text:
\section{Introduction}
\label{s1}
A \textit{gapset} is a finite set $G \subset \mathbb{N}$ satisfying the following property. Let $z \in G$ and write $z = x + y$, with $x$ and $y \in \mathbb{N}$. Then $x \in G$ or $y \in G$. This concept was formally introduced by Eliahou and Fromentin \cite{EF}. Recall that a numerical semigroup $S$ is a submonoid of the set of non-negative integers $\mathbb{N}_0$, equipped with the usual addition, such that $G(S):=\mathbb{N}_0\setminus S$, the set of {\em gaps} of $S$, is finite (the book \cite{GS-R} is an excellent background on this subject). Hence, a gapset is nothing more than the set of gaps of some numerical semigroup. Throughout this paper, we denote $[a,b] := \{x \in \mathbb{Z}: a \leq x \leq b\}$ and $[a, \infty) := \{x \in \mathbb{Z}: x \geq a\}$, for integers $a$ and $b$.

The cardinality of a gapset $G$ is called the {\em genus} of $G$. Numerical semigroup properties imply that if $S$ is a numerical semigroup with genus $g$, then $[2g, \infty) \subset S$. Hence, a non-empty gapset $G$ with genus $g$ satisfies $G \subseteq [1,2g-1]$. Some invariants play an important role in this theory. For instance, the multiplicity, the conductor, the depth and the Frobenius number of a gapset are $m(G) := \min\{s \in \mathbb{N}: s \notin G\}$, $c(G) := \min\{s \in \mathbb{N}: s + n \notin G, \forall n \in \mathbb{N}_0\}, q(G) := \lceil \frac{c(G)}{m(G)} \rceil$ and $F(G) := c(G) - 1$, respectively.

The set of gapsets is denoted by $\Gamma$ and the set of gapsets with a fixed genus $g$ by $\Gamma(g)$, which has $n_g$ elements. The first few elements of the sequence $(n_g)$ are $1, 1, 2, 4, 7, 12, 23, 39, 67, 118$ and it is registered as the sequence A007323 at OEIS (the on-line encyclopedia of integer sequences). This sequence has been deeply studied after Bras-Amor\' os \cite{Amoros1} conjectured three items about its behaviour. The only one which is still an open question is: ``is it true that $n_g + n_{g+1} \leq n_{g+2}$, for all $g$?". Even the weaker version of this conjecture, namely ``is $(n_g)$ a non-decreasing sequence?" is an open problem. The other two questions are about the asymptotic behaviour of $(n_g)$, namely $\lim_{g \to \infty} \frac{n_{g+1}}{n_g} = \frac{1+\sqrt{5}}{2}$ (golden ratio) and  $\lim_{g \to \infty} \frac{n_{g+1} + n_g}{n_{g+2}} = 1$; both were proved by Zhai \cite{Zhai}. The main tool to prove those results is the fact that most of gapsets with fixed genus have depth at most 3. More precisely:

\paragraph{Zhai's theorem.}
Let $n_g' := \#\{G \in \Gamma(g): q(G) \leq 3\} $. Then
$$\lim_{g \to \infty} \frac{n_g'}{n_g} = 1.$$
For more details on this subject, we recommend the survey written by Kaplan \cite{Kaplan2}.

Some authors studied the set of numerical semigroups with fixed genus and some other invariants. For instance, Bernardini and Torres \cite{MF} studied numerical semigroups with fixed genus and fixed number of even gaps; Blanco and Rosales \cite{BR} studied numerical semigroups with fixed genus and fixed Frobenius number; Bras-Amor\' os \cite{Amoros3} studied numerical semigroups with fixed genus and fixed ordinarization number; Kaplan \cite{Kaplan} studied numerical semigroups with fixed genus and fixed multiplicity; all of them obtained some interesting partial results.

Throughout this paper, we write a gapset $G$ with genus $g$ as $G = \{\ell_1 < \ell_2 < \ldots < \ell_g\}$ and its elements will always be enumerated in the increasing natural order. In this paper, we study gapsets with fixed genus and we focus on pure $\kappa$-sparse gapsets. A $\kappa$-sparse gapset $G$ is a gapset such that the distance between two consecutive elements (in the natural order) is at most $\kappa$, i.e., $\ell_{j+1} - \ell_j \leq \kappa$, for all $j$. A pure $\kappa$-sparse gapset $G$ is a $\kappa$-sparse gapset such that there are two consecutive elements $\ell_i$ and $\ell_{i+1}$ such that $\ell_{i+1} - \ell_i = \kappa$. By convention, we say that $\emptyset$ is a pure $0$-sparse gapset and $\{1\}$ is a pure $1$-sparse gapset. The set of pure $\kappa$-sparse gapsets with genus $g$ is denoted by $\mathcal{G}_\kappa(g)$.

Munuera, Torres and Villanueva \cite{MTV} studied the so-called sparse semigroups, which are numerical semigroups such that its pairs of consecutive gaps have distance at most 2. In particular, they prove that the set of Arf semigroups is a proper subset of the set of sparse semigroups. Tizziotti and Villanueva \cite{JG} studied $\kappa$-sparse and pure $\kappa$-sparse numerical semigroups, which naturally generalize the concept of sparse semigroups. Here, we choose using gapset theory because we use information about the set of gaps of a numerical semigroup, such as its cardinality, the maximum distance between two consecutive gaps and its depth.

Throughout, we use the following notation: if $G$ is a pure $\kappa$-sparse gapset with genus $g \geq 2$, we write $G = \{1 = \ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\}$, with the index $\alpha$ being the maximum element of the set $\{i: \ell_{i+1} - \ell_i = \kappa\}$.

Here is an outline of this paper. In section 2, we present some general properties of gapsets. We show how  the multiplicity, the conductor and the depth of a gapset are related to its genus. In section 3, we study pure $\kappa$-sparse gapsets and we check some properties in relation to the elements of a pure $\kappa$-sparse gapset. In this case, we prove that the maximum distance between the consecutive elements can be limited by its multiplicity, its genus and its conductor. In addition, when we have the hypothesis $2g \leq 3\kappa$, where $g$ is the genus of the gapset and $\kappa$ is the biggest distance between two consecutive elements, the gapset has exactly one pair of consecutive elements with distance $\kappa$ (except for one case) and it has depth at most 3. In section 4, we build an injective map $\phi$ (see equation (\ref{mapaphi}), in page 8) from the family $\mathcal{G}_\kappa(g)$ to the set of subsets of $[1, 2g + 1]$ and we look for conditions on $g$ and $\kappa$ so that the image of a gapset is again a gapset. In particular, we prove that every gapset with depth at most two and every pure $\kappa$-sparse gapset with genus $g$ such that $2g \leq 3\kappa$ satisfy this property. We emphasize that this map brings a new approach to this topic. In section 5, we prove that the application $\varphi: \mathcal{G}_\kappa(g) \to \mathcal{G}_{\kappa+1}(g+1), G \mapsto \varphi(G) = \phi(G)$ is surjective and therefore establishes a bijection between $\mathcal{G}_{\kappa}(g)$ and $\mathcal{G}_{\kappa+1}(g+1)$ whenever the condition $2g \leq 3\kappa$ is satisfied. Thus, we conclude that $\# \mathcal{G}_{k}(g) = \# \mathcal{G}_{k+n}(g+n)$, for all $n \in \mathbb{N}$. In section 6, we discuss some further questions related to this work.

\section{Some basic results on gapsets}
\label{s2}
We bring some familiar results on numerical semigroup theory to gapset theory and we keep the proofs for the sake of completeness. The next result tells us about the relation between multiplicity and genus of a gapset.

\begin{proposition}
If a non-empty gapset has genus $g$ and multiplicity $m$, then $2 \leq m \leq g+1$.
\label{mg}
\end{proposition}

\begin{proof}
If $m \geq g+2$, then $[1,g+1] \subseteq G$ and $\# G \geq g+1$, which does not occur. Hence $m \leq g+1$. On the other hand, since $G \neq \emptyset$, then $1 \in G$ and $m \geq 2$.
\end{proof}

The next result gives us information about each element of a gapset and it can be found in \cite{Gilvan}.

\begin{proposition}
Let $G = \{\ell_1 < \ell_2 < \cdots < \ell_g\}$ be a gapset with genus $g$. Then $j \leq \ell_j \leq 2j-1$.
\label{gilvan}
\end{proposition}

\begin{proof}
If $\ell_j < j$ for some $j$, then $\ell_1 < 1$, which is a contradiction. On the other hand, let $j \in [1,g]$. If $\ell_j$ is even, then $\frac{\ell_j}{2} \in G$ and for each $\ell \in [1, \frac{\ell_j}{2} - 1]$, there is at least one element of $G$ in $\{\ell, \ell_j - \ell\}$. There are $j$ elements of $G$ in $[1, \ell_j]$, including $\frac{\ell_j}{2}$ and $\ell_j$, hence $j \geq (\ell_j/2 - 1) + 2$ and we obtain $\ell_j \leq 2j - 2 < 2j - 1$.  If $\ell_j$ is odd, then for each $\ell \in [1, \frac{\ell_j-1}{2}]$, there is at least one element of $G$ in $\{\ell, \ell_j - \ell\}$. There are $j$ elements of $G$ in $[1, \ell_j]$, including $\ell_j$, hence $j \geq \frac{\ell_j-1}{2} + 1$ and we obtain $\ell_j \leq 2j - 1$.
\end{proof}

The next result tells us about the relation between the conductor and the genus of a gapset.

\begin{proposition}
If a non-empty gapset has genus $g$ and conductor $c$, then $g+1 \leq c \leq 2g$.
\label{cond}
\end{proposition}

\begin{proof}
If $c \leq g$, then $G \subseteq [1,g-1]$ and $\# G \leq g-1$, which does not occur. Hence, $c \geq g+1$. On the other hand, Proposition \ref{gilvan} ensures that $\ell_g \leq 2g - 1$ and thus $c \leq 2g$.
\end{proof}

As a consequence, we obtain that if a non-empty gapset has genus $g$ and Frobenius number $F$, then $g \leq F \leq 2g-1$. Putting together Propositions \ref{mg} and \ref{cond}, we obtain a relation between the depth and the genus of a gapset.

\begin{corollary}
If a non-empty gapset has genus $g$ and depth $q$, then $1 \leq q \leq g$.
\label{dep}
\end{corollary}

\begin{proof}
By applying Propositions \ref{mg} and \ref{cond}, we obtain
$$1 = \frac{g+1}{g+1} \leq \frac{c}{m} \leq \frac{2g}{2} = g.$$
Hence, $1 \leq q \leq g$.
\end{proof}

Next, we present three classical examples. In particular, Examples \ref{q=1} and \ref{q=g} confirm that the bounds presented in Propositions \ref{mg}, \ref{gilvan}, \ref{cond} and in Corollary \ref{dep} are accurate.

\begin{example}
The trivial gapset $\emptyset$ has genus $0$, multiplicity $1$, conductor $0$ and depth $0$. It is the unique gapset with depth 0.
\label{q=0}
\end{example}

\begin{example}
Let $g \in \mathbb{N}$. The gapset $[1,g]$ has genus $g$, multiplicity $g+1$, conductor $g+1$ and depth $1$. There are no other gapsets with depth 1. Following the numerical semigroup terminology, those gapsets are \textit{ordinary gapsets}.
\label{q=1}
\end{example}

\begin{example}
Let $g \in \mathbb{N}$. The gapset $(2\mathbb{N} + 1) \cap [1,2g-1]$ has genus $g$, multiplicity $2$, conductor $2g$ and depth $g$. There are no other gapsets with depth $g$. Following numerical semigroup terminology, those gapsets are \textit{hyperelliptic gapsets}.
\label{q=g}
\end{example}

We end up this section with the following result.

\begin{proposition}
Let $G = \{\ell_1 < \ell_2 < \ldots < \ell_g\}$ be a gapset with multiplicity $m$ and genus $g$. Then $[am + \ell_j + 1, am + \ell_{j+1} - 1] \cap G = \emptyset$ for all $a \in \mathbb{N}_0$ and for all $j \in [1,g-1]$.
\label{gilberto_partition}
\end{proposition}

\begin{proof}
Let $a \in \mathbb{N}_0$ and $j \in [1,g-1]$. If $b \in [am + \ell_j + 1, am + \ell_{j+1} - 1]$, then  $b = am + \ell_j + c$, with $c \in [1, \ell_{j+1} - \ell_j - 1]$. Since $am \notin G$ and $\ell_j + c \notin G$, then $b \notin G$.
\end{proof}

\section{Pure $\kappa$-sparse gapsets}
\label{s3}

Following the notation for numerical semigroups, a gapset $G$ is \textit{$\kappa$-sparse} if the difference between two consecutive elements of $G$ (in the natural order) is at most $\kappa$. If there are two consecutive elements $\ell_{i}$ and $\ell_{i+1} \in G$ such that $\ell_{i+1} - \ell_i = \kappa$, then we say that $G$ is a pure $\kappa$-sparse gapset. Let $\mathcal{G}_{\kappa}$ be the set of all pure $\kappa$-sparse gapsets and $\tilde{\mathcal{G}}_{\kappa}$ be the set of all $\kappa$-sparse gapsets. In particular,  $\mathcal{G}_{\kappa} \subset \tilde{\mathcal{G}_{\kappa}}$ and $\tilde{\mathcal{G}_{\kappa}} \subset \tilde{\mathcal{G}}_{\kappa+1}$. In this section, we study the set of pure $\kappa$-sparse gapsets with genus $g$, namely $\mathcal{G}_{\kappa}(g) := \Gamma(g) \cap \mathcal{G}_{\kappa}$ and its subset $\mathcal{G}_{\kappa}(g, q) := \{G \in \mathcal{G}_{\kappa}(g): q(G) = q\}$. By convention, we say that $\emptyset$ is a $0$-sparse gapset and $\{1\}$ is a $1$-sparse gapset. We recall that if $G$ is a pure $\kappa$-sparse gapset with genus $g \geq 2$, then we write $G = \{\ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\}$, where $\alpha = \max\{i: \ell_{i+1} - \ell_i = \kappa\}$.

\begin{example}
Let $G = \{\ell_1 < \ell_2 < \ldots < \ell_g\}$ be a pure $1$-sparse gapset. Then $\ell_{i+1} - \ell_i = 1$, for all $i$. Hence, $G = \{1, 2, \ldots, g\}$ is an ordinary gapset.
\end{example}

Now we obtain a relation between the multiplicity of a gapset and the maximum distance between two consecutive elements.

\begin{proposition}
Let $G$ be a pure $\kappa$-sparse gapset with multiplicity $m$. Then $\kappa \leq m$.
\label{km}
\end{proposition}

\begin{proof}
Let $\ell_i$ and $\ell_{i+1} \in G$ such that $\ell_{i+1} - \ell_i = \kappa$. If $\kappa > m$, then $\ell_i + 1, \ell_i + 2, \ldots, \ell_i + m \notin G$ ($\ell_i + m < \ell_{i+1}$). There is $a \in \mathbb{Z}$ such that $\ell_i + 1 \leq \ell_{i+1} - am \leq \ell_i + m$.  Notice that $G \ni \ell_{i+1} = (\ell_{i+1} - am) + am$. But $\ell_{i+1} - am \notin G$ and $am \notin G$, which leads to a contradiction.
\end{proof}

At first, Proposition \ref{km} together with Proposition \ref{mg} imply that a pure $\kappa$-sparse gapset with genus $g$ satisfies $\kappa \leq g + 1$. We obtain a better result in the next proposition.

\begin{proposition}
Let $G$ be a pure $\kappa$-sparse gapset with genus $g$. Then $\kappa \leq g$.
\end{proposition}

\begin{proof}
Let $\ell_i$ and $\ell_{i+1} \in G$ such that $\ell_{i+1} - \ell_i = \kappa$. If $\kappa \geq g+1$, then $\ell_i + 1, \ell_i + 2, \ldots, \ell_i + g \in [1,2g-2] \cap (\mathbb{Z} \setminus G)$ ($\ell_i + g < \ell_{i+1} \leq \ell_{g}$). Hence, \linebreak $\# G \leq (2g - 2 - g) + 1 = g - 1$, which is a contradiction.
\end{proof}

The next result improves the upper bound on Proposition \ref{cond}.

\begin{proposition}
Let $G$ be a pure $\kappa$-sparse gapset with conductor $c$ and genus $g$. Then $g + \kappa \leq c$.
\label{kF}
\end{proposition}

\begin{proof}
Suppose that $g + \kappa > c$. By writing $G = \{\ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\}$, we conclude that $\ell_g \leq g + \kappa - 2$. We have that \linebreak $I = [\ell_\alpha + 1, \ell_{\alpha+1} - 1] \cap G = \emptyset$, $\#I = \kappa-1$ and $G \subset [1, g + \kappa - 2]$. Hence, $\#G \leq (g+ \kappa -2) - (\kappa-1) = g - 1$, which is a contradiction.
\end{proof}

\begin{example}
The ordinary gapset with genus $g$ has multiplicity $g+1$, conductor $g+1$ and $\kappa = 1$. The hyperelliptic gapset  with genus $g$ has multiplicity $2$, conductor $2g$ and $\kappa = 2$. Notice that they do not attain the maximum $\kappa$ with respect to the multiplicity and the genus. The ordinary gapset attains the bound $g + \kappa = c$. The gapset $G = \{1, 2, \ldots, g-1, 2g-1\}$ has genus $g$, multiplicity $g$ and $\kappa = g$, i. e., it attains the maximum value of $\kappa$ with respect to the multiplicity and with respect to the genus.
\end{example}

\begin{remark}
If $G = \{\ell_1 < \ell_2 < \ldots < \ell_g\}$ is a pure $\kappa$-sparse gapset with genus $g$ such that $\ell_{i+1} - \ell_i = \kappa$, then $\kappa \leq i+1$. This follows from Proposition \ref{gilvan}, since $\kappa = \ell_{i+1} - \ell_i \leq 2i+1 - i = i+1$.
\end{remark}

The next result is important for the proofs of the main results of this paper. It confirms that the Frobenius number of a gapset cannot be too big with respect to its element $\ell_\alpha$.

\begin{proposition}
Let $G = \{\ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\}$ be a pure $\kappa$-sparse gapset with genus $g$, where $\alpha = \max\{i: \ell_{i+1} - \ell_i = \kappa\}$. Then $\ell_g \leq \ell_\alpha + m$.
\label{abaixo}
\end{proposition}

\begin{proof}
Suppose that $\ell_g > \ell_\alpha + m$ and let $r$ be the smallest index such that $\ell_r > \ell_\alpha + m$. By Proposition \ref{gilberto_partition}, we conclude that $\ell_r \geq \ell_{\alpha+1} + m$ and thus $\ell_r - \ell_{r-1} \geq \kappa$. Since $G$ is a pure $\kappa$-sparse gapset, we have $\ell_r - \ell_{r-1} = \kappa$ and thus $r \geq \alpha + 1$. On the other hand, $r \leq \alpha + 1$. In fact, we know that $\ell_j \leq \ell_{\alpha} < \ell_{\alpha} + m$, for all $j \in [1,\alpha]$. By the definition of $\alpha$, we know that $\alpha$ is the greatest index such that $\ell_{i+1} - \ell_i = \kappa$ and thus $r \leq \alpha +1$. Hence, the only possibility is $r = \alpha + 1$. However $\ell_{\alpha+1} = \ell_\alpha + \kappa \leq \ell_\alpha + m$ and we obtain a contradiction. Hence, $\ell_g \leq \ell_\alpha + m$.
\end{proof}

Recall that the canonical partition of a gapset $G$ with multiplicity $m$ and depth $q$ is $G_0 \cup G_1 \cup \ldots \cup G_{q-1}$, where $G_0 =[1, m-1]$ and $G_i = G \cap [im + 1, (i+1)m - 1]$ (cf. \cite{EF}). As a consequence of Proposition \ref{abaixo}, we obtain information about where the element $\ell_\alpha$ can be.

\begin{corollary}
Let $G = \{\ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\}$ be a pure $\kappa$-sparse gapset with genus $g$, where $\alpha = \max\{i: \ell_{i+1} - \ell_i = \kappa\}$, and consider its canonical partition $G_0 \cup G_1 \cup \cdots \cup G_{q-1}$. Then one of the following occurs:
  \begin{itemize}
    \item[$\bullet$] $\ell_\alpha, \ell_{\alpha+1} \in G_{q-2}$
    \item[$\bullet$] $\ell_\alpha, \ell_{\alpha+1} \in G_{q-1}$
    \item[$\bullet$] $\ell_\alpha \in G_{q-2}$ and $\ell_{\alpha+1} \in G_{q-1}$
  \end{itemize}
\label{trescasos}
\end{corollary}

\begin{proof}
From Proposition \ref{abaixo}, we conclude that $\ell_\alpha \in G_{q-2}$ or  $\ell_\alpha \in G_{q-1}$, because $\ell_g \in G_{q-1}$. For the first case, both possibilities can occur: $\ell_{\alpha+1} \in G_{q-2}$ or $\ell_{\alpha+1} \in G_{q-1}$. For the second case, we must have $\ell_{\alpha+1} \in G_{q-1}$ and we are done.
\end{proof}

The next example illustrates that the three cases obtained in Corollary \ref{trescasos} can occur. For emphasizing, the last pair of consecutive elements of a gapset that attains the maximum distance will be written in bold.

\begin{example}
For an integer $m \geq 3$, we have the following examples:
  \begin{itemize}
    \item[$\bullet$]  $\{1, \ldots, m-1, \boldsymbol{m + 1}, \boldsymbol{2m - 1}, 2m + 1\} = [1,m-1] \cup \{m+1, 2m-1\} \cup \{2m+1\}$ has $q = 3$, $\ell_\alpha = m+1 \in G_1$ and $\ell_{\alpha+1} = 2m-1 \in G_1$;
    \item[$\bullet$]  $\{1, \ldots, m-1, \boldsymbol{m + 1}, \boldsymbol{2m - 1}\} = [1,m-1] \cup \{m+1, 2m-1\}$ has $q = 2$, $\ell_\alpha = m + 1 \in G_1$ and $\ell_{\alpha+1} = 2m - 1 \in G_1$;
    \item[$\bullet$]  $\{1, \ldots, m-1, \boldsymbol{m + 1}, \boldsymbol{2m + 1}\} = [1, m-1] \cup \{m+1\} \cup \{2m + 1\}$ has $q = 3$, $\ell_\alpha = m+1 \in G_1$ and $\ell_{\alpha+1} = 2m + 1 \in G_2$.
  \end{itemize}
\end{example}

\noindent \textbf{The case $2g \leq 3\kappa$.} At this point, we focus on pure $\kappa$-sparse gapsets with genus $g$ such that $2g \leq 3\kappa$ which play an important role in the end of Section 5 and in Section 6.

\begin{proposition}
Let $G$ be a pure $\kappa$-sparse gapset with genus $g$ and depth $q$. If $2g \leq 3\kappa$, then $q \leq 3$.
\label{2g3kq3}
\end{proposition}

\begin{proof}
Since $m \geq \kappa \geq \frac{2g}{3}$ and $c \leq 2g$, we have $q = \left \lceil \frac{c}{m} \right \rceil \leq \frac{2g}{\frac{2g}{3}} = 3$.
\end{proof}

\begin{remark}
It is important to notice that the gapset $\{1, 2, 3, 4, \textbf{6}, \textbf{9}, 11\}$ has depth $3$, genus $7$ and $\kappa = 3$. In this case $2g > 3\kappa$. Hence, the converse of Proposition \ref{2g3kq3} is not true.
\end{remark}

The next proposition tells us that, if $2g \leq 3\kappa$, then the set $\{i: \ell_i - \ell_{i-1} = \kappa\}$ has only the element $\alpha$, with only one exception.

\begin{proposition}
Let $G$ be a pure $\kappa$-sparse gapset with genus $g$ such that $2g \leq 3\kappa$. Then there is an unique $i \in [1, g-1]$ such that $\ell_{i+1} - \ell_i = \kappa$, except for $G = \{1,3,5\}$.
\label{unique}
\end{proposition}

\begin{proof}
Suppose that there are $i \neq j$ such that $\ell_{i+1} - \ell_i = \ell_{j+1} - \ell_j = \kappa$. Then $\ell_g \geq 1 + 2 \cdot \kappa + (g - 3) \cdot 1 = 2\kappa + g - 2 \geq 4g/3 + g - 2 = 7g/3 - 2 \geq 2g$, if $g \geq 6$. By Proposition \ref{gilvan}, it is a contradiction. The cases $g \leq 5$ can be done by using GAP.
\end{proof}

\begin{remark}
If $2g > 3\kappa$, then it is possible to have a unique $i$ such that $\ell_{i+1} - \ell_i = \kappa$ and it is possible to have more than one index that satisfies that property. For instance, if $G_1 = \{1,2,3,5,6\}$, then $2g(G_1) = 10 > 3\kappa(G_1) = 6$ and $\kappa(G_1) = 5 - 3$; if $G_2$ is the hyperelliptic gapset with genus $g \geq 4$, then $2g(G_2) \geq 8 > 3\kappa(G_1) = 6$ and every pair of consecutive elements of $G_2$ have distance 2.
\end{remark}

For the next result we consider a pure $\kappa$-sparse gapset $G$ with genus $g$ such that $2g \leq 3\kappa$ and we use the last result about the uniqueness of a pair of consecutive elements with the greatest distance $\kappa$. If $q(G) = 3$ and $G_0 \cup G_1 \cup G_2$ is the canonical partition of $G$, then Corollary \ref{trescasos} ensures that $\ell_\alpha \in G_1$ or $\ell_a \in G_2$. The next result ensures that $\ell_\alpha \in G_1$ under the hypothesis $2g \leq 3k$.

\begin{proposition}
Let $g > 1$ and $\kappa$ be positive integers such that $2g \leq 3\kappa$ and consider $G = \{\ell_1 < \ell_2 < \ldots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \ldots < \ell_g\}$ be a pure $\kappa$-sparse gapset with multiplicity $m$,  where $\alpha = \max\{i: \ell_{i+1} - \ell_i = \kappa\}$. Then $\ell_\alpha \leq 2m - 1$.
\label{gilberto}
\end{proposition}

\begin{proof}
The hypothesis $2g \leq 3\kappa$ ensures that the depth of $G$ is at most 3. Suppose that $\ell_\alpha \geq 2m+1$. By Proposition \ref{km}, $\ell_{\alpha+1} = \ell_\alpha + \kappa \geq 2m + 1 + \kappa \geq 3\kappa +1$. Hence, $\ell_{\alpha+1} \geq 2g + 1$, which is a contradiction.
\end{proof}

\begin{remark}
This property does not hold true in the general case, even for depth at most 3. For instance, the gapset $G = \{1,2,3,4,6,7,9,\textbf{11},\textbf{14}\}$ has $\ell_\alpha = 11 > 9 = 2m-1$.
\end{remark}

\begin{remark}
Proposition \ref{gilberto} gives informations about $\ell_\alpha$, but not about $\ell_{\alpha+1}$. For instance, if $G = \{1, 2, \ldots, 10, \textbf{12}, \textbf{21}, 23\}$, then $\ell_{\alpha+1} \leq 2m-1$; if $G = \{1,2,\textbf{4},\textbf{7}\}$, then $\ell_{\alpha+1} > 2m-1$.
\label{2casos}
\end{remark}

\section{The map $\phi$}
\label{s4}

In this section, we introduce the map $\phi$ and we discuss which properties the set $\phi(G)$ has, when $G \in \mathcal{G}_\kappa(g)$. Our aim is that $\phi(G)$ is a pure $(\kappa+1)$-sparse gapset with genus $g+1$. Hence, we look for hypotheses that guarantee it.

Eliahou and Fromentin \cite{EF} introduced the notion of $m$-extension, for $m \in \mathbb{N}$. An $m$-extension $A \subset \mathbb{N}$ is a finite set containing $[1,m-1]$ that admits a partition \linebreak $A = A_0 \cup A_1 \cup \ldots \cup A_t$, for some $t \in \mathbb{N}_0$, where $A_0 = [1,m-1]$ and $A_{i+1} \subseteq m + A_i$ for all $i$. In particular, if $A$ is an $m$-extension, then $A \cap m\mathbb{N} = \emptyset$. For our approach, we deal with more general sets and we define them as follows: let $m > 2$ be an integer. We say that a finite set $M \subset \mathbb{N}$ is an $m$-set if $[1, m-1] \subset M$ and $M \cap m\mathbb{N} = \emptyset$. In particular, an $m$-extension is an $m$-set. If $c$ is the largest element of an $m$-set $M$, then its depth is defined as $\left \lceil \frac{c}{m} \right \rceil$. We denote by $\mathcal{M}_\kappa(g)$ the set of all $m$-sets with $g$ elements, such that the maximum distance between two consecutive elements (with respect to the natural order) is $\kappa$ and that lies in $[1,2g - 1]$. In some cases, we also deal with subsets of $[1, 2g - 1]$ with no specific property. We denote by $\mathcal{C}_{\kappa}(g)$ the set of those sets that have $g$ elements and such that the maximum distance between two consecutive elements (with respect to the natural order) is $\kappa$. Notice that $\mathcal{G}_\kappa(g) \subset \mathcal{M}_\kappa(g) \subset \mathcal{C}_\kappa(g)$.

Let $g$ and $\kappa$ be non-negative integers and define $\phi: \mathcal{G}_{\kappa}(g) \to \mathcal{C}_{k+1}(g+1)$, with

$$
G = \{1 = \ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\} \mapsto
$$
\begin{equation}
\label{mapaphi}
\mapsto \phi(G) = \{1 = \ell_0 < 2 = \ell_1 + 1 < \ell_2 + 1 < \cdots < \ell_\alpha + 1 < \ell_{\alpha+1} + 2 < \cdots < \ell_g + 2\},
\end{equation}
where $\alpha := \max\{i: \ell_{i+1} - \ell_i = \kappa\}$.

As a convention, $\phi(\emptyset) = \{1\}$ and $\phi(\{1\}) = \{1,3\}$.

\begin{theorem}
The map $\phi$ is well defined and it is injective. Moreover, if $G \in \mathcal{G}_\kappa(g)$ has multiplicity $m$, then

\begin{enumerate}
  \item if $q(G) = 1$, then $\phi(G)$ is a gapset of depth $2$;
  \item if $q(G) = 2$, then $\phi(G)$ is an $(m+1)$-set of depth $2$;
  \item if $q(G) = 3$, then $\phi(G)$ is an $(m+1)$-set of depth $3$, except for the case $2m + 1 \in G$ and $\ell_\alpha \geq 2m+1$.
\end{enumerate}
\label{phi}
\end{theorem}

\begin{proof}
First we prove that $\phi(G) \in \mathcal{C}_{\kappa+1}(g+1)$. The maximum distance between two consecutive elements in $\phi(G)$ is $(\ell_{\alpha+1} + 2) - (\ell_a + 1) = \kappa + 1$ and there are $g+1$ elements in $\phi(G)$. Since $\ell_g \leq 2g -1$, then $\ell_g + 2 \leq 2g+1$ and $\phi(G) \subset [1, 2(g+1) - 1]$.

Now we prove that $\phi$ is injective.  Let $G_1 = \{a_1 < a_2 < \ldots < a_{\alpha_1} < a_{\alpha_1+1} = a_{\alpha_1} + \kappa < \ldots <a_g\}$ and $G_2 =  \{b_1 < b_2 < \ldots < b_{\alpha_2} < b_{\alpha_2+1} = b_{\alpha_2} + \kappa < \ldots <b_g\} \in \mathcal{G}_{k}(g)$ such that $\phi(G_1) = \phi(G_2)$, where $\alpha_1 = \max\{i: a_{i+1} - a_{i} = \kappa\}$ and $\alpha_2 = \max\{i: b_{i+1} - b_{i} = \kappa\}$. First, we prove that $\alpha_1 = \alpha_2$. Otherwise, without loss of generality, suppose that $\alpha_1 < \alpha_2$. Then $a_n = b_n$, for all $n \in [1,\alpha_1] \cup [\alpha_2+1, g]$ and $a_{\alpha_2} + 2 = b_{\alpha_2} + 1$. Hence, $\kappa = b_{\alpha_2+1} - b_{\alpha_2} = a_{\alpha_2+1} - (a_{\alpha_2}+1)$, i.e., $\kappa+1 = a_{\alpha_2+1} - a_{\alpha_2}$ which is a contradiction. Hence, $\alpha_1 = \alpha_2$ and $G_1 = G_2$.

If $q = 1$, then $G$ is the ordinary gapset with genus $g$ and $\phi(G) = [1,g] \cup \{g+2\}$ is a gapset with depth 2.

It remains to solve the cases $q = 2$ and $q = 3$. Let $G = \{1 = \ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\}$ with $m(G) = m$; in particular, $[1,m-1] \subset G$ and $m \notin G$ implies that $[1,m] \subset \phi(G)$ and $m+1 \notin \phi(G)$.

If $q = 2$, then $m+1 \leq \ell_g \leq 2m-1$ and the maximum element of $\phi(G)$, $\ell_g + 2$, is such that $(m+1) + 1 < m+3 \leq \ell_g + 2 \leq 2m+1 = 2(m+1) - 1$. Thus, $2(m+1) \notin \phi(G)$, which guarantees that $\phi(G)$ is an $(m+1)$-set with depth 2.

If $q = 3$, then $2m+1 \leq \ell_g \leq 3m-1$ and the maximum element of $\phi(G)$, $\ell_g + 2$, is such that $2(m+1) + 1 = 2m+3 \leq \ell_g + 2 \leq 3m+1 < 3(m+1) - 1$. Thus, $3(m+1) \notin \phi(G)$. If $2m+1 \notin G$, then $2m+2 \notin \phi(G)$, since $2m$ and $2m+1 \notin G$. If $2m+1 \in G$ and $\ell_\alpha < 2m+1$, then $2m+2 \notin G$, since the corresponding element of $2m+1$ in $\phi(G)$ is $2m + 3$. Hence, $2m+2$ and $3m+3 \notin G$ and $\phi(G)$ is an $(m+1)$-set with depth 3.
\end{proof}

\begin{remark}
In general, the image of a gapset under $\phi$ is not an $m$-set. The gapset \linebreak $A =\{1,2,3,4,6,7,9,11,14\}$ has depth 3, multiplicity $5$ and its image \linebreak $\phi(A) = \{1,2,3,4,5,7,8,10,12,16\}$ is not a $6$-set. The gapset $B = \{1, 2, 4, 5, 7, 10\}$ has depth 4, multiplicity 3 and its image $\phi(B) = \{1, 2, 3, 5, 6, 8, 12\}$ is not a $4$-set.
\end{remark}

\begin{corollary}
If $G \in \mathcal{G}_\kappa(g)$ has multiplicity $m$ and $2g \leq 3\kappa$ , then $\phi(G)$ is an $(m+1)$-set.
\label{mset}
\end{corollary}

\begin{proof}
Proposition \ref{2g3kq3} ensures that the depth of $G$ is at most 3. The cases of depth 1 or 2 follow from (1) and (2) of Theorem \ref{phi}. If $q = 3$, then Proposition \ref{abaixo} ensures that $\ell_a \leq 2m-1$ and then (3) of Theorem \ref{phi} completes the proof.
\end{proof}

Eliahou and Fromentin \cite{EF} worked with the map $\sigma$ and for a gapset $G$ with canonical partition $G_0 \cup G_1 \cup G_2$ and multiplicity $m$. They define $\sigma(G) = (G_0 \cup \{m\}) \cup (G_1 + 1) \cup (G_2 + 2)$. In particular, $\sigma(G)$ is always a gapset, if its depth is at most 3. Notice that $g(\sigma(G)) = g(G) + 1$. The map $\phi$ considered in this paper is different from the map $\sigma$, besides the fact that both maps increase the genus by one. For instance, if $G = [1,9] \cup \{\textbf{11}, \textbf{19}\} \cup \{21\}$, then $\phi(G) = [1,10] \cup \{\textbf{12},\textbf{21}\} \cup \{23\}$ and $\sigma(G) = [1,10] \cup \{12,20\} \cup \{23\}$. More generally, let $g \geq 12$. If $G = [1,g-3] \cup \{g-1, 2g-5\} \cup \{2g-3\}$, then $\phi(G) = [1,g-2] \cup \{g,2g-3\} \cup \{2g-1\}$ and $\sigma(G) = [1,g-2] \cup \{g,2g-4\} \cup \{2g-1\}$.

Now we deal with pure $\kappa$-sparse gapsets with genus $g$ and depth 2 and 3, namely $\mathcal{G}_\kappa(g,2):= \{G \in\mathcal{G}_\kappa(g): q(G) = 2\}$ and $\mathcal{G}_\kappa(g,3):= \{G \in\mathcal{G}_\kappa(g): q(G) = 3\}$, respectively. Theorem \ref{phi} guarantees that gapsets with multiplicity $m$ and depth at most 3 are mapped onto $(m+1)$-sets with depth at most 3, except for the case of gapsets $G$ with depth 3, $2m + 1 \in G$ and $\ell_\alpha \geq 2m+1$. Now we look for conditions that ensure that those $(m+1)$-sets are in fact gapsets.

The next result guarantees that every $m$-set with $q = 1$ or $2$ is a gapset.

\begin{lemma}
Let $G \subseteq [1, 2m-1]$ be an $m$-set. Then $G$ is a gapset with multiplicity $m$ and depth at most $2$.
\label{q=2}
\end{lemma}

\begin{proof}
Let $z \in G$ and write $z = x + y$, with $x \leq y$. Since $z \leq 2m-1$, then $x \leq m-1$. Hence, $x \in G$ and the proof is complete.
\end{proof}

\begin{theorem}
Let $g$ and $\kappa$ be non-negative integers and $\phi_2 := \phi \vert_{\mathcal{G}_\kappa(g,2)}$. Then \linebreak $\Imm(\phi_2) \subseteq \mathcal{G}_{\kappa+1}(g+1,2)$.
\label{q2}
\end{theorem}

\begin{proof}
Let $G' \in \Imm(\phi_2)$. Then there is $G \in \mathcal{G}_{\kappa}(g)$ such that $\phi_2(G) = G'$. If $m(G) = m$ and $q(G) = 2$, then Theorem \ref{phi} guarantees that $G'$ is an $(m+1)$-set with depth 2. By Lemma \ref{q=2}, we conclude that $G'$ is a gapset and $G' \in \mathcal{G}_{\kappa+1}(g+1,2)$.
\end{proof}

\begin{remark}
The map $\phi_2$ is injective but it is not surjective, since $[1,g] \cup \{g+2\} \notin \phi_2^{-1}(\mathcal{G}_{\kappa+1}(g+1,2))$. In particular, $\#\mathcal{G}_\kappa(g,2) < \#\mathcal{G}_{\kappa+1}(g+1,2)$.
\end{remark}

\begin{example}
Table \ref{table1} shows how $\phi_2$ acts.

\begin{table}
\centering
\caption{Some examples of Theorem \ref{q2} for genus 2, 3 and 4.}
\label{table1}
\begin{tabular}{c c}
\hline\noalign{\smallskip}
$G$ & $\phi_2(G)$ \\
\noalign{\smallskip}\hline\noalign{\smallskip}
$\{\textbf{1},\textbf{3}\}$ & $\{1,\textbf{2},\textbf{5}\}$ \\
$\{1,\textbf{2},\textbf{5}\}$ & $\{1,2,\textbf{3},\textbf{7}\}$ \\
$\{1,\textbf{2},\textbf{4}\}$ & $\{1,2,\textbf{3},\textbf{6}\}$ \\
$\{1,2,\textbf{3},\textbf{7}\}$ & $\{1,2,3,\textbf{4},\textbf{9}\}$ \\
$\{1,2,\textbf{3},\textbf{6}\}$ & $\{1,2,3,\textbf{4},\textbf{8}\}$ \\
$\{1,2,\textbf{3},\textbf{5}\}$ & $\{1,2,3,\textbf{4},\textbf{7}\}$ \\
$\{1,\textbf{2},\textbf{4},5\}$ & $\{1,2,\textbf{3},\textbf{6},7\}$ \\
\noalign{\smallskip}\hline
   \end{tabular}
\end{table}
\end{example}

Next, we present a result similar to Theorem \ref{q2} with an additional condition. In this case, we consider gapsets with depth 3.

\begin{theorem}
Let $g$ and $\kappa$ be non-negative integers such that $2g \leq 3\kappa$ and $\phi_3 := \phi \vert_{\mathcal{G}_\kappa(g,3)}$. Then $\Imm (\phi_3) \subseteq$ $\mathcal{G}_{\kappa+1}(g+1,3)$.
\label{q3}
\end{theorem}

\begin{proof}
Let $G = \{\ell_1 < \ell_2 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa < \cdots < \ell_g\} \in \mathcal{G}_k(g,3)$,  where $\alpha = \max\{i: \ell_{i+1} - \ell_i = \kappa\}$ and $m(G) = m$. By Corollary \ref{mset}, $\phi(G)$ is an $m$-set that lies in $\mathcal{M}_{k+1}(g+1,3)$. It remains to prove that $\phi_3(G)$ is a gapset. Consider the canonical partition of $\phi(G)$, namely $G'_0 \cup G'_1 \cup G'_2$. Let $z \in \phi(G)$ and write $z = x + y$, with $x \leq y$. We consider three cases as follows:

\begin{enumerate}
  \item[$\bullet$] $z \in G'_0 = [1, m]$. In this case, both $x$ and $y \in G'_0$.
  \item[$\bullet$] $z \in G'_1 \subseteq [m+2, 2m+1]$. In this case, $2x \leq x + y = z \leq 2m +1$. Thus, $x \leq m$ and $x \in G'_0$.
  \item[$\bullet$] $z \in G'_2 \subseteq [2m+3, 3m+1]$. We claim that if $x \leq m$ or $y \geq 2m+1$, then $x \in G'_0$. The first case is trivial and the second one implies that $3m + 1 \geq z \geq x + 2m + 1$ and we obtain $x \leq m$. Hence it remains to show that if $x, y \in [m+1, 2m]$, with $x \leq y$, then $x$ or $y \in G'_1$.
  \begin{itemize}
  \item Consider $\ell_{\alpha+1} \leq 2m - 1$. In this case, we claim that $y-1 \leq \ell_\alpha$. In fact, if $y \geq \ell_\alpha + 2$, then $z \geq (m + 1) + (\ell_\alpha + 2) = \ell_\alpha + m + 3$. Thus, $G \ni z-2 \geq \ell_\alpha + m + 1$, which is a contradiction according to Proposition \ref{abaixo}. Thus, both $x-1$ and $y-1$ are smaller than or equal to $\ell_\alpha$ and we can write $z - 2 = (x-1) + (y-1)$. Since $z-2 \in G$ and $G$ is a gapset, we conclude that $x-1$ or $y-1 \in G$ and thus $x \in \phi(G)$ or $y \in \phi(G)$.
   \item Consider $\ell_{\alpha+1} \geq 2m + 1$.  We use the hypothesis $2g \leq 3\kappa$ in this case. By Proposition \ref{gilberto}, we have $\ell_\alpha \leq 2m - 1$ and thus $\ell_\alpha + 1 \in G'_1$. Hence, $z \geq \ell_{\alpha+1}+2$. In this case, $x \leq y \leq 2m < \ell_{\alpha+1}$ and $G \ni z - 2 = (x-1) + (y-1)$. Since $G$ is a gapset, then $x-1$ or $y-1 \in G$. Hence, $x \in \phi(G)$ or $y \in \phi(G)$.
  \end{itemize}
\end{enumerate}
\end{proof}

\begin{remark}
In general, $\phi(G)$ is not a gapset, even if the depth of $G$ is 3 and $\phi(G)$ is an $m$-set. The gapset $G =\{1,\ldots, 9, 12, 13, 14, 17, 18, 23, 28\}$ has depth 3 and multiplicity 10 and its image $\phi(G) = \{1,\ldots, 10, 13, 14, 15, 18, 19, 24, 30\}$ is an $11$-set, but it is not a gapset. There are some examples such as $G = \{1,2,4,5,7\}$ such that $\phi(G) = \{1,2,3,5,6,9\}$ is also a gapset, even when $2g > 3\kappa$.
\end{remark}

\begin{example}
Table \ref{table2} shows how $\phi_3$ acts.

\begin{table}
\centering
\caption{Some examples of Theorem \ref{q3} for genus 3, 4, 5 and 6.}
\label{table2}
  \begin{tabular}{c c}
\hline\noalign{\smallskip}
$G$ & $\phi_3(G)$ \\
\noalign{\smallskip}\hline\noalign{\smallskip}
$\{1,\textbf{3},\textbf{5}\}$ & $\{1,2,\textbf{4},\textbf{7}\}$ \\
$\{1,2,\textbf{4},\textbf{7}\}$ & $\{1,2,3,\textbf{5},\textbf{9}\}$ \\
$\{1,2,3,\textbf{5},\textbf{9}\}$ & $\{1,2,3,4,\textbf{6},\textbf{11}\}$ \\
$\{1,2,3,4,\textbf{6},\textbf{11}\}$ & $\{1,2,3,4,5,\textbf{7},\textbf{13}\}$ \\
$\{1,2,3,5,\textbf{6},\textbf{10}\}$ & $\{1,2,3,4,6,\textbf{7},\textbf{12}\}$ \\
$\{1,2,3,5,\textbf{7},\textbf{11}\}$ & $\{1,2,3,4,6,\textbf{8},\textbf{13}\}$ \\
$\{1,2,3,6,\textbf{7},\textbf{11}\}$ & $\{1,2,3,4,7,\textbf{8},\textbf{13}\}$ \\
\noalign{\smallskip}\hline
   \end{tabular}
\end{table}
\end{example}

\section{On the cardinality of $\mathcal{G}_\kappa(g)$, for $2g \leq 3\kappa$}
\label{s5}

In this section, we present the main result of this paper.

\begin{theorem}
Let $g$ and $\kappa$ be non-negative integers such that $2g \leq 3\kappa$. Then the map $\varphi: \mathcal{G}_\kappa(g) \to \mathcal{G}_{\kappa+1}(g+1), G \mapsto \varphi(G) = \phi(G)$ is bijective.
\label{main}
\end{theorem}

\begin{proof}
First of all, notice that the hypothesis $2g \leq 3\kappa$ implies that $\mathcal{G}_\kappa(g)$ has only the gapsets $\emptyset, \{1\}$ and gapsets with depth 2 or 3. Theorem \ref{phi} guarantees that $\varphi$ is injective; Theorems \ref{q2} and \ref{q3} guarantee that $\varphi$ is well defined. Hence, we only have to prove that $\varphi$ is surjective.

Let $G \in \mathcal{G}_{k+1}(g+1)$ such that $2g \leq 3\kappa$. If $G = \{1\}$, then $G' = \emptyset$ is such that $\phi(G') = G$ and if $G = \{1,3\}$, then $G' = \{1\}$  is such that $\phi(G') = G$ . Hence, we can assume that $G$ has genus $g + 1 \geq 3$.

Write $G = \{\ell_1 < \ell_2 < \ell_3 < \cdots < \ell_\alpha < \ell_{\alpha+1} = \ell_\alpha + \kappa + 1 < \ldots < \ell_{g+1}\}$,  where $\alpha = \max\{i: \ell_{i+1} - \ell_i = \kappa\}$ and assume that its multiplicity is $m$. Since $2g \leq 3\kappa$ then $2(g+1) \leq 3\kappa + 2 < 3(\kappa+1)$. In particular $G$ has depth at most 3 and thus $\ell_{g+1} \leq 3m - 1$. Proposition \ref{unique} ensures that $\alpha$ is the unique index such that $\ell_{\alpha+1} - \ell_{\alpha} = \kappa + 1$ (recall $\{1,3,5\} \notin \mathcal{G}_{\kappa+1}(g+1)$). We show that $G' = \{\ell_2 - 1 < \ell_3 - 1 < \cdots < \ell_\alpha - 1 < \ell_{\alpha+1} - 2 < \ldots < \ell_{g+1} - 2\} \in \mathcal{G}_{\kappa}(g)$ and satisfies $\varphi(G') = G$.

Naturally, $\#G' = g$, the maximum distance between two consecutive elements of $G'$ is $(\ell_{\alpha+1} - 2) - (\ell_\alpha - 1) = \kappa$ and the first positive integer that does not belong to $G'$ is $m - 1$.  Now we prove that $G'$ is an $(m-1)$-set:

\begin{itemize}
  \item[$\bullet$] $m-1 \notin G'$ \\
Since $\ell_j = j, \forall j \in [1,m-1]$, then $G' \ni \ell_j - 1 = j - 1, \forall j \in [2,m-1]$; we also have $\ell_m \geq m + 1$. We study two cases, as follows: (1) if $\ell_m \geq m+2$, then $\ell_m - 1 > \ell_m - 2 \geq m$. In this case, we obtain that $m-1 \notin G'$; (2) suppose that $\ell_m = m + 1$. If there is $j \in [m, g]$ such that $\ell_{j+1} - \ell_j \geq 2$, then $G' \ni \ell_m - 1 = m$ and thus $m-1 \notin G'$. Otherwise, we can write $\ell_j = j + 1, \forall j \in [m, g]$ and the maximum distance between two consecutive elements of $G$ is $\kappa + 1= 2$, which cannot occur ($2g > 3$). Hence, $m-1 \notin G'$.
  \item[$\bullet$] $2(m-1) \notin G'$  \\
Suppose that $2m - 2 \in G'$. In this case, we could have $2m \in G$ or $2m - 1 \in G$. The first case does not occur, because $m$ is the multiplicity of $G$. In the second case, we have  $2m - 1 \leq \ell_\alpha$ (otherwise, the corresponding element would be $(2m - 1) - 2 = 2m - 3)$. By Proposition \ref{gilberto}, we have that $\ell_\alpha = 2m - 1$. Thus, $\ell_{\alpha+1} = \ell_\alpha + (\kappa + 1) = 2m - 1 + \kappa + 1 \geq 2(\kappa+1) + \kappa = 3\kappa + 2 \geq 2(g+1)$, which is a contradiction. Hence, $2m - 2 \notin G'$.
  \item[$\bullet$] $3(m-1) \notin G'$ \\
We already know that $\ell_{g+1} \leq 3m - 1$ and thus the maximum element of $G'$ is such that $\ell_{g+1} - 2 \leq 3m - 3$. Hence, we only have to show that $\ell_{g+1} \neq 3m - 1$ which implies that $\ell_{g+1} - 2 \leq 3m - 4$. Suppose that $\ell_{g+1} = 3m - 1$. In this case, we have $2m - 1 \in G$ and we could have $\ell_\alpha < 2m - 1$ or $\ell_\alpha = 2m - 1$. The first case implies that $\ell_{\alpha+1} \leq 2m - 1$. By Proposition \ref{gilberto_partition}, we can denote by $\ell_r$ and $\ell_{r+1}$ the biggest element of $G$ that is smaller than or equal to $\ell_{\alpha} + m$ and the smallest element of $G$ that is bigger than or equal to $\ell_{\alpha+1} + m$, respectively. In this case, $\ell_{r+1} - \ell_r \geq \kappa$, which is a contradiction, according to Proposition \ref{unique} (notice that $G \neq \{1,3,5\}$). In the second case, we obtain $\ell_{\alpha+1} = \ell_\alpha + (\kappa + 1) = 2m - 1 + \kappa + 1 \geq 2(\kappa+1) + \kappa = 3\kappa + 2 \geq 2(g+1)$, which is a contradiction. Hence, $3m - 3 \notin G'$.
\end{itemize}

Hence, we can write $G' = G'_0 \cup G'_1 \cup G'_2$, with $G'_0 = [1,m-2]$, $G'_1 \subseteq [m, 2m-3]$ and $G'_2 \subseteq [2m-1, 3m-4]$. Now we prove that $G'$ is a gapset. Let $z \in G'$ and write $z = x + y $, with $x, y \in \mathbb{N}$ and $x \leq y$.

\begin{itemize}
  \item If $z \in G'_0$, then $x$ and $y \in G'_0$.
  \item If $z \in G'_1$, then $z \leq 2m - 3$ and $2x \leq 2m - 3$. Thus $x \in G'_0$.
  \item Let $z \in G'_2$. If $x \leq m - 2$, then $x \in G'_0$. If $y \geq 2m-1$, then $x \leq m - 3$ and $x \in G'_0$. Hence, we can consider $x, y \in [m, 2m - 3]$. Notice that $\ell_\alpha \leq 2m - 2$ (if $\ell_\alpha = 2m - 1$, then $2m - 2 \in G'$, which does not occur). In particular, $\ell_\alpha - 1 \leq 2m - 3$ and $z \geq \ell_{\alpha+1} - 2$. Thus $z = \ell_t - 2$, for some $t \in [\alpha+1, g+1]$. We claim that $y < \ell_\alpha$. Otherwise, $z = x + y \geq m + \ell_\alpha$ and it implies that $\ell_t > \ell_\alpha + m$, which is a contradiction according to Proposition \ref{abaixo}. Hence, $x \leq y < \ell_\alpha$. By writing $\ell_t - 2 = z = x + y$, i.e., $\ell_t = (x+1) + (y+1)$ and using that $G$ is a gapset, we conclude that $x+1 \in G$ or $y+1 \in G$. Hence, $x \in G'$ or $y \in G'$.
\end{itemize}

\end{proof}

\begin{corollary}
Let $g$ and $\kappa$ be non-negative integers such that $2g = 3\kappa$. Then $\#\mathcal{G}_{\kappa}(g) = \#\mathcal{G}_{k+n}(g+n)$, for all $n \in \mathbb{N}$.
\label{mainc}
\end{corollary}

\begin{proof}
We proceed by induction on $n$. Theorem \ref{main} ensures that $\#\mathcal{G}_{\kappa}(g) = \#\mathcal{G}_{k+1}(g+1)$, since $2g = 3\kappa$. Notice that $2(g+j) \leq 3(\kappa + j)$, for all $j \in \mathbb{N}_0$ and we can apply Theorem \ref{main} again: the map $\mathcal{G}_{\kappa + n - 1}(g + n - 1) \to \mathcal{G}_{\kappa+n}(g+n), G \mapsto \phi(G)$ is bijective, which completes the proof.
\end{proof}

Next, we present instances of pure $\kappa$-sparse gapsets, with large $\kappa$.

\begin{example}
Let $g \geq 0$. The unique pure $g$-sparse gapset with genus $g$ is $[1,g-1] \cup \{2g-1\}$.
\end{example}

\begin{example}
Let $g \geq 3$. The two pure $(g-1)$-sparse gapsets with genus $g$ are $[1,g-2] \cup \{g, 2g-1\}$ and $[1,g-1] \cup \{2g-2\}$.
\end{example}

\begin{example}
Let $g \geq 6$. The five pure $(g-2)$-sparse gapsets with genus $g$ are:
\begin{itemize}
  \item[$\bullet$] $[1,g-1] \cup \{2g-3\}$
  \item[$\bullet$] $[1,g-2] \cup \{2g-4,2g-3\}$
  \item[$\bullet$] $[1,g-3] \cup \{g-1,g, 2g-2\}$
  \item[$\bullet$] $[1,g-3]\cup \{g,g+1, 2g-1\}$
  \item[$\bullet$] $[1,g-3] \cup \{g - 1, g+1, 2g-1\}$
\end{itemize}
\end{example}

We end up this section presenting Table \ref{g,k}, which illustrates our results. The values ware obtained using the package \texttt{numericalsgps} \cite{GAP} in GAP.

\begin{landscape}
\begin{table}[h!]
\caption{A few values for $\#\mathcal{G}_{\kappa}(g)$. The entries in bold correspond to the case $2g = 3\kappa$.}
\label{g,k}
\begin{tabular}{|c|c c c c c c c c c c c c c c c c c c c c| c|}
 \hline
\diagbox[height=0.6cm]{$g$}{$\kappa$} & $0$ & $1$ & $2$ & $3$ &$4$ & $5$ & $6$ & $7$ & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & $n_g$ \\
\hline
0 & \textbf{1} &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 1 \\
\hline
1 &  & 1 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 1 \\
\hline
2 &  & 1 & 1 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 2 \\
\hline
3 &  & 1 & \textbf{2} & 1  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 4 \\
\hline
4 &  & 1 & 3 & 2 & 1 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 7 \\
\hline
5 &  & 1 & 5 & 3 & 2 & 1 &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 12 \\
\hline
6 &  & 1 & 7 & 7 & \textbf{5} & 2 & 1 &  &  &  &  &  &  &  &  &  &  &  &  &  & 23 \\
\hline
7 &  & 1 & 10 & 12 & 8 & 5 & 2 & 1 &  &  &  &  &  &  &  &  &  &  &  &  & 39 \\
\hline
8 &  & 1 & 15 & 18 & 17 & 8 & 5 & 2 & 1 &  &  &  &  &  &  &  &  &  &  &  & 67 \\
\hline
9 &  & 1 & 20 & 31 & 28 & 18 & \textbf{12} & 5 & 2 & 1 &  &  &  &  &  &  &  &  &  &  & 118 \\
\hline
10 &  & 1 & 27 & 51 & 49 & 34 & 22 & 12 & 5 & 2 & 1 &  &  &  &  &  &  &  &  &  & 204 \\
\hline
11 &  & 1 & 38 & 78 & 87 & 57 & 40 & 22 & 12 & 5 & 2 & 1 &  &  &  &  &  &  &  &  & 343 \\
\hline
12 &  & 1 & 51 & 125 & 147 & 100 & 76 & 42 & \textbf{30} & 12 & 5 &  2 & 1 &  &  &  &  &  &  &  & 592 \\
\hline
13 &  & 1 & 70 & 195 & 237 & 177 & 134 & 83 & 54 & 30 & 12 & 5 & 2 & 1 &  &  &  &   &  &   & 1001 \\
\hline
14 &  & 1 & 95 & 297 & 399 & 309 & 239 & 150 & 99 & 54 & 30 & 12 & 5 & 2 & 1 &  &  &  &  &  & 1693 \\
\hline
15 &  & 1 & 128 & 457 & 654 & 530 & 422 & 259 & 183 & 103 & \textbf{70} & 30 & 12 & 5 & 2 & 1 &  &  &  &  & 2857 \\
\hline
16 &  & 1 & 172 & 705 & 1061 & 902 & 723 & 452 & 336 & 199 & 135 & 70 & 30 & 12 & 5 & 2 & 1 &  &  &  & 4806 \\
\hline
17 &  & 1 & 230 & 1074 & 1717 & 1513 & 1248 & 811 & 590 & 363 & 243 & 135 & 70 & 30 & 12 & 5 & 2 & 1 &  &  & 8045 \\
\hline
18 &  & 1 & 309 & 1621 & 2777 & 2535 & 2148 & 1411 & 1037 & 646 & 444 & 251 & \textbf{167} & 70 & 30 & 12 & 5 & 2 & 1 &  & 13467 \\
\hline
19	  & & 1 & 413 & 2448 & 4464 & 4232 & 3636 & 2434 & 1810 & 1124 & 804 & 480 & 331 & 167 & 70 & 30 & 12 & 5 & 2 & 1 & 22464 \\
\hline
\end{tabular}
\end{table}
\end{landscape}

\section{Further questions}

A natural question that arises from Theorem \ref{main} is about the existence of an injective map from $\mathcal{G}_\kappa(g)$ to $\mathcal{G}_{\kappa+1}(g+1)$ where $\kappa$ and $g \in \mathbb{N}_0$ and with no other hypothesis. The values obtained in Table \ref{g,k} point out that it is possible that it occurs. Besides, those values indicate that it is possible to have an injective map from $\mathcal{G}_\kappa(g)$ to $\mathcal{G}_{\kappa}(g+1)$. We can write those questions as follows:

\begin{question}
Let $g$ and $\kappa$ be non-negative integers. Is there an injective map \linebreak $\mathcal{G}_\kappa(g) \to \mathcal{G}_{\kappa+1}(g+1)$?
\end{question}

\begin{question}
Let $g$ and $\kappa$ be non-negative integers. Is there an injective map \linebreak $\mathcal{G}_\kappa(g) \to \mathcal{G}_{\kappa}(g+1)$?
\end{question}

Notice that a positive answer to any of them implies that the sequence $(n_g)$ is increasing for $g \in \mathbb{N}$.

Another question that arises is about the behaviour of the sequence $(g_w)$, where $g_w := \#\mathcal{G}_{2w}(3w)$. In fact, Corollary \ref{mainc} ensures that $\#\mathcal{G}_{\kappa}(g) = \#\mathcal{G}_{\kappa+n}(g+n)$, for all $n \in \mathbb{N}$, as can be seen in Table \ref{g,k} where columns progressively stabilize below the diagonal. Here are the first few values of the sequence $(g_w)$:
$$(g_0, g_1, g_2, g_3, g_4, g_5, g_6) = (1, 2, 5, 12, 30, 70, 167).$$
Hence, it could be of interest to know information about the sequence, which was unknown at OEIS before this work, but which is now listed there as sequence A348619.

\begin{table}
\centering
\caption{Some sequences related to $(g_w)$}
\begin{tabular}{c c c c}
\hline
$w$ & $g_w$ & $\frac{g_{w}}{g_{w-1}}$ & $\frac{\sum_{i=0}^{w} g_{i}}{g_{w}}$
\\
\noalign{\smallskip}\hline\noalign{\smallskip}
0 & 1 & $-$ & 1  \\
\hline
1 & 2 & 2.000 & 1.5  \\
\hline
2 & 5 & 2.500 & 1.6 \\
\hline
3 & 12 & 2.400 & 1.66 \\
\hline
4 & 30 & 2.500 & 1.66 \\
\hline
5 & 70 & 2.333 & 1.714 \\
\hline
6 & 167 &  2.386 & 1.719 \\
\hline
7 & 395 & 2.365 & 1.727 \\
\hline
8 & 936 & 2.370 & 1.729 \\
\hline
9 & 2212 & 2.363 & 1.731 \\
\noalign{\smallskip}\hline
   \end{tabular}
\end{table}

\begin{question}
Are there remarkable properties about the sequence $(g_w)$?
\end{question}

\textbf{Acknowledgements.} The authors thank to the anonymous referee and to Nathan Kaplan for their careful corrections, suggestions and comments that allowed to improve the last version of the paper.

redpajama_set_name: RedPajamaArXiv


How do you create new products and services that deliver new value?
Saturn removed negotiations from the car buying experience.
Subway removed the traditional kitchen from the fast food restaurant.
Netflix removed the storefront from video rentals.
Little Caesar's removed the restaurant from the pizzeria.
Apple removed complexity from the user interface.
Yellow Tail removed the pretension from selecting wine.
Tagged Creativity, Innovation, new business concepts, products, services, subtractive thinking.
Where do new business concepts come from?
How did Apple come up with the idea of the iPod and later iPhone? How did Google come up with a way make money from their search engine?
While there are enough books that can tell you the story of how it happened, they won't tell you what happened inside the minds of the people or individual who came up with the idea.
New business concepts come from people. Those who have the ambition and where with all to discuss open topic issues with others, those who are willing to step outside the box of comfort and ask questions, research and try to think as if they are in the other persons shoes. Putting yourself "out there" and being focused at the same time while trying to figure out how your idea can benefit from your thoughts.
So where do new business concepts come from?
The profound insight doesn't come out of any strategic planning process nor does it come from some brainstorming session, it comes from a mix of individual desire, curiosity, ambition and need. But there also needs to be some foresight, a sense of where things might go.
New business concepts come out of a mix of unexpected problems, novel experiences, random conversations and newly discovered facts. The fact is there is no predictable and mechanical process for creating a bold new business concept.
Where do you think new business concepts come from?
Tagged Creativity, Innovation, new business concepts.


text:
Harron should face trial
By The West Virginia Record | Jan 23, 2009
West Virginia's most famous--critics say infamous-- radiologist is asking the court to let him off the hook.
Dr. Ray Harron, the man said to have made millions through his challenged asbestos illness diagnoses, wants a federal judge to dismiss a civil lawsuit against him brought by one of his alleged victims.
Rail giant CSX was typical of companies attacked by trial lawyers armed with Harron's wholesale diagnoses. But CSX turned the tables and sued the Bridgeport native in 2005, charging X-ray reports he provided to a Pittsburgh law firm -- later used to sue CSX -- were fraudulent.
Jerald Jones, Harron's lawyer, said the court should dismiss its case because CSX didn't ask for enough damages.
Federal court requires $75,000. But the x-ray in question, of alleged plaintiff Earl Baylor, couldn't have cost CSX that much money, Jones says.
Saying the client's behavior wasn't that damaging is an ironic argument for a guy the New York Times called a "vital cog" in America's multi-billion dollar absestos lawsuit machine. According to a Times report, the Bridgeport native has been paid for more than 75,000 lung injury diagnoses since the mid-1990s.
The depth of fraudulent activity may never be known. What's certain is the $70 billion -- that's $70 billion -- in litigation costs and $49 billion in damages shelled out by corporate targets of some 700,000 asbestos lawsuits over the past 30 years. And Dr. Harron apparently had his fingerprints on more than 10% of those suits.
That kind of damage to our economy, often involving the work of our West Virginia doctor, is difficult to fathom. Maybe that's why America has yet to come to grips with the enormous consequences of this asbestos lawsuit colossus; it's too much and too raw for an average person to get their arms around.
Here and elsewhere, the trial of Dr. Harron, the "vital cog," has a chance to change that. It's time for hard evidence to be aired and hard questions to be asked.
Such as who found these plaintiffs? How much was paid for the medical diagnoses? Which law firms did most of the paying? How much did those firms earn from it all?
America needs to know. And Ray Harron should help us find out.
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Kroger shopper alleges liquid on floor of self-checkout area caused fall
Former Department of Health and Human Resources employee alleges she was denied job because of harassment report
Attorney General announces prescription drug disposal site at capitol
Mechanic alleges Aero-Smith, Sylvanus Group failed to pay him
Man claims WVU Board of Governors used harassment, intimidation to prevent him from attending board meeting
Federal judge files attorney removal order for Timothy George Hayes
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redpajama_set_name: RedPajamaCommonCrawl


This is an overview of some of the important topics. You can view other topics grouped by, activity, hottest, newest, views, votes.
How Can I Propose New Features Or Sensors For PRTG?
How and where does PRTG store its data?
My SNMP sensors don't work. What can I do?
My Windows sensors do not work when using direct Performance Counter access. What can I do?


The results of the 2016 presidential election prompted many Americans to question the electoral college – a winner-takes-all system which empowers a group of 538 electors to name the next president of the United States.
Equal representation, citizen-funded elections and equal access to the ballot are the three actionable steps towards change, suggests Equal Citizens, a nonprofit founded by renowned law professor Lawrence Lessig.
Two lawsuits citing the 14th Amendment – one in a largely Republican state on behalf of Democrats and another in a majority Democratic state on behalf of Republicans – arguing that the current system disenfranchises voters, could be an effective legal strategy to spark reform.


Dubai, UAE – Electricity demand in the Middle East will triple by 2050, accelerating the growth of renewable energy and triggering the diversification of economies away from fossil fuels.These are some of the findings of DNV GL's Energy Transition Outlook 2018, which provides an outlook of the global energy landscape up to 2050.
By 2050, solar PV will be the main source of energy in the Middle East and North Africa, generating 39% of total supply. Onshore wind will be second, with a share of 28%. In the past few years the region has seen significant investments in onshore wind energy both in Oman and Jordan and according to the report, in 2030 onshore wind will start to grow rapidly, followed by solar PV.
Primary energy consumption will continue to grow moderately and increase 36% by 2050, this growth will be driven by manufacturing, buildings, and transport up to 2040. In buildings increased uptake of energy-hungry air conditioners will grow as the technologies become affordable for a larger share of the population and average temperatures in the region increase.For buildings, electrification will continue, and natural gas will be displaced by electricity in households, including for cooking so that by 2050, 70% of buildings energy use is covered by electricity.
Renewables uptake is starting to mature; Egypt aims to get 42% of its electricity from renewables by 2025 and Iran 5 GW by 2020. Saudi Arabia, with the largest petroleum reserves in the region, sees renewables as a strategic priority and is investing in 9.5 GW of solar and wind by 2023. Its goal is for 30% of electricity to come from renewables by 2030.
The rise of solar and wind will create a need for changes to the electricity market fundamentals in the region. This requires major regulatory intervention,including policy decisions that are favourable to the renewables market. Market mechanisms such as a reduction of fossil-fuel subsidies are the first step to accelerate the diversification of economies in the region.


text:
<dom-module id="paper-range-slider">
    <template>

            <paper-slider id='sliderMax' on-down="_sliderMaxDown" on-up="_sliderMaxUp" step="[[step]]" min="[[min]]" max="[[max]]" value="[[valueMax]]" secondary-progress='[[max]]' style="width:100%;"></paper-slider>

            <paper-slider id='sliderMin' on-down="_sliderMinDown" on-up="_sliderMinUp" noink step="[[step]]" min="[[min]]" max="[[max]]" value="[[valueMin]]" style="width:100%;"> </paper-slider>

    </template>

</dom-module>

redpajama_set_name: RedPajamaGithub


Pharmaceuticals play a major role in today's population health era – they can prevent and cure disease, improve or maintain wellness and slow progression of existing conditions. Yet, their promise can also be a curse if high prices limit patient access and bankrupt the healthcare providers and insurers bearing significant financial risk for patient care.
The proactive new leadership at the FDA is promoting competitive markets by combatting the abuse of well-intentioned programs and market share monopolies. Commissioner Scott Gottlieb has ramped up the FDA's efforts to prevent drug manufacturers from "gaming the system" in a number of ways.
For the first time, the FDA made publicly available a list of off-patent, off-exclusivity branded drugs without generic competition. Using the list, Premier immediately identified a number of critical drugs for patient care and has been working with manufacturers to participate in the FDA's new expedited review process. Additionally, Congress recently enacted legislation creating an expedited review process for generic drug applications when there are fewer than three manufacturers in the market for a given drug product. We strongly support and helped to champion these efforts, and are hopeful that the FDA will use this new authority to foster competition and curb price spikes and shortages in generic drugs where only a few players dominate.
We anticipate the FDA will soon release key documents to streamline the Abbreviated New Drug Application process, which will help eliminate unnecessary, duplicative procedures and greatly increase efficiency without compromising standards.
Commissioner Gottlieb has strong words for abusers of the REMs program, stating, "some manufacturers are… using the law as a way to delay the ability of generic firms to purchase the doses of a branded drug that they need to run their studies and get to market." The FDA has been taking important steps to help prevent exploitation of the loophole by releasing draft guidance to stem drug manufacturers' abuse of REMS.
We also call on Congress to level the playing field for new market entrants that foster competition in order to drive down inflated drug prices.
Policy change is urgently needed to close loopholes and get competitive entrants into the market faster. The Fair Access for Safe and Timely (FAST) Generics Act and the Creating, Restoring Equal Access to Equivalent Samples (CREATES) and Preserve Access to Affordable Generics Act are clear, bipartisan solutions to curb abusive, anticompetitive business practices that are driving up unsustainable spending within healthcare and impede patient access to generic and biosimilar medicines.
We know competition works. Our aggregated group purchasing power alone puts manufacturers through a head-to-head bidding process that brings results. Premier's drug portfolio prices have held constant at just 4.55 percent per year, well below the inflation rate of 10 percent over the same time period on all drugs with no contractual terms.
While the FDA has made significant progress toward adopting policies that push more competition, there still is a long way to go before these policies have a real, meaningful upstream effect. From reducing and prioritizing backlogged approvals, accelerating entrance of biosimilars into the market and facilitating increased competition around branded drugs, stakeholders from across healthcare, including Premier, are ready and willing to work with the FDA to bring in more manufacturers into the fray and foster healthier markets for pharmaceuticals.
Calling out the abuses is important, but we badly need action by both FDA and Congress.
Michael J. Alkire is chief operating officer of Premier Inc., a healthcare improvement company.
The shell game is profound.
Barry- Iam off this week and would have to pull out budgets at my office, but I generally think of pharmacy costs as running around 10% of total inpatient costs. That includes the cost of pharmacists and drug machines, so I would have look at our details.
Steve — Thanks as always.
Please tell me how this will be able to mitigate the traditional business model of Big Pharma and its effect on over-all healthcare that is to eventually become EQUITABLY available, ECOLOGICALLY accessible, JUSTLY efficient AND RELIABLY effective for each citizen, community by community. By traditional business model, I mean: 40% of annual cash income is equal to net income (aka profit) plus the expense of business promotion (aka marketing).


text:
Science Signaling
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Nancy R. Gough3,*
1Editor of Science's STKE and a Senior Editor of Science, American Association for the Advancement of Science, 1200 New York Avenue, N.W., Washington, DC 20005, USA.
2Associate Editor of Science's STKE and Associate Editor of Science, American Association for the Advancement of Science, 1200 New York Avenue, N.W., Washington, DC 20005, USA.
3Managing Editor of Science's STKE, American Association for the Advancement of Science, 1200 New York Avenue, N.W., Washington, DC 20005, USA.
↵*Contact information. E-mail: bray{at}aaas.org (L.B.R.), lchong{at}aaas.org (L.D.C.), ngough{at}aaas.org (N.R.G.)
Science's STKE 03 Sep 2002:
Vol. 2002, Issue 148, pp. eg10
DOI: 10.1126/stke.2002.148.eg10
L. Bryan Ray
1Editor of and a Senior Editor of , American Association for the Advancement of Science, 1200 New York Avenue, N.W., Washington, DC 20005, USA.
Lisa D. Chong
2Associate Editor of and Associate Editor of , American Association for the Advancement of Science, 1200 New York Avenue, N.W., Washington, DC 20005, USA.
Nancy R. Gough
3Managing Editor of , American Association for the Advancement of Science, 1200 New York Avenue, N.W., Washington, DC 20005, USA.
vol. 2002 no. 148 eg10
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text:
require_relative 'dataset_loader'

module CitySDK
  class StreamDatasetLoader < DatasetLoader
    def initialize(stream)
      super()
      @stream = stream
    end # def

    private

    attr_reader :stream
  end # class
end # module


Want to register a LLC. Have a few questions.
1: Can I register my house (own) as the Primary Place of Business?
2: Assuming answer to question 1 is yes, if tomorrow someone sues me and I have to pay the damages by my assets, will my house be in danger. Please not, I am trying to open an LLC.
Bottomline: I want to start a business from my house but not put the ownership of my house in danger.
Re: Want to register a LLC. Have a few questions.
The first thing to do would be to form your LLC. LLC has certain protections and tax advantages. In most cases, your personal assest's are protected. It would be a good idea to ask your lawyer and/or accountant for specifics.
Bottom line: Form your LLC first and then open/operate the business under the LLC. And, seek a professional -- they can also advice you of reporting requirements, licenses and permits required, record-keeping practices, etc.
Thanks so much for the information. Really appreciate it.
A question follow-up, so It sounds like it will be ok to call my house as a Primary place of business when I register the LLC? This is not way will be in danger if there are any legal damages against me. Thanks again.
Yes. Your place of business is where your business operates. If your residence is not owned by the LLC, it should, in most cases, be protected. Once again, seek a professional. Also, you may want to protect your privacy by setting up a P.O. Box under your LLC. Once again seek legal counsel. If you have a Registered Agent you may not need one.
A good place to go for other legal advice would be to check out SCORE. You may be able to find a Mentor in your area.


POURQUOI MON CODE RIO N'EST-IL PAS RECONNU ?
AU SECOURS : MON CODE RIO EST INVALIDE !
SOSH, LA MARQUE QUI PROPOSE DES PROMOS À GOGO !
TÉLÉPHONE PORTABLE SANS ABONNEMENT, QUEL OPÉRATEUR CHOISIR ?
LE FORFAIT SFR RED ILLIMITÉ À VIE, RÊVE OU RÉALITÉ ?


text:
The cell viability of both the PANC-1 and HepG2 cells was measured at 24 h after each treatment
The cell viability of both the PANC-1 and HepG2 cells was measured at 24 h after each treatment. of HepG2 cells was measured using ATP content-based method. (*** is used for P < 0.001).(TIF) pone.0201920.s004.tif (793K) GUID:?420C1B01-E07D-4A2F-BDEA-A23674C708F0 S5 Fig: Effects of the sonication dispersion of EGCG ABT-751 (E-7010) solution on the triple treatment-induced anticancer effects. EGCG stock solution was treated with or without the sonication dispersion, and then the solutions were used for the combined triple treatment. After treatment for 24 h or 72 h, the viability of PANC-1 cells was measured using MTT assay.(TIF) pone.0201920.s005.tif (856K) GUID:?D411AB2A-B778-405E-BC77-1F08B89BACAA S1 File: Raw data of MTT assay. (RAR) pone.0201920.s006.rar (40K) GUID:?BCC49FC0-2117-43FE-8039-A86173F4359D S2 File: Raw data of ATP-based viability assay. (RAR) pone.0201920.s007.rar (37K) GUID:?261EAF21-093C-4DFC-837A-5C3053061852 S3 File: Raw data ABT-751 (E-7010) of DHE flow cytometry. (RAR) pone.0201920.s008.rar (8.9K) GUID:?F0DA772A-7B11-4D89-A4B7-BA9D0454FD2A S4 File: Raw image of MDC staining. (RAR) pone.0201920.s009.rar (3.1M) GUID:?AE73E16A-0C7B-430A-8992-9B342F7072C0 S5 File: Raw images of PANC-1 proteins. (RAR) pone.0201920.s010.rar (1.1M) GUID:?47F8518F-A466-4368-8B9E-CEE6DE85C613 S6 File: Raw data of inhibitors. (RAR) pone.0201920.s011.rar (218K) GUID:?D6670012-E838-48D9-80EE-07D4CD4455F6 S7 File: Raw data of HepG2 proteins. (RAR) pone.0201920.s012.rar (1.0M) GUID:?75536836-ACA9-4081-B923-D59E56D3C31E S8 File: Raw data of EGCG sonication. (RAR) pone.0201920.s013.rar (7.5K) GUID:?F211D89B-0BBB-4046-B713-B9956861925A Data Availability StatementAll relevant data are within the paper and its Supporting Information files. Abstract Cancer is one of the most troublesome diseases and a leading cause of death worldwide. Recently, novel treatments have been continuously developed to improve the disadvantages of conventional therapies, such as prodigious expenses, unwanted Rabbit Polyclonal to CCRL1 side effects, and tumor recurrence. Here, we ABT-751 (E-7010) provide the first non-invasive treatment that has combined epigallocatechin gallate (EGCG), the most abundant catechin in green tea, with a low strength pulsed electric field (PEF) and a low energy ultrasound (US). It has been observed that the cell viability of human pancreatic cancer PANC-1 was decreased approximately to 20% of the control after this combination treatment for 72 h. Besides, the combined triple treatment significantly reduced the high tolerance of HepG2 cells to the EGCG-induced cytotoxicity and similarly exhibited persuasive proliferation-inhibitory effects. We also found the combined triple treatment improved the intracellular reactive oxygen varieties (ROS) and acidic vesicles, and the EGCG-induced inhibition of Akt phosphorylation was dramatically intensified. In this study, the apoptosis inhibitor Z-VAD-FMK and the autophagy inhibitor 3-MA were, respectively, shown to attenuate the anticancer effects of the triple treatment. This indicates the triple treatment-induced autophagy was switched from cytoprotective to cytotoxic, and hence, cooperatively caused cell death with the apoptosis. Since the EGCG is definitely easily accessible from the green tea and mild for any long-term treatment, and the non-invasive physical stimulations could be modified to focus on a specific location, this combined triple treatment may serve as a encouraging strategy for anticancer therapy. Introduction Cancer is the second leading cause of death worldwide and remains a major challenge for general public health study [1]. Traditional therapies such as surgery, radiation, and chemotherapy are commonly used to treat individuals diagnosed with this disease. However, individuals treated with ABT-751 (E-7010) conventional treatments still have a high risk of tumor recurrence, and many of them are refractory to treatment. Therefore, newer approaches to improve the effectiveness of a tumor therapy at an affordable cost are urgently needed. The most common methods are combination therapies that use two or more anticancer medicines, and these strategies are considered to target different pathways and to enhance their restorative effectiveness inside a synergistic or additive manner [2]. Nevertheless, combination therapies could also reduce effectiveness due to the drug competition [3]. Besides, unwanted side effects and dangerous drug relationships still exist as the potentially harmful effects. Recently, we have reported that a noninvasive low strength pulsed electric field (PEF) can enhance the epigallocatechin gallate (EGCG) to combat against the pancreatic malignancy cells [4]. It was found that the synergistic reactions in the double treatment of the EGCG and PEF disturbed the mitochondria, enhanced the intrinsic pathway transduction, and effectively induced apoptosis. On the other hand, it has been reported that EGCG is not stable and simultaneously transforms into many EGCG auto-oxidation products (EAOPs) in the cell tradition system [5, 6]. Even so, one of the EAOPs, theasinensin A, has also been shown to cause apoptotic cell death in malignancy cells [7]. Recently, certain EAOPs have been demonstrated to possess equal cytotoxic activities as EGCG and to exhibit an enhanced ability to deplete sulfhydryl group of cysteine, which is a major resource for sustaining malignancy cell malignancy [6]. Consequently, we suggested the natural products of EGCG combined with the non-invasive and moderate physical stimulations would be a encouraging strategy in the anticancer treatment. In addition to the noninvasive PEF, an alternative software of physical stimulations is the exposure of malignancy cells to the ultrasound (US), and this technique.
Posted in PKD


text:
William Lloyd Prosser ( à New Albany dans l'Indiana et décédé le ) fut le doyen de Boalt Hall de 1948 à 1961. Il est l'auteur de plusieurs éditions du Prosser on Torts (traité de droit de la responsabilité), universellement reconnu comme l'ouvrage de référence en la matière (11 éditions ont été publiées à ce jour).

Notes et références

Liens externes

Juriste américain
Étudiant de l'université du Minnesota
Naissance en mars 1898
Naissance à New Albany (Indiana)
Lieu de décès inconnu
Décès en mai 1972
Décès à 74 ans

redpajama_set_name: RedPajamaWikipedia


For today's "Where are they now?" I'd like to talk a bit about a fellow who worked with Damon Slye on games back in the 1980s and continued to develop games to this very day. From the many "Where are they now?" posts I've done so far, I've started to appreciate how rare it is for someone to be so passionate about games to continue developing them for over four decades - many would've simply retired by now. Jeff Tunnell is one of those rare individuals that has stayed in the business to this very day, but where did it all start?
Well, back in 1981, Tunnell managed to get his first Apple II (it seemed to be the game development machine of choice during the early 80s) and he started his first business selling software from his house while at the same time attempting to program games. Tunnell eventually met Damon Slye who was programming games in assembler which made the man "a God" in his eyes. Tunnell hired him to work in his store and when he realised how good he was at programming, Tunnell focused on other aspects of game development, namely "design, production and learning to run the business".
Eventually in 1984, Tunnell and Slye formed the game development company "Dynamix". Dynamix would go on to develop many simulation games in the 1980s - most I've never heard of (but that's probably because I was too young at that stage) with the exception of 1989's A-10 Tank Killer which I fondly remember. Tunnell served as an Executive Producer on that project.
Tunnell left Dynamix in 1990 to form his own company called Jeff Tunnell Productions but around that time a few noteworthy Dynamix games were released, namely the WWI flight sim Red Baron in 1990 (where Tunnell was executive producer), the 1990 cyberpunk adventure Rise of the Dragon (which Tunnell designed), the 1991 adventure game Heart of China (which Tunnell wrote and directed) and the 1991 adventure game The Adventures of Willy Beamish (which Tunnell directed).
After Tunnell left Dynamix he developed the game that he's arguably most famous for: 1992's The Incredible Machine, a game where you solve puzzles by developing Rube Goldberg-like devices. While Tunnell didn't program the game, he was the designer and producer.
Tunnell returned to Dynamix in 1995 in a leadership role and before the company disbanded in 2001, he was executive producer on the classic 1999 mecha simulation game Starsiege. When Dynamix disbanded, Tunnell formed a new company called GarageGames, an independent video game publisher which also developed the Torque Game Engine, originally developed for the 2001 game Tribes 2. Tunnell eventually left GarageGames in 2008.
In 2009, Tunnell founded the company PushButton Labs which focused on the development of social games but they eventually wound down operations in 2011 after being acquired by Disney.
In 2012, Tunnell founded Spotkin along with several partners from his previous companies and he still works there today as Creative Director and Managing Partner. In 2014, Spotkin released a spiritual successor to The Incredible Machine called Contraption Maker which has rated very well with Steam users (Very Positive rating of 94%) which just goes to show that when it comes to gameplay, it doesn't matter how old the concepts are - they can still be relevant today.
Well, it's good to see Tunnell is still involved in making games and I hope he continues to channel his enthusiasm and creativity into the field for many years to come! Thank you for all the games you made at Dynamix and The Incredible Machine series!
Dynamix Jeff Tunnell Sierra Where are they now?


I've been in a baking mood for the last week and thought I'd share another one of my favourites from the 80/20 eBook. These high protein blueberry muffins are practically a meal in itself and are lovely for breakfast on the run served with your favourite cup of tea or long black. Each muffin has just over 7 grams of protein which will help fill you up, keep blood sugars stable and support a healthy metabolism. I've made them with almonds which are kind to the digestive system and won't bloat you out.
Preheat the oven to 170°C / 300°F.
Combine the almond meal and baking powder in a large mixing bowl.
Add the eggs, oil, cinnamon and honey to form a batter.
Fold in the apple and blueberries.
Divide into paper-lined muffin tins.
Bake for 35–40 minutes until cooked through and golden.
For nut-free muffins use ¼ cup coconut flour (40 g) + 2 tablespoons of Healthy Chef Organic Pea Protein in place of the 3 cups of almond meal - reduce the baking powder to 1 good teaspoon - reduce the blueberries to 100 g. You can get coconut flour in any good health food store. If the mix looks a little wet - add a touch more coconut flour or Pea Protein. This recipe makes 6 large muffins or 12 cupcake sized muffins.
Hazelnut meal also works really well with this recipe and marries perfectly with the apple and blueberries.


text:
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text:
James Patrick Green (* 30. Mai 1950 in Philadelphia, USA) ist ein US-amerikanischer Geistlicher, römisch-katholischer Erzbischof und Diplomat des Heiligen Stuhls.

Leben
James Patrick Green empfing am 14. Juni 1969 durch Erzbischof John Joseph Kardinal Krol das Sakrament der Priesterweihe für das Erzbistum Philadelphia. Papst Johannes Paul II. verlieh ihm am 4. Mai 1988 den Ehrentitel Kaplan Seiner Heiligkeit (Monsignore).

Am 17. August 2006 ernannte ihn Papst Benedikt XVI. zum Titularerzbischof von Altinum und bestellte ihn zum Apostolischen Nuntius in Südafrika und Namibia sowie zum Apostolischen Delegaten in Botswana. Die Bischofsweihe spendete ihm Kardinalstaatssekretär Angelo Sodano am 6. September desselben Jahres; Mitkonsekratoren waren der Präfekt der Kongregation für die Evangelisierung der Völker, Ivan Kardinal Dias, und der Erzbischof von Philadelphia, Justin Francis Kardinal Rigali. James Patrick Green wurde am 6. September 2006 zudem zum Apostolischen Nuntius in Lesotho und am 23. September 2006 zum Apostolischen Nuntius in Swasiland bestellt. Am 15. Oktober 2011 wurde James Patrick Green Apostolischer Nuntius in Peru.

Papst Franziskus ernannte ihn am 6. April 2017 zum Apostolischen Nuntius in Schweden und Island. Am 13. Juni 2017 wurde er zusätzlich zum Apostolischen Nuntius in Dänemark, am 12. Oktober desselben Jahres zum Apostolischen Nuntius in Finnland und am 18. Oktober zum Apostolischen Nuntius in Norwegen ernannt. Am 30. April 2022 nahm Papst Franziskus seinen Rücktritt von diesen Ämtern an.

Weblinks

Einzelnachweise

Römisch-katholischer Bischof (21. Jahrhundert)
Römisch-katholischer Geistlicher (20. Jahrhundert)
Titularerzbischof
Apostolischer Nuntius in Botswana
Apostolischer Nuntius in Eswatini
Apostolischer Nuntius in Lesotho
Apostolischer Nuntius in Namibia
Apostolischer Nuntius in Peru
Apostolischer Nuntius in Südafrika
Botschafter in Eswatini
Botschafter in Schweden
Botschafter in Island
Botschafter in Dänemark
Person (Philadelphia)
US-Amerikaner
Geboren 1950
Mann


text:
The Netherlands' partial ban on the burqa/niqab – the first step to an Islamophobic government?
1st August 2019 saw the first partial ban on the burqa and niqab come into effect in the Netherlands. This was placed upon all women who wear the niqab or burqa, which includes prohibiting women from wearing clothing that covers their entire face. This law is applied within public places such as on public transport, within health and educational institutions and inside government buildings. Although the women are allowed to wear the burqa and niqab on public streets, if stopped by police and demanded to remove then they will be required to take it off. Refusal to do so may result in a fine up to 410 Euros.
The government describes its Dutch-law as "religion-neutral," but this ban is specifically targeted at Muslim women. An estimate of 100 – 400 women are said to wear the niqab/burqa in the European country of 17 million people. So what is the necessity of this law? Targeted at such a small minority, it brings into question exactly what problem are the Dutch government attempting to solve? With the weak argument from Cabinet ministers of the burqa not fitting in with their "open society," the Dutch parliament was actually the first European country to put forward the ban for the burqa. This idea was first proposed in 2005 by the Freedom Party far-right politician Geert Wilders who claimed this development as a "major victory." It took almost 15 years for the law to pass, resulting in the Netherlands now being another European country to have placed the ban on religious clothing, the first two being France and Belgium. France was the first country to ban the full-face Islamic veil on 11th April 2011 where, under the ban, no women, French or foreign, is able to leave their own home with their face covered behind a veil without the risk of facing a fine. The controversial ban of "burkinis," was then later imposed by French Riviera in 2016, where Prime Minister Manuel Valls called these swimsuits "the affirmation of political Islam in the public space."
To a Muslim woman, the religious garment she wears is not simply just an item of clothing to cover her face, but a symbol of empowerment. Women wear this to hide and preserve their beauty, where some see it as a part of their own identity. By enforcing this ban, this not only strips a woman of her religious freedom, but also from freedom of movement. This affects both elderly women in their everyday lives, as well as the young in their education and participating in school sports and activities.
This partial ban is, in actuality, a complete ban. There are only a limited amount of public spaces still available to women to wear face-covering clothing, which is the streets and the private sector. But these, too, are limited. As mentioned above, if a female covering her face is requested to remove her niqab, she must oblige or fear receiving an extensive fine. And, naturally, within the private sector they can have their own procedures which could possibly (and most likely) legislate against women dressed specifically in a burqa/niqab. Bans like these will also affect women as they heighten the risk of Islamophobic hate crimes. This ban compromises the safety of these women as it opens the door for people to take the law into their own hands. With this law in place, this is seen as an opportunity for the public to act as "vigilantes." This form of terror preinstalls a fear into these women and prohibits them from stepping out of their own homes. Activist Flavia Dzodan, has commented on the Dutch newspaper ADnl on its horrific suggestions on what to if they spot a face-covering women in the streets: "They suggest asking the person to leave, call the police or, alternatively, exercise the right to a citizen's arrest." Geert Wilders, who first proposed this ban, was of course in favour of this, with senator Marjolein Faber-Van de Klashorst referring to this as a "historical day as this is the first step to de-Islamise the Netherlands. This is the first and next step to close all the mosques."
JAN Trust supports the rights of Muslim women and their own decision to wear the niqab and burqa. If they have the free will to do so, they should have the right to exercise it and express their religion in the way they choose. Read more of our blogs where we address similar issues in women having to fight for their rights.
#JT30BelgiumBurqaburqa banEmpowermentfrancefreedomgeert wildersheadscarfidentityIslamIslamophobiaJAN TrustnetherlandsniqabreligionVeilWomenWomen's rights


Another update week! macOS 10.13.3 and iOS 11.2.5 dropped along with the usual other updates. Security Updates for Sierra and El Capitan bring Meltdown fixes and (maybe) APFS support for Sierra.
The end of one beta cycle is the begin of the next. macOS 10.13.4 and iOS 11.3 started in beta. Apple seems so proud of this update that they put up an iOS 11.3 preview. Also an announcement that the macOS Server application will be severely reduced in functionality in a future release.
Next Next post: And now Server.app, too!