---
author:
- Francisco Franco Munoz
title: '**On the counting function for numerical monoids**'
---

Introduction
============

The study of numerical monoids is a huge area of research (see [@RS] and references therein). A numerical monoid $M$ is a subset of ${\mathbb{N}}$ containing $0$ that is closed under addition and finite complement. In other words, a finite complement submonoid of $({\mathbb{N}},+)$.

The aim of this paper is to calculate in an elementary way the main term of the counting function of a monoid (Theorems \[thm-5\], \[thm-13\], \[thm-14\]) which is a natural object. We’ll avoid the use of more advanced treatments such as quasi-polynomial functions and integer points of polytopes ([@BLDPS]) that provide a more comprehensive view of the subject. The paper is essentially self-contained and requires no prior knowledge of the subject.

Notation
--------

${\mathbb{N}}= \{0, 1, 2, 3, ...\}$, the natural numbers\
${\mathbb{Z}}= -{\mathbb{N}}\cup {\mathbb{N}}$, the integers\
${\mathbb{N}}_{+} = \{1, 2, 3, ...\}$, the positive integers

Setting
=======

Let $M = \langle a_1, \dots, a_{n-1}, a_n \rangle$ be a submonoid of $({\mathbb{N}}, +)$. By definition $M$ is called a numerical monoid if its complement is finite. Denote by $M_{\widehat{a_n}} = \langle a_1, \dots, a_{n-1} \rangle $ the submonoid obtained by dropping the generator $a_n$.

For $x\in {\mathbb{R}}$, $a\in {\mathbb{N}}_{+}$, define $\displaystyle {\{x; a\}} = x-a{\floor*{\frac{x}{a}} } $.

With this definition, we have $\{x\} = {\{x; 1\}}$ the fractional part of $x$.

\[prop-1\] Let $x, k\in {\mathbb{Z}}$, $a, b\in {\mathbb{N}}_{>0}$

1.  ${\{x; a\}}$ is the residue of $x$ modulo $a$

2.  $0\leq {\{x; a\}} < a$

3.  ${\{x; a\}}=0 \iff a | x$.

4.  ${\{x+ka; a\}}={\{x; a\}}$

5.  ${\{x; b\}} = {\{{\{x; ab\}}; b\}}$

Immediate.

\[lem-2\] For $y\in {\mathbb{N}}$, if $y<a_n$, then $y\in M \iff y\in M_{\widehat{a_n}}$. In particular, for $x\in {\mathbb{N}}$, ${\{x; a_n\}} \in M \iff {\{x; a_n\}} \in M_{\widehat{a_n}}$.

This is clear.

We define the counting function $$T_{M}(x) = \# \{(x_1, ... , x_n) \in {\mathbb{N}}^n  \mid  x_1a_1 + ... + x_na_n = x \}$$

The Theory of Erhart polynomials [@BLDPS] provides ways to determine $T_M$. Instead we’ll address them directly in an elementary way.

$$T_{M}(x)=T_{M}(x-a_n)+T_{M_{\widehat{a_n}}}(x)$$

This is immediate since the set $ \{(x_1, ... , x_n) \in {\mathbb{N}}^n  \mid  x_1a_1 + ... + x_na_n = x \}$ can be partitioned as $ \{(x_1, ... , x_n) \in {\mathbb{N}}^n  \mid  x_1a_1 + ... + x_na_n = x \} = $ $$=\{(x_1, ... , x_n) \in {\mathbb{N}}^n  \mid  x_n > 0 , x_1a_1 + ... + x_na_n = x \}\sqcup  \{(x_1, ... , x_n) \in {\mathbb{N}}^n  \mid  x_n=0, x_1a_1 + ... + x_na_n = x\}$$ $= \{(x_1, ... , x_{n-1}, y_n) \in {\mathbb{N}}^n  \mid  y_n=x_n-1\geq 0 , x_1a_1 + ... + x_{n-1}a_{n-1}+y_na_n = x-a_n \}\sqcup \{(x_1, ... , x_{n-1}, 0) \in {\mathbb{N}}^n  \mid  x_n=0, x_1a_1 + ... + x_{n-1}a_{n-1}+0 = x\}$ and the counting for each is $T_{M}(x-a_n)+T_{M_{\widehat{a_n}}}(x)$.

\[cor-4\]

$\displaystyle T_{M} (x) = \sum_{k=0}^{{\floor*{\frac{x}{a_n}} }} T_{M_{\widehat{a_n}}}(x-ka_n) =  \sum_{l=0}^{{\floor*{\frac{x}{a_n}} }} T_{M_{\widehat{a_n}}}({\{x; a_n\}}+la_n) $

$\displaystyle T_{M} (x) - T_{M}\left(x-a_n\floor*{\frac{x}{a_n}}\right) =  \sum_{k=0}^{ {\floor*{\frac{x}{a_n}} } -1} \left[T_{M}(x-ka_n)-T_{M}(x-(k+1)a_n)\right] \\=\sum_{k=0}^{{\floor*{\frac{x}{a_n}} } - 1} T_{M_{\widehat{a_n}}}(x-ka_n)$. Also $T_{M}\left(x-a_n\floor*{\frac{x}{a_n}}\right)=T_{M}({\{x; a_n\}})=T_{M_{\widehat{a_n}}}({\{x; a_n\}}) = T_{M_{\widehat{a_n}}}\left(x-a_n\floor*{\frac{x}{a_n}}\right)$ by Lemma \[lem-2\] and we conclude that $\displaystyle T_{M} (x) = \sum_{k=0}^{{\floor*{\frac{x}{a_n}} }} T_{M_{\widehat{a_n}}}(x-ka_n)$ and making the change of variable $k={\floor*{\frac{x}{a_n}} }-l$ we get the last equality.

Dimension $n=2$
===============

$M = \langle a, b \rangle$ and assume that $\gcd(a,b)=1$ in this subsection.

\[thm-5\] We have $$T_{\langle a, b \rangle} (x) =  {\floor*{\frac{x}{ab}} } + \epsilon_{\langle a, b\rangle}({\{x; ab\}})$$ where for $0\leq y < ab$, $\epsilon(y)=\delta(y\in \langle a, b\rangle)$ is the indicator function.

Applying Corollary \[cor-4\] with $n=2$ and $a_2=b$, we have $\displaystyle T_{\langle a, b \rangle}(x) =  \sum_{l=0}^{{\floor*{\frac{x}{b}} }} T_{M_{\widehat{b}}}({\{x; b\}}+lb) = \sum_{l=0}^{{\floor*{\frac{x}{b}} }} T_{\langle a\rangle}({\{x; b\}}+lb)$. Now, it’s clear that $T_{\langle a \rangle }(y)$ is $1$ or $0$ according to whether $a|y$ or not and so $T_{\langle a\rangle}(y+ka)=T_{\langle a\rangle}(y)$. Let $y={\floor*{\frac{x}{b}} }$. By Proposition \[prop-1\], ${\floor*{\frac{{\floor*{\frac{x}{b}} }}{a}} }={\floor*{\frac{x}{ab}} }$, so $y={\{y; a\}} + a{\floor*{\frac{x}{ab}} }$. We split the sum $\displaystyle \sum_{l=0}^{{\floor*{\frac{x}{b}} }} T_{\langle a\rangle}({\{x; b\}}+lb) =  \sum_{i=0}^{{\floor*{\frac{x}{ab}} } - 1} \sum_{j=0}^{a-1} T_{\langle a\rangle}\left({\{x; b\}}+(ai+j)b\right) + \sum_{j=0}^{{\{y; a\}}} T_{\langle a\rangle}({\{x; b\}}+(a{\floor*{\frac{x}{ab}} }+j)b) =   \sum_{i=0}^{{\floor*{\frac{x}{ab}} } - 1} \sum_{j=0}^{a-1} T_{\langle a\rangle}({\{x; b\}}+jb) + \sum_{j=0}^{{\{y; a\}}} T_{\langle a\rangle}({\{x; b\}}+jb) = {\floor*{\frac{x}{ab}} }\sum_{j=0}^{a-1} T_{\langle a\rangle}({\{x; b\}}+jb) + \sum_{j=0}^{{\{y; a\}}} T_{\langle a\rangle}({\{x; b\}}+jb)$ and the proof follows from the next two lemmas.

For any $y\in {\mathbb{N}}$, $\displaystyle \sum_{j=0}^{a-1} T_{\langle a\rangle}(y+jb)=1$

Since $\gcd(a,b)=1$, there’s unique $j, 0\leq j\leq a-1$ such that $a\mid (y+jb)$.

For any $x\in {\mathbb{N}}$, then $\displaystyle \sum_{j=0}^{{\{y; a\}}} T_{\langle a\rangle}({\{x; b\}}+jb) = \epsilon_{\langle a, b\rangle}({\{x; ab\}})$ where $y={\floor*{\frac{x}{b}} }$.

Omitted.

A different proof
-----------------

\[lem-8\] If $x=\alpha a+\beta b\in M=\langle a, b\rangle$, then $T_{M}(x)={\floor*{\frac{\alpha}{b}} }+{\floor*{\frac{\beta}{a}} }+1$

Fix a representation $x=\alpha a+\beta b$. We need to count representations $x=\alpha_1 a+\beta_1 b$. But $\alpha a+\beta b=\alpha_1 a+\beta_1 b \iff (\alpha-\alpha_1)a=(\beta_1-\beta)b$ and since $\gcd(a,b)=1$ this happens $\iff a\mid (\beta_1-\beta), b\mid (\alpha-\alpha_1)$ and $\beta_1-\beta=ja, \alpha-\alpha_1 = jb$. So we need to count the possibilities for $j\in {\mathbb{Z}}$. We have $\beta_1=ja+\beta$, $\alpha_1=-jb+\alpha$. And the condition is exactly $\beta_1\geq 0$ and $\alpha_1\geq 0$, which happens $\iff ja+\beta\geq 0, -jb+\alpha\geq 0 \iff \frac{-\beta}{a}\leq j\leq \frac{\alpha}{b} \iff {\ceil*{\frac{-\beta}{a}} }\leq j\leq {\floor*{\frac{\alpha}{b}} } $. Now we have $ {\ceil*{\frac{-\beta}{a}} } = - {\floor*{\frac{\beta}{a}} }$ hence there are ${\floor*{\frac{\alpha}{b}} }+{\floor*{\frac{\beta}{a}} }+1$ such $j$.

\[cor-9\] For $x<ab$, $T_{M}(x)=1 \iff x\in M$.

It’s enough to show that if $x\in M$ and $x<ab$, then $T_{M}(x)=1$. By Lemma \[lem-8\], this holds iff ${\floor*{\frac{\alpha}{b}} }={\floor*{\frac{\beta}{a}} }=0$ where $x=\alpha a+\beta b$ is some representation. But since $x< ab$, then $\alpha a\leq x < ab$ so $\alpha<b$ and $\beta b\leq x < ab$ so $\beta<a$ so ${\floor*{\frac{\alpha}{b}} }={\floor*{\frac{\beta}{a}} }=0$.

\[lem-10\] If $x\in M$, then for all $k\in {\mathbb{N}}$, $T_{M}(x+kab)=T_{M}(x)+k$

By Lemma \[lem-8\], take $x=\alpha a+\beta b\in M$ then $x+kab\in M$ and $x+kab=(\alpha+kb) a+\beta b$ and so $T_M(x+kab)={\floor*{\frac{\alpha+kb}{b}} }+{\floor*{\frac{\beta}{a}} }+1=k+{\floor*{\frac{\alpha}{b}} }+{\floor*{\frac{\beta}{a}} }+1= T_{M}(x)+k$.

\[lem-11\] If $x\notin M$, $0\leq x\leq ab-1$, then $T_{M}(x+ab)=1$

Since $x+ab\geq ab$ and the Frobenius number of $M$ is $ab-a-b$ ([@RS]), $x+ab$ belongs to $M$, so $T_{M}(x+ab)\geq 1$. Now, as in Corollary \[cor-9\], it’s enough to show that in any representation of $x+ab=\alpha a + \beta b$, both $\alpha < b$ and $\beta < a$. If not, say $\alpha\geq b$, then $x=\alpha a + \beta b -ab = (\alpha-b)a + \beta b$ which says that $x\in M$, contradiction. In the same way, $\beta < a$ and we’re done.

For all $x\in {\mathbb{N}}, T_{M}(x+ab)=T_{M}(x)+1$ and Theorem \[thm-5\] is valid.

By contradiction, take $x$ minimal such that this formula is not valid for $x$. Notice that by Lemma \[lem-11\], we have $x\geq ab$, and by Lemma \[lem-10\], we have $x\notin M$. Let $y=x-ab < x$ then by minimality, the theorem is valid for $y$ and so $T_{M}(x)=T_{M}(y+ab)=T_{M}(y)+1\geq 1$ in particular $x\in M$, contradiction.\
Since $\forall x\in {\mathbb{N}}, T_{M}(x+ab)=T_{M}(x)+1$, by induction $\forall x, k\in {\mathbb{N}}$, $T_{M}(x+kab)=T_{M}(x)+k$, and so $T_{M}(x)=T_{M}({\{x; ab\}}+ab{\floor*{\frac{x}{ab}} })= {\floor*{\frac{x}{ab}} } + T_{M}({\{x; ab\}})= {\floor*{\frac{x}{ab}} } + \epsilon_{\langle a, b\rangle}({\{x; ab\}})$ the last equality since ${\{x; ab\}} < ab$ and using Corollary \[cor-9\]. This proves the theorem.

Dimension $n=3$
===============

Assume $\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1$. These conditions are stronger than necessary, but the theory of numerical monoids permits one to reduce to this case in any dimension [@RS].

Let $M = \langle a, b, c \rangle$, $M_{\widehat{c}}= \langle a, b \rangle$ dimension $3$ and $2$ numerical monoids.\
Recall $\displaystyle T_{M} (x) = \sum_{l=0}^{{\floor*{\frac{x}{c}} }} T_{\langle a, b\rangle}({\{x; c\}}+lc)$

Let $y={\floor*{\frac{x}{c}} }$. Then $y= {\{y; ab\}} + ab{\floor*{\frac{x}{abc}} }$. By Theorem \[thm-5\], $T_{\langle a, b\rangle}$ splits as sum, consequently we split the sum $\displaystyle  \sum_{l=0}^{{\floor*{\frac{x}{c}} }} T_{\langle a, b\rangle}({\{x; c\}}+lc) = \sum_{l=0}^{{\floor*{\frac{x}{c}} }} {\floor*{\frac{{\{x; c\}}+lc}{ab}} } + \sum_{l=0}^{{\floor*{\frac{x}{c}} }}  \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+cl; ab\}})= S1 + S2$\
$S1= \displaystyle \sum_{l=0}^{{\floor*{\frac{x}{c}} }} {\floor*{\frac{{\{x; c\}}+lc}{ab}} } =   \sum_{i=0}^{{\floor*{\frac{x}{abc}} } - 1} \sum_{j=0}^{ab-1} {\floor*{\frac{{\{x; c\}}+(abi+j)c}{ab}} } +  \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+(ab{\floor*{\frac{x}{abc}} }+j)c}{ab}} } \\= \sum_{i=0}^{{\floor*{\frac{x}{abc}} } - 1} \sum_{j=0}^{ab-1} {\floor*{\frac{{\{x; c\}}+jc}{ab}} }  +  \sum_{i=0}^{{\floor*{\frac{x}{abc}} } - 1} \sum_{j=0}^{ab-1} ic+ \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} }+ \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{x}{abc}} }c \\=  {\floor*{\frac{x}{abc}} }\sum_{j=0}^{ab-1} {\floor*{\frac{{\{x; c\}}+jc}{ab}} }  + \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right){\floor*{\frac{x}{abc}} } + \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} }+ ({\{y; ab\}}+1) {\floor*{\frac{x}{abc}} }c $

$S2= \displaystyle \sum_{l=0}^{{\floor*{\frac{x}{c}} }} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+cl; ab\}})= \sum_{i=0}^{{\floor*{\frac{x}{abc}} } - 1} \sum_{j=0}^{ab-1} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+(abi+j)c; ab\}})  +  \sum_{j=0}^{{\{y; ab\}}} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+(ab{\floor*{\frac{x}{abc}} }+j)c; ab\}}) =  \sum_{i=0}^{{\floor*{\frac{x}{abc}} } - 1} \sum_{j=0}^{ab-1} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}})  +  \sum_{j=0}^{{\{y; ab\}}} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}}) \\=  {\floor*{\frac{x}{abc}} } \sum_{j=0}^{ab-1} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}})  +  \sum_{j=0}^{{\{y; ab\}}} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}}) $\
Using Hermite reciprocity, since $\gcd(ab,c)=1$:\
$\displaystyle \sum_{j=0}^{ab-1} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } = \sum_{j=0}^{c-1} {\floor*{\frac{{\{x; c\}}+jab}{c}} } =  \sum_{j=0}^{c-1} {\floor*{\frac{{\{x; c\}}+{\{jab; c\}} + c{\floor*{\frac{jab}{c}} }   }{c}} } =  \sum_{j=0}^{c-1} {\floor*{\frac{{\{x; c\}}+{\{jab; c\}} }{c}} } + \sum_{j=0}^{c-1}{\floor*{\frac{jab}{c}} } =  \sum_{l=0}^{c-1} {\floor*{\frac{{\{x; c\}}+l }{c}} } + \sum_{j=0}^{c-1}{\floor*{\frac{jab}{c}} } =  \floor*{c\frac{{\{x; c\}}}{c}}+\frac{(ab-1)(c-1)}{2} = {\{x; c\}}+\frac{(ab-1)(c-1)}{2} $\
Hence we obtain $S1=$\
$\displaystyle \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } + {\floor*{\frac{x}{abc}} }\sum_{j=0}^{ab-1} {\floor*{\frac{{\{x; c\}}+jc}{ab}} }  + \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right){\floor*{\frac{x}{abc}} } + ({\{y; ab\}}+1) {\floor*{\frac{x}{abc}} }c \\= \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } + {\floor*{\frac{x}{abc}} }\left({\{x; c\}}+\frac{(ab-1)(c-1)}{2}\right)+ \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right){\floor*{\frac{x}{abc}} } + \\ ({\{y; ab\}}+1) {\floor*{\frac{x}{abc}} }c = \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } +  {\floor*{\frac{x}{abc}} }\left( \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right) + \\ {\{x; c\}}+\frac{(ab-1)(c-1)}{2} + \\ ({\{y; ab\}}+1)c  \right) \\=   \sum_{j=0}^{{\{y; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } + S3$, where $S3$ is the sum\
$\displaystyle S3= {\floor*{\frac{x}{abc}} }\left( \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right) + {\{x; c\}}+\frac{(ab-1)(c-1)}{2} +  ({\{y; ab\}}+1)c  \right) \\= {\floor*{\frac{x}{abc}} }\left( \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right) + {\{x; c\}}+\frac{(ab-1)(c-1)}{2} + \\ ({\{{\floor*{\frac{x}{c}} }; ab\}}+1)c  \right) \\= {\floor*{\frac{x}{abc}} }\left( \frac{abc}{2}\left({\floor*{\frac{x}{abc}} }-1\right) + x-c{\floor*{\frac{x}{c}} }+\frac{(ab-1)(c-1)}{2} +  ({\floor*{\frac{x}{c}} }-ab{\floor*{\frac{x}{abc}} }+1)c  \right) \\= {\floor*{\frac{x}{abc}} }\left( \frac{abc}{2}{\floor*{\frac{x}{abc}} }-\frac{abc}{2} + x-c{\floor*{\frac{x}{c}} }+\frac{abc-ab-c+1}{2} + c{\floor*{\frac{x}{c}} }-abc{\floor*{\frac{x}{abc}} }+c  \right) \\= {\floor*{\frac{x}{abc}} }\left(x-\frac{abc}{2}{\floor*{\frac{x}{abc}} } +\frac{c-ab+1}{2} \right) = {\floor*{\frac{x}{abc}} }\left(\frac{abc}{2}{\floor*{\frac{x}{abc}} } + {\{x; abc\}} + \frac{c-ab+1}{2} \right)$

Finally we get $\displaystyle S1=  {\floor*{\frac{x}{abc}} }\left(\frac{abc}{2}{\floor*{\frac{x}{abc}} } + {\{x; abc\}} + \frac{c-ab+1}{2} \right) +  \sum_{j=0}^{{\{{\floor*{\frac{x}{c}} }; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } \\=  \frac{abc}{2}{\floor*{\frac{x}{abc}} }^2 +{\floor*{\frac{x}{abc}} }\left( {\{x; abc\}} + \frac{c-ab+1}{2} \right) +\sum_{j=0}^{{\{{\floor*{\frac{x}{c}} }; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} }$\
Notice that the main term is $\displaystyle  \frac{abc}{2}{\floor*{\frac{x}{abc}} }^2$ which has order $\displaystyle \sim \frac{x^2}{2abc}$.\
The entire result is then $\displaystyle S1+S2 \\=  \frac{abc}{2}{\floor*{\frac{x}{abc}} }^2 +{\floor*{\frac{x}{abc}} }\left( {\{x; abc\}} + \frac{c-ab+1}{2} \right) +\sum_{j=0}^{{\{{\floor*{\frac{x}{c}} }; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } + \\ {\floor*{\frac{x}{abc}} } \sum_{j=0}^{ab-1} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}})  +  \sum_{j=0}^{{\{y; ab\}}} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}})$.\
Each of the sums have easy to find bounds:

-   $\displaystyle {\{x; abc\}} + \frac{c-ab+1}{2} < abc+\frac{c-ab+1}{2}$.

-   $\displaystyle \sum_{j=0}^{{\{{\floor*{\frac{x}{c}} }; ab\}}} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } \leq \displaystyle \sum_{j=0}^{ab-1} {\floor*{\frac{{\{x; c\}}+jc}{ab}} } = {\{x; c\}}+\frac{(ab-1)(c-1)}{2} \leq c+\frac{(ab-1)(c-1)}{2}$.

-   $\displaystyle \sum_{j=0}^{ab-1} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}}) \leq  \sum_{j=0}^{ab-1} 1= ab$

-   $\displaystyle \sum_{j=0}^{{\{y; ab\}}} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}}) \leq  \sum_{j=0}^{ab-1} \epsilon_{\langle a, b\rangle}({\{{\{x; c\}}+jc; ab\}}) \leq ab$

Hence we obtain:

\[thm-13\] $n=3$, $M = \langle a, b, c \rangle$, $\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1$. We have $$T_{M}(x) = \frac{abc}{2}{\floor*{\frac{x}{abc}} }^2 + r_{M}(x)$$ where $|r_M(x)|\leq Kx+L$ for absolute constants $K, L$ (depending only on $a, b, c$).

Dimension $n$ arbitrary
=======================

The technique above generalizes, by summing up the leading terms using Corollary \[cor-4\] obtaining:

\[thm-14\] For $M=\langle a_1, ... , a_d\rangle$, $\gcd(a_i,a_j)=1$ for all $i\neq j$. The counting function $T_{M}(x)$ has main order $$T_M(x) \sim \frac{(a_1....a_d)^{d-2}}{(d-1)!}{\floor*{\frac{x}{a_1.... a_d}} }^{d-1}$$ meaning that $T_{M}(x)=\frac{(a_1...a_d)^{d-2}}{(d-1)!}{\floor*{\frac{x}{a_1.... a_d}} }^{d-1} + r_{M}(x)$, where $|r_{M}(x)|$ is bounded by a polynomial in $x$ (depending only on $M$) of degree at most $d-2$.

[1]{}

Beck, M.; De Loera, J. A.; Develin, M.; Pfeifle, J.; Stanley, R. P. (2005), “Coefficients and roots of Ehrhart polynomials", Integer points in polyhedra- geometry, number theory, algebra, optimization, Contemp. Math., 374, Providence, RI: Amer. Math. Soc., pp. 15-36

J. C. Rosales, J. C and P. A. García-Sánchez, P. A. Numerical semigroups. Developments in Mathematics, 20. Springer, New York, 2009

Francisco Franco Munoz, <span style="font-variant:small-caps;">Department of Mathematics, University of Washington, Seattle, WA 98195</span>

*E-mail address*: `ffm1@uw.edu`
