In analogy with the Delta conjecture, we will refer to the case $m=0$ of the following conjecture simply as the \emph{Delta square conjecture}.

We formulated the following conjecture in \cite{DAdderio-Iraci-VandenWyngaerd-Delta-Square}.

\begin{conjecture}[Generalized Delta square]
For $m,n,k\in \mathbb{N}$, $m\geq 0$ and $n>k\geq 0$,
\[
\frac{[n-k]_t}{[n]_t} \Delta_{h_m}\Delta_{e_{n-k}}\omega(p_n) = \gensquare.
\]	
\end{conjecture}

\begin{remark} \label{rmk:symmetry}
Observe that $\gensquare$ is a symmetric function. Indeed, consider the expression $\sum_Pq^{\dinv(P)}t^{\area(P)}x^P$ where the sum is taken over all $P\in \PLSQE(m,n)^{\ast k}$ of a fixed \emph{shape}, i.e. a fixed underlying square path with prescribed zero valleys. From this sum we can factor $t^{\area(P)}$, as the area is the same for all such paths $P$, and $q^{a(P)}$, where $a(P)$ is the contribution to the dinv of the $0$ labels and of the negative letters of the area word (the bonus dinv): indeed this contribution does not depend on the nonzero labels, but only on the shape, so it will be the same for all our paths. What we are left with is in fact an LLT polynomial: the argument is essentially the same as in \cite[Section~6.2]{Haglund-Remmel-Wilson-2015}, so we omit it (cf also \cite[Remark~6.5]{Haglund-Book-2008}). As it is well-known that the LLT polynomials are symmetric functions (cf. \cite[Appendix]{Haglund-Haiman-Loehr}), we deduce that also $\gensquare$ is symmetric.
\end{remark}

This conjecture answers a question in \cite[Section~8.2]{Haglund-Remmel-Wilson-2015}.

\begin{remark}
Notice that the case $m=k=0$ of the generalized Delta square conjecture reduces precisely to the \emph{square conjecture} of Loehr and Warrington \cite{Loehr-Warrington-square-2007}, recently proved by Sergel \cite{Leven-2016} after the breakthrough of Carlsson and Mellit \cite{Carlsson-Mellit-ShuffleConj-2015}.
\end{remark}

% We need some symmetric functions identities if we want to include this examplet

%\begin{example}
%Using \eqref{eq:en_expansion} and \eqref{eq:p_expansion} (see Section~\ref{sec:SF} for definitions and notations about symmetric functions), it is easy to see that for $m=0$ and $k=n-1$ we get
%\begin{equation}
%\frac{[1]_t}{[n]_t} \Delta_{h_0}\Delta_{e_{1}}\omega(p_n) = [n]_qe_n.
%\end{equation}
%We leave to the reader the straightforward verification that indeed
%\begin{equation}
%\mathsf{PLSQ^E}_{\underline x,q,t}(0,n)^{\ast n-1} = [n]_qe_n,
%\end{equation}
%proving in this way our conjecture at $m=0$ and $k=n-1$.
%\end{example}

