1. Introduction
Throughout this paper,
$\mathbb {Z}$
,
$\mathbb {N}$
and
$\mathbb {Z}^{+}$
denote the set of all integers, nonnegative integers and positive integers, respectively.
A subset S of
$\mathbb {N}$
is a numerical semigroup if
$0\in S$
,
$\mathbb {N}\setminus S$
is finite and S is closed under the addition in
$\mathbb {N}$
. Given a positive integer sequence
$A=(a_1,a_2,\ldots ,a_k)$
, if
$\gcd (A)=1$
, then
$$ \begin{align*}\langle A\rangle=\{x_1a_1+x_2a_2+\cdots +x_ka_k \mid k\geq 2, x_i\in \mathbb{N}, 1\leq i\leq k \}\end{align*} $$
is a numerical semigroup (see [Reference Rosales and García-Sánchez13]) and A is a system of generators of
$S=\langle A\rangle $
. If no proper subset of A generates S, then we say that A is a minimal system of generators of S. Sylvester [Reference Sylvester16] defined the denumerant
$d(a_0;a_1,a_2,\ldots ,a_k)$
as
$$ \begin{align*}d(a_0;a_1,a_2,\ldots,a_k)=\#\{(x_1,\ldots,x_k)\mid x_1a_1+x_2a_2+\cdots +x_ka_k=a_0, x_i\in \mathbb{N} \}. \end{align*} $$
If
$\gcd (A)=1$
, then there exists a positive integer N such that
$d(a_0;a_1,\ldots ,a_k)>0$
for any integer
$a_0\geq N$
(see, for example, [Reference Ramírez Alfonsín11, Theorem 1.0.1]). The greatest integer not belonging to
$\langle A\rangle $
is the Frobenius number of A defined by
$$ \begin{align*} F(A)=\max\{a_0\in \mathbb{Z}^{+}\mid d(a_0;a_1,a_2,\ldots,a_k)=0\}.\end{align*} $$
For more descriptions and results about numerical semigroups, see [Reference Assi, D’Anna and García-Sánchez4, Reference Ramírez Alfonsín11, Reference Rosales and García-Sánchez13].
Suppose
$\langle A\rangle $
is a numerical semigroup and
$p\in \mathbb {Z}^{+}$
. The quotient of
$\langle A\rangle $
by p,
$$ \begin{align*}\frac{\langle A\rangle}{p}=\{n\in \mathbb{N} \mid pn\in \langle A\rangle\}=\{n \mid pn=x_1a_1+x_2a_2+\cdots +x_ka_k, x_i\in\mathbb{N}, 1\leq i\leq k\},\end{align*} $$
was introduced in [Reference Rosales, García-Sánchez, García-García and Urbano-Blanco14]. It is easy to verify that
${\langle A\rangle }/{p}$
is a numerical semigroup, that
$\langle A\rangle \subseteq {\langle A\rangle }/{p}$
, and that
${\langle A\rangle }/{p}=\mathbb {N}$
if and only if
$p\in \langle A\rangle $
. For example, let
$p=3$
and
$\langle A\rangle =\langle 5,6\rangle =\{0,5,6,10,11,12,15,16,17,18,20\rightarrow \}$
, where the symbol
$\rightarrow $
means that all subsequent integers are included. Then
${\langle A\rangle }/{3}=\{0,2,4\rightarrow \}=\langle 2,5\rangle $
.
Let
$a_1,a_2,p$
be pairwise relatively prime positive integers. Rosales [Reference Rosales12] obtained a system of generators for
${\langle a_1,a_2\rangle }/{2}$
and Rosales and Urbano-Blanco [Reference Rosales and Urbano-Blanco15] gave a characterisation of a system of generators for
${\langle a_1,a_2\rangle }/{p}$
by means of modular permutations and certain congruence equations. In [Reference Cabanillas6], Cabanillas discussed the minimal generators of
${\langle a_1,a_2\rangle }/{p}$
. In [Reference Moscariello10], Moscariello also gave a characterisation of the generating system of
${\langle A\rangle }/{p}$
by defining a class of partitions. There are many open problems related to
${\langle A\rangle }/{p}$
(see, for example, [Reference Delgado, García-Sánchez and Rosales7]).
The representation generating function of
${\langle A\rangle }/{p}$
is the generating function
$$ \begin{align*}\textrm{RGF}_p(x)=\sum_{n\geq 0}d(pn;a_1,a_2,\ldots,a_k)x^n.\end{align*} $$
The function
$\textrm {RGF}_p(x)$
is easily seen to be rational (see Section 2.1) and we can use it to obtain a system of generators for
${\langle A\rangle }/{p}$
. For example, let
$a_1$
and
$a_2$
be relatively prime odd positive integers. Then
$$ \begin{align*}\sum_{n\geq 0}d(n;a_1,a_2)x^n=\frac{1}{(1-x^{a_1})(1-x^{a_2})}\end{align*} $$
and the representation generating function of
$\langle a_1,a_2\rangle /2$
is determined by
$$ \begin{align*}\textrm{RGF}_2(x^2)=\frac{1}{2}\bigg(\frac{1}{(1-x^{a_1})(1-x^{a_2})}+\frac{1}{(1-(-x)^{a_1})(1-(-x)^{a_2})}\bigg) =\frac{1+x^{a_1+a_2}}{(1-x^{2a_1})(1-x^{2a_2})}.\end{align*} $$
Therefore,
${\langle a_1,a_2\rangle }/{2}=\langle a_1,a_2,{(a_1+a_2)}/{2}\rangle $
.
We use MacMahon’s partition analysis [Reference MacMahon9] to represent
$\textrm {RGF}_p(x)$
as the constant term of a rational function in a new variable
$\lambda $
. For small
$p\in \mathbb {Z}^{+}$
and
$A=(a_1,a_2,a_3)$
with
$\gcd (A)=1$
, we can calculate
$\textrm {RGF}_p(x)$
and obtain a system of generators of the quotient of the numerical semigroup
$\langle A\rangle $
by p. We give the results for
$p=2$
and 3 in Table 1. We write
$a_i=pk_i+t_i$
, where
$0\leq t_i\leq p-1$
and
$p,k_i\in \mathbb {Z}^{+}$
for
$1\leq i\leq 3$
.
A system of generators of
${\langle a_1,a_2,a_3\rangle }/{p}$
for
$p=2,3$
.
We can extend this idea to give the following simple characterisations for
${\langle A\rangle }/{p}$
.
Theorem 1.1. Suppose
$A=(a_1,a_2,\ldots ,a_n)=(pk_1+t_1,pk_2+t_2,\ldots ,pk_n+t_n)$
with
$\gcd (A)=1$
,
$p\in \mathbb {Z}^{+}$
,
$k_i\in \mathbb {N}$
,
$1\leq t_i\leq p-1$
for
$1\leq i\leq n$
and
$n\geq 2$
. Let
$$ \begin{align*}\mathcal{T}_p=\bigg\{(x_1,x_2,\ldots,x_n) \mid 0\leq x_1,x_2,\ldots,x_n\leq p-1, p\ \bigg| \ \sum_{i=1}^nx_it_i \ (\neq 0)\bigg\}.\end{align*} $$
Then a system of generators of the quotient of the numerical semigroup
$\langle A\rangle $
by p is given by
$$ \begin{align*}\frac{\langle A\rangle}{p}=\bigg\langle a_1,a_2,\ldots,a_n,\frac{1}{p}\sum_{i=1}^nx_ia_i\ \bigg|\ (x_1,x_2,\ldots,x_n)\in \mathcal{T}_p\bigg\rangle.\end{align*} $$
The paper is organised as follows. In Section 2, we introduce MacMahon’s partition analysis and the constant term method following [Reference Xin17, Reference Xin18]. We calculate
$\textrm {RGF}_p(x)$
and obtain a system of generators for
${\langle 3k_1+1,3k_2+2,3k_3+2\rangle }/{3}$
and
${\langle a,a+1\rangle }/{(a-1)}$
to illustrate how to use the method. In Section 3, we give the proof of Theorem 1.1.
2. MacMahon’s partition analysis
In algebraic combinatorics, MacMahon’s partition analysis [Reference MacMahon9] is one of the tools for solving counting problems connected to linear Diophantine equations and inequalities. Such problems can be transformed into finding the constant term of an Elliott rational function, that is, a rational function whose denominator is a product of binomials. This process has been studied by Andrews et al. using computer algebra [Reference Andrews1–Reference Andrews, Paule and Riese3]. Algorithms have been developed, such as the Omega package [Reference Andrews2], the Ell package [Reference Xin17] and the CTEuclid package [Reference Xin18]. We will work with symbolic data.
We introduce some basic definitions and results from [Reference Xin17, Reference Xin18]. We work in the field
$K={\mathbb {Q}}((\lambda ))((x))$
of double Laurent series. In this field, every rational function has a unique Laurent series expansion, so that the following definition makes sense.
Definition 2.1 [Reference Xin17]
Suppose an element in
$K={\mathbb {Q}}((\lambda ))((x))$
is written as a formal Laurent series
$\sum _{i=-\infty }^{\infty } a_{i}\lambda ^{i}$
in
$\lambda $
, where
$a_{i}$
are elements in
${\mathbb {Q}}((x))$
. Then the constant term operator
$\mathrm {CT}_{\lambda }$
acts by
$$ \begin{align*}{\mathrm{CT}}_{\lambda}\sum_{i=-\infty}^{\infty}a_{i}\lambda^{i}=a_0.\end{align*} $$
This definition is extended to
$\mathop {\mathrm {CT}}_{\Lambda }$
for a set of variables
$\Lambda =\{ \lambda _1,\lambda _2, \ldots ,\lambda _m\}$
in [Reference Xin17]. Here we only need the case
$m=1$
.
To work with rational functions in K, we need to clarify their series expansions. A monomial
$M=x^k \lambda ^\ell \neq 1$
is said to be small, denoted
$M<1$
, if
$k>0$
or if
$k=0$
and
$\ell>0$
, and is said to be large, denoted
$M>1$
, otherwise. The series expansion for
$1/(1-M)$
in K is
$$ \begin{align*} \dfrac{1}{1-M}=\left\{ \begin{array}{@{}lr} \displaystyle \sum_{k\geq 0}M^k&\ \text{if}\ M<1; \\ \dfrac{1}{-M(1-1/M)}=- \displaystyle\sum_{k\geq 0}\dfrac{1}{M^{k+1}}&\ \text{if}\ M>1. \end{array} \right. \end{align*} $$
To obtain the series expansion of an Elliott rational function E, we write E in its proper form,
$$ \begin{align*} E = \frac{L}{ \prod_{j=1}^n (1-M_j)}=L\prod_{j=1}^n \bigg(\sum_{k\ge 0} (M_j) ^k\bigg), \end{align*} $$
where L is a Laurent polynomial and each monomial
$M_j$
is small. Note that the proper form of E is not unique. For instance,
$1/(1-x)=(1+x)/(1-x^2)$
are both proper forms.
2.1. Extracting the constant term
Consider the
$\textrm {RGF}_p(x)$
of a numerical semigroup
${\langle A\rangle }/{p}$
, where
$\langle A\rangle =\langle a_1,a_2,\ldots ,a_k\rangle $
,
$\gcd (A)=1$
and
$p\in \mathbb {Z}^{+}$
. We introduce a new variable
$\lambda $
to replace the linear constraint
$pn=c_1a_1+c_2a_2+\cdots +c_ka_k$
, so that
$$ \begin{align} \textrm{RGF}_p(x)&=\sum_{n\geq 0}d(pn;a_1,a_2,\ldots,a_k)x^n\notag \\&=\sum_{n\geq 0, c_i\geq 0}{\mathrm{CT}}_{\lambda}\,\lambda^{c_1a_1+c_2a_2+\cdots +c_ka_k-pn}x^n\notag \\&={\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^p}) (1-\lambda^{a_1})(1-\lambda^{a_2})\cdots (1-\lambda^{a_k})}. \end{align} $$
In the third line, we used the sum of a geometric series and the linearity of the
$\mathop {\mathrm {CT}}$
operator. This gives a power series in x with the powers of
$\lambda $
ranging from
$-\infty $
to
$\infty $
. Thus, we have represented
$\textrm {RGF}_p(x)$
as the constant term of an Elliott rational function. It follows that
$\textrm {RGF}_p(x)$
is also an Elliott rational function, since by [Reference Xin17, Theorem 3.2], the constant term of an Elliott rational function is still Elliott rational.
Remark 2.2. By definition, the Frobenius number of
${\langle A\rangle }/{p}$
is the greatest integer m with
$\textrm {RGF}_p^{(m)}(0)=0$
, that is,
$$ \begin{align*}F\bigg(\frac{\langle A\rangle}{p}\bigg)=\max\{n\in \mathbb{N}\mid d(pn;a_1,a_2,\ldots,a_k)=0\} =\max\{m \in\mathbb{N}\mid \textrm{RGF}_p^{(m)}(0)=0\}.\end{align*} $$
To extract the constant term, we use partial fraction decompositions of univariate rational functions, from which the constant term can be read off. To this end, we write
$$ \begin{align} E=\dfrac{L(\lambda)}{\prod_{i=1}^n(1-u_i\lambda^{a_i})}, \end{align} $$
where
$L(\lambda )$
is a Laurent polynomial, the
$u_i$
are free of
$\lambda $
and the
$a_i$
are positive integers for all i. Note that we might have
$u_i\lambda ^{a_i}=x^{-1}\lambda ^2>1$
, so that (2.2) is not a proper form.
Proposition 2.3 [Reference Xin18]
Suppose that the partial fraction decomposition of E is given by
$$ \begin{align} E=P(\lambda)+\dfrac{p(\lambda)}{\lambda^k}+\sum_{i=1}^n\dfrac{A_i(\lambda)}{1-u_i\lambda^{a_i}}, \end{align} $$
where the
$u_i$
are free of
$\lambda $
,
$P(\lambda ), p(\lambda )$
and
$A_i(\lambda )$
are all polynomials,
$\deg _p(\lambda )<k$
, and
$\deg A_{i}(\lambda )<a_i$
for all i. Then
$$ \begin{align*}{\mathrm{CT}}_{\lambda}\,E=P(0)+\sum_{u_i\lambda^{a_i}<1}A_i(0),\end{align*} $$
where the sum ranges over all i such that
$u_i\lambda ^{a_i}$
is small in
${\mathbb {Q}}((\lambda ))((x))$
.
We can see that the proposition holds by direct series expansion:
$$ \begin{align*} \dfrac{A_i(\lambda)}{1-u_i\lambda^{a_i}}=\left\{ \begin{array}{@{}lr} \dfrac{A_i(\lambda)}{1-u_i\lambda^{a_i}} \stackrel{\mathop{\mathrm{CT}}_{\lambda}}{\longrightarrow}A_i(0)& \text{if } u_i\lambda^{a_i}<1; \\ \dfrac{A_i(\lambda)}{-u_i\lambda^{a_i}(1-{1}/{u_i\lambda^{a_i}})} =\dfrac{\lambda^{-a_i}A_i(\lambda)}{-u_i(1-{1}/{u_i\lambda^{a_i}})} \stackrel{\mathop{\mathrm{CT}}_{\lambda}}{\longrightarrow}0& \text{if } u_i\lambda^{a_i}>1. \end{array} \right. \end{align*} $$
For clarity, we have written the rational function in its proper form before applying the operator
$\mathop {\mathrm {CT}}_\lambda $
.
Theorem 2.4 [Reference Xin18]
Let E be as in (2.3). Then
$A_s(\lambda )$
is uniquely characterised by
$$ \begin{align} A_s(\lambda)\equiv E\cdot (1-u_s\lambda^{a_s}) \,\mod\langle 1-u_s\lambda^{a_s}\rangle,\quad deg_{\lambda}A_s<a_s, \end{align} $$
where
$\langle 1-u_s\lambda ^{a_s}\rangle $
denotes the ideal generated by
$1-u_s\lambda ^{a_s}$
.
To compute
$\mathop {\mathrm {CT}}_{\lambda }E$
for E as in (2.2) in K, we need to compute
$$ \begin{align*}A_s(0):=\mathcal{A}_{1-u_s\lambda^{a_s}}E=\mathcal{A}_{1-(u_s\lambda^{a_s})^{-1}}E,\end{align*} $$
where
$A_s(\lambda )$
is characterised by (2.4). In this new notation, Proposition 2.3 reads
$$ \begin{align*} {\mathrm{CT}}_{\lambda}\,E=P(0) +\sum_{i}\chi(u_i\lambda^{a_i}<1)\mathcal{A}_{1-u_i\lambda^{a_i}}E, \end{align*} $$
where
$\chi (\varepsilon )=1$
if the proposition
$\varepsilon $
is true and
$\chi (\varepsilon )=0$
if
$\varepsilon $
is false.
Theorem 2.5 [Reference Xin18]
Let E be as in (2.2). If E is proper in
$\lambda $
, that is, the degree in the numerator is less than the degree in the denominator, then
$$ \begin{align} {\mathrm{CT}}_{\lambda}\,E =\sum_{i=1}^n\chi(u_i\lambda^{a_i}<1)\mathcal{A}_{1-u_i\lambda^{a_i}}E. \end{align} $$
If
$E|_{\lambda =0}=\lim _{\lambda \rightarrow 0}E$
exists, then
$$ \begin{align} {\mathrm{CT}}_{\lambda}\,E =E|_{\lambda=0}-\sum_{i=1}^n\chi(u_i\lambda^{a_i}>1)\mathcal{A}_{1-u_i\lambda^{a_i}}E. \end{align} $$
Equation (2.6) is a kind of dual of (2.5). Because of these two formulae, it is convenient to call the denominator factor
$1-u_i\lambda ^{a_i}$
contributing if
$u_i\lambda ^{a_i}$
is small and dually contributing if
$u_i\lambda ^{a_i}$
is large. We also write
$$ \begin{align*}{\mathrm{CT}}_{\lambda}\,\underline{\dfrac{1}{1-u_s\lambda^{a_s}}}E(1-u_s\lambda^{a_s})=\mathcal{A}_{1-u_s\lambda^{a_s}}E=A_s(0).\end{align*} $$
For this notation, we allow
$a_s<0$
. One can think that only the single underlined factor of the denominator contributes when taking the constant term in
$\lambda $
.
Lemma 2.6. If E given by (2.2) is proper in
$\lambda $
, that is, the degree in the numerator is less than the degree in the denominator, and
$E|_{\lambda =0}=0$
, then
$$ \begin{align*} \sum_{s=1}^n \mathcal{A}_{1-u_s\lambda^{a_s}}E =0.\end{align*} $$
2.2. Two examples
In this section, we obtain systems of generators for the numerical semigroups
${\langle 3k_1+1,3k_2+2,3k_3+2\rangle }/{3}$
and
${\langle a, a+1\rangle }/{(a-1)}$
by calculating their representation generating functions
$\textrm {RGF}_p(x)$
.
Proposition 2.7. Let
$A=(a_1,a_2,a_3)=(3k_1+1,3k_2+2,3k_3+2)$
,
$k_1,k_2,k_3\in \mathbb {N}$
, with
$\gcd (A)=1$
. A system of generators of
${\langle A\rangle }/{3}$
is given by
$$ \begin{align} \frac{\langle A\rangle}{3}=\bigg\langle a_1,a_2,a_3,\frac{a_1+a_2}{3},\frac{a_1+a_3}{3},\frac{2a_2+a_3}{3},\frac{2a_3+a_2}{3}\bigg\rangle. \end{align} $$
Proof. The right-hand side of (2.7) is easily seen to be contained in the left-hand side. To show that the left-hand side is contained in the right-hand side, we compute as follows. By (2.1),
$$ \begin{align} \textrm{RGF}_3(x)&={\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{3}})(1-\lambda^{3k_1+1}) (1-\lambda^{3k_2+2})(1-\lambda^{3k_3+2})}\notag \\&={\mathrm{CT}}_{\lambda}\,\frac{-1}{\underline{(1-{x}/{\lambda^{3}})} (1-\lambda^{3k_1+1})(1-\lambda^{3k_2+2})(1-\lambda^{3k_3+2})}\quad \text{(by Theorem }2.5)\notag \\&={\mathrm{CT}}_{\lambda}\,\frac{-1}{\underline{(1-{x}/{\lambda^{3}})} (1-x^{k_1}\lambda)(1-x^{k_2}\lambda^{2})(1-x^{k_3}\lambda^{2})}\quad \text{(by Theorem }2.4)\notag \\&={\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{3}}) \underline{(1-x^{k_1}\lambda)(1-x^{k_2}\lambda^{2})(1-x^{k_3}\lambda^{2})}}\quad \text{(by Lemma }2.6)\notag \\&=\frac{1}{(1-x^{3k_1+1})(1-x^{k_2-2k_1})(1-x^{k_3-2k_1})}\notag \\&\quad +{\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{3}}) (1-x^{k_1}\lambda)\underline{(1-x^{k_2}\lambda^{2})}(1-x^{k_3}\lambda^{2})} \notag \\&\quad +{\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{3}}) (1-x^{k_1}\lambda)(1-x^{k_2}\lambda^{2})\underline{(1-x^{k_3}\lambda^{2})}}. \end{align} $$
The second term of (2.8) is
$$ \begin{align*} &{\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{3}}) (1-x^{k_1}\lambda)\underline{(1-x^{k_2}\lambda^{2})}(1-x^{k_3}\lambda^{2})} \\&\quad={\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x^{k_2+1}}/{\lambda}) (1-x^{k_1}\lambda)\underline{(1-x^{k_2}\lambda^{2})}(1-x^{k_3-k_2})}\quad \text{(by Theorem }2.4) \\&\quad={\mathrm{CT}}_{\lambda}\,\frac{-\lambda x^{-k_2-1}}{(1-{\lambda}/{x^{k_2+1}}) (1-x^{k_1}\lambda)\underline{(1-x^{k_2}\lambda^{2})}(1-x^{k_3-k_2})} \\&\quad={\mathrm{CT}}_{\lambda}\,\frac{\lambda x^{-k_2-1}}{\underline{(1-{\lambda}/{x^{k_2+1}}) (1-x^{k_1}\lambda)}(1-x^{k_2}\lambda^{2})(1-x^{k_3-k_2})}\quad \text{(by Lemma }2.6) \\&\quad=\frac{1}{(1-x^{k_1+k_2+1})(1-x^{3k_2+2})(1-x^{k_3-k_2})}+\frac{x^{-k_1-k_2-1}}{(1-x^{-k_1-k_2-1})(1-x^{k_2-2k_1})(1-x^{k_3-k_2})}. \end{align*} $$
Similarly, the third term of (2.8) is
$$ \begin{align*} &{\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x^{k_3+1}}/{\lambda}) (1-x^{k_1}\lambda)(1-x^{k_2-k_3})\underline{(1-x^{k_3}\lambda^2)}}\quad \text{(by Theorem }2.4) \\&\quad={\mathrm{CT}}_{\lambda}\,\frac{\lambda x^{-k_3-1}}{\underline{(1-{\lambda}/{x^{k_3+1}}) (1-x^{k_1}\lambda)}(1-x^{k_2-k_3})(1-x^{k_3}\lambda^2)}\quad \text{(by Lemma }2.6) \\&\quad=\frac{1}{(1-x^{k_1+k_2+1})(1-x^{k_3-k_2})(1-x^{3k_3+2})}+\frac{x^{-k_1-k_3-1}}{(1-x^{-k_1-k_3-1})(1-x^{k_3-k_2})(1-x^{k_3-2k_1})}. \end{align*} $$
Therefore, we obtain the representation generating function in the form
$$ \begin{align*} &\textrm{RGF}_3(x)\\&\hspace{-3pt}=\frac{1+(x^{k_1+1}+x^{k_2+k_3+2})(x^{k_2}+x^{k_3})+x^{2k_1+2}(x^{2k_2}+x^{2k_3}) +x^{k_1+k_2+k_3}(x^{k_1+2}+x^{k_2+k_3+3})}{(1-x^{3k_1+1})(1-x^{3k_2+2})(1-x^{3k_3+2})} \\&\hspace{-3pt}=\frac{1+x^{({a_1+a_2})/{3}}+x^{({a_1+a_3})/{3}}+x^{({2a_2+a_3})/{3}}+x^{({2a_3+a_2})/{3}} +x^{{2(a_1+a_2)}/{3}}+x^{{2(a_1+a_3)}/{3}}+x^{({2a_1+a_2+a_3})/{3}} +x^{({a_1+2a_2+2a_3})/{3}}}{(1-x^{a_1})(1-x^{a_2})(1-x^{a_3})}\!. \end{align*} $$
Since
${2a_1+a_2+a_3}={(a_1+a_2)}+{(a_1+a_3)}$
and
${a_1+2a_2+2a_3}={(a_1+a_2)}+(2a_3+ a_2)$
, the power of each term in the series expansion of
$\textrm {RGF}_3(x)$
is contained in the right-hand side. This completes the proof.
Proposition 2.8. Let
$A=(a,a+1)$
,
$a\in\mathbb{Z}^+$
,
$a\geq 3$
. A system of generators of
${\langle a,a+1\rangle }/{(a-1)}$
is given by
$$ \begin{align*}\frac{\langle a,a+1\rangle}{a-1}=\begin{cases} \bigg\langle \dfrac{a+1}{2},\dfrac{a+3}{2},\ldots,a-1,a\bigg\rangle & \text{if } a\ \text{is odd}; \\ \bigg\langle \dfrac{a}{2}+1,\dfrac{a}{2}+2,\ldots,a,a+1\bigg\rangle & \text{if } a\ \text{is even}. \end{cases}\end{align*} $$
Proof. As in the previous proof, we only need to show that the left-hand side is contained in the right-hand side. By (2.1),
$$ \begin{align*} \textrm{RGF}_{a-1}(x)&={\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{p}})(1-\lambda^{a}) (1-\lambda^{a+1})} \\&={\mathrm{CT}}_{\lambda}\,\frac{-1}{\underline{(1-{x}/{\lambda^{a-1}})} (1-\lambda^{a})(1-\lambda^{a+1})}\quad \text{(by Theorem }2.5) \\&={\mathrm{CT}}_{\lambda}\,\frac{-1}{\underline{(1-{x}/{\lambda^{a-1}})} (1-x\lambda)(1-x\lambda^{2})}\quad \text{(by Theorem }2.4) \\&={\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{a-1}}) \underline{(1-x\lambda)(1-x\lambda^2)}}\quad \text{(by Lemma }2.6) \\&=\frac{1}{(1-x^{a})(1-x^{-1})} +{\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x}/{\lambda^{a-1}}) (1-x\lambda)\underline{(1-x\lambda^{2})}}. \end{align*} $$
The computation of the second term depends on the parity of a. If a is odd, it is
$$ \begin{align*} &{\mathrm{CT}}_{\lambda} \frac{1}{(1-x^{{(a+1)}/{2}})(1-x\lambda)\underline{(1-x\lambda^{2})}} =\frac{1}{(1-x^{{(a+1)}/{2}})}\bigg(1-{\mathrm{CT}}_{\lambda}\,\frac{1}{\underline{(1-x\lambda)}(1-x\lambda^2)}\bigg) \\ &\quad=\frac{1}{(1-x^{{(a+1)}/{2}})}\bigg(1-\frac{1}{1-x^{-1}}\bigg) =\frac{1}{(1-x)(1-x^{{(a+1)}/{2}})}. \end{align*} $$
Thus, we obtain
$$ \begin{align*} \textrm{RGF}_{a-1}(x)&=\frac{1-x-x^a+x^{{(a+3)}/{2}}}{(1-x)(1-x^a)(1-x^{{(a+1)}/{2}})} =\frac{1+x^{{(a+3)}/{2}}+x^{{(a+5)}/{2}}+\cdots +x^{a-1}}{(1-x^a)(1-x^{{(a+1)}/{2}})}. \end{align*} $$
If instead a is even, the second term is
$$ \begin{align*} {\mathrm{CT}}_{\lambda}\,\frac{1}{(1-{x^{{a}/{2}}}/{\lambda})(1-x\lambda)\underline{(1-x\lambda^{2})}} &={\mathrm{CT}}_{\lambda}\,\frac{-({\lambda}/{x^{a/2}})}{(1-{\lambda}/{x^{a/2}})(1-x\lambda)\underline{(1-x\lambda^2)}} \\ &={\mathrm{CT}}_{\lambda}\,\frac{{\lambda}/{x^{a/2}}}{\underline{(1-{\lambda}/{x^{a/2}})(1-x\lambda)}(1-x\lambda^2)} \\ &=\frac{1}{(1-x^{{(a+2)}/{2}})(1-x^{a+1})}+\frac{x}{(1-x^{{(a+2)}/{2}})(1-x)}. \end{align*} $$
Thus, we obtain
$$ \begin{align*} \textrm{RGF}_{a-1}(x)&=\frac{-x}{(1-x)(1-x^{a})}+\frac{1}{(1-x^{{(a+2)}/{2}})(1-x^{a+1})}+\frac{x}{(1-x^{{(a+2)}/{2}})(1-x)}\\[4pt] &=\frac{1+(x^{{(a+2)}/{2}}+x^{a+2})(1+x+x^2+\cdots +x^{{a}/{2}-2})}{(1-x^a)(1-x^{a+1})}. \end{align*} $$
The proposition then follows.
Note that
${\langle a,a+1\rangle }/({a-1)}$
is a half-line numerical semigroup (see [Reference Bras-Amorós5]). Therefore, its Frobenius number is given by
$$ \begin{align*}F\bigg(\frac{\langle a,a+1\rangle}{a-1}\bigg)=\begin{cases} \dfrac{a-1}{2} &\ \text{if}\ a\ \text{is odd}, \\[2 ex] \dfrac{a}{2} &\ \text{if}\ a\ \text{is even}. \end{cases}\end{align*} $$
3. A system of generators of
${\langle A\rangle }/{p}$
Let
$A=(a_1,a_2)=(pk_1+t_1,pk_2+t_2)$
,
$0\leq t_i\leq p-1$
,
$p,k_1,k_2\in \mathbb {Z}^{+}$
and
$\gcd (A)=1$
. We can compute
$\textrm {RGF}_p(x)$
for
$p=2,3,4,5$
as in the proof of Proposition 2.7 and obtain a system of generators of
${\langle a_1,a_2\rangle }/{p}$
. The results agree with those in [Reference Rosales and Urbano-Blanco15, Proposition 17].
Similarly, for
$A=(a_1,a_2,a_3)$
, we can obtain
$\textrm {RGF}_{p}(x)$
for
$p=2,3$
. The corresponding systems of generators are given in Table 1. This table illustrates a pattern summarised in Theorem 1.1. The theorem has a simple direct proof.
Proof of Theorem 1.1
Let
$\overline {B}:=\langle a_1,a_2,\ldots ,a_n, \sum _{i=1}^nx_ia_i/p \mid (x_1,x_2,\ldots ,x_n)\in \mathcal {T}_p\rangle $
. This is well defined since
$(x_1a_1+x_2a_2+\cdots +x_na_n)/p\in \mathbb {Z}^+$
by definition of
$\mathcal {T}_p$
. Let
$\overline {A}:={\langle A\rangle }/{p}$
. The containment
$\overline {A}\supseteq \overline {B}$
is obvious and we need to show that
$\overline {A}\subseteq \overline {B}$
.
If
$x\in \overline {A}={\langle A\rangle }/{p}$
, then
$xp=y_1a_1+y_2a_2+\cdots +y_na_n$
for some
$y_1,y_2,\ldots ,y_n\in \mathbb {N}$
. Each
$y_i$
is uniquely written as
$y_i=m_ip+r_i$
for some
$m_i\geq 0$
and
$0\leq r_i\leq p-1$
. Then we have
$x=m_1a_1+m_2a_2+\cdots +m_na_n+(r_1a_1+r_2a_2+\cdots +r_na_n)/p$
and we have
$p\mid (r_1a_1+r_2a_2+\cdots +r_na_n)$
. By the definition of
$\mathcal {T}_p$
,
$(r_1a_1+r_2a_2+\cdots +r_na_n)/p$
is either
$0$
or an element in
$\{(x_1a_1+x_2a_2+\cdots +x_na_n)/p\ |\ (x_1,x_2,\ldots ,x_n)\in \mathcal {T}_p\}$
. In either case,
$x\in \overline {B}$
. Therefore,
$\overline {A}\subseteq \overline {B}$
.
Corollary 3.1. Suppose that
$A=(a_1,a_2,a_3)=(pk_1+t_1,pk_2+t_2,pk_3+t_3)$
with
$p\in \mathbb {Z}^{+}$
,
$k_i\in \mathbb {N}$
and
$1\leq t_i\leq p-1$
for
$1\leq i\leq 3$
. If
$\gcd (A)=1$
, then a system of generators of the quotient of the numerical semigroup
$\langle A\rangle $
by p is given by
$$ \begin{align*}\frac{\langle A\rangle}{p}=\bigg\langle a_1,a_2,a_3,\frac{1}{p}(x_1a_1+x_2a_2+x_3a_3)\ \bigg|\ (x_1,x_2,x_3)\in \mathcal{T}_p\bigg\rangle,\end{align*} $$
where
$$ \begin{align*}\mathcal{T}_p=\{(x_1,x_2,x_3) \mid 0\leq x_1,x_2,x_3\leq p-1, p\mid (t_1x_1+t_2x_2+t_3x_3) ( \neq 0)\}.\end{align*} $$
We observe that Theorem 1.1 can be strengthened in the following sense. Suppose that
$A=(a_1,\ldots ,a_e,a_{e+1},\ldots ,a_n)$
,
$p\mid a_i$
for
$1\leq i\leq e$
and
$p\nmid a_j$
for
$e+1\leq j\leq n$
. Then
$$ \begin{align} \frac{\langle A\rangle}{p}=\bigg\langle \frac{a_1}{p},\frac{a_2}{p},\ldots,\frac{a_e}{p},\mathcal{L}_p\bigg\rangle, \end{align} $$
where
$\mathcal {L}_p$
is a system of generators of
$\langle a_{e+1},\ldots ,a_n\rangle /p$
. We only explain why the left-hand side is contained in the right-hand side because the other containment is trivial. For any
$x\in {\langle A\rangle }/{p}$
, there exists
$xp=y_1a_1+y_2a_2+\cdots +y_na_n$
for some
$y_1,y_2,\ldots ,y_n\in \mathbb {N}$
and
$$ \begin{align*}x=y_1\frac{a_1}{p}+\cdots +y_e\frac{a_e}{p}+\frac{1}{p}(y_{e+1}a_{e+1}+\cdots +y_na_n).\end{align*} $$
Therefore,
$x\in \langle {a_1}/{p},{a_2}/{p},\ldots ,{a_e}/{p},\mathcal {L}_p\rangle $
.
Remark 3.2. Theorem 1.1 only gives a system of generators of
${\langle A\rangle }/{p}$
, rather than a minimal system of generators. A minimal system of generators of
${\langle a_1,a_2\rangle }/{p}$
is given in [Reference Cabanillas6].
Combining (3.1) and Theorem 1.1, we reobtain the following result.
Corollary 3.3 [Reference Rosales and Urbano-Blanco15, Corollary 18]
Let
$a_1,a_2,k_1,k_2\in \mathbb {Z}^{+}$
and
$\gcd (a_1,a_2)=1$
. Then
$$ \begin{align*}\frac{\langle a_1,a_2\rangle}{2}=\begin{cases} \bigg\langle \dfrac{a_1}{2},a_2\bigg\rangle & \mbox{if } a_1=2k_1, a_2=2k_2+1, \\ \bigg\langle a_1,a_2,\dfrac{a_1+a_2}{2}\bigg\rangle & \mbox{if }a_1=2k_1+1, a_2=2k_2+1. \end{cases} \end{align*} $$
Another consequence of Theorem 1.1 is the following result.
Corollary 3.4 [Reference Rosales and Urbano-Blanco15, Corollary 19]
Let
$a_1,a_2,k_1,k_2\in \mathbb {Z}^{+}$
and
$\gcd (a_1,a_2)=1$
. If
$a_1=3k_1+1$
,
$a_2=3k_2+1$
, or
$a_1=3k_1+2$
,
$a_2=3k_2+2$
, then
$$ \begin{align*}\frac{\langle a_1,a_2\rangle}{3}=\bigg\langle a_1,a_2,\frac{2a_1+a_2}{3},\frac{2a_2+a_1}{3}\bigg\rangle.\end{align*} $$
If
$a_1=3k_1+1$
,
$a_2=3k_2+2$
, then
$$ \begin{align*}\frac{\langle a_1,a_2\rangle}{3}=\bigg\langle a_1,a_2,\frac{a_1+a_2}{3}\bigg\rangle.\end{align*} $$
Proof. If
$(t_1,t_2)=(1,1)$
or
$(t_1,t_2)=(2,2)$
, then
$\mathcal {T}_p=\{(1,2),(2,1)\}$
. If
$(t_1,t_2)=(1,2)$
, then
$\mathcal {T}_p=\{(1,1),(2,2)\}$
. This completes the proof.
4. Future work
Let
$s\in \mathbb {N}$
,
$A=(a_1,a_2,\ldots ,a_k)$
and
$\gcd (A)=1$
. Komatsu [Reference Komatsu8] introduced the s-numerical semigroup defined by
$\langle A;s\rangle =\{n\in \mathbb {N} \mid d(n;a_1,a_2,\ldots ,a_k)\geq s+1\} \cup \{0\}$
and considered its Frobenius number, called the s-Frobenius number
$F_s(A)$
of A. In other words,
$F_s(A)$
is the largest number N satisfying
$d(N;a_1,\ldots ,a_k)\leq s$
. These concepts reduce to the classical one when
$s=0$
. It would be of interest to see if our methods can be used to compute these more general Frobenius numbers.
Acknowledgements
The author would like to thank the referee for helpful comments and suggestions. The author thanks his advisor Guoce Xin for guidance and support.
