Infinite sequences of almost symmetric non-Weierstrass numerical semigroups

Abstract

Let u be any positive integer. We construct infinite sequences of almost symmetric non-Weierstrass numerical semigroups whose conductors are the genera double minus $2u-1$. Moreover, let v be any non-negative integer. We give an infinite number of non-Weierstrass numerical semigroups whose conductors are the genera double minus 2v.

1 Introduction

Let ${\mathbb {N}}_0$ be the additive monoid of non-negative integers. A submonoid H of ${\mathbb {N}}_0$ is called a numerical semigroup if the complement ${\mathbb {N}}_0\backslash H$ is a finite set. The cardinality of ${\mathbb {N}}_0\backslash H$ is called the genus of H, denoted by g(H). Let (C, P) be a pointed curve of genus g where a curve means a complete non-singular irreducible algebraic curve over an algebraically closed field k of characteristic 0 in this article. We set

$$\begin{aligned} H(P)=\{n\in {\mathbb {N}}_0\mid \text{ there } \text{ exists } f\in k(C) \text{ such } \text{ that } (f)_{\infty }=nP\} \end{aligned}$$

where k(C) denotes the field of rational functions on C. Then H(P) becomes a numerical semigroup of genus g. A numerical semigroup H is said to be Weierstrass if there exists a pointed curve (C, P) with $H(P)=H$. Hurwitz [2] posed the following problem in 1893: Is every numerical semigroup Weierstrass? This is a long-standing problem. In 1980 Buchweitz [1] gave a computable necessary condition for a numerical semigroup to be Weierstrass. Using this criterion he finally found an example of a non-Weierstrass numerical semigroup. About ten years later after Buchweitz gave the non-Weierstrass numerical semigroup, Torres [13] showed that the criterion given by Buchweitz is not a sufficient condition for a numerical semigroup to be Weierstrass. So the following problem is still unsolved:

Give a computable necessary and sufficient condition on a numerical semigroup to be Weierstrass.

Let H be a numerical semigroup. We set $c(H)=\min \{c\in {\mathbb {N}}_{0}\mid c+{\mathbb {N}}_{0}\subseteqq H\}$, which is called the conductor of H. The integer $c(H)-1$ is called the Frobenius number of H. We have $c(H)\leqq 2g(H)$. A numerical semigroup H is said to be symmetric if the equality $c(H)=2g(H)$ holds. A numerical semigroup H is said to be quasi-symmetric if the equality $c(H)=2g(H)-1$ holds. We denote by PF(H) the set

$$\begin{aligned} \{f\in {\mathbb {N}}_0\backslash H\mid f+h\in H \text{ for } \text{ all } h \in H \text{ with } h> 0\}. \end{aligned}$$

Since $c(H)-1$ belongs to PF(H), the set PF(H) is non-empty. An element of PF(H) is called a pseudo-Frobenius number of H. The cardinality of the set PF(H) is called the type of H, which is denoted by t(H). We have $c(H)+t(H)\leqq 2g(H)+1$ (see Proposition 2.2 in [10]). A numerical semigroup H is said to be almost symmetric if $c(H)+t(H)=2g(H)+1$. Hence, any symmetric numerical semigroup is almost symmetric. Let l be an integer with $l\geqq 2$. For any numerical semigroup H, we set

$$ d_l(H)=\{h'\in {\mathbb {N}}_0\mid lh'\in H\}, $$

which is also a numerical semigroup. In Sect. 2, using the result in Torres [13], we give infinite sequences

$$\begin{aligned} H_{0,0}=H_0 \xleftarrow {d_2} H_1 \xleftarrow {d_2} H_2 \xleftarrow {d_2} \cdots \xleftarrow {d_2} H_{i-1} \xleftarrow {d_2} H_i \xleftarrow {d_2} \cdots \end{aligned}$$

of symmetric non-Weierstrass numerical semigroups. In Sect. 3, for any positive integer t and any almost symmetric numerical semigroup $H_0$ of type t, we construct infinite sequences

$$\begin{aligned} H_0 \xleftarrow {d_3} H_1 \xleftarrow {d_3} H_2 \xleftarrow {d_3} \cdots \xleftarrow {d_3} H_{i-1} \xleftarrow {d_3} H_i \xleftarrow {d_3} \cdots \end{aligned}$$

of almost symmetric numerical semigroups of type t, hence $c(H_i)=2g(H_i){-}(t{-}1)$. Oliveira and Stöhr [11] constructed quasi-symmetric non-Weierstrass numerical semigroups $H_{0,1}$. In Sect. 4 for any integer u with $u\geqq 2$, we show that there are almost symmetric non-Weierstrass numerical semigroups $H_{0,2u-1}$ of type 2u, hence $c(H_{0,2u-1})=2g(H_{0,2u-1})-(2u-1)$. Thereafter for any $u\geqq 1$, we construct infinite sequences of almost symmetric non-Weierstrass numerical semigroups whose starting points are the above numerical semigroups $H_{0,2u-1}$ as follows:

$$\begin{aligned} H_{0,2u-1}=H_0 \xleftarrow {d_3} H_1 \xleftarrow {d_3} H_2 \xleftarrow {d_3} \cdots \xleftarrow {d_3} H_{i-1} \xleftarrow {d_3} H_i \xleftarrow {d_3} \cdots \end{aligned}$$

with $c(H_i)=2g(H_i)-(2u-1)$. In Sect. 5 we study numerical semigroups whose conductors are even. Specifically, we give an infinite number of non-Weierstrass numerical semigroups H with $c(H)=2g(H)-2v$ for any $v\geqq 1$.

2 Symmetric non-Weierstrass numerical semigroups

For a numerical semigroup H we use the following notation: We set

$$\begin{aligned} m=m(H)=\min \{h\in H\mid h>0\}, \end{aligned}$$

which is called the multiplicity of H. Let

$$\begin{aligned} s_i=\min \{h\in H\mid h\equiv i \text{ mod } m\} \end{aligned}$$

for any i with $1\leqq i\leqq m-1$. The numerical semigroup H is generated by $s_1,\ldots ,s_{m-1}$. The set $\{m,s_1,\ldots ,s_{m-1}\}$ is called the standard basis for H, which is denoted by S(H). We set

$$\begin{aligned} s_{max}=\max \{s_i\mid i=1,2,\ldots , m-1\}. \end{aligned}$$

We note that $c(H)=s_{max}-m+1$. To construct infinite sequences of symmetric non-Weierstrass numerical semigroups we need the following lemma:

Lemma 1

Let H be a symmetric numerical semigroup with multiplicity m and n be an odd number with $n\geqq c(H)+m-1$. We set

$$\begin{aligned} \tilde{H}=2H+n{\mathbb {N}}_0:=\{2h+nl\mid h\in H, l\in {\mathbb {N}}_0\}. \end{aligned}$$

Then $\tilde{H}$ is a symmetric numerical semigroup whose genus is $2g(H)+\dfrac{n-1}{2}$.

Proof

Using Proposition 2.1 in [8] we get the description of $g(\tilde{H})$. We note that the proof of Proposition 2.1 in [8] works well even if $n\geqq c(H)+m-1$. Then we have

$$\begin{aligned} c(\tilde{H})=n+2s_{max}-2m+1=n-1+2(s_{max}-m+1)=n-1+2c(H). \end{aligned}$$

Since H is symmetric, we obtain

$$\begin{aligned} c(\tilde{H})=n-1+4g(H)=2g(\tilde{H}), \end{aligned}$$

which implies that $\tilde{H}$ is symmetric. $\square$

Remark 1

The semigroup $\tilde{H}$ in Lemma 1 is obtained by “gluing” and the Frobenius numbers are easily computed (see Theorem 9.2 and Proposition 9.11 in [12]). Using the above we can also prove that $\tilde{H}$ is symmetric.

The following is useful for constructing infinite sequence of non-Weierstrass numerical semigroups from a non-Weierstrass numerical semigroup:

Remark 2

Let H be a non-Weierstrass numerical semigroup and $\tilde{H}$ be a numerical semigroup with $d_2(\tilde{H})=H$. If $g(\tilde{H})\geqq 6g(H)+1$, then $\tilde{H}$ is non-Weierstrass.

Proof

Assume that $\tilde{H}$ is Weierstrass, i.e., there exists a pointed curve (C, P) such that $H(P)=\tilde{H}$. If $g(\tilde{H})\geqq 6g(H)+4$, $g(\tilde{H})=6g(H)+3$ or $6g(H)+2$, and $g(\tilde{H})=6g(H)+1$, by Corollary 2.10 in [13], Theorem B in [3] and Main Theorem in [9] respectively, C is a double cover of some curve $C'$ such that P is its ramification point. Let $\pi :C\longrightarrow C'$ be the double covering. Then we have $H=H(\pi (P))$ by Lemma 2 in [5], which implies that H is Weierstrass. This is a contradiction. $\square$

Theorem 1

Let $H_0$ be a symmetric non-Weierstrass numerical semigroup. Then we have infinite sequences

$$\begin{aligned} H_0 \xleftarrow {d_2} H_1 \xleftarrow {d_2} H_2 \xleftarrow {d_2} \cdots \xleftarrow {d_2} H_{i-1} \xleftarrow {d_2} H_i \xleftarrow {d_2} \cdots \end{aligned}$$

of symmetric non-Weierstrass numerical semigroups.

Proof

Let H be a symmetric non-Weierstrass numerical semigroup with multiplicity m. For any odd $n\geqq c(H)+m-1$ the numerical semigroup $\tilde{H}=2H+n{\mathbb {N}}_0$ is symmetric from Lemma 1. If $n\geqq 8g(H)+3$, then we have $g(\tilde{H})=2g(H)+\dfrac{n-1}{2}\geqq 6g(H)+1$. Hence by Remark 2 the numerical semigroup $\tilde{H}$ is non-Weierstrass. Repeating this process we get the desired infinite sequences. $\square$

3 Infinite sequences of almost symmetric numerical semigroups

Proposition 1

Let H be a numerical semigroup with multiplicity m and n be an integer not divisible by 3 with

$$\begin{aligned} n\geqq \max \{c(H)+m-1,3m+1\}. \end{aligned}$$

Then we have $g(3H+n{\mathbb {N}}_0)=3g(H)+n-1$.

Proof

Since $n\geqq 3m$, $3H+n{\mathbb {N}}_0$ is a $3m-$semigroup. Moreover, the standard basis $S(3H+n{\mathbb {N}}_0)$ is

$$\begin{aligned}&\left\{ 3m,3s_1,\ldots ,3s_{m-1}\}\cup \{n,n+3s_1,\ldots ,n+3s_{m-1}\} \cup \{2n,2n+3s_1,\right. \\&\quad \left. \ldots ,2n+3s_{m-1}\right\} , \end{aligned}$$

because $n\geqq c(H)+m-1=s_{max}$.

We put $H_1=3H+n{\mathbb {N}}_0$. We will count $\sharp ({\mathbb {N}}\backslash H_1)$.

• (a) For a non-negative integer y we have $x=3y\not \in H_1$ if and only if $y\not \in H$. We have $\sharp \{y\in {\mathbb {N}}_0\mid y\not \in H\}=g(H)$.

• (b) For an integer y we have $x=n+3y\in {\mathbb {N}}_0\backslash H_1$ if and only if $y\in {\mathbb {N}}_0\backslash H$ or $0< n+3y<n$. We have

  $$\begin{aligned} \sharp \{y\in {\mathbb {Z}}\mid y\in {\mathbb {N}}_0\backslash H \text{ or } 0< n+3y<n\}=g(H)+ \left[ \dfrac{n}{3}\right] . \end{aligned}$$

• (c) For an integer y we have $x=2n+3y\in {\mathbb {N}}_0\backslash H_1$ if and only if $y\in {\mathbb {N}}_0\backslash H$ or $0< 2n+3y<2n$. We have

  $$\begin{aligned} \sharp \{y\in {\mathbb {Z}}\mid y\in {\mathbb {N}}_0\backslash H \text{ or } 0<2 n+3y<2n\}=g(H)+\left[ \dfrac{2n}{3}\right] . \end{aligned}$$

Thus, we obtain

$$\begin{aligned} g(H_1)=\sharp ({\mathbb {N}}_0\backslash H_1)=3g(H) +\left[ \dfrac{n}{3}\right] +\left[ \dfrac{2n}{3}\right] =3g(H)+n-1. \end{aligned}$$

$\square$

We will show from now on that there is an infinite sequence of almost symmetric numerical semigroups whose starting point is any almost symmetric numerical semigroup. For that purpose we need to check whether a given numerical semigroup is almost symmetric. To show that a numerical semigroup H is almost symmetric we investigate the set PF(H) of pseudo-Frobenius numbers. The following lemma is useful for describing the set PF(H):

Lemma 2

Let H be a numerical semigroup with multiplicity m. Then we have

$$\begin{aligned} PF(H)=\{s-m\mid s\in SPF_+(H)\} \end{aligned}$$

where we set

$$\begin{aligned} SPF_+(H)=\{s\in S(H)\mid s+s'\not \in S(H) \text{ for } \text{ all } s'\in S(H)\} \end{aligned}$$

Proof

Let $s\in SPF_+(H)$. Take $h\in H$ with $h>0$. Then $h=s'+lm$ for some $s'\in S(H)$ with $l\in {\mathbb {N}}_0$. We have

$$\begin{aligned} s-m+h=s+s'+(l-1)m=s''+l'm+(l-1)m=s''+(l'+l-1)m \end{aligned}$$

for some $s''\in S(H)$ with a positive integer $l'$. Hence we get $s-m+h\in H$. Thus, we have $s-m\in PF(H)$.

Let $f\in PF(H)$. Then we have $f=s-m$ with $s\in S(H)$, because $f\not \in H$ and $f+m\in H$. Moreover, we obtain $H\ni f+s'=s-m+s'$ for any $s'\in S(H)$, which implies that $s+s'=h+m$ with $h\in H$. Hence, we get $s+s'\not \in S(H)$. $\square$

We will show how to construct an almost symmetric numerical semigroup $\tilde{H}$ from any almost symmetric numerical semigroup H satisfying $d_3(\tilde{H})=H$. We denote by $\langle a_1,\ldots ,a_l\rangle$ the additive monoid generated by positive integers $a_1,\ldots ,a_l$. This notation is useful for describing a numerical semigroup.

Theorem 2

Let t be a positive integer. Let H be an almost symmetric numerical semigroup with multiplicity m and $c(H)=2g(H)-(t-1)$. Let n be an integer with $n\geqq \max \{c(H)+m-1,3m+1\}$ which is not divisible by 3. We set

$$\begin{aligned} \tilde{H}=3H+\langle n,2n+3f_1,\ldots ,2n+3f_{t-1}\rangle \end{aligned}$$

where $PF^*(H)=PF(H)\backslash \{c(H)-1\}=\{f_1,\ldots ,f_{t-1}\}$, because H is almost symmetric. Then we have

$$\begin{aligned} g(\tilde{H})=3g(H)+n-1-(t-1) \text{ and } c(\tilde{H})=2g(\tilde{H})-(t-1) . \end{aligned}$$

Moreover, we obtain $t(\tilde{H})=t$, which implies that the numerical semigroup $\tilde{H}$ is almost symmetric.

Proof

It follows from Lemma 2 that

$$\begin{aligned}&S(\tilde{H})=\{3m,3s_1,\ldots ,3s_{m-1},n,n+3s_1,\ldots ,n+3s_{m-1}\}\\&\quad \cup \{2n+3f_1,\ldots ,2n+3f_{t-1}\}\cup \{2n+3s\mid s\in S(H) \text{ with } s-m\not \in PF(H)\}. \end{aligned}$$

Using Proposition 1 and $f_i\in PF(H)$ for any i with $1\leqq i\leqq t-1$, we get

$$\begin{aligned} g(\tilde{H})=3g(H)+n-1-(t-1) \end{aligned}$$

Since $2n+3s_{max}$ is the maximum element in $S(\tilde{H})$, we have $2n+3s_{max}-3m \in PF(\tilde{H})$. Moreover, we have

$$\begin{aligned}&c(\tilde{H})=2n+3s_{max}-3m+1=3(s_{max}-m+1)+2(n-1)\\&\quad =3c(H)+2(n-1)\\&\quad =3(2g(H)-(t-1))+2(n-1)\\&\quad =2(3g(H)+n-t)-t+1=2g(\tilde{H})-(t-1). \end{aligned}$$

We set $s^{(i)}=f_i+m$ for all i. Then

$$\begin{aligned} (n+3s^{(i)})+n=(2n+3f_i)+3m\not \in S(\tilde{H}). \end{aligned}$$

Moreover, for any $s\in S(H)$ we have

$$\begin{aligned} (n+3s^{(i)})+3s=n+3(s^{(i)}+s)=n+3(s'+\nu m) \end{aligned}$$

with a positive integer $\nu$ from Lemma 2, because $f_i\in PF(H)$. Using Lemma 2 again we obtain $n+3f_i\in PF(\tilde{H})$ for all $i=1,\ldots ,t-1$. Thus, we have $t(\tilde{H})\geqq t$. Moreover, the equality $c(\tilde{H})=2g(\tilde{H})-(t-1)$ implies that $t(\tilde{H})\leqq t$. Hence, we get $t(\tilde{H})=t$. Thus, $\tilde{H}$ is almost symmetric. $\square$

Here, we give an example of an almost symmetric numerical semigroup H with $c(H)=2g(H)-(t-1)$ for any positive integer t.

Example 1

Let t be a positive integer. We set $m=t+1$. Let $H=\langle m,m+1,\ldots ,m+m-1\rangle$. By Lemma 2 we have $t(H)=m-1=t$. Moreover, we obtain

$$\begin{aligned} c(H)=m=2(m-1)-(m-2)=2g(H)-(t-1). \end{aligned}$$

Hence, the numerical semigroup H is almost symmetric.

Using Theorem 2 repeatedly we get infinite sequences of almost symmetric numerical semigroups.

Corollary 1

Let t be a positive integer. Let H be any almost symmetric numerical semigroup with $c(H)=2g(H)-(t-1)$. Then there exists an infinite sequences

$$\begin{aligned} H_0=H \xleftarrow {d_3} H_1 \xleftarrow {d_3} H_2 \xleftarrow {d_3} \cdots \xleftarrow {d_3} H_{i-1} \xleftarrow {d_3} H_i \xleftarrow {d_3} \cdots \end{aligned}$$

of almost symmetric numerical semigroups whose conductors are equal to the genera double minus $t-1$.

4 Almost symmetric non-Weierstrass numerical semigroups with odd conductor

To find almost symmetric non-Weierstrass numerical semigroups with odd conductor we use the construction in the following lemma whose origin is Scholium 3.5 in [13] and Theorem 5.1 in [11].

Lemma 3

Let u be a positive integer and H be a numerical semigroup with multiplicity $m\geqq 2$. Let g be an integer with $g\not \equiv u \text{ mod } 3$ and

$$\begin{aligned} g>\max \{4u-3,2u+2c(H)+m-4\}. \end{aligned}$$

We set

$$\begin{aligned} \tilde{H}=3H\cup (g+2u+3{\mathbb {N}}_0)\cup \{2g-2u-3r\mid r \in {\mathbb {Z}}\backslash H\}. \end{aligned}$$

Then the following hold:

• (i) $\tilde{H}$ is a numerical semigroup of genus g.

• (ii) We have $c(\tilde{H})=2g-(2u-1)$, hence the maximum element in $S(\tilde{H})$ is $2g-2u+3m$.

• (iii) We obtain

  $$\begin{aligned} PF(\tilde{H})\subseteqq \{g+2u+3l-3m\mid l=0,1,\ldots ,m-1\}\cup \{2g-2u+3m-3m\}, \end{aligned}$$

  which implies that $t(\tilde{H})\leqq m+1$.

Proof

• (i) First, we will prove that $\tilde{H}$ becomes a submonoid of the additive monoid ${\mathbb {N}}_0$. We set

  $$\begin{aligned} A=3H, B=g+2u+3{\mathbb {N}}_0 \,\, \text{ and } \,\, C=\{2g-2u-3r\mid r\in {\mathbb {Z}}\backslash H\}. \end{aligned}$$

  We note that the minimum element of C is $2g-2u-3(c(H)-1)$, which is non-negative, because we have

  $$\begin{aligned} 2g-2u-3(c(H)-1)\geqq & {} 2(2u+2c(H)+m-3)-2u-3(c(H)-1)\\= & {} 2u+c(H)+2m-3\geqq 2u+c(H)+1>0. \end{aligned}$$

  It is trivial that

  $$\begin{aligned} A+A\subseteqq A, A+B\subseteqq B \text{ and } A+C\subseteqq C. \end{aligned}$$

  Let $l\in {\mathbb {N}}_0$ and $l'\in {\mathbb {N}}_0$. We have

  $$\begin{aligned} (g+2u+3l)+(g+2u+3l')= & {} 2g-2u+6u+3(l+l')\\= & {} 2g-2u-3(-2u-l-l'), \end{aligned}$$

  which implies that $B+B\subseteqq C$, because $-2u-l-l'<0$.

  Let $l\in {\mathbb {N}}_0$ and $r\in {\mathbb {Z}}\backslash H$. We have

  $$\begin{aligned} (g+2u+3l)+(2g-2u-3r)=3(g+l-r). \end{aligned}$$

  Since $g\geqq 2u+2c(H)+m-3>2u+2c(H)-2$, we get

  $$\begin{aligned} g+l-r\geqq & {} g-r> 2u+2c(H)-2-r\geqq 2+2c(H)-2-r\\= & {} 2c(H)-r\geqq 2c(H)-(c(H)-1)=c(H)+1, \end{aligned}$$

  because $r\leqq c(H)-1$. Hence, we get $g+l-r\in H$, which implies that $B+C\subseteqq A$.

  Let $r\in {\mathbb {Z}}\backslash H$ and $r'\in {\mathbb {Z}}\backslash H$. Then we have

  $$\begin{aligned} (2g-2u-3r)+(2g-2u-3r')=g+2u+3(g-2u-r-r'). \end{aligned}$$

  It follows from $g>\max \{4u-3,2u+2c(H)+m-4\}$ that

  $$\begin{aligned} g-2u-r-r'\geqq & {} g-2u+2-2c(H)\\\geqq & {} 2u+2c(H)+m-3-2u+2-2c(H)=m-1>0, \end{aligned}$$

  which implies that $C+C\subseteqq B$. Thus, $\tilde{H}$ becomes a submonoid of ${\mathbb {N}}_0$. By the definition of $\tilde{H}$ it is a numerical semigroup.

  Secondly, we will show that the genus of the numerical semigroiup $\tilde{H}$ is equal to g. We have

  $$\begin{aligned} {\mathbb {N}}_0\backslash \tilde{H}= & {} 3{\mathbb {N}}_0\backslash 3H \cup (3{\mathbb {N}}_0+a)\backslash \{g+2u+3{\mathbb {N}}_0\}\\&\cup (3{\mathbb {N}}_0+b)\backslash \{2g-2u-3r\mid r \in {\mathbb {Z}}\backslash H\} \end{aligned}$$

  where $g+2u\equiv a \text{ mod } 3$ and $2g-2u\equiv b \text{ mod } 3$ with $\{a,b\}=\{1,2\}$, because $g\not \equiv u \text{ mod } 3$. We obtain that

  $$\begin{aligned} \sharp (3{\mathbb {N}}_0\backslash 3H)=g(H) \text{ and } \sharp ((3{\mathbb {N}}_0+a)\backslash \{g+2u+3{\mathbb {N}}_0\}) =\frac{g+2u-a}{3}. \end{aligned}$$

  Moreover, the maximum element in $(3{\mathbb {N}}_0+s)\backslash \{2g-2u-3r\mid r\in {\mathbb {Z}}\backslash H\}$ is $2g-2u$, which implies that

  $$\begin{aligned} \sharp ((3{\mathbb {N}}_0+b)\backslash \{2g-2u-3r\mid r \in {\mathbb {Z}}\backslash H\})=\frac{2g-2u+3-b}{3}-g(H). \end{aligned}$$

  Combining the above results about the cardinalities of the three sets we get

  $$\begin{aligned} g(\tilde{H})=\sharp ({\mathbb {N}}_0\backslash \tilde{H})=\frac{3g-a-b+3}{3}=g. \end{aligned}$$

• (ii) We have $3(c(H)-1)\not \in 3H$ and $3l\in 3H$ for any $l\geqq c(H)$. Moreover, it is clear that $g+2u-3\not \in g+2u+3{\mathbb {N}}_0$ and $g+2u+3l\in g+2u+3{\mathbb {N}}_0$ for all $l\geqq 0$. It follows that $2g-2u\not \in C$ and $2g-2u+3l'\in C$ for all $l'>0$. Since we have the inequalities

  $$\begin{aligned}&2g-2u-(g+2u-3)=g-(4u-3)>0 \text{ and } \\&2g-2u-(3c(H)-3)=2\left( g-\left( u+\frac{3}{2}(c(H)-1)\right) \right. \\&\quad \geqq 2\left( 2u+2c(H)+m-3-u-\frac{3}{2}(c(H)-1)\right) \\&\quad =2\left( u+\frac{1}{2}c(H)+m-\frac{3}{2}\right) >0, \end{aligned}$$

  we obtain $c(\tilde{H})=2g-(2u-1)$. Hence, the maximal element in $S(\tilde{H})$ is equal to

  $$\begin{aligned} c(\tilde{H})-1+3m= 2g-(2u-1)-1+3m=2g-2u+3m. \end{aligned}$$

• (iii) We obtain

  $$\begin{aligned} S(\tilde{H})= & {} \{3m,3s_1,\ldots ,3s_{m-1}\}\cup \{g+2u+3l\mid l=0,1,\ldots ,m-1\}\\&\cup \{2g-2u-3(s_i-m)\mid i=1,\ldots ,m-1\}\cup \{2g-2u+3m\}. \end{aligned}$$

  Moreover, we have

  $$\begin{aligned} 3s_i+2g-2u-3(s_i-m)=2g-2u+3m \end{aligned}$$

  for all $i=1,\ldots ,m-1$. By Lemma 2 we get the inclusion in (iii).

$\square$

Proposition 2

Let u, H, g and $\tilde{H}$ be as in Lemma 3. Moreover, assume that $m\leqq 2u-1$. Then we get $t(\tilde{H})=m+1$. If $m=2u-1$, then $\tilde{H}$ is an almost symmetric numerical semigroup with $c(\tilde{H})=2g(\tilde{H})-(2u-1)$.

Proof

For $l\geqq 0$ we have

$$\begin{aligned} (g+2u+3l)+3s_i\geqq g+2u+3s_i\geqq g+2u+3(m+1)>g+2u+3(m-1). \end{aligned}$$

Moreover, we obtain

$$\begin{aligned}&((g+2u+3l)+(2g-2u-3(s_i-m)))-3s_{max}\geqq 3g-3s_i+3m-3s_{max}\\&\quad \geqq 3(2u+2c(H)+m-3)-3s_i+3m-3s_{max}\\&\quad \geqq 3(2u+2c(H)-1) -3s_i+3m-3s_{max}\\&\quad =3(2u+c(H)-s_{max}+m-1+c(H)-s_i)=3(2u+c(H)-s_i)\\&\quad =3(2u+s_{max}-m+1-s_i)\\&\quad \geqq 3(2u-m+1)\geqq 3(2u-(2u-1)+1)= 6>0, \end{aligned}$$

because $m\leqq 2u-1$.

Lastly, for $l\geqq 0$ and $l'\geqq 0$ we get

$$\begin{aligned}&(g+2u+3l)+(g+2u+3l')-(2g-2u+3m)\\&\quad \geqq 2g+4u-(2g-2u+3m)=3(2u-m)\geqq 3>0 \end{aligned}$$

by the assumption $m\leqq 2u-1$. By Lemmas 2 and 3(iii) we obtain

$$\begin{aligned} PF(\tilde{H})= \{g+2u+3l-3m\mid l=0,1,\ldots ,m-1\}\cup \{2g-2u+3m-3m\}. \end{aligned}$$

Thus, we get $t(\tilde{H})=m+1$.

If $m=2u-1$, then it follows from Lemma 3(ii) that

$$\begin{aligned}&c(\tilde{H})+t(\tilde{H})=2g(\tilde{H})-(2u-1)+m+1\\&\quad =2g(\tilde{H})-(2u-1)+(2u-1)+1=2g(\tilde{H})+1, \end{aligned}$$

which implies that $\tilde{H}$ is almost symmetric. $\square$

The following result is useful for getting non-Weierstrass numerical semigroups:

Remark 3

Let H be a numerical semigroup with $g(H)\geqq 15g(d_3(H))+11$. If H is Weierstrass, then so is $d_3(H)$.

Proof

By Corollaries 2.6 and 2.7 in [14] we can apply to H the proof of Theorem 5.1 in [11]. Hence we can see that $d_3(H)$ is Weierstrass. $\square$

We find almost symmetric non-Weierstrass numerical semigroups $\tilde{H}$ with an odd conductor $c(\tilde{H})=2g(\tilde{H})-(2u-1)$ for any positive integer u using Proposition 2.

Theorem 3

Let $u\geqq 7$. Then there exists an almost symmetric non-Weierstrass numerical semigroup $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-(2u-1)$.

Proof

Let m be an odd number $\geqq 13$. We set $m=2u-1$, which implies that $u=\dfrac{m+1}{2}\geqq 7$. Then there exists a non-Weierstrass m-semigroup H (for example, see [6]). Let g be an integer as in Lemma 3. Moreover, we assume that $g\geqq 15g(H)+11$. We set

$$\begin{aligned} \tilde{H}=3H\cup (g+2u+3{\mathbb {N}}_0)\cup \{2g-2u-3r\mid r \in {\mathbb {Z}}\backslash H\}. \end{aligned}$$

By Remark 3 the numerical semigroup $\tilde{H}$ is non-Weierstrass, because so is $H=d_3(\tilde{H})$. Moreover, it follows from Proposition 2 that $\tilde{H}$ is an almost symmetric numerical semigroup with $c(\tilde{H})=2g(\tilde{H})-(2u-1)$. $\square$

Theorem 4

Let $u=3,5$ or 6. Then there exists an almost symmetric non-Weierstrass numerical semigroup $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-(2u-1)$.

Proof

We set $m=2u+2$. Let H, g and $\tilde{H}$ be as in Lemma 3. By Lemma 3(iii) we have

$$\begin{aligned} PF(\tilde{H})\subseteqq \{g+2u+3l-3m\mid l=0,1,\ldots ,m-1\}\cup \{2g-2u+3m-3m\}. \end{aligned}$$

We obtain

$$\begin{aligned}&(g+2u)+(g+2u+3\cdot 2)=(g+2u+3\cdot 1)+(g+2u+3\cdot 1)\\&\quad =2g-2u+3(2u+2)=2g-2u+3m \end{aligned}$$

from $m=2u+2$. Hence, by Lemma 2 we get

$$\begin{aligned} PF(\tilde{H})\subseteqq \{g+2u+3l-3m\mid l=3,\ldots ,m-1\}\cup \{2g-2u+3m-3m\}. \end{aligned}$$

We have

$$\begin{aligned} (g+2u+3l)+3s_i>g+2u+3(m-1) \end{aligned}$$

for $l\geqq 0$. Moreover, for $l\geqq 3$ we obtain

$$\begin{aligned} \begin{aligned}&((g+2u+3l)+(2g-2u-3(s_i-m)))-3s_{max} \geqq 3g+9+3m-3(s_i+s_{max})\\&\quad \geqq 3(2u+2c(H)+m-3)+9+3m-3(s_i+s_{max})\\&\quad =3(2u-m+1)+9-9-3+6c(H)+9m-3(s_i+s_{max})\\&\quad =3(2u-m+1)+9+6(c(H)-1+m)-3(s_i+s_{max})-3+3m-3\\&\quad \geqq 3(2u-m+1)+9+3(m-2)\geqq 3(m-2-m+1)+9=6>0 \end{aligned}\end{aligned}$$

from $m=2u+2$. Lastly, for $l\geqq 3$ and $l'\geqq 0$ we get

$$\begin{aligned}&(g+2u+3l)+(g+2u+3l')-(2g-2u+3m)\\&\quad \geqq 2g+4u-(2g-2u+3m)+9=3(2u-m)+9\\&\quad =3(m-2-m)+9= 3>0 \end{aligned}$$

from $m=2u+2$. Thus, we obtain

$$\begin{aligned} PF(\tilde{H})= \{g+2u+3l-3m\mid l=3,\ldots ,m-1\}\cup \{2g-2u+3m-3m\}, \end{aligned}$$

which implies that $t(\tilde{H})=m-2=2u$. Since we have $c(\tilde{H})=2g(\tilde{H})-(2u-1)$ from Lemma 3(ii), the numerical semigroup $\tilde{H}$ is almost symmetric.

For $m=8,12,14$, i.e., $u=3,5,6$, respectively there exists a non-Weierstrass m-semigroup from [6, 7]. Assume that $g\geqq 15g(H)+11$. By Remark 3 the numerical semigroup $\tilde{H}$ is non-Weierstrass. $\square$

Theorem 5

There exists an almost symmetric non-Weierstrass numerical semigroup $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-7$.

Proof

Let $u=4$ and $m=2u=8$. Let $H=2\langle 4,6,9,11\rangle +\langle n,n+4\rangle$ where n is an odd integer with $n\geqq 45$. Let g and $\tilde{H}$ be as in Lemma 3. By Lemma 3(iii) we have

$$\begin{aligned} \begin{aligned} PH(\tilde{H})\subseteqq \{g+8+3l-24\mid l=0,1,\ldots , 7\}\cup \{2g-8+24-24\}. \end{aligned} \end{aligned}$$

It follows from $m=2u=8$ that $g+8+g+8=2g-8+24$. Hence, we get

$$\begin{aligned} \begin{aligned} PH(\tilde{H})\subseteqq \{g+8+3l-24\mid l=1,\ldots , 7\}\cup \{2g-8+24-24\}. \end{aligned}\end{aligned}$$

We obtain

$$\begin{aligned}&(g+2u)+(2g-2u-3(s_i-m))-3s_{max}\\&\quad \geqq 3(2u+2c(H)+5)-3s_i+3m-3s_{max}> 6u+6c(H)+3m-6s_{max}\\&\quad =6u+6c(H)+3m-6(c(H)+m-1)=6u-3m+6=24-24+6>0 \end{aligned}$$

and

$$\begin{aligned} (g+2u+3l)+3s_i>g+2u+3(m-1). \end{aligned}$$

Moreover, for any $l\geqq 1$ and $l'\geqq 0$ we have

$$\begin{aligned}&(g+2u+3l)+(g+2u+3l')-(2g-2u+3m)\\&\quad \geqq 2g+4u-(2g-2u+3m)+3=3(2u-m)+3= 3>0 \end{aligned}$$

from $m=2u$.

Thus, we obtain

$$\begin{aligned} \begin{aligned} PH(\tilde{H})=\{g+8+3l-24\mid l=1,\ldots , 7\}\cup \{2g-8+24-24\}, \end{aligned} \end{aligned}$$

which implies that $t(\tilde{H})=8$. By Lemma 3 we have $c(\tilde{H})=2g(\tilde{H})-7$. Hence, $\tilde{H}$ is almost symmetric. By [7, 9, 13] the numerical semigroup H is non-Weierstrass. Assum that $g\geqq 15g(H)+11$. Since $d_3(\tilde{H})=H$, it follows from Remark 3 that the numerical semigroup $\tilde{H}$ is non-Weierstrass. $\square$

Theorem 6

There exists an almost symmetric non-Weierstrass numerical semigroup $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-3$.

Proof

Let $u=2$ and $m=2u+4=8$. Let $H=2\langle 4,6,9,11\rangle +\langle n,n+4\rangle$ where n is an odd integer with $n\geqq 45$. Let g and $\tilde{H}$ be as in Lemma 3. By Lemma 3 we have

$$\begin{aligned} \begin{aligned} PH(\tilde{H})\subseteqq \{g+4+3l-24\mid l=0,1,\ldots , 7\}\cup \{2g-4+24-24\}. \end{aligned}\end{aligned}$$

We get

$$\begin{aligned} (g+2u)+(g+2u+3\cdot 4)= & {} (g+2u+3\cdot 1)+(g+2u+3\cdot 3)\\= & {} (g+2u+3\cdot 2)+(g+2u+3\cdot 2)\\= & {} 2g+4u+12=2g+8+12=2g-4+24, \end{aligned}$$

which implies that

$$\begin{aligned} PH(\tilde{H})\subseteqq \{g+4+3l-24\mid l=5,6,7\}\cup \{2g-4+24-24\}. \end{aligned}$$

We obtain

$$\begin{aligned}&(g+2u)+(2g-2u-3(s_i-m))-3s_{max}\\&\quad \geqq 3(2u+2c(H)+5)-3s_i+3m-3s_{max}\\&\quad =3(2u+2s_{max}-2m+2+5)-3s_i+3m-3s_{max}\\&\quad> 6u-3m+21=12-24+21=9>0 \end{aligned}$$

and

$$\begin{aligned} (g+2u+3l)+3s_i>g+2u+3(m-1). \end{aligned}$$

Moreover, for any $l\geqq 5$ and $l'\geqq 0$ we have

$$\begin{aligned}&(g+2u+3l)+(g+2u+3l')-(2g-2u+3m)\\&\quad \geqq 2g+4u-(2g-2u+3m)+15\\&\quad =3(2u-m)+15=3(2u-2u-4)+15 =3>0 \end{aligned}$$

from $m=2u+4.$ Hence, we obtain

$$\begin{aligned} PH(\tilde{H})=\{g+4+3l-24\mid l=5,6,7\}\cup \{2g-4+24-24\}, \end{aligned}$$

which implies that $t(\tilde{H})=4$. By Lemma 3 we have $c(\tilde{H})=2g(\tilde{H})-3$. Hence, $\tilde{H}$ is almost symmetric. Assume that $g\geqq 15g(H)+11$. By the same way as the proof of Theorem 5 the numerical semigroup $\tilde{H}$ is non-Weierstrass. $\square$

By Theorem 5.1 in [11] we obtain the following:

Remark 4

There exists a quasi-symmetric non-Weierstrass numerical semigroup $\tilde{H}$, i.e., $c(\tilde{H})=2g(\tilde{H})-1$.

By Theorems  3, 4, 5, 6 and Remark 4 we obtain the following:

Theorem 7

For any positive integer u there exists an almost symmetric non-Weierstrass numerical semigroup $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-(2u-1)$.

By Theorems 2, 7 and [11] we obtain the following:

Corollary 2

For any $u\geqq 1$ we have infinite sequences

$$\begin{aligned} \tilde{H}=H_0 \xleftarrow {d_3} H_1 \xleftarrow {d_3} H_2 \xleftarrow {d_3} \cdots \xleftarrow {d_3} H_{i-1} \xleftarrow {d_3} H_i \xleftarrow {d_3} \cdots \end{aligned}$$

of almost symmetric non-Weierstrass numerical semigroups whose conductors are genera double minus $2u-1$.

5 Non-Weierstrass numerical semigroups with even conductor

First, we give non-Weierstrass numerical semigroups $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-2v$ where v is an odd number in the following Proposition:

Proposition 3

Let m and l be integers larger than 1. We set

$$\begin{aligned}&\quad H=\langle 2m,2m+2\cdot 1,\ldots ,2m+2\cdot (m-1),2lm+2 \cdot 0+1,\\&\quad \ldots ,2lm+2\cdot (m-1)+1\rangle . \end{aligned}$$

For any odd integer n with $n\geqq 2c(H)+2m-1$ we set $\tilde{H}=2H+\langle n,n+4\rangle$.

• (i) We have $g(H)=m(l+1)-1$, $c(H)=2g(H)-2(m-1)$ and $t(H)=2m-1$, hence H is almost symmetric.

• (ii) We have $g(\tilde{H})=2g(H)+\dfrac{n-1}{2}-1$, $c(\tilde{H})=2g(\tilde{H})-2(2m-3)$ and $t(\tilde{H})=2m-1$.

• (iii) If $n\geqq 8g(H)+11$, then $\tilde{H}$ is non-Weierstrass.

Proof

• (i) We have

  $$\begin{aligned} g(H)=m-1+lm=m(l+1)-1 \end{aligned}$$

  and

  $$\begin{aligned} c(H)=2lm+2\cdot (m-1)+1-2m+1=2lm=2g(H)-2(m-1). \end{aligned}$$

  Moreover, it follows from Lemma 2 that $t(H)=2m-1$.

• (ii) Since $2=2m+2\cdot 1-2m$, the genus $g(\tilde{H})$ of $\tilde{H}$ equals to $2g(H)+\dfrac{n-1}{2}-1$. We have

  $$\begin{aligned} c(\tilde{H})= & {} n+2(2lm+2\cdot (m-1)+1)-4m+1=n+4lm-1\\= & {} 2\left( 2(m(l+1)-1)+\frac{n-1}{2}-1\right) -2(2m-3)=2g(\tilde{H})-2(2m-3). \end{aligned}$$

  Moreover, we have $t(\tilde{H})=2m-1$. As a matter of fact, we obtain that

  $$\begin{aligned} (n+4)+2(2lm+1)=n+2(2lm+2\cdot 1+1)\in S(\tilde{H}), \end{aligned}$$

  which implies that $n+4-4m\not \in PF(\tilde{H})$. Hence, the set $PF(\tilde{H})$ is equal to

  $$\begin{aligned}&\{2(2m+2)-4m,n+2s_1-4m,n+2s_3-4m,n+2s_4-4m,\\&\quad \ldots ,n+2s_{2m-1}-4m\} \end{aligned}$$

  where $S(H)=\{2m,s_1,s_2,s_3,\ldots ,s_{2m-1}\}$ is the standard basis for H, i.e., $s_{2i}=2m+2i$ for $i=1,\ldots ,m-1$ and $s_{2i+1}=2lm+2i+1$ for $i=0,1,\ldots ,m-1$.

• (iii) We will show that the numerical semigroup $\tilde{H}$ is non-Weierstrass. Assume that $\tilde{H}$ is Weierstrass. Since we have

  $$\begin{aligned} g(\tilde{H})=2g(H)+\dfrac{n-1}{2}-1\geqq 2g(H) +\frac{8g(H)+11-1}{2}-1=6g(H)+4, \end{aligned}$$

  it follows from [13] that there is a double covering $\pi :{\tilde{C}}\longrightarrow C$ with a ramification point $\tilde{P}$ over P such that $H(\tilde{P})=\tilde{H}$ and $H(P)=H$. Moreover, there is a divisor $D=\dfrac{n+1}{2}P-Q$ with a point Q of C distinct form P such that $h^0(2P+Q)=2$. Thus, C is a trigonal curve with the point P with

  $$\begin{aligned} H(P)= & {} \langle 2m,2m+2\cdot 1,\ldots ,2m+2\cdot (m-1),2lm\\&+2 \cdot 0+1,\ldots ,2lm+2\cdot (m-1)+1\rangle . \end{aligned}$$

  By Theorem on the page 172 in [4] we must have

  $$\begin{aligned} {\mathbb {N}}_0\backslash H= & {} \{1,2,\ldots ,2m-1,2m+1,2(m+1) +1,\ldots ,2(ml-1)+1\}\\= & {} \{1,2,\ldots ,2n+1,2(n+1)+1,\ldots ,2(g-n-1)+1\} \end{aligned}$$

  with $\dfrac{g-1}{3}\leqq n\leqq \dfrac{g}{2}$ where we set $m=n+1$. Here, we note that $g=g(H)=m(l+1)-1$ and $g-n-1=ml-1$. We must have

  $$\begin{aligned} m-1=n\geqq \dfrac{g-1}{3}=\dfrac{m(l+1)-1-1}{3}, \end{aligned}$$

  which implies that $l\leqq 2-\dfrac{1}{m}<2$. This contradicts $l\geqq 2$. Thus, $\tilde{H}$ should be non-Weierstrass.

$\square$

We prove the following lemma which is used at the construction of non-Weierstrass numerical semigroups $\tilde{H}$ with $c(\tilde{H})=2g(\tilde{H})-2v$ where v is an even number:

Lemma 4

Let v be a positive integer. Let H be a numerical semigroup of type t with multiplicity m and $c(H)=2g(H)-2v$. We take a subset $\{f_1,\ldots ,f_q\}$ of $PF^*(H)=PF(H)\backslash \{c(H)-1\}$ with $0\leqq q\leqq t-1$. For an odd integer n with $n\geqq 2c(H)+2m-1$ we set

$$\begin{aligned} \tilde{H}=2H+\langle n,n+2f_1,\ldots ,n+2f_q\rangle . \end{aligned}$$

Then we have $c(\tilde{H})=2g(\tilde{H})-2(v+v-q)$ and $t(\tilde{H})\geqq t-q$.

Proof

Since we have $PF^*(H)\supseteq \{f_1,\ldots ,f_q\}$, we get

$$\begin{aligned} g(\tilde{H})=2g(H)+\frac{n-1}{2}-q \end{aligned}$$

and

$$\begin{aligned} c(\tilde{H})= & {} n+2s_{max}-2m+1=2(s_{max}-m+1)+n-1=2c(H)+n-1\\= & {} 2(2g(H)-2v)+n-1=2\left( 2g(H)+\frac{n-1}{2}-q\right) \\&-4v+2q=2g(\tilde{H})-2(v+v-q). \end{aligned}$$

We set

$$\begin{aligned} \begin{aligned} S(H)=\{m,s^{(1)},\ldots ,s^{(q)},\ldots ,s^{(t-1)},s^{(t)}, \ldots ,s^{(m-1)}=s_{max}\} \end{aligned} \end{aligned}$$

such that $f_i=s^{(i)}-m$ for any i with $1\leqq i\leqq q$ and

$$\begin{aligned} PF^*(H)=\{f_1,\ldots ,f_q,s^{(q+1)}-m,\ldots ,s^{(t-1)}-m\}. \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} S(\tilde{H})=&{} \{2m,2s^{(1)},\ldots ,2s^{(q)},\ldots ,2s^{(m-1)}\} \cup \{n,n+2f_1,\ldots ,n+2f_{q}\}\\&\cup \{n+2s^{(q+1)},\ldots ,n+2s^{(t-1)},n+2s^{(t)},\ldots ,n\\&+2s^{(m-2)},n+2s^{(m-1)}=n+2s_{max}\}. \end{aligned}\end{aligned}$$

It follows from the assumption $n\geqq 2c(H)+2m-1$ that

$$\begin{aligned} n+(n+2s^{(j)})> 2c(H)+2m-2+n+2s^{(j)} =2s_{max}+n+2s^{(j)}>n+2s_{max}. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} PF(\tilde{H})\supseteqq \{n+2s^{(q+1)},\ldots ,n+2s^{(t-1)},n+2s_{max}\}. \end{aligned}$$

Thus, we get $t(\tilde{H})\geqq t-q$. $\square$

Combining Proposition 3 with Lemma 4 we can find non-Weierstrass numerical semigroups $\tilde{H}$ with an even conductor $c(\tilde{H})=2g(\tilde{H})-2u$ for any positive integer u.

Theorem 8

Let v be a positive integer. Then there exists an infinite number of non-Weierstrass numerical semigroups H with $c(H)=2g(H)-2v$.

Proof

By Proposition 3 for any odd number $v\geqq 1$ we get an infinite number of non-Weierstrass numerical semigroups H with $c(H)=2g(H)-2v$ and $t(H)=v+2$ because l in Proposition 3 is any integer larger than 1. We set $q=v-1$ in Lemma 4. Let $\tilde{H}$ be the numerical semigroup as in Lemma 4. Then it follows from Lemma 4 that

$$\begin{aligned} c(\tilde{H})=2g(\tilde{H})-2(v+v-(v-1))=2g(\tilde{H})-2(v+1). \end{aligned}$$

If $n\geqq 8g(H_0)+2v+1$, then we have

$$\begin{aligned} g(\tilde{H})= & {} 2g(H_0)+\frac{n-1}{2}-q\geqq 2g(H_0) +\frac{8g(H_0)+2v+1-1}{2}-q\\= & {} 6g(H_0)+v-q=6g(H_0)+1, \end{aligned}$$

which implies that $\tilde{H}$ is a non-Weierstrass numerical semigroup with $c(\tilde{H})=2g(\tilde{H})-2(v+1)$ by Remark 2, because $H=d_2(\tilde{H})$ is a non-Weierstrass numerical semigroup. Since v is any odd number larger than or equal to 1, any positive even numbers can be obtained by $v+1$. $\square$