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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 14, Issue 1

# On the saturated numerical semigroups

Sedat Ilhan
/ Meral Süer
Published Online: 2016-11-03 | DOI: https://doi.org/10.1515/math-2016-0074

## 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.

MSC 2010: 20M14

## 1 Introduction

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, bSa + bS,

• (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. $S=〈a1,a2,...ar〉={∑k=1tciai,...,cr∈N}$ where a1, a2, ... ,arS and r ≥ 1. In this case, {a1, a2, ... , ar} is a minimal system of generators if no proper subset of {a1, a2, ... , ar} generates S. The numbers e(S) = r and m(S) = min{aS : a > 0} are called the embedding dimension and multiplicity of S respectively. In general, it holds that e(S) ≤ m(S). We say that S has maximal embedding dimension if e(S) = m(S) (see [6]).

We define the following invariants of numerical semigroups: $F(S)=max{x:x∈Z∖S}$ and $n(S)=|{0,1,2,...,F(S)}∩S|.$

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

We can write $S=〈a1,a2,...,ar〉=〈s0=0,s1,s2,...,sn−1,sn=F(S)+1,→...〉$ where si < si+1 and n = n(S). The arrow means that every integer greater than F(S) + 1 belongs to S, for i = 1, 2, ... , n = n(S) (see [2]).

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 xH(S) is called a fundamental gap of S if 2x, 3xS. The set of all the fundamental gaps of S is denoted by FH(S), i.e. $FH(S)={x∈H(S):2x,3x∈S}.$

An element x ∈ ℤ is called a Pseudo-Frobenius number of S if xS and x + sS, for sS \ {0}. We denote by PF(S) the set of all Pseudo-Frobenius numbers of S, i.e. $PF(S)={x∈Z∖S:x+s∈S,foralls∈S∖{0}}$ (see [7]). Given a numerical semigroup S and xS \ {0}, we define the Apery set of x in S as Ap(S, x) = {sS : sxS} (for details see [9]).

If a numerical semigroup S satisfies the condition x + yzS, for every x, y, zS such that xyz, 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 + c1s1 + c2s2 + ... + ckskS, where s, siS and ci ∈ ℤ such that c1s1 + c2s2 + ... + cksk ≥ 0 and sis for i = 1, 2, ... , k. Also, all saturated numerical semigroup are Arf. However an Arf numerical semigroup need not be to be saturated. The numerical semigroup $S=〈7,12,15,16,17,18,20〉$ is Arf, but it is not saturated since 12 + (−5).7 + 3.12 = 13 ∉ S.

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(mod4) 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.

## 2 Main results

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 + dS (a)S for all aS, a > 0 where dS (a) = gcd{xS : xa }.

• (iii)

a + kdS (a)S for all aS, 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, r1 and r ∈ ℤ. Thus, we have $S=〈4,k,k+2,k+3〉={0,4,8,⋯,k−7,k−3,k→⋯,}={0,4,8,⋯,4r−4,4r,4r+3,→⋯}.$

In this case,

• (a)

If a < 4r + 3, then dS (a) = 1. So, we find that a + dS (a)S since a + dS (a) = a + 14r + 4S, for all aS, a > 0.

• (b)

If a ≥ 4r + 3, then dS (a) = 4. So, we have a + dS (a) = a + 4 ∈ S, for all aS, 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 $S=〈4,k,k+t,k+t+2〉={0,4,8,...,k−6,k−2,k,k+2,...,k+t−3,k+t−1,→...,}={0,4,8,...,4r−4,4r,4r+2,4r+4,...,4r+t−1,4r+t+1,→...,}.$

In this case,

• (i)

If a > 4r + t + 1, then dS (a) = 1. So, we obtain a + dS (a)S from the inequality a + dS (a) = a + 1 ≥ 4r + t + 2 ∈ S, for all aS, a > 0.

• (ii)

If 4ra ≤ 4r + t + 1, then dS (a) = 2. So, we obtain a + dS (a) = a + 2 ∈ S from the inequality 4ra ≤ 4r + t + 1, for all aS, a > 0.

• (iii)

If a < 4r, then dS (a) = 4. So, we obtain a + dS (a) = a + 4 ∈ S since a + 4r < 4r + 4, for all aS, 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 {n1 < n2 < ... < nr}. Then S has maximal embedding dimension if and only if Ap(S, n1) = {0,n2, n3,...,nr}.

([6]).Let S be a numerical semigroup minimally generated by {n1 < n2 < ... < nr}. Then the following conditions are true:

• (1)

If S has maximal embedding dimension, then F(S) = nrn1.

• (2)

S has maximal embedding dimension if and only if $g(S)=n2+n3+...+nrn1−n1−12.$

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, ... , 4r − 3, 4r − 2, 4r − 1, 4r + 1, 4r + 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, ... , 4r − 4, 4r, 4r − 3, → ... ,}.

Let S= 〈4, k, k + t, k + t + 2be 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 $n\left(S\right)=\left(k+t-2\right)+1-\frac{3k+2t-4}{4}=\frac{k+2t}{4}$ from g(S) = F(S) + 1 − n(S).

• (e)

We observe that H(S) = {1, 2, 3, 5, 6, 7, ... , 4r − 3, 4r − 2, 4r − 1, 4r − 1, 4r + 3, 4r + 5, ... , 4r + 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 + t1, → ... ,} = {0, 4, 8, ... ,4r − 4, 4r, 4r + 2, 4r + 4, ... , 4rt − 1, 4r + 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)

$g\left(S\right)=\frac{3k-1}{4}=\frac{45-1}{4}=11$,

• (c)

PF(S) = {k − 4, k − 2, k − 1} = {11, 13, 14},

• (d)

$n\left(S\right)=\frac{k+1}{4}=\frac{15+1}{4}=4$,

• (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)

$g\left(S\right)=\frac{3k+2t-4}{4}=\frac{64}{4}=16$,

• (c)

PF(S) = {k − 4, k + t − 4, k + t − 2} = {10, 23, 25},

• (d)

$n\left(S\right)=\frac{k+2t}{4}=\frac{40}{4}=10$,

• (e)

H(S) = {1, 2, 3, 5, 6, 7, ... , k − 5, k − 4, k + 3, k − 1, k − 1, k +3, ... , kt − 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}.

## Acknowledgement

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

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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,

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