On certain families of sparse numerical semigroups with Frobenius number even
This paper is about sparse numerical semigroups and applications in the Weierstrass semigroups theory. We describe and find the genus of certain families of sparse numerical semigroups with Frobenius number even and we also study the realization of the elements on these families as Weierstrass semig...
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Tizziotti, G. Villanueva, J. 2023-02-28T19:21:21Z 2023-02-28T19:21:21Z 2019 On certain families of sparse numerical semigroups with Frobenius number even / G. Tizziotti, J. Villanueva // Algebra and Discrete Mathematics. — 2019. — Vol. 27, № 1. — С. 99–116. — Бібліогр.: 20 назв. — англ. 1726-3255 2010 MSC: 20M10, 20M14, 14H55 https://nasplib.isofts.kiev.ua/handle/123456789/188426 This paper is about sparse numerical semigroups and applications in the Weierstrass semigroups theory. We describe and find the genus of certain families of sparse numerical semigroups with Frobenius number even and we also study the realization of the elements on these families as Weierstrass semigroups. Partially supported by CNPq 446913/2014-6 and FAPEMIG APQ-01607-14. Partially supported by PNPD/CAPES, by Programa de Pós-doutorado da FAMAT, Câmpus Santa Mônica, Universidade Federal de Uberlândia. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics On certain families of sparse numerical semigroups with Frobenius number even Article published earlier |
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On certain families of sparse numerical semigroups with Frobenius number even |
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On certain families of sparse numerical semigroups with Frobenius number even Tizziotti, G. Villanueva, J. |
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On certain families of sparse numerical semigroups with Frobenius number even |
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On certain families of sparse numerical semigroups with Frobenius number even |
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On certain families of sparse numerical semigroups with Frobenius number even |
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On certain families of sparse numerical semigroups with Frobenius number even |
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on certain families of sparse numerical semigroups with frobenius number even |
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This paper is about sparse numerical semigroups and applications in the Weierstrass semigroups theory. We describe and find the genus of certain families of sparse numerical semigroups with Frobenius number even and we also study the realization of the elements on these families as Weierstrass semigroups.
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On certain families of sparse numerical semigroups with Frobenius number even / G. Tizziotti, J. Villanueva // Algebra and Discrete Mathematics. — 2019. — Vol. 27, № 1. — С. 99–116. — Бібліогр.: 20 назв. — англ. |
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“adm-n1” — 2019/3/22 — 12:03 — page 99 — #107
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 27 (2019). Number 1, pp. 99–116
c© Journal “Algebra and Discrete Mathematics”
On certain families of sparse numerical
semigroups with Frobenius number even
Guilherme Tizziotti1 and Juan Villanueva2
Communicated by V. Mazorchuk
Abstract. This paper is about sparse numerical semigroups
and applications in the Weierstrass semigroups theory. We describe
and find the genus of certain families of sparse numerical semigroups
with Frobenius number even and we also study the realization of
the elements on these families as Weierstrass semigroups.
Introduction
Let Z be the set of integers numbers and N0 be the set of non-negative
integers. A subset H =
{
0 = n0(H) < n1(H) < · · ·
}
of N0 is a numerical
semigroup if its is closed respect to addition and its complement N0 \H is
finite. The cardinality of the set Gaps(H) := N0 \H is called genus of the
numerical semigroup H and is denoted by g = g(H). Note that g(H) = 0
if and only if H = N0. If g(H) > 0 the elements of Gaps(H) are called
gaps. The smallest integer c = c(H) such that c+ h ∈ H, for all h ∈ N0
is called the conductor of H. The least positive integer n1 = n1(H) ∈ H
is called the multiplicity of H. As N0 \ H is finite, the set Z \ H has
a maximum, which is called Frobenius number and will be denoted by
ℓg = ℓg(H). A property known of this number is that ℓg(H) 6 2g − 1, see
1Partially supported by CNPq 446913/2014-6 and FAPEMIG APQ-01607-14.
2Partially supported by PNPD/CAPES, by Programa de Pós-doutorado da FAMAT,
Câmpus Santa Mônica, Universidade Federal de Uberlândia.
2010 MSC: 20M10, 20M14, 14H55.
Key words and phrases: Arf numerical semigroup, numerical semigroup, sparse
numerical semigroup, Weierstrass semigroup, weight of numerical semigroup.
“adm-n1” — 2019/3/22 — 12:03 — page 100 — #108
100 On certain families of sparse numerical semigroups
[15]. In particular, H = N0 if and only if −1 is the Frobenius number
of H. Aa a consequence of this fact, from now on we use the notation
ℓ0 = ℓ0(H) := −1, for all numerical semigroup H . When g > 0, we denote
Gaps(H) =
{
1 = ℓ1(H) < · · · < ℓg(H)
}
. So, c = ℓg(H) + 1 and is clear
that c = nc−g(H). For simplicity of notation we shall write ℓi for ℓi(H)
and nk for nk(H), for all integers i, k such that 0 6 i 6 g and k > 0, when
there is no danger of confusion. More details about numerical semigroups
theory, see e.g. [17].
Currently, there are several families that have been of interest in the
literature due to their properties and applications. Examples of such
families are the sparse semigroups, which were introduced in [14]. A
numerical semigroup H = {0 = n0 < n1 < · · · } of genus g > 0 with
Gaps(H) = {ℓ1 < · · · < ℓg} is called sparse numerical semigroup if
ℓi − ℓi−1 6 2, for all integer i such that 1 6 i 6 g, or equivalently
ni − ni−1 > 2, for all integer i such that 1 6 i 6 c − g, where c is the
conductor of H. For convenience, we considerer the numerical semigroup
N0 as sparse. Thus, the concept of sparse numerical semigroups it’s in a
way a generalization of the concept of Arf numerical semigroups, which
was introduced in [1].
Among other applications, the study of numerical semigroup is related
to Algebraic Geometry in the treatment of algebraic curves and their
Weierstrass semigroups. More explicitly, given a numerical semigroup H,
does it exist a curve X such that for some point P ∈ X has H = H(P )?,
where H(P ) is the Weierstrass semigroup of X at P . If the answer is yes,
we say that the numerical semigroup H is Weierstrass. Studies to answer
this question have been done for decades, see e.g. [4], [12], [15] and [19].
From a geometrical point of view, sparse numerical semigroups are closely
related to Weierstrass semigroups arising in double covering of curves, cf.
[19]. Its arithmetical structure is strongly influenced by the parity of ℓg.
In this work, we study certain families of sparse numerical semigroups
which are examples of Weierstrass semigroups. In addition, we manage
to describe and find the genus of the semigroups on these families. These
aspects are very important in the study of numerical semigroups theory.
It is important to note that in [6] the authors find an upper bound for the
genus of sparse numerical semigroups with Frobenius number even. Here,
in this paper, we get a better bound for the genus of the semigroups on
the families of sparse numerical semigroups studied.
This paper is organized as follows. Section 1, contains basic concepts
about numerical semigroups and backgrounds for the next sections. In
Section 2, we study certain families of sparse numerical semigroups with
“adm-n1” — 2019/3/22 — 12:03 — page 101 — #109
G. Tizziotti, J. Villanueva 101
Frobenius number even. As the main results, we describe and find the
genus of the semigroups on these families of sparse numerical semigroups.
Finally, in Section 3 we study the realization of the sparse numerical
semigroups determined in the previous section as Weierstrass semigroups.
1. Preliminaries
1.1. Basic concepts
Let H =
{
0 = n0 < n1 < · · ·
}
be a numerical semigroup of genus
g > 0 and Gaps(H) = {ℓ1 < · · · < ℓg}. For each 1 6 i 6 g, the ordered
pair (ℓi−1, ℓi) will be called leap on H (or simply leap). The set of leaps
on H will be denoted by
V = V(H) :=
{
(ℓi−1, ℓi) : 1 6 i 6 g
}
.
Note that |V | = g. The ordered pair (ℓi−1, ℓi) will be called single leap if
ℓi − ℓi−1 = 1 and double leap if ℓi − ℓi−1 = 2.
Based on this set, for a positive integer m, let us define the subset
Vm = Vm(H) :=
{
(ℓi−1, ℓi) : ℓi − ℓi−1 = m, 1 6 i 6 g
}
and for an interval [a, b], with −1 6 a < b 6 ℓg, let us define the subset
V[a,b] =
{
(ℓi−1, ℓi) : ℓi−1, ℓi ∈ [a, b], 1 6 i 6 g
}
.
For convenience, we define Vm(N0) := ∅, for all positive integer m. To
simplify the notation we will denote the cardinality of the set Vm(H) by
vm(H), that is, vm = vm(H) := |Vm(H) |.
1.2. Arf numerical semigroup
A numerical semigroup H = {0 = n0 < n1 < · · · } is called Arf
numerical semigroup if
ni + nj − nk ∈ H, (1)
for all integers i, j, k such that 0 6 k 6 j 6 i, or equivalently ni+nj−nk ∈
H, for all integers i, j, k such that 0 6 k 6 j 6 i 6 c − g, where c is
the conductor and g is the genus of H, respectively. The Arf numerical
semigroups was introduced in [1]. For more details about this family of
numerical semigroups, see e.g. [2], [5], [16] and [18].
If g > 0 and Gaps(H) = {ℓ1 < · · · < ℓg}, the Arf property (1) implies
that
ℓi − ℓi−1 6 2, (2)
“adm-n1” — 2019/3/22 — 12:03 — page 102 — #110
102 On certain families of sparse numerical semigroups
for all integer i such that 1 6 i 6 g, or equivalently ni − ni−1 > 2, for all
integer i such that 1 6 i 6 c− g (see [14, Corollary 1] and [20, Corollary
2.1.4]). Using this property is not difficult to see that if H is an Arf
numerical semigroup, then H is a sparse numerical semigroup.
For each non-negative integer g, let
Ng := {0} ∪ {n ∈ N : n > g + 1}
(in the case g = 0, it notation is itself the N0). It is clear that Ng is a
numerical semigroup of genus g. The semigroup Ng is called ordinary nu-
merical semigroup and is a canonical example of Arf numerical semigroups
and, consequently, a example of sparse numerical semigroups.
1.3. Sparse numerical semigroups
Theorem 1.1 ([18], Theorem 2.2). Let H be a numerical semigroup of
genus g.
(1) H is an sparse numerical semigroup if and only if v1(H)+v2(H) = g.
In this case, vm(H) = 0, for all positive integer m > 3.
(2) If H is an sparse numerical semigroup, then the Frobenius number of
H is ℓg = v1(H) + 2v2(H)− 1.
(3) If g > 0 and ℓg(H) = 2g −K, for some positive integer K, then H
is an sparse numerical semigroup if and only if v1(H) = K − 1 and
v2(H) = g −K + 1.
For a numerical semigroup H = {0 = n0 < n1 < · · · }, define M =
M(H) := n1 − 1. The parameter M was introduced in [14], where if
ℓg(H) = 2g −K, for some positive integer K, we have that
0 6 M 6 K. (3)
If g > 1 and Gaps(H) = {1 = ℓ1 < · · · < ℓg}, define
SM = SM (H) :=
∣
∣
{
(ℓi−1, ℓi) : ℓi − ℓi−1 = 1, M + 1 6 i 6 g
}∣
∣ . (4)
Note that SM < v1, since
{
(ℓi−1, ℓi) : ℓi− ℓi−1 = 1, M +1 6 i 6 g
}
( V1.
Before the next result, let us remember that a semigroup H is called
γ-hyperelliptic if it has exactly γ even gaps. If γ = 0, H is called simply
hyperelliptic.
Proposition 1.2. Let H = {0 = n0 < n1 < · · · } be a numerical semi-
group of genus g and M = M(H) = n1 − 1. Then:
(1) 0 6 M 6 g;
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G. Tizziotti, J. Villanueva 103
(2) M(H) = 0 if and only if H = N0;
(3) M(H) = g if and only if H = Ng;
(4) M(H) = 1 if and only if H is hyperelliptic;
(5) If g > 0 and Gaps(H) = {ℓ1 < · · · < ℓg}, then M is the largest integer
such that ℓM = M .
Proof. The assertions (1), (2) and (3) are clear. Item (4), follows from the
fact that M = 1 if and only if n1 = 2. Finally, item (5) follows directly
from the fact that all integers belong to interval [1, n1 − 1] are gaps.
Corollary 1.3. Let H be a sparse numerical semigroup of genus g > 1
with Frobenius number ℓg = 2g − K, for some positive integer K, and
Gaps(H) = {ℓ1 < · · · < ℓg}. Let M = M(H) and SM = SM (H) be
defined in (4). If M < g, then SM = K −M.
Proof. By the definition of SM , since M < g, we have that SM 6= 0 and
SM = v1(H) − (M − 1). Now, by Theorem 1.1 item (2), follows that
SM = K − 1− (M − 1) = K −M .
In Proposition 1.2, we present the semigroups for the cases M = 0, 1
and M = g. The case M = 2 is treated in [14, Proposition 3]. The
following result treat of the case M = K.
Theorem 1.4. Let H be a numerical semigroup of genus g > 0 with
Frobenius number ℓg = 2g−K, for some positive integer K. The following
statements are equivalent:
(1) H is an sparse numerical semigroup and M = K;
(2) H = NK or H = {K + 2i− 1 : 1 6 i 6 g −K} ∪N2g−K ;
(3) v1(H) = M − 1 and v2(H) = g −M + 1.
In this case, Gaps(H) = {1, . . . , ℓK , . . . , ℓg}, where ℓi = 2i − K, for all
integer i such that K 6 i 6 g.
Proof. We prove that (1) ⇒ (2) ⇒ (3) ⇒ (1). By [14, Theorem 1 (2)]
follows that (1) ⇒ (2). Now, suppose (2). If H = NK , then M = K
and by Theorem 1.1 (3) follows that v1(H) = M − 1 and v2(H) =
g − M + 1. If H = {K + 2i − 1 : 1 6 i 6 g − K} ∪ N2g−K , then
Gaps(H) = {1, . . . ,K, ℓK+1, . . . , ℓg}, where ℓK+i = K + 2i, for all integer
i such that 1 6 i 6 g − K. So, M = K, v1(H) = K − 1 = M − 1 and
v2(H) = g−K+1 = g−M +1. Thus, we have (2) ⇒ (3). The implication
(3) ⇒ (1), follows directly from the Theorem 1.1 items (1) and (3).
The assertion on the gaps of H follows as in the proof of the implication
(2) ⇒ (3).
“adm-n1” — 2019/3/22 — 12:03 — page 104 — #112
104 On certain families of sparse numerical semigroups
2. Sparse numerical semigroups with Frobenius number
even
For each positive integer k, consider the family H sfe
k of sparse numerical
semigroups H with genus g = g(H) and Frobenius number even of the
form 2g − 2k. That is,
H
sfe
k := {H : H is a sparse numerical semigroup
with genus g and ℓg = 2g − 2k}.
In this section, we will study the classification of the elements of
some proper subsets in H sfe
k as well as the cardinality of these subsets.
If H ∈ H sfe
k in [14, Theorem 2] is proved that g(H) 6 6k − 3 and that
the family H sfe
k is finite. In [20, Question 2.3.10], was conjectured that
g(H) 6 4k − 1. This conjecture was proved by Contiero, Moreira and
Veloso in [6, Corollary 3.7]. In Theorem 2.8 and Theorem 2.13, we get a
better bound for g(H) for certain H ∈ H sfe
k .
In the previous section, in particular, we describe all the sparse nu-
merical semigroups belongs to H sfe
k with M = 0, 1 and M = 2k. The next
result describe the case M = 3.
Lemma 2.1. Let k > 2 be an integer and H be a numerical semigroup
of genus g and Frobenius number ℓg = 2g − 2k. If M = 3, the following
statements are equivalent:
(1) H is be an Arf numerical semigroup;
(2) H is be an sparse numerical semigroup;
(3) H = 〈4, 4k − 1, 4k + 1, 4k + 2〉;
(4) H is k-hyperelliptic.
In this case, g = 3k − 1 and
Gaps(H) = {1, . . . , 2g − 2k} \
{
4i : i ∈ N, 1 6 i 6 2g−2k−2
4
}
.
Proof. We prove that (1) ⇒ (2) ⇒ (3) ⇒ (4) ⇒ (3) ⇒ (1). Implication
(1) ⇒ (2) is always true. Before show the next implications, since M = 3,
we have that the multiplicity of H is n1 = 4. Let G := {1, . . . , 2g − 2k} \
{
4i : i ∈ N, 1 6 i 6 2g−2k−2
4
}
. It is clear that Gaps(H) is a subset of G.
Since 4 ∈ H and ℓg = 2g− 2k is even, we have that ℓg = 4i0 +2, for some
i0 ∈ N. Note that if ℓ ∈ G is even, then ℓ = 4i+ 2, for some i ∈ N0 with
i 6 i0. So, ℓ ∈ G and ℓ even implies that ℓ /∈ H, because otherwise we
will have that ℓg ∈ H, since ℓ = 4i+ 2, 4 ∈ H, ℓg = 4i0 + 2 and i 6 i0.
“adm-n1” — 2019/3/22 — 12:03 — page 105 — #113
G. Tizziotti, J. Villanueva 105
Now, suppose (2). Let ℓ ∈ G. If ℓ is even, we have seen above that
ℓ ∈ Gaps(H). If ℓ ≡ 1 (mod 4), then ℓ − 1 ∈ H and we have that
ℓ ∈ Gaps(H), since H is sparse. If ℓ ≡ 3 (mod 4), then ℓ + 1 ∈ H and
we also have that ℓ ∈ H, since H is sparse. Thus, Gaps(H) = G, and so
H = 〈4, 4k − 1, 4k + 1, 4k + 2〉. Thus, we have (2) ⇒ (3).
Now, we prove the implication (3) ⇒ (4). If H = 〈4, 4k−1, 4k+1, 4k+
2〉, then 2 ∈ Gaps(H) and the number of even gaps on the interval [4, 4k]
is equal to 4k−4
4 = k − 1. So, H is k-hyperelliptic.
To prove the implication (4) ⇒ (3), first note that G∩2N = Gaps(H)∩
2N. Suppose that H is k-hyperelliptic. Then, G ∩ 2N = Gaps(H) ∩ 2N
and |G ∩ 2N | = g − k − 2g−2k−2
4 implies that g − k − 2g−2k−2
4 = k, that
is, g = 3k − 1. Since M = 3, g = 3k − 1 and ℓg = 2g − 2k = 4k − 2, we
conclude that Gaps(H) = [1, 4k − 2] \ {4i : i ∈ N, 1 6 i 6 k − 1}, and
then H = 〈4, 4k − 1, 4k + 1, 4k + 2〉.
Finally, we prove the implication (3) ⇒ (1). If H = 〈4, 4k − 1, 4k +
1, 4k + 2〉, then H = 4N0 ∪ {n ∈ N : n > 4k − 1}. Suppose that
H = {0 = n0 < 4 = n1 < n2 < · · · } and let i, j, s integers such that
0 6 s 6 j 6 i. If ni, nj or ns belongs to {n ∈ N : n > 4k − 1}, then it is
clear that ni + nj − ns ∈ {n ∈ N : n > 4k − 1} and so ni + nj − ns ∈ H.
If ni, nj , ns ∈ 4N0 then it is also clear that ni + nj − ns ∈ 4N0 and so
ni + nj − ns ∈ H. Therefore, H is Arf.
For H ∈ H sfe
k , by Equation (3) and Proposition 1.2, follows that
2 6 M(H) 6 2k. (5)
For each integer J such that 0 6 J 6 2k − 2, consider the set Hk
2+J of
sparse numerical semigroups H ∈ H sfe
k such that M(H) = 2 + J . That is,
H
k
2+J =
{
H ∈ H
sfe
k : M(H) = 2 + J
}
. (6)
Since
H
sfe
k =
2k−2
⋃
·
J=0
H
k
2+J ,
in order to study the cardinality of H sfe
k it is enough to study the cardinality
of Hk
2+J , for all integer J such that 0 6 J 6 2k − 2. Next we will study
the cases: J = 2k − 2, J = 2k − 3 and J = 2k − 4.
The following result shows that all numerical semigroups H belongs
to Hk
2k are Arf and (g(H)− k)-hyperelliptic.
“adm-n1” — 2019/3/22 — 12:03 — page 106 — #114
106 On certain families of sparse numerical semigroups
Proposition 2.2. Let H be a numerical semigroup of genus g > 0 with
Frobenius number ℓg = 2g − 2k, for some positive integer k. If M = 2k,
then the following statements are equivalent:
(1) H is an Arf numerical semigroup;
(2) H is an sparse numerical semigroup;
(3) H = N2k or H = {2k + 2i− 1 : 1 6 i 6 g − 2k} ∪N2g−2k;
(4) v1(H) = 2k − 1 and v2(H) = g − 2k + 1;
(5) H is (g − k)-hyperelliptic.
In this case, 2k 6 g 6 3k and Gaps(H) = {1, . . . , 2k − 1, ℓ2k, . . . , ℓg},
where ℓi = 2(i− k), for all integer i such that 2k 6 i 6 g.
Proof. We prove that (1) ⇒ (2) ⇒ (3) ⇔ (4), (3) ⇔ (5) and (3) ⇒ (1).
The implication (1) ⇒ (2) is always true. The implication (2) ⇒ (3), the
equivalence (3) ⇔ (4) and the assertion on the gaps of H is a particular
case from Theorem 1.4, by taking M = K = 2k.
Now, suppose (3). Then, ni = 2k + 2i− 1, for all integer i such that
1 6 i 6 g − 2k + 1, Let i, j integers such that 0 6 j 6 i 6 g − 2k + 1. If
j = 0 is clear that 2ni − nj ∈ H ; other case 2ni − nj = 2k+4i− 2j − 1 >
2k + 2i − 1 > 2k + 1 > ℓ2k. Since in the interval [ℓ2k, ℓg] the gaps ℓi’s
are even numbers and 2ni − nj is odd, follow that 2ni − nj ∈ H. This
shows that (3) ⇒ (1). Also, by Theorem 1.4, (3) implies that Gaps(H) =
{1, . . . , 2k − 1, ℓ2k, . . . , ℓg}, where ℓi = 2(i− k), for all integer i such that
2k 6 i 6 g, and thus (3) ⇒ (5). Finally, suppose (5). Since in the interval
[2, 2g − 2k] there are g − k elements even, these elements should be gaps
and so (5) ⇒ (3).
Note that, since n1 = 2k + 1, from item (3) follows that 2(2k + 1) >
ℓg + 2 = (2g − 2k) + 2. Therefore, we have that g 6 3k.
Theorem 2.3. For all integer k > 1,
H
k
2k = {N2k} ∪
{
H(k,r) : r ∈ N, 1 6 r 6 k
}
,
where
H(k,r) = {2k + 2i− 1 : i ∈ N, 1 6 i 6 r} ∪N2k+2r,
for 1 6 r 6 k. In addition, g
(
H(k,r)
)
= 2k+r. In particular, the cardinality
of Hk
2k is
∣
∣Hk
2k
∣
∣ = k + 1.
Proof. For each integer r such that 1 6 r 6 k, let H(k,r) = {2k+2i−1 : i ∈
N, 1 6 i 6 r}∪N2k+2r. By the definition, is clear that H(k,r) is a numerical
semigroup of genus g
(
H(k,r)
)
= 2k + r and ℓg
(
H(k,r)
)
= 2g
(
H(k,r)
)
− 2k.
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G. Tizziotti, J. Villanueva 107
So, by Proposition 2.2, we have that H(k,r) belongs to H k
2k, for all integer
r such that 1 6 r 6 k. Also, by Proposition 2.2, N2k is contained in H k
2k.
Now, let H ∈ H k
2k of genus g and H 6= N2k. By Proposition 2.2, we have
2k+ 1 6 g 6 3k and H = {2k+ 2i− 1 : i ∈ N, 1 6 i 6 g− 2k} ∪N2g−2k.
Let r := g − 2k. Then, 1 6 r 6 k and so H = H(k,r). This completes the
proof.
A numerical semigroup H of genus g > 0 is called quasi-symmetric
if the Frobenius number of H is ℓg = 2g − 2. In [2] and [3], has been
proved that there are only two Arf numerical semigroups quasi-symmetric:
〈3, 4, 5〉 and 〈3, 5, 7〉. In [14, Example 4] and [20, Example 2.3.13], was
proved that Arf numerical semigroups quasi-symmetric is equivalent to
sparse numerical semigroups quasi-symmetric. This result can be obtained
directly from the previous theorem by taking k = 1. That is, the family
of sparse numerical semigroups quasi-symmetric is
H
sfe
1 =
{
N2, {3} ∪N4
}
= H
1
2 .
The following is a further demonstration of the [14, Example 5] about the
classification of sparse numerical semigroups of genus g with Frobenius
number ℓg = 2g − 4.
Corollary 2.4. The family of sparse numerical semigroups of genus g
and Frobenius number 2g − 4 is
H
sfe
2 =
{
{3, 6} ∪N8, {3, 6, 9} ∪N10, {4} ∪N6,N4, {5} ∪N6, {5, 7} ∪N8
}
.
In this case,
H
2
2 =
{
{3, 6} ∪N8, {3, 6, 9} ∪N10
}
,
H
2
3 =
{
{4} ∪N6
}
,
H
2
4 =
{
N4, {5} ∪N6, {5, 7} ∪N8
}
.
Proof. First, note that H sfe
2 = H2
2 ∪ H2
3 ∪ H2
4 . From [14, Proposition
3], we have that H2
2 =
{
{3, 6} ∪N8, {3, 6, 9} ∪N10
}
. From the item (3)
of Lemma 2.1 follows that H2
3 =
{
{4} ∪ N6
}
and taking k = 2 in the
previous theorem follows that H2
4 =
{
N4, {5} ∪N6, {5, 7} ∪N8
}
.
Lemma 2.5. Let k > 2 be an integer. For each pair of integers r and αr
such that 1 6 r 6 k − 1 and 1 6 αr 6 r, let
Hαr
(k,r) := {2(k + µ− 1) : 1 6 µ 6 αr}∪
{2k + 2ν + 1 : αr 6 ν 6 r − 1} ∪N2k+2r.
“adm-n1” — 2019/3/22 — 12:03 — page 108 — #116
108 On certain families of sparse numerical semigroups
Then, Hαr
(k,r) ∈ H k
2k−1 and g
(
Hαr
(k,r)
)
= 2k + r, for all pairs (r, αr).
Proof. Let r, αr and Hαr
(k,r) be as above. By the definition, it is clear that
Hαr
(k,r) is a numerical semigroup with set of gaps given by
Gaps
(
Hαr
(k,r)
)
= {i : 1 6 i 6 2k − 1} ∪· {2k + 2µ− 1 : 1 6 µ 6 αr}∪·
{2k + 2ν : αr 6 ν 6 r}.
In particular, g
(
Hαr
(k,r)
)
= (2k − 1) + αr + (r − αr + 1) = 2k + r. Also,
M
(
Hαr
(k,r)
)
= 2k − 1 and ℓg
(
Hαr
(k,r)
)
= 2k + 2r = 2g − 2k. Therefore,
Hαr
(k,r) ∈ H k
2k−1, for all pairs (r, αr) as above.
Henceforth Hαr
(k,r) is the sparse numerical semigroup defined in the
Lemma 2.5.
Remark 2.6. Let k > 2 be an integer and (r, αr) be as defined in the
previous lemma. Then,
Gaps
(
Hαr
(k,r)
)
=
{
ℓ1
(
Hαr
(k,r)
)
, . . . , ℓ2k+r
(
Hαr
(k,r)
)}
,
where
ℓi
(
(Hαr
(k,r)
)
=
i, if 1 6 i 6 2k − 1;
2(i− k) + 1, if 2k 6 i 6 2k − 1 + αr;
2(i− k), if 2k + αr 6 i 6 2k + r.
The following theorem will give us a new bound for the genus of
semigroups in Hk
2k−1. Before, we will make an observation that will be
very useful in the following.
Remark 2.7. Let H be a sparse numerical semigroup of genus g > 1
with Frobenius number ℓg = 2g − K, for some positive integer K, and
Gaps(H) = {ℓ1 < · · · < ℓg}. Let r and s integers such that 1 6 r < s 6 g.
If V[ℓr,ℓs] ∩ V1 =
{
(ℓj−1, ℓj)
}
, for some j ∈ {r + 1, . . . , s}, then
ℓv = ℓj−1 − 2(j − v − 1),
for all integer v such that r 6 v 6 j − 1, and
ℓw = ℓj + 2(w − j),
for all integer w such that j 6 w 6 s. In particular, ℓj−1 ≡ ℓv (mod 2),
for all integer v such that r 6 v 6 j − 1, and ℓj ≡ ℓw (mod 2), for all
integer w such that j 6 w 6 s.
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G. Tizziotti, J. Villanueva 109
Theorem 2.8. Let k > 2 be an integer. If H ∈ Hk
2k−1, then
2k + 1 6 g(H) 6 3k − 1.
Proof. Let k > 2 be an integer and let H ∈ Hk
2k−1 with g(H) = g and
Gaps(H) = {1 = ℓ1 < · · · < ℓg}. Since H is a sparse semigroup and
M = 2k − 1, we have that ℓM+1 = 2k + 1. If g 6 2k, then ℓg 6 2k, a
contradiction with the value of ℓM+1. So, g > 2k + 1. Now, suppose that
g = 3k + s, for some integer s > 0. So, ℓg = 4k + 2s. By the Corollary
1.3, SM = 1, that is, there exist a unique single leap (ℓj−1, ℓj), for some
integer j such that 2k+1 = M +2 6 j 6 g. Then, by the Remark 2.7, all
the even numbers greater than 2k+ 1 and smaller than ℓj−1 are non-gaps.
Moreover all the even numbers from ℓj to ℓg are gaps. More precisely, we
have ℓg = ℓj + 2(g − j). So, ℓj = (4k + 2s)− 2(3k + s− j) = 2j − 2k. We
conclude the proof by studying the following cases separately: j < 3k and
j > 3k. Firstly, assume that j < 3k. Then, ℓj = 2j − 2k < 4k 6 4k + 2s.
So, 4k is a gap, a contradiction since 2k is a non-gap. Now, suppose that
j > 3k. Then, 2k 6 2j − 4k < 2j − 2k − 1 = ℓj−1. Thus, 2j − 4k ∈ H.
Therefore, ℓj = 2j − 2k = 2k + (2j − 4k) ∈ H, a contradiction, and the
proof is complete.
Theorem 2.9. For all integer k > 2,
H
k
2k−1 =
{
Hαr
(k,r) : (r, αr) ∈ N
2, 1 6 r 6 k − 1, 1 6 αr 6 r
}
.
Proof. Firstly, by the Lemma 2.5, we have that
{
Hαr
(k,r) : (r, αr) ∈ N
2, 1 6 r 6 k − 1, 1 6 αr 6 r
}
⊂ H
k
2k−1.
Now, let H ∈ Hk
2k−1 with g(H) = g. By Theorem 2.8, we have 2k + 1 6
g 6 3k − 1. Let r := g − 2k. So, 1 6 r 6 k − 1. Since ℓg = 2g − 2k, we
have that ℓg = ℓ2k+r = 2k + 2r. Let Gaps(H) = {1 = ℓ1 < · · · < ℓg}. By
Corollary 1.3, SM = 1, that is, there exist a unique single leap (ℓj−1, ℓj),
for some integer j such that 2k + 1 = M + 2 6 j 6 g = 2k + r. Let
αr := j − 2k. By Remark 2.7, ℓv = ℓ2k + 2(v − 2k), for all integer v such
that 2k 6 v 6 j − 1 and ℓw = ℓ2k+r + 2
[
w − (2k + r)
]
, for all integer w
such that j 6 w 6 2k+ r. Since M = 2k− 1, we have ℓ2k−1 = 2k− 1 and
ℓ2k = 2k + 1. Thus, we concluded that ℓv = 2(v − k) + 1, for all integer v
such that 2k 6 v 6 j− 1 = 2k− 1+αr, and ℓw = 2(w− k), for all integer
w such that 2k+αr = j 6 w 6 2k+ r = g. Therefore, by Remark 2.6, we
have that H = Hαr
(k,r) and follows the result.
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110 On certain families of sparse numerical semigroups
Corollary 2.10. For all integer k > 2,
∣
∣Hk
2k−1
∣
∣ =
(
k
2
)
.
Proof. By the Theorem 2.9, we have that
H
k
2k−1 =
k−1
⋃
·
r=1
{
Hαr
(k,r) : αr ∈ N, 1 6 αr 6 r
}
.
Therefore,
∣
∣
∣
H
k
2k−1
∣
∣
∣
=
k−1
∑
r=1
r =
(
k
2
)
.
Lemma 2.11. Let k > 3 be an integer. Let r, s and αs be a triple of
integers such that 1 6 r 6 k − 2, 1 6 s 6 r and 1 6 αs 6 r − s+ 1. For
r ∈ {k, k + 1}, let s = 1 and αs = k. For each triple (r, s, αs), consider
H
(s,αs)
(k,r) :={2k + 2λ− 3 : 1 6 λ 6 s}∪
{2(k + s+ µ− 1) : 1 6 µ 6 αs − 1}∪
{2k + 2ν + 2s− 1 : αs 6 ν 6 r − s} ∪N2k+2r.
Then, H
(s,αs)
(k,r) ∈ Hk
2k−2 and g
(
H
(s,αs)
(k,r)
)
= 2k + r, for all triples (r, s, αs).
Proof. Let r, s, αs and H
(s,αs)
(k,r) be as above. By the definition, it is clear
that H
(s,αs)
(k,r) is a numerical semigroup with set of gaps given by
Gaps
(
H
(s,αs)
(k,r)
)
= {i : 1 6 i 6 2k − 2} ∪· {2(k + λ− 1) : 1 6 λ 6 s}∪·
{2k + 2µ+ 2s− 3 : 1 6 µ 6 αs}∪·
{2(k + ν + s) : αs − 1 6 ν 6 r − s}.
In particular, g
(
H
(s,αs)
(k,r)
)
= (2k−2)+s+αs+(r−s−αs+2) = 2k+r. Also,
M
(
H
(s,αs)
(k,r)
)
= 2k − 2 and ℓg
(
H
(s,αs)
(k,r)
)
= 2k + 2r = 2g − 2k. Therefore,
H
(s,αs)
(k,r) ∈ Hk
2k−2, for all triple (r, s, αs) as above.
Henceforth H
(s,αs)
(k,r) is the sparse numerical semigroup defined in the
Lemma 2.11.
Remark 2.12. Let k > 3 be an integer and (r, s, αs) be as defined in the
previous lemma. Then,
Gaps
(
H
(s,αs)
(k,r)
)
=
{
ℓ1
(
H
(s,αs)
(k,r)
)
, . . . , ℓ2k+r
(
H
(s,αs)
(k,r)
)}
,
“adm-n1” — 2019/3/22 — 12:03 — page 111 — #119
G. Tizziotti, J. Villanueva 111
where
ℓi
(
H
(s,αs)
(k,r)
)
=
i, if 1 6 i 6 2k − 2;
2(i− k + 1), if 2k − 1 6 i 6 2k + s− 2;
2(i− k) + 1, if 2k + s− 1 6 i 6 2k + s+ αs − 2;
2(i− k), if 2k + s+ αs − 1 6 i 6 2k + r.
Theorem 2.13. Let k > 3 be an integer. If H ∈ Hk
2k−2, then 2k + 1 6
g(H) 6 3k + 1 and g(H) 6= 3k − 1.
Proof. Let k > 3 be an integer and let H ∈ Hk
2k−2 with g(H) = g and
Gaps(H) = {1 = ℓ1 < · · · < ℓg}. Since M = M(H) = 2k − 2 and H is a
sparse semigroup, we have that ℓ2k−1 = ℓM+1 = 2k and, by the Corollary
1.3, SM = 2. If g < 2k, then ℓg < 2k, a contradiction with the value
of ℓM+1. If g = 2k, then ℓg = 2k. So, g = M + 1 = 2k − 1 = g − 1, a
contradiction. Therefore, g > 2k+1. It is clear that g 6= 3k− 1, because if
g = 3k−1, then ℓg = 4k−2 ∈ Gaps(H), a contradiction, since 2k−1 ∈ H .
Now, suppose that H has genus g = 3k + s, for some integer s > 2. So,
ℓg = 4k+2s and follows that ℓg/2 = 2k+ s ∈ Gaps(H). Since 2k−1 ∈ H ,
we have that ℓ := 2k + 2s+ 1 ∈ Gaps(H). We prove that 2k + 2s ∈ H.
In fact, suppose that 2k + 2s ∈ Gaps(H). Thus, (ℓ − 1, ℓ) is a single
leap. If ℓ2k = 2k + 1, then (2k, 2k + 1) is a single leap. Note that, if
ℓ2k = 2k+2, then 2k+1 ∈ H and so 2k+2s−1 ∈ Gaps(H) (since s > 2),
or equivalently, (ℓ − 2, ℓ − 1) is a single leap. On the other hand, since
SM = 2, by Remark 2.7, follows that (2k + 2s+ 1) ≡ (4k + 2s) (mod 2),
a contradiction. Therefore, 2k + 2s ∈ H. Then, 2k + 2s − 1 ∈ Gaps(H)
and 4k+2s− 1 = (2k− 1)+ (2k+2s) ∈ H . Thus, ℓg−1 = 4k+2s− 2 and
follows that 2k+ s− 1 ∈ Gaps(H). So, (2k+ s− 1, 2k+ s) is a single leap.
Therefore, since 2k + 2s+ 1 is odd and 4k + 2s− 2 is even, we have that
∣
∣V[2k+2s+1,4k+2s−2] ∩ V1(H)
∣
∣ = 1. We conclude the proof by studying the
following cases separately: s = 2 and s > 3. Firstly, assume that s = 2. In
this case, (2k + 1, 2k + 2) and (2k + 2, 2k + 3) are both single leaps. So,
SM = 3, a contradiction. Now, suppose that s > 3. Then, 2k + s− 2 ∈ H
and consequently 4k+2s− 4 ∈ H . So, ℓg−2 = 4k+2s− 3, a contradiction,
since 4k + 2s− 3 = (2k − 1) + (2k + 2s− 2) ∈ H.
Theorem 2.14. For all integer k > 3,
H
k
2k−2 =
{
H
(s,αs)
(k,r) : (r, s, αs) ∈ N
3, 1 6 r 6 k − 2, 1 6 s 6 r,
1 6 αs 6 r − s+ 1
}
∪
{
H
(1,k)
(k,r) : r ∈ {k, k + 1}
}
.
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112 On certain families of sparse numerical semigroups
Proof. Firstly, by the Lemma 2.11, we have that
{
H
(s,αs)
(k,r) : (r, s, αs) ∈ N
3, 1 6 r 6 k− 2, 1 6 s 6 r, 1 6 αs 6 r− s+1
}
∪
{
H
(1,k)
(k,r) : r ∈ {k, k + 1}
}
⊂ Hk
2k−2.
Now, let H ∈ Hk
2k−2 with g(H) = g. By Theorem 2.13, we have
2k + 1 6 g 6 3k + 1 and g 6= 3k − 1. Let r := g − 2k, then 1 6 r 6 k − 2
or r ∈ {k, k + 1}. Since ℓg = 2g − 2k, we have that ℓg = ℓ2k+r = 2k + 2r.
Let Gaps(H) = {1 = ℓ1 < · · · < ℓg} be the gaps set of H. By Corollary
1.3, SM = 2. That is, there exists exactly two single leaps (ℓi−1, ℓi) and
(ℓj−1, ℓj), for some integers i and j such that 2k 6 i < j 6 g = 2k + r.
We affirm that if r ∈ {k, k + 1}, then i = 2k and j = 3k. In fact, first
we will prove that (2k, 2k + 1) is a single leap. Indeed, since M = 2k − 2,
we have that ℓ2k−2 = 2k − 2 and ℓ2k−1 = 2k. In particular, 2k − 1 ∈ H.
Suppose that 2k + 1 ∈ H. Thus, we have the following: if r = k, then
ℓg = 4k = (2k − 1) + (2k + 1) ∈ H; and if r = k + 1, then ℓg = 4k + 2 =
2(2k + 1) ∈ H. Therefore, in both cases, we get a contradiction. Thus,
(2k, 2k + 1) is a single leap and so i = 2k. Now, we will prove that j = 3k.
Since, 4k − 2 = 2(2k − 1) ∈ H, follows that 4k − 1 is a gap. Then,
by Remark 2.7, 2k + 2 ∈ H, since (2k + 2) 6≡ (4k − 1) (mod 2). Thus,
4k + 1 = (2k − 1) + (2k + 2) ∈ H and so 4k is a gap. Consequently,
(4k − 1, 4k) is a single leap. Note that, if r = k, then ℓ3k = ℓg = 4k, and
if r = k + 1, then ℓ3k = ℓg−1 = 4k. Therefore, in both cases, j = 3k.
For r ∈ {1, . . . , k−2}∪{k, k+1}, let s := i−2k+1 and αs := j−i. Note
that if r ∈ {1, . . . , k− 2}, then 1 6 s 6 r and 1 6 αs = (j− 2k)− s+1 6
r − s+ 1, and if r ∈ {k, k + 1}, then s = 1 and αs = k.
We will prove that H = H
(s,αs)
(k,r) , with s and αs as above. In fact, by
Remark 2.7, we have that ℓu = ℓi−1 − 2(i− u− 1), for all integer u such
that 2k − 1 6 u 6 i − 1, ℓv = ℓi + 2(v − i), for all integer v such that
i 6 v 6 j−1, and ℓw = ℓj+2(w−j), for all integer w such that j 6 w 6 g.
Since ℓg = 2g−2k, it is not hard conclude that ℓj = 2(j−k) and, therefore,
ℓj−1 = 2(j−k)−1. In the same way we can conclude that ℓi = 2(i−k)+1
and ℓi−1 = 2(i− k), and that ℓu = 2(u− k+1), for all integer u such that
2k − 1 6 u 6 i− 1 = 2k + s− 2, ℓv = 2(v − k) + 1, for all integer v such
that 2k + s− 1 = i 6 v 6 j − 1 = 2k + s+ αs − 2 and ℓw = 2(w − k), for
all integer w such that 2k + s+ αs − 1 = j 6 w 6 g = 2k + r. Therefore,
by Remark 2.12, we have that H = H
(s,αs)
(k,r) .
Corollary 2.15. For all integer k > 3,
∣
∣Hk
2k−2
∣
∣ =
(
k
3
)
+ 2.
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G. Tizziotti, J. Villanueva 113
Proof. By the Theorem 2.14, we have that
H
k
2k−2 =
k−2
⋃
·
r=1
r
⋃
·
s=1
{
H
(s,αs)
(k,r) : αs ∈ N, 1 6 αs 6 r − s+ 1
}
∪·
{
H
(1,k)
(k,r) : r ∈ {k, k + 1}
}
.
Therefore,
∣
∣
∣
H
k
2k−2
∣
∣
∣
=
k−2
∑
r=1
r
∑
s=1
(r − s+ 1) + 2 =
k−2
∑
r=1
(
r + 1
2
)
+ 2 =
(
k
3
)
+ 2.
3. On Weierstrass semigroup
Let X be a non-singular, projective, irreducible, algebraic curve of
genus g > 1 over a field K. Let K(X ) be the field of rational functions on
X and for f ∈ K(X ), (f)∞ will denote the divisor of poles of f . Let P be
a point on X . The set
H(P ) := {n ∈ N0 : there exist f ∈ K(X ) with (f)∞ = nP},
is a numerical semigroup called Weierstrass semigroup of X at P .
Given a numerical semigroup H , does it exist a curve X such that for
some point P ∈ X has H = H(P )? If the answer is yes, we say that the
numerical semigroup H is Weierstrass. Studies to answer this question
have been done for decades, see e.g. [4], [8], [10], [11], [12], [13], [15] and [19].
In addition to the genus g(H), Frobenius number ℓg(H) and multiplicity
n1(H), an important concept in this study is the weight of a numerical
semigroup H. If Gaps(H) = {1 = ℓ1 < · · · < ℓg} be the gaps set of H,
the weight of H is
w(H) =
g
∑
i=1
(ℓi − i).
As a particular result, a numerical semigroup H is Weierstrass if the
following condition hold:
either w(H) 6 g(H)/2, or g(H)/2 < w(H) 6 g(H)− 1 and 2n1(H) > ℓg(H) (∗)
(see Eisenbud-Harris [7], Komeda [9]).
Next, we will see which of the semigroups in the families studied in
the previous section are Weierstrass.
“adm-n1” — 2019/3/22 — 12:03 — page 114 — #122
114 On certain families of sparse numerical semigroups
Lemma 3.1. Let k > 2 be an integer. If H = Hαr
(k,r) ∈ Hk
2k−1, then
w(H) = αr +
r(r+1)
2 and 2n1(H) > ℓg(H).
Proof. Since H = Hαr
(k,r) ∈ Hk
2k−1, by Remark 2.6, we have that
w(H) =
2k−1+αr
∑
i=2k
(i−2k+1)+
2k+r
∑
i=2k+αr
(i−2k) =
r
∑
i=1
i+αr =
r(r + 1)
2
+αr.
On the other hand, by Lemma 2.5, ℓg(H) = 2k + 2r 6 4k − 2 < 4k =
2(2k) = 2n1(H), since r 6 k − 1.
Proposition 3.2. Let k > 2 be an integer and H = Hαr
(k,r) ∈ Hk
2k−1. If
αr +
r(r−1)
2 6 2k − 1, then H is Weierstrass. In particular, if k > 3 and
r ∈
{
1, . . . ,
⌊
−1+
√
16k−7
2
⌋}
, then H is Weierstrass.
Proof. By previous lemma, 2n1(H) > ℓg(H). So, from the condition (∗)
above, follows that H is Weierstrass if w(H) 6 g(H)−1. On the other hand,
by Lemma 2.5, g(H) = 2k + r, and, by Lemma 3.1, w(H) = αr +
r(r+1)
2 .
Thus, H is Weierstrass if αr +
r(r+1)
2 6 2k+ r− 1, that is, if αr +
r(r−1)
2 6
2k − 1.
Now, if r ∈
{
1, . . . ,
⌊
−1+
√
16k−7
2
⌋}
, then r + r(r−1)
2 6 2k − 1. Also,
since k > 3, we have that r 6 k − 1. So, since αr 6 r, the required result
follows.
Lemma 3.3. Let k > 3 be an integer. If H = H
(s,αs)
(k,r) ∈ Hk
2k−2, then
w(H) = 2s−1+αs+
r(r+1)
2 . In addition, if r 6∈ {k, k+1}, then 2n1(H) >
ℓg(H).
Proof. Since H = H
(s,αs)
(k,r) ∈ Hk
2k−2, by Remark 2.12, we have that
w(H) =
2k+s−2
∑
i=2k−1
(i− 2k + 2) +
2k+s+αs−2
∑
i=2k+s−1
(i− 2k + 1) +
2k+r
∑
i=2k+s+αs−1
(i− 2k)
=
r
∑
i=−1
i+ 2s+ αs
= −1 +
r(r + 1)
2
+ 2s+ αs.
On the other hand, by Lemma 2.11, ℓg(H) = 2k + 2r 6 4k − 2 =
2(2k − 1) = 2n1(H), since r 6 k − 2.
“adm-n1” — 2019/3/22 — 12:03 — page 115 — #123
G. Tizziotti, J. Villanueva 115
Proposition 3.4. Let k > 3 be an integer and H = H
(s,αs)
(k,r) ∈ Hk
2k−2.
If r 6∈ {k, k + 1} and 2s + αs +
r(r−1)
2 6 2k, then H is Weierstrass.
In particular, if k > 4 and r ∈
{
1, . . . ,
⌊
−3+
√
16k+1
2
⌋}
, then H is
Weierstrass.
Proof. Using the lemmas 2.11 and 3.3, the proof of the first statement is
analogous to the proof of the Proposition 3.2.
Now, if r ∈
{
1, . . . ,
⌊
−3+
√
16k+1
2
⌋}
, then 2r + r(r−1)
2 6 2k − 1. Also,
since k > 4, we have that r 6 k − 2. So, since s 6 r and αs 6 r − s+ 1,
the required result follows.
We observe that, for r ∈ {k, k+ 1}, the condition (∗) above cannot be
used to conclude that the semigroups H
(1,k)
(k,r) ’s are Weierstrass.
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Contact information
G. Tizziotti Universidade Federal de Uberlândia, Câmpus
Santa Mônica, Faculdade de Matemática, Av.
João Naves de Ávila 2.160, Santa Mônica,
38.408-100, Uberlândia - MG, Brasil
E-Mail(s): guilhermect@ufu.br
J. Villanueva Universidade Federal de Mato Grosso, Câmpus
Universitário do Araguaia, Instituto de Ciências
Exatas e da Terra, Av. Senador Valdon Varjão
6.390, Setor Industrial, 78.600-000, Barra do
Garças - MT, Brasil
E-Mail(s): vz_juan@yahoo.com.br
Received by the editors: 20.07.2017
and in final form 22.11.2017.
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