:HypDet}, too. Hence, $g(l-2;i,j)=f(l;i,j)$. Andrews \cite[Theorem~1 \& Lemma~2]{And87} evaluated the determinant $D_n(a) \coloneqq \det_{0 \le i,j \le n-1} \left( g(a;i,j) + \delta_{i,j} \right)$ for $a \ge 0$ and proved that

\begin{align}
\frac{D_{2n}(a)}{D_{2n-1}(a)} &= 2^n \prod_{k=1}^{n} \frac{2\left(n+k\right)+a-1}{n+k},\label{eq:AndrewsD1}\\
\frac{D_{2n+1}(a)}{D_{2n}(a)} &= 2^{n+1} \prod_{k=1}^{n} \frac{2\left(n+k\right)+a-1}{n+k}.\label{eq:AndrewsD2}
\end{align}

Since $D_1(a)=2$, \eqref{eq:AndrewsD1} and \eqref{eq:AndrewsD2} imply Theorem~\ref{thm:2Enum}.\qedhere
\end{proof}

The determinant $D_n(a)$ originally emerged in \cite{MRR87} as a generating function of a differently weighted enumeration of column strict shifted plane partitions of class $2a$ with at most $n$ parts in the first row.

Note that $\mathcal{Z}_{\AST}(n,1;x,1,1,1)$ is not equivalent to the $x$-enumeration of quasi alternating sign triangles since we altered the weight in the case $l=1$. However, computer experiments suggest that neither the $2$-enumeration $2$, $5$, $22$, $188$, $3152$, $104704$, $6905856,\dots$ nor the $3$-enumeration $2$, $5$, $24$, $252$, $5832$, $301077$, $34720812,\dots$ of quasi alternating sign triangles yields round numbers. Computer experiments suggest that the $3$-enumeration of $(n,l)$-alternating sign trapezoids does not provide round numbers either unless $l=3$. Furthermore, note that for $l=2$, the $2$-enumeration coincides with the total homology of free $2$-step nilpotent Lie algebras \cite[Theorem~1.1]{GKT02} as well as with the number of rhombus tilings of half-hexagons with glued sides \cite[(3.5)]{DFZJZ05}.

\bibliography{ASM}
\bibliographystyle{halpha}

\textsc{Universit{\"a}t Wien, Fakult{\"a}t f{\"u}r Mathematik, Oskar-Morgenstern-Platz 1, 1090 Wien, Austria}

\textit{E-mail address:} \href{mailto:hans.hoengesberg@univie.ac.at}{hans.hoengesberg@univie.ac.at}

\end{document}

config: arxiv


TITLE: $p$ prime and $m_p$ a proper divisor of $p−1$. Is it $\sigma(m_p)-\nu(m_p)<p-2$ for every $p$ and $m_p$?
QUESTION [1 upvotes]: (This is a sharper version of this other question of mine.)
Along the problem I'm facing, I've come to the following lemma (if it is true):
Let $p$ be a prime and $m_p$ a proper divisor of $p-1$. Then the difference between the sum of all the divisors of $m_p$, say $\sigma(m_p)$, and their number, say $\nu(m_p)$, is less than $p-2$.
Examples:

$p=13$; then, for $m_{13}=6$: $1+2+3+6-4=8<11$, and likewise for $m_{13}=4,3,2,1$;
$p=31$; then for $m_{31}=15$ we have: $1+3+5+15-4=20<29$, or for $m_{31}=10$: $1+2+5+10-4=14<29$, and likewise for $m_{31}=5,3,2,1$;
$p=37$; then for $m_{37}=18$ we have: $1+2+3+6+9+18-6=33<35$, and likewise for $m_{37}=9,6,3,2,1$;
$p=101$; then for $m_{101}=50$ we have: $1+2+5+10+25+50-6=87<99$, and likewise for $m_{101}=25,20,10,5,2,1$.

Is it true for every $p$ and $m_p$?

REPLY [1 votes]: here are the counterexamples from $m=30$ to $m = 210,$ those being consecutive primorials. The primorial thing seems to die out after
m: 2310   p:   4621    sig:   6912 nu: 32  diff:  6880  WOW 2310 =  2 3 5 7 11

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
m:   30   p:     61    sig:     72 nu: 8  diff:  64  WOW 30 =  2 3 5
m:   36   p:     73    sig:     91 nu: 9  diff:  82  WOW 36 =  2^2 3^2
m:   48   p:     97    sig:    124 nu: 10  diff:  114  WOW 48 =  2^4 3
m:   54   p:    109    sig:    120 nu: 8  diff:  112  WOW 54 =  2 3^3
m:   56   p:    113    sig:    120 nu: 8  diff:  112  WOW 56 =  2^3 7
m:   78   p:    157    sig:    168 nu: 8  diff:  160  WOW 78 =  2 3 13
m:   90   p:    181    sig:    234 nu: 12  diff:  222  WOW 90 =  2 3^2 5
m:   96   p:    193    sig:    252 nu: 12  diff:  240  WOW 96 =  2^5 3
m:  114   p:    229    sig:    240 nu: 8  diff:  232  WOW 114 =  2 3 19
m:  120   p:    241    sig:    360 nu: 16  diff:  344  WOW 120 =  2^3 3 5
m:  138   p:    277    sig:    288 nu: 8  diff:  280  WOW 138 =  2 3 23
m:  140   p:    281    sig:    336 nu: 12  diff:  324  WOW 140 =  2^2 5 7
m:  156   p:    313    sig:    392 nu: 12  diff:  380  WOW 156 =  2^2 3 13
m:  168   p:    337    sig:    480 nu: 16  diff:  464  WOW 168 =  2^3 3 7
m:  174   p:    349    sig:    360 nu: 8  diff:  352  WOW 174 =  2 3 29
m:  176   p:    353    sig:    372 nu: 10  diff:  362  WOW 176 =  2^4 11
m:  186   p:    373    sig:    384 nu: 8  diff:  376  WOW 186 =  2 3 31
m:  198   p:    397    sig:    468 nu: 12  diff:  456  WOW 198 =  2 3^2 11
m:  200   p:    401    sig:    465 nu: 12  diff:  453  WOW 200 =  2^3 5^2
m:  204   p:    409    sig:    504 nu: 12  diff:  492  WOW 204 =  2^2 3 17
m:  210   p:    421    sig:    576 nu: 16  diff:  560  WOW 210 =  2 3 5 7

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
Alright, noticed that the above counterexamples all have the minimal prime $p = 1 + 2m.$   It is also possible to have counterexamples with $p=1+3m,$  listed below, but $p \geq 1+4m$  seems to allow no counterexamples at all.
m:  540   p:   1621    sig:   1680 nu: 24  diff:  1656  WOW 540 =  2^2 3^3 5  THREE
m:  720   p:   2161    sig:   2418 nu: 30  diff:  2388  WOW 720 =  2^4 3^2 5  THREE
m:  840   p:   2521    sig:   2880 nu: 32  diff:  2848  WOW 840 =  2^3 3 5 7  THREE
m: 1620   p:   4861    sig:   5082 nu: 30  diff:  5052  WOW 1620 =  2^2 3^4 5  THREE
m: 2040   p:   6121    sig:   6480 nu: 32  diff:  6448  WOW 2040 =  2^3 3 5 17  THREE
m: 2160   p:   6481    sig:   7440 nu: 40  diff:  7400  WOW 2160 =  2^4 3^3 5  THREE
m: 2184   p:   6553    sig:   6720 nu: 32  diff:  6688  WOW 2184 =  2^3 3 7 13  THREE
m: 2376   p:   7129    sig:   7200 nu: 32  diff:  7168  WOW 2376 =  2^3 3^3 11  THREE
m: 2520   p:   7561    sig:   9360 nu: 48  diff:  9312  WOW 2520 =  2^3 3^2 5 7  THREE
m: 2700   p:   8101    sig:   8680 nu: 36  diff:  8644  WOW 2700 =  2^2 3^3 5^2  THREE
m: 2772   p:   8317    sig:   8736 nu: 36  diff:  8700  WOW 2772 =  2^2 3^2 7 11  THREE
m: 2880   p:   8641    sig:   9906 nu: 42  diff:  9864  WOW 2880 =  2^6 3^2 5  THREE
m: 3000   p:   9001    sig:   9360 nu: 32  diff:  9328  WOW 3000 =  2^3 3 5^3  THREE
m: 3300   p:   9901    sig:  10416 nu: 36  diff:  10380  WOW 3300 =  2^2 3 5^2 11  THREE
m: 3720   p:  11161    sig:  11520 nu: 32  diff:  11488  WOW 3720 =  2^3 3 5 31  THREE
m: 3900   p:  11701    sig:  12152 nu: 36  diff:  12116  WOW 3900 =  2^2 3 5^2 13  THREE
m: 4032   p:  12097    sig:  13208 nu: 42  diff:  13166  WOW 4032 =  2^6 3^2 7  THREE
m: 4140   p:  12421    sig:  13104 nu: 36  diff:  13068  WOW 4140 =  2^2 3^2 5 23  THREE
m: 4200   p:  12601    sig:  14880 nu: 48  diff:  14832  WOW 4200 =  2^3 3 5^2 7  THREE
m: 4284   p:  12853    sig:  13104 nu: 36  diff:  13068  WOW 4284 =  2^2 3^2 7 17  THREE
m: 4560   p:  13681    sig:  14880 nu: 40  diff:  14840  WOW 4560 =  2^4 3 5 19  THREE
m: 5040   p:  15121    sig:  19344 nu: 60  diff:  19284  WOW 5040 =  2^4 3^2 5 7  THREE
m: 5220   p:  15661    sig:  16380 nu: 36  diff:  16344  WOW 5220 =  2^2 3^2 5 29  THREE
m: 5292   p:  15877    sig:  15960 nu: 36  diff:  15924  WOW 5292 =  2^2 3^3 7^2  THREE
m: 5460   p:  16381    sig:  18816 nu: 48  diff:  18768  WOW 5460 =  2^2 3 5 7 13  THREE

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
Correction, ratio 4 is possible, just need to keep looking at larger and larger $m$
m:60480   p: 241921    sig: 243840 nu: 112  diff:  243728  WOW 60480 =  2^6 3^3 5 7  FOUR OR MORE
m:83160   p: 332641    sig: 345600 nu: 128  diff:  345472  WOW 83160 =  2^3 3^3 5 7 11  FOUR OR MORE
m:128520   p: 514081    sig: 518400 nu: 128  diff:  518272  WOW 128520 =  2^3 3^3 5 7 17  FOUR OR MORE
m:184800   p: 739201    sig: 749952 nu: 144  diff:  749808  WOW 184800 =  2^5 3 5^2 7 11  FOUR OR MORE
m:211680   p: 846721    sig: 861840 nu: 144  diff:  861696  WOW 211680 =  2^5 3^3 5 7^2  FOUR OR MORE
m:231840   p: 927361    sig: 943488 nu: 144  diff:  943344  WOW 231840 =  2^5 3^2 5 7 23  FOUR OR MORE
m:240240   p: 960961    sig: 999936 nu: 160  diff:  999776  WOW 240240 =  2^4 3 5 7 11 13  FOUR OR MORE
m:272160   p:1088641    sig: 1100736 nu: 144  diff:  1100592  WOW 272160 =  2^5 3^5 5 7  FOUR OR MORE
m:277200   p:1108801    sig: 1199328 nu: 180  diff:  1199148  WOW 277200 =  2^4 3^2 5^2 7 11  FOUR OR MORE
m:289800   p:1159201    sig: 1160640 nu: 144  diff:  1160496  WOW 289800 =  2^3 3^2 5^2 7 23  FOUR OR MORE
m:292320   p:1169281    sig: 1179360 nu: 144  diff:  1179216  WOW 292320 =  2^5 3^2 5 7 29  FOUR OR MORE
m:299880   p:1199521    sig: 1200420 nu: 144  diff:  1200276  WOW 299880 =  2^3 3^2 5 7^2 17  FOUR OR MORE
m:347760   p:1391041    sig: 1428480 nu: 160  diff:  1428320  WOW 347760 =  2^4 3^3 5 7 23  FOUR OR MORE
m:383040   p:1532161    sig: 1584960 nu: 168  diff:  1584792  WOW 383040 =  2^6 3^2 5 7 19  FOUR OR MORE
m:411840   p:1647361    sig: 1664208 nu: 168  diff:  1664040  WOW 411840 =  2^6 3^2 5 11 13  FOUR OR MORE
m:428400   p:1713601    sig: 1798992 nu: 180  diff:  1798812  WOW 428400 =  2^4 3^2 5^2 7 17  FOUR OR MORE
m:436800   p:1747201    sig: 1763776 nu: 168  diff:  1763608  WOW 436800 =  2^6 3 5^2 7 13  FOUR OR MORE

$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$


text:
\section{Introduction}
\label{section:intro}

Bayesian optimization (\bo) is a powerful technique for optimizing expensive-to-evaluate black-box functions.
Important applications include materials design \citep{zhang2020bayesian}, drug discovery \citep{griffiths2020constrained}, machine learning hyperparameter tuning \citep{snoek2012practical}, neural architecture search \citep{kandasamy2018neural, zhang2019d}, etc.
\bo operates by constructing a \emph{surrogate model} for the target function, typically a Gaussian process (\gp), and then using a
cheap-to-evaluate \emph{acquisition function} (\af) to guide iterative queries of the target function until a predefined budget is expended.

We refer to \cite{shahriari2016taking} for a literature survey.

Most of the existing acquisition policies are only one-step optimal, that is, optimal if the decision horizon were one. An example is \emph{expected improvement} (\ei) \citep{movckus1974bayesian}, arguably the most widely used policy. Such \emph{myopic} policies only consider the immediate utility of the decision, ignoring the long-term impact of exploration. Despite the sub-optimal balancing of exploration and exploitation, they are widely used in practice due to their simplicity and computational efficiency.

When the query budget is explicitly considered, \bo can be formulated as a Markov decision process (\mdp), whose optimal policy maximizes the expected utility of the terminal dataset \citep{lam2016bayesian, jiang2019binoculars}.

However, solving the \mdp is generally intractable due to the set-based, uncountable state space, uncountable action space, and potentially long decision horizon. There has been recent interest in developing nonmyopic policies \citep{gonzalez2016glasses, lam2016bayesian, wu2019practical, yue2019why}, but these policies are often heuristic in nature or computationally expensive. A recent work known as \binoc \citep{jiang2019binoculars} achieved both efficiency and a certain degree of nonmyopia by maximizing a lower bound of the multi-step expected utility.
However, a general implementation of multi-step lookahead for \bo has, to our knowledge, not been attempted before.

\textbf{Main Contributions.} Our work makes progress on the intractable multi-step formulation of \bo through the following methodological and empirical contributions:
\begin{itemize}[leftmargin=*]
    \item \textit{One-shot multi-step trees.} We introduce a novel, scenario tree-based acquisition function for \bo that performs an approximate, multi-step lookahead. Leveraging the reparameterization trick, we propose a way to jointly optimize all decision variables in the multi-step tree in a \emph{one-shot} fashion, without resorting to explicit dynamic programming recursions involving nested expectations and maximizations. Our tree formulation is fully differentiable, and we compute gradients using auto-differentiation, permitting the use of gradient-based optimization.

    \item \textit{Fast-fantasies and parallelism.}
    Our multi-step scenario tree is built by recursively sampling from the \gp posterior and conditioning on the sampled function values (``fantasies'').
    This tree grows exponentially in size with the number of lookahead steps.
    While our algorithm cannot negate this reality, our novel efficient linear algebra methods for conditioning the \gp model combined with a highly parallelizable implementation on accelerated hardware allows us to tackle the problem at practical wall times for moderate lookahead horizons (less than 5).

    \item \textit{Improved optimization performance.} Using our method, we are able to achieve significant improvements in optimization performance over one-step \ei and \binoc on a range of benchmarks, while maintaining competitive wall times. To further improve scalability, we study two special cases of our general framework which are of linear growth in the lookahead horizon. We empirically show that these alternatives perform surprisingly well in practice.
\end{itemize}
We set up our problem setting in Section \ref{sec:bayesoptimalpolicy} and propose the one-shot multi-step trees in Section \ref{sec:OST}. We discuss how we achieve fast evaluation and optimization of these trees in Section \ref{sec:fastdifffantasies}. In Section \ref{sec:pseudo}, we show some notable instances of multi-step trees and make connections to related work in Section \ref{section:related}. We present our empirical results in Section \ref{sec:exper} and conclude in Section \ref{section:conclusion}.


TITLE: Distribution of a function of Brownian motion
QUESTION [0 upvotes]: Q) Let $W = \int_{0}^{t}B_sds$, $B$ is a Brownian motion. Find $EW$ and $EW^2$.
My attempt:
$B_s \sim N(0,s) $
$$W = \int_{0}^{t}\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}ds , EW = \int_{-\infty}^{\infty}x(\int_{0}^{t}\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}ds)dx$$
I am not sure how to find $\int_{0}^{t}\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}ds$? Integration by reduction makes sense in the case of $\int x^ne^{ax}dx$ if n is a positive integer, where in if $I_n = \int x^ne^{ax}dx$, then $I_n = \frac{1}{a}(e^{ax}x^{n+1}-aI_{n-1})$, $I_0$ can be computed and all $I_n$ recursively.
Thanks and appreciate a hint.

REPLY [2 votes]: This is relatively straightforward if you can use Ito's Lemma and Ito's Isometry.
If you are allowed to use those result than $\mathbb E W$ and $\mathbb E W^2$ can be computed in the following way. Start by applying Ito's Lemma to $t B_t$ to obtain
$$ d(tB_t)=B_tdt+tdB_t$$
From this we obtain that
$$\int_0^t B_sds = t B_t - \int_0^t s dB_s$$
It is immediate form the above expression that $\mathbb E W =0$. To compute $\mathbb E W^2$, note that the expression for $\int_0^t B_sds$ derived above implies that
$$ \int_0^t B_sds = t \int_0^t \left(1-\frac{s}{t}\right)dB_s$$ Therefore,
$$ \mathbb E W^2 = t^2\mathbb E\left[\left(\int_0^t \left(1-\frac{s}{t}\right)dB_s\right)^2\right] = t^2 \int_0^t \left(1-\frac{s}{t}\right)^2ds$$ by Ito Isometry. Computing this integral, we note that $\mathbb E W^2=t^3/3$.