数理解析研究所講究録
第 1915 巻 2014 年 53-57

53The proportion of numerical semigroups with no descendantor an infinite number of descendants 1
神奈川工科大学・基礎・教養教育センター

米田二良

Jiryo Komeda
Center for Basic Education and Integrated Learning
Kanagawa Institute of Technology

Abstract

Let be the map between the sets of numerical semigroups sending a numerical semigroup to the one whose genus is decreased by 1. We give many examples of numericalsemigroups $H$ with. We investigate the density of some kinds of numerical$H$$p^{-1}(H)=\emptyset$semigroups
within the whole set of numerical semigroups. Moreover,
we determine the numerical semigroups $H$ withfor any$p$$p^{-1}(H)=\emptyset$$p^{-n}(H)\neq\emptyset$

1$n.$

The conductor and descendants

Letis called
be the additive monoid of non-negative integers. A submonoid $H$ of
a numerical semigroup if the complement
isis finite. The cardinality ofcalled the genus of $H$ , denoted by $g(H)$ . In this section $H$ stands for a numericalsemigroup of genus . We set$\mathbb{N}_{0}$$\mathbb{N}_{0}$$\mathbb{N}_{0}\backslash H$$\mathbb{N}_{0}\backslash H$$g$$m(H)= \min\{h\in H|h>0\},$which is called the multiplicity of $H$ . We set$c(H)= \min\{c\in \mathbb{N}_{0}|c+\mathbb{N}_{0}\subseteqq H\},$

which is called the conductor of $H$ . Then we have $g+1\leqq c(H)\leqq 2g$ . We note that$c(H)-1\not\in H$ . We set $p(H)=H\cup\{c(H)-1\}$ , which is a numerical semigroup ofgenus $g-1$ . The numerical semigroup $p(H)$ is called the parent of $H$ . The numericalsemigroup $H$ is called a child of $p(H)$ . Let $M(H)$ be the minimal set of generators
for $H$ . For $\mu\in M(H)$ with $\mu>f(H)$ , which is called an eﬀective generator of $H$ , we
, which is a child of $H$ , and vice versa. A numerical semigroup $H’$ isset
called a descendant of $H$ if there exists $i\geqq 1$ such that $p^{i}(H’)=H$ . A child of $H$ isa descendant of $H$ . In this paper we are interested in numerical semigroups $H$ whichhave either no descendant, i.e., no child or an infinite number of descendants.
$H_{\mu}=H\backslash \{\mu\}$

Proposition 1.1. Suppose that $c(H)=g+1$ . Then we have $H=\langle g+1arrow 2g+1\rangle,$which has an infinite number of descendants. In fact, for any $i\geqq 1$ we have$p^{i}(\langle g+1+iarrow 2g+1+i\rangle)=H.$

1This paper is an extended abstract and the details will appear elsewhere.

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Proposition 1.2. Suppose that $c(H)=g+2$ . Then $H$ has an infinite number ofdescendants.

Proof. We have $c(H)-1-g=1$ . Since we have
10 in [1]$H$

$g.c.m.(\lambda_{0}, \lambda_{1})=\lambda_{1}>1$, by Theorem

has an infinite number of descendants.

$\square$

We set $\alpha_{i}=l_{i}-i$ for $i=1$ , . . . , where
the Schubert index of $H$ . Then we have$\mathbb{N}_{0}\backslash H=\{l_{1}< .

$g$$\alpha(H)=(\alpha_{1}, \ldots, \alpha_{g})$Proposition 1.3. Assume that $c(H)=2g$ . If

Proof. Assume that$c(H)=2g$ , we have

$H$.

.

<l_{g}\}$. We call.

$\alpha(p(H))=(\alpha_{1}, \ldots, \alpha_{g-1})$$H\neq\langle 2,$ $2g+1\rangle$, then $H$ has no child.

, i.e., $p(\tilde{H})=H$ . Since $H$ is symmetric, i.e.,. Hence we get $\alpha_{g+1}=g-1$ oris symmetric. Since $2g-1\not\in\tilde{H}$ , we have $\tilde{H}\ni 2(g+$has a child$\tilde{H}$$\alpha(\tilde{H})=(\alpha_{1}, \ldots, \alpha_{g-1}, g-1, \alpha_{g+1})$

Case 1: $\alpha_{g+1}=g$ . Then1) $-1-(2g-1)=2$ , which implies that$\tilde{H}$$\tilde{H}=\langle 2,$$g.$$2(g+1)+1\rangle$. Hence, we get$H=p(\tilde{H})=\langle 2,$ $2g+1\rangle.$is quasi-symmetric. Since $2g-1\not\in\tilde{H}$ , we have$\tilde{H}\ni 2(g+1)-2-(2g-1)=1$ , which implies that

Case 2: $\alpha_{g+1}=g-1$ . Then$\tilde{H}$$\tilde{H}=N_{0}.$$\square$Proposition 1.4. Assume that $c(H)=2g-1$ . If $H$ is diﬀerent from 3, $g+2,$
mod 3 and $\langle garrow 2g-3,$ $2g-1\rangle$ , then $H$ has no child.
with$\langle$$2g+1\rangle$$g\not\equiv 1$

Proof. For the proof see Theorem 3.9 in [4].

口The proportion of certain kinds of numerical semi-2groupsLet$\epsilon$

be a fixed positive number. Let$\gamma=\frac{5+\sqrt{5}}{10}=\frac{\phi}{\sqrt{5}}$where$\phi$is the golden ratio.

For a non-negative integer let $NS(g)$ be the set of numerical semigroups of genus
We set $\Phi S_{\epsilon}(g)=\{H\in NS(g)|(\gamma-\epsilon)g<m(H)<(\gamma+\epsilon)g\}.$$g$

Remark 2.1. ([3])) We have$g.$$\lim_{garrow\infty}\frac{\#\Phi S_{\epsilon}(g)}{\# NS(g)}=1.$

For any positive integer $n\geqq 2$ we set$L_{n}(H)=\{l_{1}+\cdots+l_{n}|l_{i}\in \mathbb{N}_{0}\backslash H$, all$i\}.$and $m\geqq 420$ . Assume that $m=m(H)$ and$(2-\epsilon)m<c(H)-1<(2+\epsilon)m$ . If $\# L_{n}(H)\geqq(2n-1)(g-1)-19$ with some $n\geqq 2,$
then we have $g<1.38175m.$

Key Lemma 2.2. Let$0< \epsilon<\frac{1}{21}$For the proof see [5].
Theorem 2.3. We set$BS(-19,9)=\{H\in NS(g)|\# L_{n}(H)\geqq(2n-1)(g-1)-19$ for some $n\geqq 2\}.$

Then we obtain$\lim_{garrow\infty}\frac{\# BS(-19,g)}{\# NS(g)}=0.$

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For the proof see [5].
Remark 2.4. ([7])) Assume that $c(H)\neq 2g$ . Then we have$L_{2}(H)\supseteqq\{2$, 3, 4, 5, . . . , $2g\}.$Using Remark 2.4 we get the following:
Key Lemma 2.5. Assume that $c(H)=2g-i$ with $1\leqq i\leqq g-1$ . Then we have$\# L_{2}(H)\geqq 3g-3-(i-1)$ .

For the proof see [5].

Main Theorem 2.6. We set$CS(20, g)=\{H\in NS(g)|2g-20\leqq c(H)\}.$Then we obtain$\lim_{garrow\infty}\frac{\# CS(20_{)}g)}{\# NS(g)}=0.$

For the proof see [5].
Corollary 2.7. We have $\lim_{garrow\infty}\frac{\#\{H\in NS(g)|c(H)=2g\}}{\# NS(g)}=0.$
Corollary 2.8. We have $\lim_{garrow\infty}\frac{\#\{H\in NS(g)|c(H)=2g-1\}}{\# NS(g)}=0.$

Problem 1. Assume that $c(H)\leqq 2g-21$ . What kind of numerical semigroup $H$ hasa child?
Problem 2.
$\lim_{garrow\infty}\frac{\#\{H\in NS(g)|Hhasnochild\}}{\# NS(g)}=0$3?

Numerical semigroups with an infinite number ofdescendants

We are interested in numerical semigroups which have infinite numbers of descendants.
Such a numerical semigroup is said to be $IND$. We set $d_{2}(H)=\{h’\in \mathbb{N}_{0}|2h’\in H\}$
which is also a numerical semigroup. $n(H)$ stands for the minimum odd number in $H.$}

Theorem 3.1. Assume that $n(H)\geqq 2c(d_{2}(H))+1$ . Then the following are equivalent:i) $H$ is $IND.$ii) $H=2d_{2}(H)+\langle n,$ $n+2$ , . . . , $n+2(m’-1)\rangle$ where $n=n(H)$ and $m’=m(d_{2}(H))$ .

For the proof see [6].
Example 3.1. Let $t\geqq 1$ . We set $H=2\langle 2,$ $2t+1\rangle+\langle 4t+1,$ $4t+3\rangle$ . Then we have$n(H)=4t+1,$ $d_{2}(H)=\langle 2,$
and $c(d_{2}(H))=2t$ . Hence, $H$ is IND. In fact, when$2t+1\rangle+\langle 4t+1+2i,$ $4t+3+2i\rangle$ , we obtain $p^{i}(H_{i})=H$ for $i\geqq 1.$
we set$2t+1\rangle$$H_{i}=2\langle 2,$

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Theorem 3.2. Let $H$ be a numerical semigroup and $m’=m(d_{2}(H))$ . For an odd
number we set $H=2d_{2}(H)+\langle n,$ $n+2$ , . . . , $n+2(m’-1$i) If $n\geqq 2c(d_{2}(H))+1$ , then $H$ is $IND.$ii) If $g(d_{2}(H))\geqq 1$ and $n=2c(d_{2}(H))-1$ , then $H$ is $IND.$iii) If $n=n(H)$ and $n\leqq 2c(d_{2}(H))-5$ , then $H$ is not $IND.$$n$

For the proof see [6].
Theorem 3.3. Let $H$ be a numerical semigroup, $m’=m(d_{2}(H))$ , $g’=g(d_{2}(H))\geqq 2$$andd=c(d_{2}(H))$ . We $setH=2d_{2}(H)+\langle 2d-3,$ $2d-3+2$ , . . . , $2d-3+2(m’-1$i) If $d_{2}(H)$ is not $IND$, then neither is $H.$ii) Assume that $d_{2}(H)$ is $IND$ . Then $H$ is $IND$ if and only if we have$(\lambda_{0}’, \lambda_{1}’, \ldots, \lambda_{d-1-g’}’, 2c’-3)>1$

where $d_{2}(H)=\{\lambda_{0}’<\lambda_{1}’<\cdots<\lambda_{c-1-g}’, <\cdots\}.$For the proof see [6].

Theorem 3.4. Assume that $n(H)\leqq 2d-1$ where $d=c(d_{2}(H))$ . If $H$ is $IND$, thenthere exists
such that $p^{i}(H)$ is one of the following:$d_{2}(p^{i}(H))+\langle2c^{(i)}-1,$ $2c^{(i)}+1$ , . . . , $2c^{(i)}+2m^{(i)}-3\rangle$ where $c^{(i)}=c(d_{2}(p^{i}(H)))$ andi) 2$i\geqq 0$$m^{(i)}=m(d_{2}(p^{i}(H)))$ii) 2 $d_{2}(p^{i}(H))+\langle 2c^{(i)}-3,$ $2c^{(i)}-1$ , . . . , $2c^{(i)}+2m^{(i)}-5\rangle$ with$(\lambda_{0}^{(i)},$$\lambda_{1}^{(i)}$,...,$\lambda_{c^{(i)}-1-g^{(:\rangle}}^{(i)},$$2c^{(i)}-$$3)>1$ where $g^{(i)}=g(d_{2}(p^{i}(H)))$ and $d_{2}(p^{i}(H))=\{\lambda_{0}^{(i)}<\lambda_{1}^{(i)}<\cdots\}.$For the proof see [6].
Remark 3.5. The converse of Theorem 3.4 does not hold. In fact, let$H=\langle 10, 15, 17, 18, 21, 22, 23, 24, 26, 29\rangle.$

Then we have $c(H)=20,$ $g(H)=15$ and $c(H)-1-g(H)=4$ . It follows from $H=$$\{0<10<15<17<18<\cdots\}$ and $(0,10,15,17,18)=1$ that $H$ is not IND. Moreover,. Then we obtain 2 $c(d_{2}(H))-3=2\cross 9-3=15,$
we have $d_{2}(H)=\langle 5$ , 9, 11, 12,$m(d_{2}(H))=5,$ $2c(d_{2}(H))+2m(d_{2}(H))-5=23$ and$13\rangle$$p(H)=2\langle 5, 9, 11, 12, 13\rangle+\langle 15, 17, 19, 21, 23\rangle.$We note that $d_{2}(p(H))=\langle 5$ , 9, 11, 12,$13\rangle,$$c(d_{2}(H))-1-g(d_{2}(H))=9-1-7=1$ and$(0,5)=5>1.$

On the other hand we consider$H’=\langle 10, 15, 18, 19, 21, 22, 23, 24, 26, 27\rangle.$

Since $g(H’)=15$ and $c(H’)=18$ , we obtain $c(H’)-g(H’)-1=2$ . It follows from$(0,10,15)=5>1$ that $H’$ is IND. Moreover, we have $p(H)=p(H’)$ .

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