## Abstract

In this study, we characterize all families of saturated numerical semigroups with multiplicity four. We also present some results about invariants of these semigroups.

Show Summary Details# On the saturated numerical semigroups

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## Abstract

## 1 Introduction

## 2 Main results

## Acknowledgement

## References

## About the article

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In this study, we characterize all families of saturated numerical semigroups with multiplicity four. We also present some results about invariants of these semigroups.

KeyWords: Saturated numerical semigroup; Frobenius number; Gaps

MSC 2010: 20M14

Let ℕ = {1, 2, ..., *n*, ..} and ℤ be the set of integers. A subset *S* of the set ℕ of nonnegative integers is called a numerical semigroup if it satisfies the following conditions:

- (i)
0 ∈

*S*, - (ii)
*a, b*∈*S*⇒*a*+*b*∈*S*, - (iii)
ℕ \

*S*has a finite number of elements.

Condition (iii) is equivalent to *gcd*(*S*) = 1 (Here, *gcd*(*S*) is the greatest common divisor of the element of *S*).

All numerical semigroups are finitely generated, i.e.
$$S=\u3008{a}_{1},{a}_{2},...{a}_{r}\u3009\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\{\sum _{k=1}^{t}{c}_{i}{a}_{i},...,{c}_{r}\in \mathbb{N}\}$$
where *a*_{1}*, a*_{2}*, ... ,a _{r}* ∈

We define the following invariants of numerical semigroups: $$F(S)=max\{x\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}x\in \mathbb{Z}\mathrm{\setminus}S\}$$ and $$n(S)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left|\{0,1,2,...,F(S)\}\cap S\right|.$$

*F* ( *S* ) and *n* ( *S* ) are called the Frobenius number of *S* and the number determiner of *S*, respectively.

We can write
$$S=\u3008{a}_{1},{a}_{2},...,{a}_{r}\u3009=\u3008{s}_{0}=0,{s}_{1},{s}_{2},...,{s}_{n-1},{s}_{n}=F(S)+1,\to ...\u3009$$
where *s _{i}* <

The set ℕ \ *S* is the gap of *S*, and the set of gaps of *S* is denoted by *H*(*S*). *g*(*S*) = |*H*(*S*)| is called the genus of *S*. It is clear that *g*(*S*) = *F*(*S*) + 1 − *n*(*S*). An element *x* ∈ *H*(*S*) is called a fundamental gap of *S* if 2*x*, 3*x* ∈ *S*. The set of all the fundamental gaps of *S* is denoted by *FH*(*S*), i.e.
$$FH(S)=\{x\in H(S):2x,3x\in S\}.$$

An element *x* ∈ ℤ is called a Pseudo-Frobenius number of *S* if *x* ∉ *S* and *x* + *s* ∈ *S*, for *s* ∈ *S* \ {0}. We denote by *PF*(*S*) the set of all Pseudo-Frobenius numbers of *S*, i.e.
$$PF(S)=\{x\phantom{\rule{thinmathspace}{0ex}}\in \mathbb{Z}\mathrm{\setminus}S:x+s\in S,\phantom{\rule{thinmathspace}{0ex}}\text{for}\phantom{\rule{thinmathspace}{0ex}}\text{all}\phantom{\rule{thinmathspace}{0ex}}s\phantom{\rule{thinmathspace}{0ex}}\in \phantom{\rule{thinmathspace}{0ex}}S\mathrm{\setminus}\{0\}\}$$
(see [7]). Given a numerical semigroup *S* and *x* ∈ *S* \ {0}, we define the Apery set of *x* in *S* as *Ap*(*S, x*) = {*s* ∈ *S* : *s* − *x* ∉ *S*} (for details see [9]).

If a numerical semigroup *S* satisfies the condition *x* + *y* − *z* ∈ *S*, for every *x, y, z* ∈ *S* such that *x* ≥ *y* ≥ *z*, then *S* is called Arf. If *S* is an Arf numerical semigroup, then *S* has maximal embedding dimension.

The investigation of combinatorial properties of semigroups is very important, because they often occur in applications ([1, 3, 5]) and are related to automata theory (see [4]). A numerical semigroup *S* is saturated if *s* + *c*_{1}*s*_{1} + *c*_{2}*s*_{2} + ... + *c _{k}*

In this study, we show that all families of numerical semigroups with multiplicity four are saturated numerical semigroups; these are numerical semigroups of the form *S* = 〈4, *k, k* + 2, *k* + 3〉, for *k* = 3(*mod*4) and *k* ≥ 7 and *S* = 〈4, *k, k* + *t, k* + *t* + 2〉, for *k* = 2(*mod* 4) and *k* ≥ 6 and *t* an odd integer. We also give the formulae for *F*(*S*), *n*(*S*), *PF*(*S*), *g*(*S*), *H*(*S*) and *FH*(*S*) of these numerical semigroups.

In this section we provide some results for numerical semigroups with multiplicity four; i.e. numerical semigroups of the form *S* = 〈4, *k, k* + 2, *k* + 3〉 (for *k* ≡ 3(*mod* 4) and *k* ≥ 7) and *S* = 〈4, *k, k* + *t, k* + *t* + 2〉, (for *k* ≡ 2(*mod* 4) and *k* ≥ 6 and *t* an odd integer).

([8]).*Let S be a numerical semigroup, then the following conditions are equivalent*:

- (i)
*S is a saturated numerical semigroup*. - (ii)
*a*+*d*_{S}*(a)*∈*S for all a*∈*S*,*a*> 0*where d*_{S}*(a)*=*gcd*{*x*∈*S*:*x*≤*a*}. - (iii)
*a*+*kd*_{S}*(a)*∈*S for all a*∈*S*,*a*> 0*and k*∈ ℕ.

([10]). *If S* = 〈4,*k, k* + *1, k* + 2 〉 *, then S is a saturated numerical semigroup, for k* ≡ 1(*mod* 4) *and k* ≥ 5.

*Let S* = 〈4, *k, k* + 2, *k* + 3〉 *be numerical semigroup, where k* ≡ 3( *mod* 4) *and k* ≥ 7. *Then S is saturated*.

Let *S* = 〈4,*k, k* + 2, *k* + 3〉 be numerical semigroup, where *k* ≡ 3(*mod* 4) and *k* ≥ 7. We note that *k* = *4r* + *3*, *r* ≥ *1* and *r* ∈ ℤ. Thus, we have
$$\begin{array}{c}S=\u30084,k,k+2,k+3\u3009=\{0,4,8,\cdots ,k-7,k-3,k\to \cdots ,\}\\ =\{0,4,8,\cdots ,4r-4,4r,4r+3,\to \cdots \}.\end{array}$$

In this case,

- (a)
If

*a*< 4*r*+ 3, then*d*_{S}*(a)*=*1*. So, we find that*a*+*d*_{S}*(a)*∈*S*since*a*+*d*_{S}*(a)*=*a*+*1*≥*4r*+*4*∈*S*, for all*a*∈*S*,*a*> 0. - (b)
If

*a*≥ 4*r*+ 3, then*d*_{S}*(a)*= 4. So, we have*a*+*d*_{S}*(a)*=*a*+ 4 ∈*S*, for all*a*∈*S*,*a*> 0.

In view of Proposition 2.1, we find that *S* is saturated a numerical semigroup. □

*Let S* = 〈4,*k, k* + *t, k* + *t* + 2〉 *be numerical semigroup, where k* ≡ 2(*mod* 4), *k* ≥ 6, *and t is an odd integer. Then S is saturated*.

It is trivial that *gcd* {4,*k, k* + *t, k* + *t* + 2} = 1 since *k* is even and *t* is an odd integer. If we put *k* = *4r* + 2, *r* ≥ 1 and *r* ∈ ℤ, then we have
$$\begin{array}{c}S=\u30084,k,k+t,k+t+2\u3009=\{0,4,8,...,k-6,k-2,k,k+2,...,k+t-3,k+t-1,\to ...,\}\\ =\{0,4,8,...,4r-4,4r,4r+2,4r+4,...,4r+t-1,4r+t+1,\to ...,\}.\end{array}$$

In this case,

- (i)
If

*a*> 4*r*+*t*+ 1, then*d*_{S}*(a)*= 1. So, we obtain*a*+*d*_{S}*(a)*∈*S*from the inequality*a*+*d*_{S}*(a)*=*a*+ 1 ≥*4r*+*t*+ 2 ∈*S*, for all*a*∈*S*,*a*> 0. - (ii)
If 4

*r*≤*a*≤ 4*r*+*t*+ 1, then*d*_{S}*(a)*= 2. So, we obtain*a*+*d*_{S}*(a)*=*a*+ 2 ∈*S*from the inequality*4r*≤*a*≤ 4*r*+*t*+ 1, for all*a*∈*S*,*a*> 0. - (iii)
If

*a*< 4*r*, then*d*_{S}*(a)*= 4. So, we obtain*a*+*d*_{S}*(a)*=*a*+ 4 ∈*S*since*a*+ 4*r*< 4*r*+ 4, for all*a*∈*S*,*a*> 0.

In view of Proposition 2.1, we have that *S* is a saturated numerical semigroup. □

([6]).*Let S be a numerical semigroup minimally generated by* {*n*_{1} < *n*_{2} < ... < *n _{r}*}.

([6]).*Let S be a numerical semigroup minimally generated by* {*n*_{1} < *n*_{2} < ... < *n _{r}*}.

- (1)
*If S has maximal embedding dimension, then F*(*S*) =*n*−_{r}*n*_{1}. - (2)
*S has maximal embedding dimension if and only if*$$g(S)=\frac{{n}_{2}+{n}_{3}+...+{n}_{r}}{{n}_{1}}-\frac{{n}_{1}-1}{2}.$$

*If S* = 〈4, *k k* + 2, *k* + 3〉 *is a numerical semigroup, where k* ≡ 3(*mod* 4) *and k* ≡ 7. *Then we obtain following equalities*:

- (a)
*F*(*S*) =*k*− 1, - (b)
$g\left(S\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{3k\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}{4}$,

- (c)
*PF*(*S*) = {*k*− 4,*k*− 2,*k*− 1}, - (d)
$n\left(S\right)\phantom{\rule{thinmathspace}{0ex}}=\frac{k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}1}{4}$,

- (e)
*H*(*S*) = {1, 2, 3, 5, 6, 7, ... ,*k*− 6,*k*− 5,*5k*− 4,*k*− 2,*k*− 1}.

We have *Ap*(*S*, 4) = {0, *k*, *k* − 2, *k* − 3} since *S* has maximal embedding dimension. Thus,

- (a)
We have

*F*(*S*) = (*k*+ 3) − 4 =*k*− 1 from Corollary 2.6 (1). - (b)
We obtain $g\left(S\right)\phantom{\rule{thinmathspace}{0ex}}=\frac{k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}3}{4}\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\frac{4\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}{2}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{3k\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}{4}$ from Corollary 2.6 (2).

- (c)
It is obvious that

*PF*(*S*) = {*k*− 4,*k*+*2*− 4,*k*+ 3 − 4}. So we find*PF*(*S*) = {*k*− 4,*k*− 2,*k*− 1}. - (d)
We have $n\left(S\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left(k\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1\right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\frac{3k\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}{4}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\frac{k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}1}{4}$ from

*g*(*S*) =*F*(*S*) + 1 −*n*(*S*). - (e)
We find that

*H*(*S*) = {1, 2, 3, 5, 6, 7, ... , 4*r*− 3, 4*r*− 2, 4*r*− 1, 4*r*+ 1, 4*r*+*2*} = {1, 2, 3, 5, 6, 7, ... ,*k*− 6,*k*− 5,*k*− 4,*k*− 2,*k*− 1} from the equality*S*= < 4,*k*,*k*+ 2,*k*+ 3 > {0, 4, 8, ... ,*k*− 7,*k*− 3,*k*, → ... ,} = {0, 4, 8, ... , 4*r*− 4, 4*r*, 4*r*− 3, → ... ,}.

*Let S*= 〈4, *k*, *k* + *t, k* + *t* + *2*〉 *be a numerical semigroup, where k* ≡ 2(*mod* 4), *k* ≡ 6*, and t is an odd integer. Then, we have following equalities*:

- (a)
*F*(*S*) =*k*+*t*− 2, - (b)
$g\left(S\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{3k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2t\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}4}{4}$,

- (c)
*PF*(*S*) = {*k*− 4,*k*+*t*− 4,*k*+*t*− 2}, - (d)
$n\left(S\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2t}{4}$,

- (e)
*H*(*S*) = {1, 2, 3, 5, 6, 7, ... ,*k*− 5,*k*− 4,*k*− 3,*k*− 1,*k*− 1,*k*+ 3, ... ,*k*+*t*− 2}.

We have *Ap*(*S*, 4) = {0, *k, k* + *t, k* + *t* + 2} since *S* has maximal embedding dimension. Thus,

- (a)
We have

*F*(*S*) = (*k*+*t*+ 2) − 4 =*k*+*t*+ 2 from Corollary 2.6 (1). - (b)
We obtain $g\left(S\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2}{4}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{4\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}1}{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{3k\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}2t\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}4}{4}$ from Corollary 2.6 (2).

- (c)
It is obvious that

*PF*(*S*) = {*k*− 4,*k*+*t*− 4,*k*+*t*− 2 − 4}. So, we find*PF*(*S*) = {*k*− 4,*k*+*t*− 4,*k*+*t*− 2}. - (d)
We have $n(S)=(k+t-2)+1-\frac{3k+2t-4}{4}=\frac{k+2t}{4}$ from

*g*(*S*) =*F*(*S*) + 1 −*n*(*S*). - (e)
We observe that

*H*(*S*) = {1, 2, 3, 5, 6, 7, ... , 4*r*− 3, 4*r*− 2, 4*r*− 1, 4*r*− 1, 4*r*+ 3, 4*r*+ 5, ... , 4*r*+*t*} = {1, 2, 3, 5, 6, 7, ... ,*k*− 5,*k*− 4,*k*− 3,*k*− 1,*k*+ 1,*k*+ 3, ... ,*k*+*t*+ 2} since*S*= 〈4,*k, k*+*t, k*+*t*+ 2〉 = {0, 4, 8, ... ,*k*− 6,*k*− 2,*k, k*+ 2, ... ,*k*+*t*− 3,*k*+*t*−*1*, → ... ,} = {0, 4, 8, ... ,4*r*− 4, 4*r*, 4*r*+ 2, 4*r*+ 4, ... , 4*r*−*t*− 1, 4*r*+*t*+*1*, → ... ,}.

*Consider the numerical semigroup S*= 〈4, *k, k* + 2, *k* + 3〉. *If we put k* = 15 *then we have that S* = 〈4, *k, k* + 2, *k* + 3〉 = 〈4, 15, 17, 18〉 = {0, 4, 8, 12, 15, → ... ,} *is saturated. Hence, we find that*

- (a)
*F*(*S*) =*k*− 1 = 15 − 1 = 14, - (b)
$g(S)=\frac{3k-1}{4}=\frac{45-1}{4}=11$,

- (c)
*PF*(*S*) = {*k*− 4,*k*− 2,*k*− 1} = {11, 13, 14}, - (d)
$n(S)=\frac{k+1}{4}=\frac{15+1}{4}=4$,

- (e)
*H*(*S*) = {1, 2, 3, 5, 6, 7, ... ,*k*− 6,*k*− 5,*k*− 4,*k*− 2,*k*− 1} = {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14}*and Ap*(*S*, 4) = {0,*k, k*+ 2,*k*+ 3} = {0, 15, 17, 18}.

*Consider the numerical semigroup S*= 〈4, *k, k* + *t, k* + *t* + 2〉. *If we put k* = 14 *and t* = *13 then we find that S*= 〈4, *k, k* + *t, k* + *t* + 2〉 = 〈4, 14, 27, 29〉 = {0, 4, 8, 12, 14, 16, 18, 20, 22, 24, 26, → ... ,} *is saturated. Thus, we observe that*

- (a)
*F*(*S*) =*k*+*t*− 2 = 14 + 13 − 2 = 25, - (b)
$g(S)=\frac{3k+2t-4}{4}=\frac{64}{4}=16$,

- (c)
*PF*(*S*) = {*k*− 4,*k*+*t*− 4,*k*+*t*− 2} = {10, 23, 25}, - (d)
$n(S)=\frac{k+2t}{4}=\frac{40}{4}=10$,

- (e)
*H*(*S*) = {1, 2, 3, 5, 6, 7, ... ,*k*− 5,*k*− 4,*k*+ 3,*k*− 1,*k*− 1,*k*+3, ... ,*k*−*t*− 2} = {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 15, 17, 19, 21, 23, 25}*and Ap*(*S*, 4) = {0,*k, k*+*t, k*+*t*+ 2} = {0, 14, 27, 29}.

The authors thank the anonymous referee for his/her remarks which helped them to improve the presentation of the paper.

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**Received**: 2016-08-04

**Accepted**: 2016-09-23

**Published Online**: 2016-11-03

**Published in Print**: 2016-01-01

**Citation Information: **Open Mathematics, Volume 14, Issue 1, Pages 827–831, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0074.

© Ilhan and Süer, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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