Second-Level Numerical Semigroups

Abstract

Let S be a numerical semigroup with multiplicity $m\left(S\right)$. Then, S is called a second-level numerical semigroup if $x+y+z-m\left(S\right)\in S$ for every $\{x,y,z\}\subseteq S\setminus \left\{0\right\}$. In this paper, we present some algorithms to compute all the second-level numerical semigroups with multiplicity, genus, and a Frobenius fixed number. For m and r, which are positive integers, such that $m<r$ and $gcd(m,r)=1$, we show that there exists the minimal second-level numerical semigroup with multiplicity m containing r. We solve the Frobenius problem for these semigroups and show that they satisfy Wilf’s conjecture.

1. Introduction

Let $\mathbb{Z}$ be the set of integer numbers, and $\mathbb{N}=\{x\in \mathbb{Z}\mid x\ge 0\}$. A submonoid of $(\mathbb{N},+)$ is a subset of a closed $\mathbb{N}$ under addition and contains 0. A numerical semigroup S is a submonoid of $(\mathbb{N},+)$, such that $\mathbb{N}\setminus S=\{n\in \mathbb{N}\mid n\notin S\}$ is finite.

If S is a numerical semigroup, then $m\left(S\right)=min(S\setminus \{0\left\}\right)$, $F\left(S\right)=max(\mathbb{Z}\setminus S)$, and $g\left(S\right)$, the cardinal of the set $\mathbb{N}\setminus S$, are three important invariants of S called the multiplicity, the Frobenius number, and the genus of S, respectively.

If A is a non-empty subset of $\mathbb{N}$, then we denote by $\langle A\rangle$ the submonoid of $(\mathbb{N},+)$ generated by A, i.e., $\langle A\rangle =\{{\lambda }_1{a}_1+\cdots +{\lambda }_n{a}_n\mid n\in \mathbb{N}\setminus \left\{0\right\},\{a_1,\dots ,a_n\}\subseteq A,and\{{\lambda }_1,\dots ,{\lambda }_n\}\subseteq \mathbb{N}\}$. In ([1], Lemma 2.1), it is shown that $\langle A\rangle$ is a numerical semigroup if and only if $gcd\left(A\right)=1$.

If M is a submonoid of $(\mathbb{N},+)$ and $M=\langle A\rangle$, then we say that A is a system of generators of M. Furthermore, if $M\ne \langle B\rangle$ for all $B⊊A$, then we say that A is a minimal system of generators of M. In ([1], Corollary 2.8), it is shown that for any M submonoid of $(\mathbb{N},+)$, there exists a unique minimal system of generators, and it is finite. We denote this minimal system of generators as $msg\left(M\right)$. Its cardinality is called the embedding dimension ofM; we will denote it as $e\left(M\right)$.

The Frobenius problem for numerical semigroups (see [2]) is to find formulas to compute the Frobenius number and the genus of a numerical semigroup using its minimal system of generators. This problem was solved in [3] for numerical semigroups with an embedding dimension equal to 2. Today, the problem is still open for numerical semigroups with a higher embedding dimension.

An unfamiliar reader with the numerical semigroup can find these names a bit strange; however, there exists a large number of publications devoted to the study of analytically irreducible unidimensional local domains using the semigroups of values, which is a numerical semigroup and where the reader can find the connection and the interpretation of these invariants (see, for example, [4,5,6,7,8]). Following this question, there also appear interesting classes of numerical semigroups as numerical semigroups having a maximal embedding dimension, called MED-semigroups (see [4,9,10]).

In ([1], Proposition 2.10), it is shown that for any numerical semigroup S, we have $e\left(S\right)\le m\left(S\right)$. We say that S is a maximal embedding dimension semigroup if $e\left(S\right)=m\left(S\right)$. In ([1], Proposition 3.12), it is shown that a numerical semigroup has a maximal embedding dimension if and only if $x+y-m\left(S\right)\in S$ for every $\{x,y\}\subseteq S\setminus \left\{0\right\}$.

Following the terminology used in [11], a numerical semigroup is second-level (or a second-level semigroup) if $x+y+z-m\left(S\right)\in S$, for every $\{x,y,z\}\subseteq S\setminus \left\{0\right\}$. Note that every numerical semigroup with a maximal embedding dimension is a second-level numerical semigroup.

Let m be an integer greater than or equal to 2. We denote

$${\mathcal{L}}_2\left(m\right)=\{S\mid S\mathrm{is}\mathrm{a} \mathrm{second-level}\mathrm{semigroup}\mathrm{with}m\left(S\right)=m\}.$$

In Section 2, we show that ${\mathcal{L}}_2\left(m\right)$ is a Frobenius pseudo-variety. Using several results from [12], we deduce an algorithm to obtain all the elements in ${\mathcal{L}}_2\left(m\right)$ for a fixed genus. Let f be a positive integer. We denote ${\mathcal{L}}_2(f,m)=\{S\in {\mathcal{L}}_2\left(m\right)\mid F\left(S\right)=f\}$. In Section 3, we show that ${\mathcal{L}}_2(f,m)$ is a ratio-covariety. The results in [13] allow us to present an algorithm to compute all the elements in ${\mathcal{L}}_2(f,m)$.

A ${\mathcal{L}}_2\left(m\right)$-monoid is a submonoid of $(\mathbb{N},+)$, which can be written as the intersection of elements in ${\mathcal{L}}_2\left(m\right)$. For $A\subseteq \{m,\to \}$, we denote by ${\mathcal{L}}_2\left(m\right)\left[A\right]$ the intersection of all elements in ${\mathcal{L}}_2\left(m\right)$ containing A. It is clear that ${\mathcal{L}}_2\left(m\right)\left[A\right]$ is the smallest ${\mathcal{L}}_2\left(m\right)$-monoid containing A. In Section 4, we give an algorithm to compute ${\mathcal{L}}_2\left(m\right)\left[A\right]$. We also show that ${\mathcal{L}}_2\left(m\right)\left[A\right]$ belongs to ${\mathcal{L}}_2\left(m\right)$ if and only if $gcd(A\cup \{m\left\}\right)=1$.

If $M={\mathcal{L}}_2\left(m\right)\left[A\right]$, then we say that A is a ${\mathcal{L}}_2\left(m\right)$-system of generators for M. Moreover, if $M\ne {\mathcal{L}}_2\left(m\right)\left[B\right]$ for every $B⊊A$, then we say that A is a minimal${\mathcal{L}}_2\left(m\right)$-system of generators for M. In Section 4, we show that every ${\mathcal{L}}_2\left(m\right)$-monoid M has a unique minimal ${\mathcal{L}}_2\left(m\right)$-system of generators, which will be denoted as ${\mathcal{L}}_2\left(m\right)$-$msg\left(M\right)$. We define the ${\mathcal{L}}_2\left(m\right)$-rank of M as the cardinality of ${\mathcal{L}}_2\left(m\right)$-$msg\left(M\right)$.

In Section 5, we focus on the elements of ${\mathcal{L}}_2\left(m\right)$ with the ${\mathcal{L}}_2\left(m\right)$-rank being equal to 1. These semigroups will be ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ for some $r\in \mathbb{N}$, such that $m<r$ and $gcd(m,r)=1$. We provide the minimal system of generators for ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ and solve the Frobenius problem for these semigroups.

Given S, a numerical semigroup, and x, an integer number, we say that x is a pseudo-Frobenius number of S if $x\notin S$ and $x+s\in S$ for every $s\in S\setminus \left\{0\right\}$. The set of all pseudo-Frobenius numbers will be denoted as $PF\left(S\right)$. We define the type of S as the cardinality of $PF\left(S\right)$, and we will denote it as $t\left(S\right)$.

In Section 5, we show that $t\left(S\right)+1=e\left(S\right)$ for any $S\in {\mathcal{L}}_2\left(m\right)$ with the ${\mathcal{L}}_2\left(m\right)$-rank being equal to 1.

For a numerical semigroup S, we denote the set of small numbers by $N\left(S\right)=\{s\in S\mid s<F(S\left)\right\}$, and the cardinality of N(S) is denoted as $n\left(S\right)$. H.S. Wilf conjectured in [14] that $F\left(S\right)+1\le e\left(S\right)n\left(S\right)$. Today, this question is still open, and it has become one of the most important challenges in the theory of numerical semigroups. This conjecture has been proved for a large number of families of numerical semigroups. Readers interested in this problem may refer to [15,16,17,18,19,20].

In Section 5, we will show that the elements in ${\mathcal{L}}_2\left(m\right)$ with the ${\mathcal{L}}_2\left(m\right)$-rank being equal to 1 verify Wilf’s conjecture.

2. Elements in ${\mathcal{L}}_{\mathbf{2}}\left(\mathbf{m}\right)$ for a Fixed Genus

In this section, m will be an integer greater than or equal to 2.

Let g be a positive integer. We denote ${\mathcal{L}}_2(m,g)=\{S\in {\mathcal{L}}_2\left(m\right)\mid g\left(S\right)=g\}$. The aim of this section is to obtain an algorithm to compute ${\mathcal{L}}_2(m,g)$.

The following result is easy to check:

Lemma 1. 

• $\Delta \left(m\right)=\{0,m,\to \}$ is the maximum of ${\mathcal{L}}_2\left(m\right)$ with respect to the inclusion order, where $\{a_1,\dots ,a_n,\to \}$ means that every integer bigger than $a_n$ belongs to the set.
• For $k\in \mathbb{N}$ such that $k\ge m$, we have $\langle m\rangle \cup \{k,\to \}\in {\mathcal{L}}_2\left(m\right)$.

If $q\in \mathbb{Q}$, then $⌊q⌋=max\{z\in \mathbb{Z}\mid z\le q\}$. For $\{a,b\}\subseteq \mathbb{Z}$ with $b\ne 0$, we denote $amodb=a-⌊{\displaystyle \frac{a}{b}}⌋b$.

Proposition 1. 

Take $g\in \mathbb{N}$. Then, ${\mathcal{L}}_2(m,g)\ne \varnothing$ if and only if $g\ge m-1$.

Proof. 

Necessity. If $S\in {\mathcal{L}}_2(m,g)$, then $S\in {\mathcal{L}}_2\left(m\right)$ and $g\left(S\right)=g$. Applying Lemma 1, we have $S\subseteq \Delta \left(m\right)$ so $g=g\left(S\right)\ge g(\Delta (m\left)\right)=m-1$.

Sufficiency. If $g\ge m-1$, it is clear that $⌊{\displaystyle \frac{g}{m-1}}⌋+g+1\ge m$. Applying Lemma 1, we have $S=\langle m\rangle \cup \{⌊{\displaystyle \frac{g}{m-1}}⌋+g+1,\to \}\in {\mathcal{L}}_2\left(m\right)$. On the other hand, the number of positive elements in $\langle m\rangle$ smaller than $⌊{\displaystyle \frac{g}{m-1}}⌋+g$ is $\lfloor {\displaystyle \frac{⌊\frac{g}{m-1}⌋+g}{m}}\rfloor =\lfloor {\displaystyle \frac{\frac{g}{m-1}+g}{m}}\rfloor =\lfloor {\displaystyle \frac{g+(m-1)g}{(m-1)m}}\rfloor =\lfloor {\displaystyle \frac{mg}{m(m-1)}}\rfloor =⌊{\displaystyle \frac{g}{m-1}}⌋$, so $g\left(S\right)=g$, and $S\in {\mathcal{L}}_2(m,g)$.    □

A Frobenius pseudo-variety $\mathcal{P}$ is a non-empty family of numerical semigroups satisfying the following properties:

• The family $\mathcal{P}$ has a maximum element (with respect to the inclusion order);
• If $\{S,T\}\subseteq \mathcal{P}$, then $S\cap T\in \mathcal{P}$;
• If $S\in \mathcal{P}$ and $S\ne max\left(\mathcal{P}\right)$, then $S\cup \{F(S\left)\right\}\in \mathcal{P}$.

The following result can be deduced from [11], Proposition 2.6 and Corollary 3.4.

Proposition 2. 

${\mathcal{L}}_2\left(m\right)$ is a Frobenius pseudo-variety and $\Delta \left(m\right)$ is its maximum.

A graph G is a pair $(V,E)$, where V is a non-empty set and $E\subseteq \left\{\right(u,v)\in V\times V\mid u\ne v\}$. The elements in V are called the vertices of G, and the elements in E are called the edges of G. A path of length n connecting the vertices x and y of G is a sequence of different elements in E, $(v_0,v_1),(v_1,v_2),\dots ,(v_{n-1},v_n)$, where $v_0=x$ and $v_n=y$.

A graph G is called a tree if there exists a vertex r named the root of G, such that for every x of G there exists a unique path joining x and r. Finally, if $(x,y)$ is an edge of a tree, we say that x is a child of y.

We define the graph $G\left({\mathcal{L}}_2\left(m\right)\right)$ as follows: The vertices of this graph are the elements of ${\mathcal{L}}_2\left(m\right)$, and $(S,T)\in {\mathcal{L}}_2\left(m\right)\times {\mathcal{L}}_2\left(m\right)$ is an edge if and only if $T=S\cup \{F(S\left)\right\}$.

As a consequence of Proposition 2, ([12], Lemmas 11 and 12, and Theorem 3), we can obtain the following result:

Theorem 1. 

$G\left({\mathcal{L}}_2\left(m\right)\right)$ is a tree whose root is $\Delta \left(m\right)$. Moreover, the set of children of a vertex S of $G\left({\mathcal{L}}_2\left(m\right)\right)$ is $\{S\setminus \left\{x\right\}\mid x\in msg\left(S\right)withx>F\left(S\right)andS\setminus \left\{x\right\}\in {\mathcal{L}}_2\left(m\right)\}$.

The following result is easy to check:

Lemma 2. 

Let S be a numerical semigroup, and let $x\in S$. Then, $S\setminus \left\{x\right\}$ is a numerical semigroup if and only if $x\in msg\left(S\right)$. Moreover, $g(S\setminus \{x\left\}\right)=g\left(S\right)+1$.

Let S be a numerical semigroup. We consider $msg\left(S\right)=\{n_1,\dots ,n_e\}$. If $s\in S$, then we denote $L_S\left(s\right)=max\{a_1+\cdots +a_e\mid (a_1,\dots ,a_e)\in {\mathbb{N}}^e \mathrm{and} a_1{n}_1+\cdots +a_e{n}_e=s\}$.

Lemma 3 

([11], Proposition 3.5). Let $S\in {\mathcal{L}}_2\left(m\right)$, and let $x\in msg\left(S\right)\setminus \left\{m\right\}$. Then, we have $S\setminus \left\{x\right\}\in {\mathcal{L}}_2\left(m\right)$ if and only if $L_{S\setminus \left\{x\right\}}(x+m)\le 2$.

A tree can be recurrently built starting from the root and by connecting each vertex with its children using an edge. Also, note that elements in ${\mathcal{L}}_2\left(m\right)$ with genus g are children of elements in ${\mathcal{L}}_2\left(m\right)$ with genus $g-1$. Following this idea and the above results, we present Algorithm 1 to compute ${\mathcal{L}}_2(m,g)$.

Algorithm 1

Input: Integers m and g, such that $1\le m-1\le g$. Output: ${\mathcal{L}}_2(m,g)$. $\hspace{1em}A=\{\Delta (m\left)\right\}$, $i=m-1$;  If $i=g$, then return A, and the algorithm is finished;  For every $S\in A$ compute $B_S=\{x\in msg\left(S\right)\setminus \left\{m\right\}\mid x>F\left(S\right)and{L}_{S\setminus \left\{x\right\}}(x+m)\le 2\}$; $\hspace{1em}A:={\cup }_{S\in A}\{S\setminus \left\{x\right\}\mid x\in B_S\}$; $i:=i+1$ and go back to (2).

Example 1. 

Using Algorithm 1, we are going to compute ${\mathcal{L}}_2(4,5)$.

• $A=\{\langle 4,5,6,7\rangle \}$, $i=3$;
• $B_{\langle 4,5,6,7\rangle }=\{5,6,7\}$;
• $A=\{\langle 4,6,7,9\rangle ,\langle 4,5,7\rangle ,\langle 4,5,6\rangle \}$, $i=4$;
• $B_{\langle 4,6,7,9\rangle }=\{6,7,9\}$; $B_{\langle 4,5,7\rangle }=\left\{7\right\}$; $B_{\langle 4,5,6\rangle }=\varnothing$;
• $A=\{\langle 4,7,9,10\rangle ,\langle 4,6,9,11\rangle ,\langle 4,6,7\rangle ,\langle 4,5,11\rangle \}$, $i=5=g$ and the algorithm is finished.

Algorithm 1 returns ${\mathcal{L}}_2(4,5)=\{\langle 4,7,9,10\rangle ,\langle 4,6,9,11\rangle ,\langle 4,6,7\rangle ,\langle 4,5,11\rangle \}$.

3. Elements in ${\mathcal{L}}_{\mathbf{2}}\left(\mathbf{m}\right)$ for a Fixed Frobenius Number

In this section, m will be an integer greater than or equal to 2.

Let f be a positive integer. We consider ${\mathcal{L}}_2(f,m)=\{S\in {\mathcal{L}}_2\left(m\right)\mid F\left(S\right)=f\}$. The objective of this section is to obtain an algorithm to compute ${\mathcal{L}}_2(f,m)$.

Proposition 3. 

Take $f\in \mathbb{N}\setminus \left\{0\right\}$. Then, ${\mathcal{L}}_2(f,m)\ne \varnothing$ if and only if $m-1\le f$ and $m\nmid f$.

Proof. 

Necessity. If $S\in {\mathcal{L}}_2(f,m)$, then $m-1\notin S$, so $m-1\le f$. Moreover, since $f\notin S$, then $f\notin \langle m\rangle$. Hence, $m\nmid f$.

Sufficiency. Observe that $\langle m\rangle \cup \{f+1,\to \}\in {\mathcal{L}}_2(f,m)$.    □

If $f=m-1$ and $S\in {\mathcal{L}}_2(f,m)$, then $S=\{0,f+1,\to \}$. In the rest of this section, we assume $\{m,f\}\subset \mathbb{N}$, such that $2\le m<f$ and $m\nmid f$.

Let S be a numerical semigroup, such that $S\ne \mathbb{N}$. The ratio of S is $r\left(S\right)=min\{s\in S\mid m(S)\nmid s\}$. Note that $r\left(S\right)=min(msg(S)\setminus \{m\left(S\right)\left\}\right)$.

A ratio-covariety is a family $\mathcal{R}$ of numerical semigroups satisfying the following properties:

• The family $\mathcal{R}$ has a minimum element (with respect to the inclusion order);
• If $\{S,T\}\subseteq \mathcal{R}$, then $S\cap T\in \mathcal{R}$;
• If $S\in \mathcal{R}$ and $S\ne min\left(\mathcal{R}\right)$, then $S\setminus \{r(S\left)\right\}\in \mathcal{R}$.

The following result is easy to prove:

Lemma 4. 

Let $S,T$ be two numerical semigroups. Then, $S\cap T$ is also a numerical semigroup, and $F(S\cap T)=max\{F(S),F(T\left)\right\}$.

We denote $\Delta (f,m)=\langle m\rangle \cup \{f+1,\to \}$.

Proposition 4. 

${\mathcal{L}}_2(f,m)$ is a ratio-covariety and $\Delta (f,m)$ is its minimum.

Proof. 

It is clear that $\Delta (f,m)$ is the minimum of ${\mathcal{L}}_2(f,m)$. If $\{S,T\}\subseteq {\mathcal{L}}_2(f,m)$, then using Lemma 4, we deduce $S\cap T\in {\mathcal{L}}_2(f,m)$. To finish the proof, we see that if $S\in {\mathcal{L}}_2(f,m)$ such that $S\ne \Delta (f,m)$, then $S\setminus \{r\left(S\right)\}\in {\mathcal{L}}_2(f,m)$. As $S\ne \Delta (f,m)$, then $r\left(S\right)<f$, and we obtain that $S\setminus \{r(S\left)\right\}$ is a numerical semigroup with multiplicity m and a Frobenius number f using Lemma 2. On the other hand, if $\{x,y,z\}\subseteq S\setminus \{r(S),0\}$, then $x+y+z-m\in S$. Furthermore, $x+y+z-m\ne r\left(S\right)$, so $x+y+z-m\in S\setminus \{r(S\left)\right\}$. From this, we have $S\setminus \{r\left(S\right)\}\in {\mathcal{L}}_2(f,m)$.    □

Let S be a numerical semigroup, and $x\in \mathbb{Z}$. We say that x is a special gap of S if $x\notin S$ and $S\cup \left\{x\right\}$ is a numerical semigroup. We denote by $SG\left(S\right)$ the set of special gaps of S.

We define the graph $G\left({\mathcal{L}}_2(f,m)\right)$ as follows: The set of vertices is ${\mathcal{L}}_2(f,m)$, and $(S,T)\in {\mathcal{L}}_2(f,m)\times {\mathcal{L}}_2(f,m)$ is an edge if and only if $T=S\setminus \{r(S\left)\right\}$.

Using Proposition 4, [13], and Propositions 3 and 4, we obtain the following result:

Theorem 2. 

$G\left({\mathcal{L}}_2(f,m)\right)$ is a tree whose root is $\Delta (f,m)$. Moreover, the set of children of a vertex S is $\{S\cup \left\{x\right\}\mid x\in SG\left(S\right),m<x<r\left(S\right),andS\cup \left\{x\right\}\in {\mathcal{L}}_2(f,m)\}$.

The following result is Proposition 3.1 of [11].

Proposition 5. 

Let S be a numerical semigroup with multiplicity m. Then, $S\in {\mathcal{L}}_2\left(m\right)$ if and only if $a+b+c-m\in S$ for every $\{a,b,c\}\subseteq msg\left(S\right)\setminus \left\{m\right\}$.

Corollary 1. 

Consider $S\in {\mathcal{L}}_2(f,m)$, and $x\in SG\left(S\right)$, such that $m<x<f$. Then, $S\cup \left\{x\right\}\in {\mathcal{L}}_2(f,m)$ if and only if $\{3x-m,2x+a-m,x+a+b-m\}\subseteq S$ for every $\{a,b\}\subseteq msg\left(S\right)\setminus \left\{m\right\}$.

Proof. 

To apply Proposition 5, we need to observe that $msg(S\cup \{x\left\}\right)\subseteq msg\left(S\right)\cup \left\{x\right\}$.    □

Let S be a numerical semigroup, and $n\in S\setminus \left\{0\right\}$, we define (in honor of [21]) the Apéry set of n in S as $Ap(S,n)=\{s\in S\mid s-n\notin S\}$. The following result appears in [1], Lemma 2.4:

Lemma 5. 

Let S be a numerical semigroup and $n\in S\setminus \left\{0\right\}$. Then, $Ap(S,n)=\left\{w\right(0)=0,w(1),\dots ,w(n-1\left)\right\}$, where $w\left(i\right)$ is the smallest element in S congruent with i modulo n.

Remark 1. 

Let S be a numerical semigroup and assume that $Ap(S,m(S\left)\right)=\left\{w\right(0),w(1),\dots$, $w(m(S)-1)\}$ is known; then, we have the following:

1. 

We can solve the problem of belonging to S as follows: Any integer x belongs to S if and only if $x\ge w(xmodm(S\left)\right)$.

2. 

We can obtain the special gaps of S. From ([22], Remark 1), we have an easy way to compute them as $SG\left(S\right)=\{x\in PF(S)\mid 2x\notin PF(S\left)\right\}$ [22], Lemma 3.4, where $PF\left(S\right)=\{w-m\left(S\right)\mid w\in {Maximals}_{\le S}Ap(S,m\left(S\right))\}$ [23], Lemma 10. Here, ${\le }_S$ is the partial order defined on $\mathbb{Z}$ as $a{\le }_{S}b$, if and only if $b-a\in S$.

3. 

From ([22], Lemma 3.5), we have $Ap(S\cup \{x\},m(S\left)\right)=(Ap(S,m\left(S\right))\setminus \{x+m\left(S\right)\left\}\right)\cup \left\{x\right\}$, for every $x\in SG\left(S\right)$.

4. 

Finally, as $msg\left(S\right)\setminus \{m\left(S\right)\}=\{w\in Ap(S,m\left(S\right))\setminus \left\{0\right\}\mid w-w^{\prime }\notin Ap(S,m\left(S\right))\forall w^{\prime }\in Ap(S,m\left(S\right))\setminus \{0,w\}\}$, we can obtain $msg(S\cup \{x\left\}\right)$ from (3).

Using the above ideas, we present Algorithm 2. It allow to compute ${\mathcal{L}}_2(f,m)$ being f and m integers under certain conditions.

Algorithm 2

Input: Integers f and m such that $2\le m<f$ and $m\nmid f$. Output: ${\mathcal{L}}_2(f,m)$. $\hspace{1em}{\mathcal{L}}_2(f,m)=\{\Delta (f,m)\}$ and $B=\{\Delta (f,m\left)\right\}$.  Compute $Ap(\Delta (f,m),m)$.  For every $S\in B$, compute $\theta \left(S\right)=\{x\in SG(S)\mid m<x<r(S),x\ne f,\{3x-m,2x+a-m,x+a+b-m\}\subseteq Sforevery\{a,b\}\subseteq msg(S)\setminus \{m\left\}\right\}$.  If ${\cup }_{S\in B}\theta \left(S\right)=\varnothing$, then return ${\mathcal{L}}_2(f,m)$, and the algorithm is finished. $\hspace{1em}C={\cup }_{S\in B}\{S\cup \left\{x\right\}\mid x\in \theta \left(S\right)\}$. $\hspace{1em}{\mathcal{L}}_2(f,m):={\mathcal{L}}_2(f,m)\cup C$ and $B:=C$.  For every $S\in B$, compute $Ap(S,m)$ and go back to (3).

Note that the computation of $Ap(S,m)$ in line 7 is used to compute the set $SG\left(S\right)$ in a new loop, as we point out in Remark 1.

Example 2. 

Using Algorithm 2, we compute ${\mathcal{L}}_2(7,4)$.

First loop of the algorithm is:

(1) 

${\mathcal{L}}_2(7,4)=\{\Delta (7,4)\}$ and $B=\{\Delta (7,4\left)\right\}$.

(2) 

$Ap(\Delta (7,4),4)=\{0,9,10,11\}$.

Therefore, $Maximals(Ap(\Delta (7,4),4\left)\right)=\{9,10,11\}$. $PF(\Delta (7,4\left)\right)=\{5,6,7\}$.

Moreover, $SG(\Delta (\mathbf{7},\mathbf{4}\left)\right)=\{\mathbf{5},\mathbf{6},\mathbf{7}\}$.

Meanwhile, $msg(\Delta (7,4\left)\right)=\{4,9,10,11\}$ and $r(\Delta (7,4\left)\right)=9$.

Then, we have

(3) 

$\theta (\Delta (7,4\left)\right)=\{5,6\}$, as $7=f$.

(5) 

$C=\{\Delta (7,4)\cup \{5\},\Delta (7,4)\cup \{6\left\}\right\}$.

(6) 

${\mathcal{L}}_2(7,4)=\{\Delta (7,4),\Delta (7,4)\cup \left\{5\right\},\Delta (7,4)\cup \left\{6\right\}\}$ and $B=\{\Delta (7,4)\cup \{5\},\Delta (7,4)\cup \{6\left\}\right\}$.

(7) 

$Ap(\Delta (7,4)\cup \{5\},4)=\{0,5,10,11\}$ and $Ap(\Delta (7,4)\cup \{6\},4)=\{0,6,9,11\}$.

Therefore,

(a) 

$Maximals(Ap(\Delta (7,4)\cup \left\{5\right\},4\left)\right)=\{10,11\}$. $PF(\Delta (7,4)\cup \{5\left\}\right)=\{6,7\}$.

Moreover, $SG(\Delta (\mathbf{7},\mathbf{4})\cup \{\mathbf{5}\left\}\right)=\{\mathbf{6},\mathbf{7}\}$.

Meanwhile, $msg(\Delta (7,4)\cup \{5\left\}\right)=\{4,5,11\}$ and $r(\Delta (7,4)\cup \{5\left\}\right)=5$.

(b) 

$Maximals(Ap(\Delta (7,4)\cup \left\{6\right\},4\left)\right)=\{6,9,11\}$. $PF(\Delta (7,4)\cup \{6\left\}\right)=\{2,5,7\}$.

Moreover, $SG(\Delta (\mathbf{7},\mathbf{4})\cup \{\mathbf{6}\left\}\right)=\{\mathbf{2},\mathbf{5},\mathbf{7}\}$.

Meanwhile, $msg(\Delta (7,4)\cup \{6\left\}\right)=\{4,6,9,11\}$ and $r(\Delta (7,4)\cup \{6\left\}\right)=6$.

Then, second loop is

(3) 

$\theta (\Delta (7,4)\cup \{5\left\}\right)=\varnothing$, as $r(\Delta (7,4)\cup \{5\left\}\right)<6<7$

and $\theta (\Delta (7,4)\cup \{6\left\}\right)=\left\{5\right\}$, as $2<m<5<r(\Delta (7,4)\cup \{6\left\}\right)<7$.

(5) 

$C=\{\Delta (7,4)\cup \{5,6\left\}\right\}$.

(6) 

${\mathcal{L}}_2(7,4)=\{\Delta (7,4),\Delta (7,4)\cup \left\{5\right\},\Delta (7,4)\cup \left\{6\right\},\Delta (7,4)\cup \{5,6\}\}$ and $B=\{\Delta (7,4)\cup \{5,6\left\}\right\}$.

(7) 

$Ap(\Delta (7,4)\cup \{5,6\left\}\right),4)=\{0,5,6,11\}$.

Therefore, $Maximals(Ap(\Delta (7,4)\cup \{5,6\},4\left)\right)=\left\{11\right\}$. $PF(\Delta (7,4)\cup \{5,6\left\}\right)=\left\{7\right\}$.

Moreover, $SG(\Delta (\mathbf{7},\mathbf{4})\cup \{\mathbf{5},\mathbf{6}\left\}\right)=\left\{\mathbf{7}\right\}$.

Meanwhile, $msg(\Delta (7,4)\cup \{5,6\left\}\right)=\{4,5,6\}$ and $r(\Delta (7,4)\cup \{5,6\left\}\right)=5$.

Then, the third loop is

(3) 

$\theta (\Delta (7,4)\cup \{5,6\left\}\right)=\varnothing$, as $7=f$.

(4) 

${\mathcal{L}}_2(7,4)=\{\Delta (7,4),\Delta (7,4)\cup \left\{5\right\},\Delta (7,4)\cup \left\{6\right\},\Delta (7,4)\cup \{5,6\}\}$ and the algorithm is finished.

Algorithm 2 returns ${\mathcal{L}}_2(7,4)=\{\Delta (7,4),\Delta (7,4)\cup \left\{5\right\},\Delta (7,4)\cup \left\{6\right\},\Delta (7,4)\cup \{5,6\}\}$.

4. ${\mathcal{L}}_{\mathbf{2}}\left(\mathbf{m}\right)$-Systems of Generators

It is clear that ${\mathcal{L}}_2\left(1\right)=\left\{\mathbb{N}\right\}$ and ${\mathcal{L}}_2\left(2\right)=\{\langle 2,2k+1\rangle \mid k\in \mathbb{N}\setminus \left\{0\right\}\}$. In this section, m will be an integer greater than or equal to 3.

From Proposition 2, we know that the intersection of a finite number of elements in ${\mathcal{L}}_2\left(m\right)$ is again an element in ${\mathcal{L}}_2\left(m\right)$. However, when we consider an infinite number of them, the intersection is generally not an element in ${\mathcal{L}}_2\left(m\right)$, as shown in the following example:

Example 3. 

For any $k\in \{m,\to \}$, we consider $S\left(k\right)=\langle m\rangle \cup \{k,\to \}$. Clearly, $S\left(k\right)\in {\mathcal{L}}_2\left(m\right)$. However, ${\cap }_{k\in \{m,\to \}}S\left(k\right)=\langle m\rangle \notin {\mathcal{L}}_2\left(m\right)$.

The finite or infinite intersection of elements in ${\mathcal{L}}_2\left(m\right)$ is always a submonoid of $(\mathbb{N},+)$. This fact allows us to give the following definition. A ${\mathcal{L}}_2\left(m\right)$-monoid is a submonoid of $(\mathbb{N},+)$ that can be written as an intersection of elements in ${\mathcal{L}}_2\left(m\right)$.

The following result is easy to check:

Lemma 6. 

The intersection of ${\mathcal{L}}_2\left(m\right)$-monoids is again a ${\mathcal{L}}_2\left(m\right)$-monoid.

Consider $A\subseteq \Delta \left(m\right)$. We denote by ${\mathcal{L}}_2\left(m\right)\left[A\right]$ the intersection of all the ${\mathcal{L}}_2\left(m\right)$-monoids containing A. As a consequence of Lemma 6, we find that ${\mathcal{L}}_2\left(m\right)\left[A\right]$ is the smallest (with respect to the inclusion) ${\mathcal{L}}_2\left(m\right)$-monoid containing A.

We can easily reach the following result:

Lemma 7. 

Let A be a subset of $\Delta \left(m\right)$. Then, ${\mathcal{L}}_2\left(m\right)\left[A\right]$ is equal to the intersection of all the elements in ${\mathcal{L}}_2\left(m\right)$ containing A.

If $M={\mathcal{L}}_2\left(m\right)\left[A\right]$, then we will say that A is a ${\mathcal{L}}_2\left(m\right)$-system of generators of M. Furthermore, if $M\ne {\mathcal{L}}_2\left(m\right)\left[B\right]$ for any $B⊊A$, then A is called a minimal${\mathcal{L}}_2\left(m\right)$-system of generators of M.

The following result is easy to deduce from Proposition 2 and [12], Corollary 1.

Proposition 6. 

Every ${\mathcal{L}}_2\left(m\right)$-monoid has a unique minimal ${\mathcal{L}}_2\left(m\right)$-system of generators. Furthermore, such a minimal ${\mathcal{L}}_2\left(m\right)$-system of generators is finite.

Let M be a ${\mathcal{L}}_2\left(m\right)$-monoid. We denote ${\mathcal{L}}_2\left(m\right)$-$msg\left(M\right)$ as the minimal ${\mathcal{L}}_2\left(m\right)$-system of generators of M. We define the ${\mathcal{L}}_2\left(m\right)$-rank of M as the cardinality of this minimal ${\mathcal{L}}_2\left(m\right)$-system of generators.

Proposition 7. 

Let M be a submonoid of $(\mathbb{N},+)$. Then, the following statements are equivalent.

• M is a ${\mathcal{L}}_2\left(m\right)$-monoid.
• $m=min(M\setminus \{0\left\}\right)$ and $x+y+z-m\in M$ for every $\{x,y,z\}\subseteq M\setminus \left\{0\right\}$.
• $m=min(M\setminus \{0\left\}\right)$ and $x+y+z-m\in M$ for every $\{x,y,z\}\subseteq msg\left(M\right)\setminus \left\{m\right\}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$. If M is a ${\mathcal{L}}_2\left(M\right)$-monoid, then there exists a family ${\left\{S_i\right\}}_{i\in I}$ of elements in ${\mathcal{L}}_2\left(m\right)$, such that $M={\cap }_{i\in I}{S}_i$. Hence, $min(M\setminus \{0\left\}\right)=m$. On the other hand, if $\{x,y,z\}\subseteq M\setminus \left\{0\right\}$, then $\{x,y,z\}\subseteq S_i\setminus \left\{0\right\}$ for every $i\in I$, so $x+y+z-m\in S_i$ for every $i\in I$, and then $x+y+z-m\in M$.

$\left(2\right)\Rightarrow \left(3\right)$. Trivial.

$\left(3\right)\Rightarrow \left(1\right)$. For every $k\in \{m,\to \}$, we consider $S\left(k\right)=M\cup \{k,\to \}$. From Proposition 5, we easily deduce that $S\left(k\right)\in {\mathcal{L}}_2\left(m\right)$ for every $k\in \{m,\to \}$. It is clear that $M={\cap }_{k\in \{m,\to \}}S\left(k\right)$, so M is a ${\mathcal{L}}_2\left(m\right)$-monoid.    □

Proposition 7 allows us to present Algorithm 3 which compute ${\mathcal{L}}_2\left(m\right)\left[A\right]$ for any $A\subseteq \Delta \left(m\right)$.

Algorithm 3

Input: A finite subset A of $\Delta \left(m\right)$. Output: ${\mathcal{L}}_2\left(m\right)\left[A\right]$ $\hspace{1em}M=\langle A\cup \left\{m\right\}\rangle$. $\hspace{1em}B=\{x+y+z-m\mid \{x,y,z\}\subseteq msg(M)\setminus \{m\}andx+y+z-m\notin M\}$.  If $B=\varnothing$, then return M, and the algorithm is finished. $\hspace{1em}M:=\langle msg\left(M\right)\cup B\rangle$ and go back to (2).

Now, we present two different examples to show how Algorithm 3 works. In the first example, we calculate ${\mathcal{L}}_2\left(5\right)\left[\left\{7\right\}\right]$, and we will see that ${\mathcal{L}}_2\left(5\right)\left[\left\{7\right\}\right]$ belongs to ${\mathcal{L}}_2\left(5\right)$. In the second example, we calculate ${\mathcal{L}}_2\left(8\right)\left[\left\{10\right\}\right]$, and we will show that ${\mathcal{L}}_2\left(8\right)\left[\left\{10\right\}\right]$ does not belong to ${\mathcal{L}}_2\left(8\right)$. This raises the following question: What is the condition for ${\mathcal{L}}_2\left(m\right)\left[A\right]$ to belong to ${\mathcal{L}}_2\left(m\right)$?

Example 4. 

Let us calculate ${\mathcal{L}}_2\left(5\right)\left[\left\{7\right\}\right]$ using Algorithm 3.

• $M=\langle 5,7\rangle$.
• $B=\left\{16\right\}$.
• $M=\langle 5,7,16\rangle$.
• $B=\varnothing$, and the algorithm is finished.

Algorithm 3 returns ${\mathcal{L}}_2\left(5\right)\left[\left\{7\right\}\right]=\langle 5,7,16\rangle$.

Example 5. 

Let us calculate ${\mathcal{L}}_2\left(8\right)\left[\left\{10\right\}\right]$ using Algorithm 3.

• $M=\langle 8,10\rangle$.
• $B=\left\{22\right\}$.
• $M=\langle 8,10,22\rangle$.
• $B=\varnothing$, and the algorithm is finished.

Algorithm 3 returns ${\mathcal{L}}_2\left(8\right)\left[\left\{10\right\}\right]=\langle 8,10,22\rangle$. Since $gcd\{8,10,22\}\ne 1$, then $\langle 8,10,12\rangle$ is not a numerical semigroup; therefore, $\langle 8,10,22\rangle \notin {\mathcal{L}}_2\left(8\right)$.

Proposition 8. 

Consider $A\subseteq \Delta \left(m\right)$. Then, ${\mathcal{L}}_2\left(m\right)\left[A\right]\in {\mathcal{L}}_2\left(m\right)$ if and only if $gcd(A\cup \{m\left\}\right)=1$.

Proof. 

Necessity. If $gcd(A\cup \{m\left\}\right)=d\ne 1$, then using Proposition 7, we deduce $M=\left\{0\right\}\cup \{m+kd\mid k\in \mathbb{N}\}$ is a ${\mathcal{L}}_2\left(m\right)$-monoid. It is clear that $A\subseteq M$ and ${\mathcal{L}}_2\left(m\right)\left[A\right]\subseteq M$. From this, we have that ${\mathcal{L}}_2\left(m\right)\left[A\right]$ is not a numerical semigroup. Consequently, ${\mathcal{L}}_2\left(m\right)\left[A\right]\notin {\mathcal{L}}_2\left(m\right)$.

Sufficiency. We have that ${\mathcal{L}}_2\left(m\right)\left[A\right]$ is a submonoid of $(\mathbb{N},+)$ and $A\cup \left\{m\right\}\subseteq {\mathcal{L}}_2\left(m\right)\left[A\right]$. As $gcd(A\cup \{m\left\}\right)=1$, we deduce $gcd\left({\mathcal{L}}_2\left(m\right)\left[A\right]\right)=1$, and then ${\mathcal{L}}_2\left(m\right)\left[A\right]$ is a numerical semigroup. Consequently, ${\mathcal{L}}_2\left(m\right)\left[A\right]$ belongs to ${\mathcal{L}}_2\left(m\right)$. □

5. Elements in ${\mathcal{L}}_{\mathbf{2}}\left(\mathbf{m}\right)$ with ${\mathcal{L}}_{\mathbf{2}}\left(\mathbf{m}\right)$-Rank Equal to 1

In this section, m denotes an integer greater than or equal to 3.

Proposition 9. 

Consider $S\in {\mathcal{L}}_2\left(m\right)$. Then, the following statements are equivalent:

1. 

The ${\mathcal{L}}_2\left(m\right)$-rank of S is equal to 1.

2. 

$S={\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ for some integer r, such that $m<r$ and $gcd(m,r)=1$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$. If ${\mathcal{L}}_2\left(m\right)$-rank $\left(S\right)=1$, then there exists $r\in \Delta \left(m\right)$ such that ${\mathcal{L}}_2\left(m\right)$-msg $\left(S\right)=\left\{r\right\}$. Hence, $r\ge m$ and $S={\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$. Using Proposition 8, we have $gcd(m,r)=1$, which implies $m<r$.

$\left(2\right)\Rightarrow \left(1\right)$. It is clear that $\left\{r\right\}={\mathcal{L}}_2\left(m\right)$-msg $\left(S\right)$. Then, ${\mathcal{L}}_2\left(m\right)$-rank $\left(S\right)=1$. □

In the rest of this section, r will be an integer greater than m such that $gcd(m,r)=1$. We also denote $A(m,r)=\left\{\right(2k+1)r-km\mid k\in \mathbb{N}\}$.

Proposition 10. 

${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]=\langle A(m,r)\cup \left\{m\right\}\rangle$.

Proof. 

First of all, we will show, by induction on k, that $(2k+1)r-km\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$. For $k=0$, the result is clear because $(2·0+1)r-0·m=r\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$. We assume the statement to be true for k; then, we prove it for $k+1$. As $\{(2k+1)r-km,r\}\subseteq {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\setminus \left\{0\right\}$ and ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\in {\mathcal{L}}_2\left(m\right)$, then $(2k+1)r-km+r+r-m\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ and so $(2(k+1)+1)r-(k+1)m\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$.

Therefore, we have that $A(m,r)\cup \left\{m\right\}\subseteq {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ and consequently $\langle A(m,r)\cup \left\{m\right\}\rangle \subseteq {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$.

Now, we will see that ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\subseteq \langle A(m,r)\cup \left\{m\right\}\rangle$. For this, we only need to prove $\langle A(m,r)\cup \left\{m\right\}\rangle \in {\mathcal{L}}_2\left(m\right)$.

Consider $\{i,j,k\}\subseteq \mathbb{N}$. Then, $(2i+1)r-im+(2j+1)r-jm+(2k+1)r-km-m\in \langle A(m,r)\cup \left\{m\right\}\rangle$ as $(2i+1)r-im+(2j+1)r-jm+(2k+1)r-km-m=\left(2\right(i+j+k+1)+1)r-(i+j+k+1)m\in A(m,r)$. Then, Proposition 5 ensures that $\langle A(m,r)\cup \left\{m\right\}\rangle \in {\mathcal{L}}_2\left(m\right)$.

Consequently, we have ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]=\langle A(m,r)\cup \left\{m\right\}\rangle$. □

As a consequence of Proposition 10, we have $Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)\subseteq \langle A(m,r)\rangle$. We also have ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\in {\mathcal{L}}_2\left(m\right)$. Then, $x+y+z-m\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ for every $\{x,y,z\}\subseteq A(m,r)$. Then, we have the following result.

Lemma 8. 

If $w\in Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$, then we have one of the following asserts:

1. 

$w=(2i+1)r-im$ for some $i\in \mathbb{N}$.

2. 

$w=(2i+1)r-im+(2j+1)r-jm$ for some $\{i,j\}\subseteq \mathbb{N}$.

As $gcd(m,r)=1$, we have

$$\left\{\right(1·r)modm,(2·r)modm,\dots ,((m-1)·r)modm\}=\{1,2,\dots ,m-1\}.$$

Theorem 3. 

If $Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)=\{w\left(0\right),w\left(1\right),\dots ,w(m-1)\}$, then

$$w(krmodm)=\left\{\begin{array}{cc}kr-{\displaystyle \frac{k-1}{2}}m& ifkisodd\\ kr-{\displaystyle \frac{k-2}{2}}m& ifkiseven\end{array}\right.$$

for every $k\in \{1,\dots ,m-1\}$.

Proof. 

If k is odd, then there exists $s\in \mathbb{N}$ such that $k=2s+1$, so $kr-{\displaystyle \frac{k-1}{2}}m=(2s+1)r-sm\in A(m,r)\subseteq {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$. However, if k is even, then $k=2s+2$ for some $s\in \mathbb{N}$, and we have $kr-{\displaystyle \frac{k-2}{2}}m=(2s+2)r-sm=(2s+1)r-sm+r\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ because $\left\{\right(sk+1)r-sm,r\}\subseteq A(m,r)$.

Now, we will first see that in the case in which k is odd, we have $kr-{\displaystyle \frac{k-1}{2}}m-m\notin {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$. Assume the contrary. By Lemma 8, we consider two cases:

• There exist $t\in \mathbb{N}\setminus \left\{0\right\}$ and $i\in \mathbb{N}$ such that $kr-{\displaystyle \frac{k-1}{2}}m-tm=(2i+1)r-im$. Then, $kr\equiv (2i+1)r(modm)$. However, since $gcd(m,r)=1$, we have $k\equiv 2i+1(modm)$. Hence, there exists $l\in \mathbb{N}$ such that $2i+1=k+lm$. Consequently, $kr-{\displaystyle \frac{k-1}{2}}m-tm=(k+lm)r-{\displaystyle \frac{k+lm-1}{2}}m$ and then $\left({\displaystyle \frac{k+lm-1}{2}}-{\displaystyle \frac{k-1}{2}}-t\right)m=lmr$. Therefore, ${\displaystyle \frac{lm}{2}}-t=lr$. Hence, ${\displaystyle \frac{lm}{2}}>lr$, and ${\displaystyle \frac{m}{2}}>r$, which is a contradiction, since $r>m$. Observe that $l\ne 0$. Otherwise, we have $t=0$, which is impossible.
• There exist $t\in \mathbb{N}\setminus \left\{0\right\}$ and $\{i,j\}\subseteq \mathbb{N}$, such that $kr-{\displaystyle \frac{k-1}{2}}m-tm=(2i+1)r-im+(2j+1)r-jm$. Then, $k\equiv (2i+2j+2)(modm)$. Hence, $2i+2j+2=k+lm$ for some $l\in \mathbb{N}$. Therefore, $kr-{\displaystyle \frac{k-1}{2}}m-tm=(2i+2j+2)r-(i+j)m=(k+lm)r-{\displaystyle \frac{k+lm-2}{2}}m$. Then, following the same reasoning as in (1), we have ${\displaystyle \frac{lm-1}{2}}-t=lr$. Hence, ${\displaystyle \frac{lm}{2}}>lr$. Consequently, ${\displaystyle \frac{m}{2}}>r$, which is again a contradiction. Observe that $l\ne 0$. Otherwise, we have $t={\displaystyle \frac{1}{2}}$, which is impossible.

We then focus on the case where k is even and try to prove that $kr-{\displaystyle \frac{k-2}{2}}m-m\notin {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$. As above, we assume the contrary, and using Lemma 8 we consider two cases:

• There exist $t\in \mathbb{N}\setminus \left\{0\right\}$ and $i\in \mathbb{N}$ such that $kr-{\displaystyle \frac{k-2}{2}}m-tm=(2i+1)r-im$. Then, $kr\equiv (2i+1)r(modm)$, since $gcd(m,r)=1$ we have $k\equiv 2i+1(modm)$, and so there exists $l\in \mathbb{N}$ such that $2i+1=k+lm$. Then, $kr-{\displaystyle \frac{k-2}{2}}m-tm=(k+lm)r-{\displaystyle \frac{k+lm-1}{2}}m$ and then $\left({\displaystyle \frac{k+lm-1}{2}}-{\displaystyle \frac{k-2}{2}}-t\right)m=lmr$. Therefore, ${\displaystyle \frac{lm+1}{2}}-t=lr$, which implies that ${\displaystyle \frac{lm}{2}}>lr$ and ${\displaystyle \frac{m}{2}}>r$, and we obtain the same contradiction. As we have commented in the above paragraph, l must be different from 0. Otherwise, $t={\displaystyle \frac{1}{2}}$.
• There exist $t\in \mathbb{N}\setminus \left\{0\right\}$ and $\{i,j\}\subseteq \mathbb{N}$ such that $kr-{\displaystyle \frac{k-2}{2}}m-tm=(2i+1)r-im+(2j+1)r-jm$. Then, $k\equiv (2i+2j+2)(modm)$, and we can write $2i+2j+2=k+lm$ for some $l\in \mathbb{N}$. Therefore, $kr-{\displaystyle \frac{k-2}{2}}m-tm=(2i+2j+2)r-(i+j)m=(k+lm)r-{\displaystyle \frac{k+lm-2}{2}}m$. Consequently, ${\displaystyle \frac{lm}{2}}-t=lr$. Then, ${\displaystyle \frac{lm}{2}}>lr$, allowing us to obtain ${\displaystyle \frac{m}{2}}>r$, which is a contradiction. Again, we can ensure $l\ne 0$ since $t\ne 0$.

□

We give the following example to illustrate Theorem 3:

Example 6. 

Using Theorem 3, we calculate $Ap({\mathcal{L}}_2\left(10\right)\left[\left\{13\right\}\right],10)$.

$w(1·13mod10)=w\left(3\right)=1·13-{\displaystyle \frac{1-1}{2}}·10=13$	$w(2·13mod10)=w\left(6\right)=2·13-{\displaystyle \frac{2-2}{2}}·10=26$
$w(3·13mod10)=w\left(9\right)=3·13-{\displaystyle \frac{3-1}{2}}·10=29$	$w(4·13mod10)=w\left(2\right)=4·13-{\displaystyle \frac{4-2}{2}}·10=42$
$w(5·13mod10)=w\left(5\right)=5·13-{\displaystyle \frac{5-1}{2}}·10=45$	$w(6·13mod10)=w\left(8\right)=6·13-{\displaystyle \frac{6-2}{2}}·10=58$
$w(7·13mod10)=w\left(1\right)=7·13-{\displaystyle \frac{7-1}{2}}·10=61$	$w(8·13mod10)=w\left(4\right)=8·13-{\displaystyle \frac{8-2}{2}}·10=74$
$w(9·13mod10)=w\left(7\right)=9·13-{\displaystyle \frac{9-1}{2}}·10=77$

The following result is easy to prove:

Lemma 9. 

Let S be a numerical semigroup, and $n\in S\setminus \left\{0\right\}$. Then, $S=\langle (Ap(S,n)\setminus \{0\left\}\right)\cup \left\{n\right\}\rangle$.

Lemma 10. 

${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]=\langle \left\{m\right\}\cup \{(2k+1)r-km\mid k\in \{0,1,\dots ,⌊{\displaystyle \frac{m-2}{2}}⌋\}\}\rangle$.

Proof. 

Applying Lemma 9 and Theorem 3, we deduce that ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]=\langle \left\{m\right\}\cup \{jr-{\displaystyle \frac{j-1}{2}}m\mid j\in \{1,\dots ,m-1\}andjodd\}\cup \{jr-{\displaystyle \frac{j-2}{2}}m\mid j\in \{1,\dots ,m-1\}andjeven\}\rangle$. However, when j is even, we have $jr-{\displaystyle \frac{j-2}{2}}m=(j-1)r-{\displaystyle \frac{(j-1)-1}{2}}m+r$, so $jr-{\displaystyle \frac{j-2}{2}}m\notin msg\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)$. Therefore, ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]=\langle \left\{m\right\}\cup \{jr-{\displaystyle \frac{j-1}{2}}m\mid j\in \{1,\dots ,m-1\}andjodd\}\rangle =\langle \left\{m\right\}\cup \{(2k+1)r-km\mid k\in \{0,1,\dots ,⌊{\displaystyle \frac{m-2}{2}}⌋\}\}\rangle$. □

Proposition 11. 

The embedding dimension of ${\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ is $⌊{\displaystyle \frac{m-2}{2}}⌋+2$. Moreover, $msg\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)=\left\{m\right\}\cup \{(2k+1)r-km\mid k\in \{0,1,\dots ,⌊{\displaystyle \frac{m-2}{2}}⌋\}\}$.

Proof. 

Suppose, on the contrary, that $(2k+1)r-km\notin msg\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)$ for some $k\in \{0,1,\dots ,⌊{\displaystyle \frac{m-2}{2}}⌋\}$. Then, there exists some $g\in msg\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)$ such that $(2k+1)r-km-g\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\setminus \left\{0\right\}$. Theorem 3 ensures that $(2k+1)r-km\in Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$ for our k, which ensures that $g\ne m$. Then, using Lemma 10, we can assume $g=(2i+1)r-im$ for some $i<k$. Therefore, we have the following: $(2k+1)r-km-((2i+1)r-im)=2(k-i)r-(k-i)m\in {\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]$ and we also have $2(k-i)r-(k-i)m\equiv 2(k-i)r(modm)$. Furthermore, we have $2(k-i)<2(k+1)\le m$, since $k\le ⌊{\displaystyle \frac{m-2}{2}}⌋$. So, by Theorem 3, we can take $w(2(k-i)rmodm)\in Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$. Thus, using Theorem 3, we have $2(k-i)r-(k-i)m\ge w(2(k-i)rmodm)=2(k-i)r-{\displaystyle \frac{2(k-i)-2}{2}}m$. However, this implies $2(k-i)-2\ge 2(k-i)$, which is absurd, and the proof is finished. □

Example 7. 

Using Proposition 11, we can compute $msg\left({\mathcal{L}}_2\left(10\right)\left[\left\{13\right\}\right]\right)=\left\{10\right\}\cup \{(2k+1)13-k10\mid k\in \{0,1,2,3,4\}\}=\{10,13,29,45,61,77\}$.

Now, we are going to solve the Frobenius problem for elements in ${\mathcal{L}}_2\left(m\right)$ with the ${\mathcal{L}}_2\left(m\right)$-rank being equal to 1. We will use Theorem 3 and the following result from [24]:

Lemma 11. 

Let S be a numerical semigroup and $n\in S\setminus \left\{0\right\}$. Then, we have the following:

1. 

$F\left(S\right)=max(Ap(S,n\left)\right)-n$.

2. 

$g\left(S\right)={\displaystyle \frac{1}{n}}\left({\sum }_{w\in Ap(S,n)}w\right)-{\displaystyle \frac{n-1}{2}}$.

Proposition 12. 

$$F\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)=\left\{\begin{array}{cc}(m-1)r-{\displaystyle \frac{m^2}{2}}& ifmiseven\\ (m-1)r-{\displaystyle \frac{m(m-1)}{2}}& ifmisodd\end{array}\right.$$

Proof. 

Using Theorem 3, we deduce the following:

$$max(Ap\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right),m)=\left\{\begin{array}{cc}(m-1)r-{\displaystyle \frac{(m-2)m}{2}}& ifm-1isodd\\ (m-1)r-{\displaystyle \frac{(m-3)m}{2}}& ifm-1iseven\end{array}\right.$$

Moreover, with Lemma 11, we finish the proof. □

Example 8. 

1. 

$F\left({\mathcal{L}}_2\left(10\right)\left[\left\{11\right\}\right]\right)=9·11-{\displaystyle \frac{10·10}{2}}=49$.

2. 

$F\left({\mathcal{L}}_2\left(9\right)\left[\left\{13\right\}\right]\right)=8·13-{\displaystyle \frac{9·8}{2}}=68$.

Lemma 12. 

If $a_j=⌊{\displaystyle \frac{j-1}{2}}⌋$ for all $j\in \{1,\dots ,m-1\}$, then we have

$$(a_1,\dots ,a_{m-1})=\left\{\begin{array}{cc}(0,0,1,1,\dots ,{\displaystyle \frac{m-3}{2}},{\displaystyle \frac{m-3}{2}})& ifmisodd\\ (0,0,1,1,\dots ,{\displaystyle \frac{m-4}{2}},{\displaystyle \frac{m-4}{2}},{\displaystyle \frac{m-2}{2}})& ifmiseven\end{array}\right.$$

Proposition 13. 

$$g\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)=\left\{\begin{array}{cc}{\displaystyle \frac{(m-1)r}{2}}-{\displaystyle \frac{{(m-1)}^2}{4}}& ifmisodd\\ {\displaystyle \frac{(m-1)r}{2}}-{\displaystyle \frac{{(m-1)}^2+1}{4}}& ifmiseven\end{array}\right.$$

Proof. 

By Theorem 3, we can write the elements in $Ap\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)$ as $w(jrmodm)=jr-⌊{\displaystyle \frac{j-1}{2}}⌋m$ for all $j\in \{1,\dots ,m-1\}$. Moreover, from Lemmas 11 and 12, we have

• If m is odd: $g\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)={\displaystyle \frac{1}{m}}\left(r{\sum }_{j=1}^{m-1}j-2m{\sum }_{i=1}^{{\displaystyle \frac{m-3}{2}}}i\right)-{\displaystyle \frac{m-1}{2}}={\displaystyle \frac{(m-1)r}{2}}-{\displaystyle \frac{{(m-1)}^2}{4}}$.
• If m is even: $g\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)={\displaystyle \frac{1}{m}}\left(r{\sum }_{j=1}^{m-1}j-2m{\sum }_{i=1}^{{\displaystyle \frac{m-4}{2}}}i-{\displaystyle \frac{m-2}{2}}m\right)-{\displaystyle \frac{m-1}{2}}={\displaystyle \frac{(m-1)r}{2}}-{\displaystyle \frac{{(m-1)}^2+1}{4}}$.

□

Example 9. 

1. 

$g\left({\mathcal{L}}_2\left(10\right)\left[\left\{11\right\}\right]\right)={\displaystyle \frac{9·11}{2}}-{\displaystyle \frac{9·9+1}{4}}={\displaystyle \frac{99}{2}}-{\displaystyle \frac{41}{2}}={\displaystyle \frac{58}{2}}=29$.

2. 

$g\left({\mathcal{L}}_2\left(9\right)\left[\left\{13\right\}\right]\right)={\displaystyle \frac{8·13}{2}}-{\displaystyle \frac{8·8}{4}}=52-16=36$.

As we pointed out in Remark 1, the next result appears in [23], Lemma 10.

Lemma 13. 

Let S be a numerical semigroup, and $n\in S\setminus \left\{0\right\}$. Then,

$$PF\left(S\right)=\{w-n\mid w\in {Maximals}_{{\le }_S}(Ap(S,n))\}$$

Proposition 14. 

${Maximals}_{{\le }_{{\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]}}(Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m))=\{w(rjmodm)\mid jiseven\}\cup \{F\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)+m\}$. For $j\in \{1,\dots ,m-1\}$.

Proof. 

If $j<m$ is odd, then $w(jrmodm)+r=jr-{\displaystyle \frac{j-1}{2}}m+r=(j+1)r-{\displaystyle \frac{(j+1)-2}{2}}m=w((j+1)rmodm)$, so $w(jrmodm)$ is not a maximal element in $Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$.

On the other hand, if $j<m$ is even, from Proposition 11, we have that every element in $msg\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)\setminus \left\{m\right\}$ can be written as $w(irmodm)$ for any i odd number, such that $i<m$. Therefore, using Theorem 3, we have $w(jrmodm)+w(irmodm)=jr-{\displaystyle \frac{j-2}{2}}m+ir-{\displaystyle \frac{i-1}{2}}m=(i+j)r-{\displaystyle \frac{(j-2)+(i-1)}{2}}m=(i+j)r-{\displaystyle \frac{(i+j-3)}{2}}m\notin Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$, which is also stated in Theorem 3. Furthermore, $w(jrmodm)+m=jr-{\displaystyle \frac{j-2}{2}}m+m=jr-{\displaystyle \frac{j-4}{2}}m\notin Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$ for every $j<m$, such that j is even. Therefore, $w(jrmodm)$ is maximal in $Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$ when j is even. Clearly, $F\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)+m$ is also maximal in $Ap({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right],m)$. □

Corollary 2. 

$PF\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)=\{k(2r-m)\mid k\in \{1,\dots ,⌊{\displaystyle \frac{m-2}{2}}⌋\}\}\cup \{F\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)\}$.

Example 10. 

Using Corollary 2, we can compute $PF\left({\mathcal{L}}_2\left(10\right)\left[\left\{13\right\}\right]\right)=\{16k\mid k\in \{1,2,3,4\}\}\cup \{F\left({\mathcal{L}}_2\left(10\right)\left[\left\{13\right\}\right]\right)\}=\{16,32,48,64,67\}$.

We have obtained in Examples 6, 7 and 10 the Apéry set of the multiplicity, the minimal system of generators, and the pseudo-Frobenius numbers of ${\mathcal{L}}_2\left(10\right)\left[\left\{13\right\}\right]$. Now, we are going to change the parity of m to show an example where m is odd.

Example 11. 

Using Theorem 3, Proposition 11, and Corollary 2, we are going to compute the Apéry set of the multiplicity, the minimal system of generators, and the pseudo-Frobenius numbers of ${\mathcal{L}}_2\left(11\right)\left[\left\{13\right\}\right]$.

The Apéry set $Ap({\mathcal{L}}_2\left(11\right)\left[\left\{13\right\}\right],11)$ is

$w(1·13mod11)=w\left(3\right)=1·13-{\displaystyle \frac{1-1}{2}}·11=13$	$w(2·13mod11)=w\left(6\right)=2·13-{\displaystyle \frac{2-2}{2}}·11=26$
$w(3·13mod11)=w\left(9\right)=3·13-{\displaystyle \frac{3-1}{2}}·11=28$	$w(4·13mod11)=w\left(2\right)=4·13-{\displaystyle \frac{4-2}{2}}·11=41$
$w(5·13mod11)=w\left(5\right)=5·13-{\displaystyle \frac{5-1}{2}}·11=43$	$w(6·13mod11)=w\left(8\right)=6·13-{\displaystyle \frac{6-2}{2}}·11=56$
$w(7·13mod11)=w\left(1\right)=7·13-{\displaystyle \frac{7-1}{2}}·11=58$	$w(8·13mod11)=w\left(4\right)=8·13-{\displaystyle \frac{8-2}{2}}·11=71$
$w(9·13mod11)=w\left(7\right)=9·13-{\displaystyle \frac{9-1}{2}}·11=73$	$w(10·13mod11)=w\left(4\right)=10·13-{\displaystyle \frac{10-2}{2}}·11=86$

From the above, we have $msg\left({\mathcal{L}}_2\left(11\right)\left[\left\{13\right\}\right]\right)=\{11,13,28,43,58,73\}$, and $PF\left({\mathcal{L}}_2\left(11\right)\left[\left\{13\right\}\right]\right)$$=\{15,30,45,60,75\}$.

Corollary 3. 

$t\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)={\displaystyle \frac{m}{2}}$ if m is even or $t\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)={\displaystyle \frac{m-1}{2}}$ if m is odd. Summarizing $t\left({\mathcal{L}}_2\left(m\right)\left[\left\{r\right\}\right]\right)=⌊{\displaystyle \frac{m}{2}}⌋$.

The following result comes from Proposition 11 and Corollary 3:

Corollary 4. 

We have $e\left(S\right)=t\left(S\right)+1$ for every $S\in {\mathcal{L}}_2\left(m\right)$ with the ${\mathcal{L}}_2\left(m\right)$-rank being equal to 1.

On page 15 of [1], it is shown that $F\left(S\right)+1\le (t(S)+1)n\left(S\right)$ for any numerical semigroup S (it was earlier obtained in [7]). Then, using Corollary 4, we achieve the following result:

Proposition 15. 

Every element in ${\mathcal{L}}_2\left(m\right)$ with the ${\mathcal{L}}_2\left(m\right)$-rank being equal to 1 satisfies Wilf’s conjecture.