---
abstract: 'In this paper we study numerical semigroups containing a given positive integer and closed with respect to the action of an affine map. For such semigroups we find a minimal set of generators, their embedding dimension, their genus and their Frobenius number.'
address: 'Dipartimento di Matematica, Università degli studi di Trento, Via Sommarive 14, I-38123 (Italy)'
author:
- 'S. Ugolini'
bibliography:
- 'Refs.bib'
title: On numerical semigroups closed with respect to the action of affine maps
---

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

A numerical semigroup $G$ is a subsemigroup of the semigroup of non-negative integers $({\text{$\mathbb{N}$}}, +)$ containing $0$ and such that ${\text{$\mathbb{N}$}}\backslash G$ is finite. A comprehensive introduction to numerical semigroups is given in [@ros_ns]. Nevertheless, for the reader’s convenience we recall some basic notions, we will make use of in the current paper.

A set $S \subseteq {\text{$\mathbb{N}$}}$ generates a numerical semigroup $G$, namely $G = \langle S \rangle$, if and only if $$\gcd(S) = 1,$$ where $\gcd(S)$ is the greatest common divisor of the elements contained in $S$.

Any numerical semigroup $G$ has a unique finite minimal set of generators, whose cardinality is the embedding dimension $e(G)$ of $G$.

The cardinality of ${\text{$\mathbb{N}$}}\backslash G$ is called the genus of $G$ and is denoted by $g(G)$, while the integer $$F(G) := \max \{ x : x \in {\text{$\mathbb{Z}$}}\backslash G \}$$ is called the Frobenius number of $G$.

If $n \in G \backslash \{ 0 \}$, then the set $$\operatorname{Ap}(G,n) = \{ s \in G: s - n \not \in G \}$$ is called the Apéry set of $G$ with respect to $n$.

For any $a \in {\text{$\mathbb{N}$}}^* := {\text{$\mathbb{N}$}}\backslash \{ 0 \}$ and $b \in {\text{$\mathbb{N}$}}$ we define the affine map $$\begin{array}{rcl}
{\vartheta}_{a,b} : {\text{$\mathbb{N}$}}& \to & {\text{$\mathbb{N}$}}\\
x & \mapsto & ax+b.
\end{array}$$

We give the following definition.

\[tab\_def\] A subsemigroup $G$ of $({\text{$\mathbb{N}$}}, +)$ containing $0$ is a ${\vartheta}_{a,b}$-semigroup if ${\vartheta}_{a,b}(y) \in G$ for any $y \in G \backslash \{ 0 \}$.

The problem we deal with in the paper consists in finding the smallest [${\vartheta}_{a,b}$]{}-semigroup $G_{a,b} (c)$ containing a given integer $c \in {\text{$\mathbb{N}$}}\backslash \{0, 1 \}$, once two positive integers $a$ and $b$ such that $\gcd(b,c)=1$ are chosen. We notice that under such hypotheses $G_{a,b} (c)$ is a numerical semigroup, while the same does not hold if $\gcd(b,c)>1$. Indeed, if $d:=\gcd(b,c) > 1$, then all elements in $G$ are divisible by $d$ and $G$ is not co-finite. The existence and the structure of $G_{a,b} (c)$ are dealt with in Theorem \[t\_g\_is\_h\].

In literature some special cases of [${\vartheta}_{a,b}$]{}-semigroups have been studied.

In [@ros] the authors studied Thabit numerical semigroups, namely numerical semigroups defined for any $n \in {\text{$\mathbb{N}$}}^*$ as $$T(n) := \langle \left\{ 3 \cdot 2^{n+i} - 1 :  i \in {\text{$\mathbb{N}$}}\right\} \rangle.$$ Indeed, if we set $c:=3 \cdot 2^n-1$, then $T(n) = G_{2,1} (c)$.

Also Mersenne numerical semigroups [@ros_m], namely numerical semigroups defined for any $n \in {\text{$\mathbb{N}$}}^*$ as $$M(n) := \langle \left\{ 2^{n+i} - 1 :  i \in {\text{$\mathbb{N}$}}\right\} \rangle,$$ are ${\vartheta}_{2,1}$-semigroups. In this case, setting $c:=2^n-1$, we have that $M(n) = G_{2,1} (c)$.

In [@ros_rep], for a given integer $b \in {\text{$\mathbb{N}$}}\backslash \{0, 1 \}$ and a given positive integer $n$, the authors defined $$M(b,n) := \langle \left\{ b^{n+i} - 1 :  i \in {\text{$\mathbb{N}$}}\right\} \rangle$$ as a submonoid of $({\text{$\mathbb{N}$}}, +)$. If we set $c := b^n-1$, then $M(b,n) = G_{b,b-1} (c)$. We notice that this latter is not a numerical semigroup. Indeed, we have that $\gcd(b-1,c) \not = 1$.

Synopsis of the paper {#synopsis-of-the-paper .unnumbered}
---------------------

The paper is organized as follows.

-   In Section \[notations\] we introduce some notations.

-   In Section \[main\_results\] we present (omitting the proofs) the main results of the paper (Theorem \[t\_g\_is\_h\], Theorem \[t\_m\_gen\], Corollary \[c\_e\_f\] and Theorem \[gen\]). Some examples of [${\vartheta}_{a,b}$]{}-semigroups follow.

-   Section \[background\] and \[proofs\] contain all the necessary background and proofs supporting the results presented in Section \[main\_results\]. In particular, Section \[proofs\] consists of the proofs of Theorem \[t\_g\_is\_h\], Theorem \[t\_m\_gen\] and Theorem \[gen\].

Definitions and notations {#notations}
=========================

Let $\{a, b, c \} \subseteq {\text{$\mathbb{N}$}}^*$.

-   If $y$ and $z$ are two non-negative integers such that $y < z$, then $$\begin{aligned}
    [y, z[ & :=  \{x \in {\text{$\mathbb{N}$}}: y \leq x < z \}; \\
    \left[y, z \right] & :=  \{x \in {\text{$\mathbb{N}$}}: y \leq x \leq z \}; \\
    \left[y,+\infty \right[ & :=  \{x \in {\text{$\mathbb{N}$}}: y \leq x  \}.\end{aligned}$$

-   If $k \in {\text{$\mathbb{N}$}}$, then we define $$s_k(a) := 
    \begin{cases}
    0 & \text{if $k=0$,}\\
    \sum_{i=0}^{k-1} a^i & \text{otherwise,}
    \end{cases}$$ and accordingly $$t_k(a,b,c) := a^k c + b \cdot s_k(a).$$ Moreover, we define the set $$S (a,b,c) := \lbrace t_k(a,b,c) : k \in {\text{$\mathbb{N}$}}\rbrace$$ and, for any non-negative integer $\tilde{k}$, $$S_{\tilde{k}} (a,b,c) := \lbrace t_k(a,b,c) : k \in {\text{$\mathbb{N}$}}\text{ and } 0 \leq k \leq \tilde{k} \rbrace.$$

-   We denote by $H(a,b,c)$ the semigroup generated by $S(a,b,c)$, namely $$H(a,b,c) := \langle S(a,b,c) \rangle.$$ If $\gcd(b,c) = 1$, then $\gcd(c,ac+b)=1$ too. Therefore, since $\{ c, ac+b \} \subseteq S(a,b,c)$, we have that $\gcd(S(a,b,c))=1$ and $S(a,b,c)$ generates a numerical semigroup.

-   If $K$ is a non-empty finite subset of ${\text{$\mathbb{N}$}}$ and $\tilde{k} := \max \{k \in K \}$, then we say that a set $\{ j_i \}_{i \in K}$ of non-negative integers is $a$-reduced if

    -   $j_i \in [0,a]$ for any $i \in K \backslash \{ 0 \}$;

    -   $j_{\tilde{k}} \not = 0$;

    -   if $j_k = a$ for some $k \in K \backslash \{ 0 \}$, then $j_i=0$ for any $i \in K \backslash \{ 0 \}$ such that $i < k$.

-   If $\{ j_i \}_{i \in K_1}$ and $\{ \tilde{j}_i \}_{i \in K_2}$ are two $a$-reduced sets of integers indexed on two subsets $K_1$ and $K_2$ of ${\text{$\mathbb{N}$}}^*$ such that $$\begin{aligned}
    k_1 & :=  \max \{k \in K_1 \},\\
    k_2 & :=  \max \{k \in K_2 \},\end{aligned}$$ then we say that

    -   $\{ j_i \}_{i \in K_1} = \{ \tilde{j}_i \}_{i \in K_2}$ if and only if $K_1 = K_2$ and $j_i = \tilde{j}_i$ for any $i \in K_1$;

    -   $\{ j_i \}_{i \in K_1} \prec \{ \tilde{j}_i \}_{i \in K_2}$ if and only if $k_1 < k_2$ or $k_1 = k_2$ and $j_M < \tilde{j}_M$, where $M := \max \{ k \in K_1 : j_k \not =  \tilde{j}_k \}$.

Main results and examples {#main_results}
=========================

In this and the following sections $\{a, b, c \}$ is a subset of ${\text{$\mathbb{N}$}}^*$, where $c \geq 2$ and $\gcd(b,c) = 1$.

The following holds.

\[t\_g\_is\_h\] We have that $G_{a,b}(c) = H(a,b,c)$.

In the following theorem a minimal set of generators for $G_{a,b} (c)$ is provided.

\[t\_m\_gen\] Let $\tilde{k} := \min \{k \in {\text{$\mathbb{N}$}}: s_k(a) > c - 1  \}$.

Then $S_{\tilde{k}-1} (a,b,c)$ is a minimal set of generators for $G_{a,b}(c)$.

As an immediate consequence of Theorem \[t\_m\_gen\] we obtain the embedding dimension of $G_{a,b} (c)$, since for each $\tilde{k} \in {\text{$\mathbb{N}$}}$ we have that $|S_{\tilde{k}} (a,b,c)| = \tilde{k}+1$.

\[c\_e\_f\] Let $\tilde{k} := \min \{k \in {\text{$\mathbb{N}$}}: s_k(a) > c - 1  \}$.

Then $e (G_{a,b}(c)) = \tilde{k}$.

In the following theorem we determine the Frobenius number $F(G_{a,b} (c))$ and the genus $g(G_{a,b} (c))$ of $G_{a,b} (c)$.

\[gen\] For any $l \in [1,c-1]$ there exists and is unique an $a$-reduced set of integers $\{ j_i^{(l)} \}_{i=1}^{k_l}$, for some positive integer $k_l$, such that $$l = \sum_{i=1}^{k_l} j_i^{(l)} \cdot s_i(a).$$

Moreover, if we define $$x_l :=
\begin{cases}
0 & \text{if $l=0$,}\\
\sum_{i=1}^{k_l} j_i^{(l)} \cdot t_i (a,b,c) & \text{if $l \in [1,c-1]$,}
\end{cases}$$ then the following hold:

1.  $x_l  =  \min \{x \in G_{a,b} (c) : x \equiv b l \pmod{c}  \}$;

2.  $\operatorname{Ap}(G_{a,b} (c),c)  = \{x_l : l \in [0,c-1] \}$;

3.  $F(G_{a,b} (c))  = x_{c-1} - c$;

4.  $g(G_{a,b} (c))  = \frac{1}{c} \cdot \sum_{l=1}^{c-1} x_l - \frac{c-1}{2}$.

As a by-product of Theorem \[gen\] we get the following membership criterion: if $n \in {\text{$\mathbb{N}$}}$ and $n \equiv x_l \pmod{c}$ for some $l \in [0,c-1]$, then $n \in G_{a,b} (c)$ if and only if $n \geq x_l$.

\[exm\_1\] In this example we study the semigroup $G_{3,1} (3)$.

Adopting the notations introduced above we have that $$a = 3, \quad b = 1, \quad c = 3.$$ Moreover, $$\begin{aligned}
x_0 & =  0,\\
x_1 & =  1 \cdot t_1 (3,1,3) =  1 \cdot (3 \cdot 3+1) = 10,\\
x_2 & =  2 \cdot t_1 (3,1,3) = 20.\end{aligned}$$ Therefore, $$\begin{aligned}
F(G_{3,1} (3)) & =  17,\\
g(G_{3,1} (3)) & =  9,\end{aligned}$$ according to Theorem \[gen\].

Since $$\min \{ k \in {\text{$\mathbb{N}$}}: s_k(3) > 2 \} = 2,$$ we have that $$G_{3,1} (3) = \langle S_1 (3,1,3) \rangle$$ according to Theorem \[t\_m\_gen\].

The non-negative integers smaller than $21$ belonging to $G_{3,1} (3)$ are listed in the following table (the numbers in bold are the elements of $S_{1} (3,1,3)$):

   0   **3**   6     9      12   15   18  
  --- ------- --- -------- ---- ---- ---- --
                   **10**   13   16   19  
                                      20  

\[exm\_2\] In this example we study the semigroup $G_{3,1} (5)$.

We have that $$a = 3, \quad b = 1,  \quad c = 5.$$

Moreover, $$\begin{aligned}
x_0 & =  0,\\
x_1 & =  1 \cdot t_1 (3,1,5) = 1 \cdot (3 \cdot 5 + 1) = 16,\\
x_2 & =  2 \cdot t_1 (3,1,5) = 32,\\
x_3 & =  3 \cdot t_1 (3,1,5) = 48,\\
x_4 & =  1 \cdot t_2 (3,1,5) = 3^2 \cdot 5+4 = 49.\end{aligned}$$ Therefore, $$\begin{aligned}
F(G_{3,1} (5)) &  =  44,\\
g(G_{3,1} (5)) & =  27,\end{aligned}$$ according to Theorem \[gen\].

Since $$\min \{ k \in {\text{$\mathbb{N}$}}: s_k(3) > 4 \} = 3,$$ we have that $$G_{3,1} (5) = \langle S_2 (3,1,5) \rangle$$ according to Theorem \[t\_m\_gen\].

The non-negative integers smaller than $50$ belonging to $G_{3,1} (5)$ are listed in the following table (the numbers in bold are the elements of $S_{2} (3,1,5)$):

   0   **5**   10     15     20   25   30   35   40     45
  --- ------- ---- -------- ---- ---- ---- ---- ---- --------
                    **16**   21   26   31   36   41     46
                                       32   37   42     47
                                                        48
                                                      **49**

\[exm\_3\] In this example we study the semigroup $G_{2,3} (4)$.

We have that $$a = 2, \quad b = 3,  \quad c = 4.$$ Moreover, $$\begin{aligned}
x_0 & =  0,\\
x_1 & =  1 \cdot t_1 (2,3,4) = 1 \cdot (2 \cdot 4 + 3) = 11,\\
x_2 & =  2 \cdot t_1 (2,3,4) = 22,\\ 
x_3 & =  1 \cdot t_2 (2,3,4) = 1 \cdot (4 \cdot 4 + 3 \cdot 3) = 25.\end{aligned}$$ Therefore, $$\begin{aligned}
F(G_{2,3} (4)) & = 21,\\
g(G_{2,3} (4)) & = 13,\end{aligned}$$ according to Theorem \[gen\].

Since $$\min \{ k \in {\text{$\mathbb{N}$}}: s_k(2) > 3 \} = 3,$$ we have that $$G_{2,3} (4) = \langle S_2 (2,3,4) \rangle$$ according to Theorem \[t\_m\_gen\].

The non-negative integers smaller than $28$ belonging to $G_{2,3} (4)$ are listed in the following table (the numbers in bold are the elements of $S_{2} (2,3,4)$):

   0   **4**     8      12   16   20     24
  --- ------- -------- ---- ---- ---- --------
                                       **25**
                                  22     26
               **11**   15   19   23     27

Background
==========

In this section we prove some technical lemmas which we will repeatedly use in Section \[proofs\].

\[l\_pre\_0\] We have that $S(a,b,c) \subseteq G_{a,b} (c)$.

We prove by induction on $k \in {\text{$\mathbb{N}$}}$ that any $t_k (a,b,c)$ belongs to $G_{a,b}(c)$.

If $k = 0$, then $t_0(a,b,c) = c \in G_{a,b} (c)$.

Suppose now that $t_k (a,b,c) \in G_{a,b} (c)$ for some non-negative integer $k$. Then $$t_{k+1} (a,b,c) = a \cdot t_k(a,b,c) + b = {\vartheta}_{a,b} (t_k(a,b,c)) \in G_{a,b} (c). \qedhere$$

\[l\_pre\_1\] $H(a,b,c)$ is a subsemigroup of $({\text{$\mathbb{N}$}}, +)$ closed with respect to the action of the map ${\vartheta}_{a,b}$.

By definition, $H(a,b,c)$ is a subsemigroup of $({\text{$\mathbb{N}$}}, +)$.

We prove that $H(a,b,c)$ is closed with respect to the action of the map [${\vartheta}_{a,b}$]{}.

Consider an element $$y = \sum_{k \in K} j_k \cdot t_k(a,b,c) \in H(a,b,c),$$ where $K$ is a non-empty finite subset of ${\text{$\mathbb{N}$}}$ and $\{ j_k \}_{k \in K}$ is a set of positive integers. Let $\tilde{k}$ be a chosen element of $K$. Then

$$\begin{aligned}
a y + b & =  \sum_{k \in K} a j_k \cdot t_k(a,b,c) + b \\
& =  \sum_{\substack{k \in K \\ k \not = \tilde{k}} } a j_k \cdot t_k(a,b,c) + a (j_{\tilde{k}}-1) \cdot t_{\tilde{k}}(a,b,c) + a \cdot t_{\tilde{k}} (a, b, c) + b \\
& =  \sum_{\substack{k \in K \\ k \not = \tilde{k}} } a j_k \cdot t_k(a,b,c) + a (j_{\tilde{k}}-1) \cdot t_{\tilde{k}}(a,b,c) + t_{\tilde{k}+1} (a, b, c).\end{aligned}$$

Since this latter is a linear combination of elements in $S(a,b,c)$ with coefficients in ${\text{$\mathbb{N}$}}$, we conclude that $ay+b \in H(a,b,c)$.

\[l\_pre\_2\] For any $k \in {\text{$\mathbb{N}$}}$ we have that the set $$I_k (a,b,c) := \{a^kc+bi: i \in [0, s_k(a)] \}$$ is contained in $H(a,b,c)$.

We prove the claim by induction on $k$.

Proving the base step is trivial, since $c \in H(a,b,c)$ and $$I_0(a,b,c) = \{ c \}.$$

Suppose now that $I_k(a,b,c) \subseteq H(a,b,c)$ for some $k \in {\text{$\mathbb{N}$}}$.

For any $r \in [0, s_k(a)[$ and any $j \in [0,a]$ we have that $$a^{k+1} c + b (a r + j) \in H(a,b,c).$$ In fact, $$a^{k+1} c + b (a r + j) = (a-j) \cdot (a^k c+br ) +j \cdot (a^k c+b(r+1)),$$ where $$\begin{aligned}
\{a^k c + br, a^k c+b(r+1) \} \subseteq  I_k(a,b,c).\end{aligned}$$

Therefore, $$\{a^{k+1} c + b i : i \in [0,a \cdot s_k(a)] \} \subseteq H(a,b,c).$$

Finally, $$\begin{aligned}
t_{k+1} (a, b, c) & =  {\vartheta}_{a,b} (t_k(a,b,c)) \in  H(a,b,c).\end{aligned}$$ Hence, $I_{k+1} (a,b,c) \subseteq H(a,b,c)$ and the inductive step is proved.

\[l\_pre\_3\] Let $k$ be a non-negative integer such that $s_k(a) \geq c-1$. Then $$[t_k (a,b,c), + \infty[ \subseteq H(a,b,c).$$

Let $y \in [t_k(a,b,c), + \infty[$. Then $$y \equiv r \pmod{c}$$ for some $r \in [0,c-1]$.

Since $\gcd(b,c) = 1$, there exists an integer $i \in [0,c-1] \subseteq [0, s_k(a)]$ such that $$a^k c + bi \equiv r \pmod{c}.$$ Therefore, $$y-a^k c - bi \equiv 0 \pmod{c},$$ namely $$y = a^k c + bi + cq$$ for some non-negative integer $q$. Since $$\{ a^k c + bi, c \} \subseteq H(a,b,c)$$ according to Lemma \[l\_pre\_2\], we conclude that $y \in H(a,b,c)$ and the result follows.

\[l\_pre\_5\] If $k$ is a positive integer such that $s_k(a) \leq c-1$, then $$t_k(a,b,c) \not \in \langle S_{k-1} (a,b,c) \rangle.$$

Suppose by contradiction that $$\sum_{i \in K} j_i \cdot t_i(a,b,c) = t_k(a,b,c)$$ for some positive integers $\{ j_i \}_{i \in K}$ indexed on a non-empty set $K \subseteq [0, k-1]$. Before proceeding we define $K^*:=K \backslash \{ 0 \}$.

We distinguish three different cases.

-   *Case 1:* $\sum_{i \in K} j_i \cdot a^i \leq a^k$. We distinguish four subcases.

    -   *Subcase 1:* $a = 1$. Then $$\sum_{i \in K} j_i 1^i \leq 1^k.$$ This latter is possible only if $|K| = 1$ and the only integer $j_i$ is equal to $1$. Therefore, $K = \{ r \}$ for some $r \in [0,k-1]$ and $j_r = 1$. Hence, $$\sum_{i \in K} j_i \cdot t_i (a,b,c) = c + b \cdot s_r(1),$$ while $$t_k (a,b,c) = c + b \cdot s_k (1).$$ Since $s_r(1) < s_k(1)$, we conclude that $\sum_{i \in K}  j_i \cdot t_i (a,b,c) \not = t_k(a,b,c)$.

    -   *Subcase 2:* $a > 1$ and $\sum_{i \in K^*} j_i =0$. Then $K = \{ 0 \}$ and $$\sum_{i \in K} j_i \cdot t_i (a,b,c) = j_0 \cdot c \equiv 0 \pmod{c}.$$ Since $$t_k(a,b,c) \equiv b \cdot s_k(a) \pmod{c}$$ and $$b \cdot s_k(a) \not \equiv 0 \pmod{c}$$ because $\gcd(b,c) = 1$ and $1 \leq s_k(a) \leq c-1$, we conclude that $\sum_{i \in K}  j_i \cdot t_i (a,b,c) \not = t_k(a,b,c)$.

    -   *Subcase 3:* $a > 1$ and $\sum_{i \in K^*} j_i =1$. Then either $K = \{ r \}$ or $K= \{ 0, r \}$ for some positive integer $r$. In both cases, $j_r = 1$.

        If $K = \{ r \}$, then $$\sum_{i \in K} j_i \cdot t_i (a,b,c) = a^r c + b \cdot s_r(a)  < t_{k} (a,b,c).$$

        If $K= \{ 0, r \}$, then $$\sum_{i \in K} j_i \cdot t_i (a,b,c) = j_0 \cdot c + a^r c + b \cdot s_r(a).$$

        We notice that $$\begin{array}{rcll}
        t_k (a,b,c) & \equiv & b \cdot s_k (a)  & \pmod{c},\\
        \sum_{i \in K} j_i \cdot t_i (a,b,c) & \equiv & b \cdot s_r (a)&  \pmod{c}.
        \end{array}$$ Since $\gcd (b,c) = 1$ and $$0 < s_k (a) - s_r(a) < c-1,$$ we conclude that $$t_k (a,b,c) \not \equiv \sum_{i \in K} j_i \cdot t_i (a,b,c) \pmod{c},$$ and consequently $t_k(a,b,c) \not = \sum_{i \in K} j_i \cdot t_i (a,b,c)$.

    -   *Subcase 4:* $a > 1$ and $\sum_{i \in K^*} j_i \geq 2$. Then $$\begin{aligned}
        \sum_{i \in K} j_i \cdot t_i(a,b,c) & =  \sum_{i \in K} j_i \cdot a^i c + b \cdot \sum_{i \in K^*} j_i \cdot  s_i(a) \\
        & \leq  a^k \cdot c + b \cdot  \left( \sum_{i \in K^*} j_i \cdot \frac{a^i-1}{a-1} \right) \\
        & =  a^k \cdot c + \frac{b}{a-1} \cdot \left( \sum_{i \in K^*} j_i \cdot a^i - \sum_{i \in K^*} j_i \right) \\
        & \leq  a^k \cdot c + \frac{b}{a-1} \cdot \left( a^k - \sum_{i \in K^*} j_i \right) \\
        & \leq  a^k \cdot c + \frac{b}{a-1} \cdot \left( a^k - 2 \right)\\
        & <  a^k \cdot c + b \cdot \frac{a^k-1}{a-1} = t_k (a,b,c).\end{aligned}$$

-   *Case 2:* $\sum_{i \in K} j_i \cdot a^i > a^k$ and $\sum_{i \in K} j_i \cdot s_i(a) > s_k(a)$. Then $$\sum_{i \in K} j_i \cdot t_i(a,b,c) > t_k(a,b,c),$$ in contradiction with the initial assumption.

-   *Case 3:* $\sum_{i \in K} j_i \cdot a^i > a^k$ and $\sum_{i \in K} j_i \cdot s_i(a) \leq s_k(a)$. Since $$\left(\sum_{i \in K} j_i a^i - a^k \right) \cdot c = b \cdot \left( s_k(a) - \sum_{i \in K} j_i s_i (a) \right)$$ and $$\left(\sum_{i \in K} j_i a^i - a^k \right) \cdot c > 0,$$ we get that $$0 < s_k(a) - \sum_{i \in K} j_i s_i (a) \leq c-1.$$ This latter fact implies that $$b \cdot \left( s_k(a) - \sum_{i \in K} j_i s_i (a) \right) \not \equiv 0 \pmod{c},$$ in contradiction with the fact that $$\left(\sum_{i \in K} j_i a^i - a^k \right) \cdot c \equiv 0 \pmod{c}.$$ Hence, also in this case $\sum_{i \in K} j_i \cdot t_i(a,b,c) \not = t_k(a,b,c)$.

\[l\_pre\_5\_a\] Let $k$ be a positive integer.

If $x$ is a positive integer such that $$s_k(a) \leq x < s_{k+1}(a),$$ then there exists and is unique an $a$-reduced set of integers $\{ j_i \}_{i=1}^k$ such that $$x = \sum_{i=1}^k j_i \cdot s_i(a).$$

We prove the claim by induction on $k \in {\text{$\mathbb{N}$}}^*$.

If $k=1$, then $s_1(a) \leq x < s_2(a) = 1 + a$. Therefore $x = j_1 \cdot s_1(a)$, where $j_1 = x$.

Suppose that $k > 1$ and $s_k (a) \leq x < s_{k+1} (a)$. Then there exist and are unique two non-negative integers $q$ and $r$ such that $$\begin{cases}
x = q  s_k(a) + r \\
0 \leq r < s_k(a)
\end{cases}$$ and $q \leq a$.

If $r =0$, then $q \geq 1$. The result follows setting $j_k := q$ and $j_i := 0$ for any $i < k$.

If $r > 0$, then $1 \leq q < a$ and $s_{\tilde{k}} (a) \leq r < s_{\tilde{k}+1} (a)$ for some positive integer $\tilde{k}$. By inductive hypothesis we have that $$r = \sum_{i=1}^{\tilde{k}} j_i \cdot s_i(a)$$ for some $a$-reduced set of integers $\{ j_i \}_{i=1}^{\tilde{k}}$. Therefore the result follows setting $j_k := q$ and $j_i := 0$ for any $i \in [\tilde{k}+1, k-1]$.

\[l\_pre\_5\_b\] If $$x = \sum_{i \in K} j_i \cdot t_i(a,b,c),$$ where $\{ j_i \}_{i \in K}$ is a set of positive integers indexed on a finite subset $K$ of ${\text{$\mathbb{N}$}}$, then $$x = \sum_{i \in \tilde{K}} \tilde{j}_i \cdot t_i(a,b,c)$$ for some $a$-reduced set of integers $\{ \tilde{j}_i \}_{i \in \tilde{K}}$ indexed on a finite subset $\tilde{K}$ of ${\text{$\mathbb{N}$}}$.

We notice that $$a \cdot t_{i_2} (a,b,c) + t_{i_1} (a,b,c) = t_{i_2+1} (a,b,c) + a \cdot t_{i_1-1} (a,b,c)$$ for any choice of positive integers $i_1$ and $i_2$ such that $i_1 \leq i_2$.

We define $l:=0$, $K(l) := K$ and $\tilde{j}_i := j_i$ for any $i \in K(l)$.

Then we enter the following iterative procedure.

1.  If $\{ \tilde{j}_i \}_{i \in K(l)}$ is $a$-reduced, then we break the procedure, else we define $$\begin{aligned}
    M(l) & :=  \max \{ i \in K(l) : \tilde{j}_i \geq a \},\\
    m(l) & :=  \min \{ i \in K(l) \backslash \{ 0 \} : \tilde{j}_i \not = 0 \}.\end{aligned}$$

2.  We set $$\begin{aligned}
    \tilde{j}_{m(l)-1} & :=  \tilde{j}_{m(l)-1} + a,\\
    \tilde{j}_{m(l)} & :=  \tilde{j}_{m(l)} - 1,\\
    \tilde{j}_{M(l)} & :=  \tilde{j}_{M(l)} - a,\\
    \tilde{j}_{M(l)+1} & :=  \tilde{j}_{M(l)+1} + 1.\\\end{aligned}$$

3.  We set $$\begin{aligned}
    K(l+1) & :=  K(l) \cup \{ m(l)-1, M(l)+1 \},\\
    l & :=  l+1,\end{aligned}$$ and go to step (1).

We notice that for each $l$ we have that $${\displaystyle\sum}_{i \in K(l+1)} j_i = {\displaystyle\sum}_{i \in K(l)} j_i$$ and at least one of the following holds: $$m(l+1) = m(l) - 1  \quad \text{ or } \quad {\displaystyle\sum}_{i \in K(l+1) \backslash \{ 0 \}} j_i < {\displaystyle\sum}_{i \in K(l) \backslash \{ 0 \}} j_i.$$ In particular, when $m(l) = 1$ for some integer $l$, we have that $${\displaystyle\sum}_{i \in K(l+1) \backslash \{ 0 \}} j_i < {\displaystyle\sum}_{i \in K(l) \backslash \{ 0 \}} j_i.$$ Therefore, after some iterations the procedure breaks.

Hence, $$x = \sum_{i \in \tilde{K}} \tilde{j}_i \cdot t_i(a,b,c),$$ where $\tilde{K} := K(l)$.

Suppose that $$a = 2, \quad b = 3, \quad c = 4.$$

Let $K = \{ 1, 2, 4 \}$ and $$j_1 = 2, \quad j_2 = 4, \quad j_4 = 3.$$

We have that $$\begin{aligned}
t_0 (2,3,4) & =  4,\\
t_1 (2,3,4) & =  11,\\
t_2 (2,3,4) & =  25,\\
t_3 (2,3,4) & =  53,\\
t_4 (2,3,4) & =  109,\\
t_5 (2,3,4) & =  221.\end{aligned}$$

Let $$x = \sum_{i \in K} j_i \cdot t_i (2,3,4) = 449.$$

We use the iterative procedure described in the proof of Lemma \[l\_pre\_5\_b\] with the aim to write $$x = \sum_{i \in \tilde{K}} \tilde{j}_i \cdot t_i (2,3,4)$$ for some $a$-reduced set of integers $\{ \tilde{j}_i \}_{i \in \tilde{K}}$.

We set $l:=0$, $K(0):=K$ and $\tilde{j}_i = j_i$ for any $i \in K(0)$.

Since $\{ \tilde{j}_i \}_{i \in K(0)}$ is not $a$-reduced, we define $$M(0) := 4, \quad m(0) := 1.$$

Then we set $$\tilde{j}_{0} := 2, \quad \tilde{j}_{1} := 1, \quad \tilde{j}_{2} :=4, \quad \tilde{j}_{4} :=1, \quad \tilde{j}_{5} := 1,$$ and $$\begin{aligned}
K(1) & :=  \{0,1,2,4,5 \},\\
l & :=   1.\end{aligned}$$

Since $\{ \tilde{j}_i \}_{i \in K(1)}$ is not $a$-reduced, we define $$M(1) := 2, \quad m(1) := 1.$$

Then we set $$\tilde{j}_{0} := 4, \quad \tilde{j}_{1} := 0, \quad \tilde{j}_{2} :=2, \quad \tilde{j}_3 := 1, \quad \tilde{j}_{4} :=1, \quad \tilde{j}_{5} := 1,$$ and $$\begin{aligned}
K(2) & :=  \{0,1,2,3,4,5 \},\\
l & :=   2.\end{aligned}$$

We notice that $\{ \tilde{j}_i \}_{i \in K(2)}$ is $a$-reduced and define $\tilde{K} := K(2)$.

We have that $$x = \sum_{i \in \tilde{K}} \tilde{j}_i \cdot t_i (2,3,4).$$

\[l\_pre\_6\] Let $k$ be a positive integer.

If $\{ j_i \}_{i=1}^k$ is an $a$-reduced set of integers, then $${\displaystyle\sum}_{i=1}^k j_i \cdot t_i(a,b,c) < t_{k+1} (a,b,c).$$

We prove the claim by induction on $k$.

If $k=1$, then $${\displaystyle\sum}_{i=1}^1 j_i \cdot t_i(a,b,c) \leq a \cdot t_1 (a,b,c) = t_2(a,b,c) - b < t_2 (a,b,c).$$

Let $k > 1$. We distinguish two cases.

-   If $j_{k} = a$, then $${\displaystyle\sum}_{i=1}^k j_i \cdot t_i(a,b,c) = a \cdot t_k (a,b,c) < t_{k+1} (a,b,c).$$

-   If $j_k \leq a-1$, then, by inductive hypothesis, we have that $$\begin{aligned}
    {\displaystyle\sum}_{i=1}^k j_i \cdot t_i(a,b,c) & =  {\displaystyle\sum}_{i=1}^{k-1} j_i \cdot t_i(a,b,c) + j_k \cdot t_k (a,b,c)\\
    & <  t_{k} (a,b,c) + (a-1) \cdot t_k (a,b,c) < t_{k+1} (a,b,c). \qedhere\end{aligned}$$

\[l\_pre\_7\] Suppose that

-   $k_1$ and $k_2$ are two positive integers such that $k_1 \leq k_2$;

-   $\{ j_i \}_{i=1}^{k_1}$ and $\{ \tilde{j}_i \}_{i=1}^{k_2}$ are two different $a$-reduced sets of integers;

-   $x$ and $y$ are two integers such that $$\begin{aligned}
    x & =  \sum_{i=1}^{k_1} j_i \cdot t_i (a,b,c),\\
    y & =  \sum_{i=1}^{k_2} \tilde{j}_i \cdot  t_i (a,b,c).\end{aligned}$$

The following hold.

-   If $\{ j_i \}_{i=1}^{k_1} \prec \{ \tilde{j}_i \}_{i=1}^{k_2}$, then $x < y$.

-   If $\{ \tilde{j}_i \}_{i=1}^{k_2} \prec \{ j_i \}_{i=1}^{k_1}$, then $y < x$.

Without loss of generality we suppose that $\{ j_i \}_{i=1}^{k_1} \prec \{ \tilde{j}_i \}_{i=1}^{k_2}$.

If $k_1 < k_2$, then $$x = \sum_{i=1}^{k_1} j_i \cdot t_i (a,b,c) < t_{k_1+1} (a,b,c)  \leq t_{k_2} (a,b,c) \leq y$$ according to Lemma \[l\_pre\_6\].

If $k_1 = k_2$, then we define $M:= \max \{i \in [1,k_2]: j_i \not= \tilde{j}_i \}$.

Since $\{ j_i \}_{i=1}^{k_1} \prec \{ \tilde{j}_i \}_{i=1}^{k_2}$ we have that $j_M < \tilde{j}_M$. Therefore, $$\begin{aligned}
x & =  \sum_{i=1}^{k_1} j_i \cdot  t_i (a,b,c) =  \sum_{i=1}^{k_2} j_i \cdot t_i (a,b,c)  \\
& =  \sum_{i=1}^{M-1} j_i \cdot t_i (a,b,c) + j_M \cdot t_M (a,b,c) + \sum_{i=M+1}^{k_2} j_i \cdot t_i (a,b,c) \\
& <  t_M (a,b,c) + j_M \cdot t_M (a,b,c) + \sum_{i=M+1}^{k_2} \tilde{j}_i \cdot t_i (a,b,c) \\
& \leq  \tilde{j}_M \cdot t_M (a,b,c) + \sum_{i=M+1}^{k_2} \tilde{j}_i \cdot t_i (a,b,c)\\
& \leq  \sum_{i=1}^{k_2} \tilde{j}_i \cdot  t_i (a,b,c) = y. \qedhere\end{aligned}$$

In analogy with Lemma \[l\_pre\_6\] and Lemma \[l\_pre\_7\] we state two more lemmas, whose proofs (which are omitted) follow the same lines as the proofs of the two lemmas above.

\[l\_pre\_8\] If $\{ j_i \}_{i=1}^k$ is an $a$-reduced set of integers for some positive integer $k$, then $${\displaystyle\sum}_{i=1}^k j_i \cdot s_i(a) < s_{k+1} (a).$$

\[l\_pre\_9\] Suppose that

-   $k_1$ and $k_2$ are two positive integers such that $k_1 \leq k_2$;

-   $\{ j_i \}_{i=1}^{k_1}$ and $\{ \tilde{j}_i \}_{i=1}^{k_2}$ are two different $a$-reduced sets of integers;

-   $x$ and $y$ are two integers such that $$\begin{aligned}
    x & =  \sum_{i=1}^{k_1} j_i \cdot  s_i (a),\\
    y & =  \sum_{i=1}^{k_2} \tilde{j}_i \cdot  s_i (a).\end{aligned}$$

The following hold.

-   If $\{ j_i \}_{i=1}^{k_1} \prec \{ \tilde{j}_i \}_{i=1}^{k_2}$, then $x < y$.

-   If $\{ \tilde{j}_i \}_{i=1}^{k_2} \prec \{ j_i \}_{i=1}^{k_1}$, then $y < x$.

Proofs
======

Proof of Theorem \[t\_g\_is\_h\]
--------------------------------

First, we notice that $H(a,b,c) \subseteq G_{a,b} (c)$ according to Lemma \[l\_pre\_0\].

According to Lemma \[l\_pre\_1\], $ H(a,b,c)$ is a subsemigroup of $({\text{$\mathbb{N}$}}, +)$ closed with respect to the action of the map ${\vartheta}_{a,b}$.

Moreover, ${\text{$\mathbb{N}$}}\backslash  H(a,b,c)$ is finite. In fact, according to Lemma \[l\_pre\_3\] we have that $$[t_k (a,b,c), + \infty[ \subseteq H(a,b,c)$$ for any positive integer $k$ such that $s_k(a) \geq c-1$.

Hence, $H(a,b,c)$ is a numerical semigroup and $G_{a,b} (c) = H(a,b,c)$.

Proof of Theorem \[t\_m\_gen\]
------------------------------

According to Lemma \[l\_pre\_5\] we have that $$t_k(a,b,c) \not \in \langle S_{k-1} (a,b,c) \rangle$$ for any positive integer $k < \tilde{k}$ . In fact, for any such $k$ we have that $s_k(a) \leq c-1$. Nevertheless, $$t_{\tilde{k}} (a,b,c)  \in \langle S_{\tilde{k}-1} (a,b,c) \rangle.$$ The latter assertion holds since $$\begin{cases}
t_{\tilde{k}} (a,b,c) - a^{\tilde{k}} c = b(q c +r) \\
0 \leq r < c
\end{cases}$$ for some non-negative integers $q$ and $r$. Since $$r \leq c-1 < s_{\tilde{k}} (a),$$ we have that $$a^{\tilde{k}} c + b r \in G_{a,b} (c)$$ according to Lemma \[l\_pre\_2\]. Therefore, $$t_{\tilde{k}} (a,b,c) \in \langle S_{\tilde{k}-1} (a,b,c) \rangle,$$ namely $S_{\tilde{k}-1} (a,b,c)$ is a minimal set of generators for $G_{a,b} (c)$.

Proof of Theorem \[gen\]
------------------------

We prove separately the $4$ assertions.

1.  Let $l \in [1,c-1]$. According to Lemma \[l\_pre\_5\_a\] there exists and is unique an $a$-reduced set $\{ j_i^{(l)} \}_{i=1}^{k_l}$ such that $$l = \sum_{i=1}^{k_l} j_i^{(l)} \cdot s_i(a).$$

    Moreover, $$\begin{aligned}
    x_l & =  \sum_{i=1}^{k_l} j_i^{(l)} \cdot t_i(a,b,c) \\
    & =  \sum_{i=1}^{k_l} j_i^{(l)} \cdot a^i c + b \cdot \sum_{i=1}^{k_l} j_i^{(l)} \cdot s_i(a)\\
    & \equiv  b l \pmod{c}.\end{aligned}$$

    Let $x \in G_{a,b} (c)$. Since $$\{bl: l \in [0,c-1] \}$$ is a set of representatives of the residue classes in ${\text{$\mathbb{Z}$}}/ c {\text{$\mathbb{Z}$}}$, we can say that $$x \equiv b l \pmod{c}$$ for some $l \in [0,c-1]$.

    If $x \not \equiv 0 \pmod{c}$, then $$x = \sum_{i \in \tilde{K}} \tilde{j}_i \cdot t_i(a,b,c)$$ for some $a$-reduced set of integers $\{ \tilde{j}_i \}_{i \in \tilde{K}}$, according to Lemma \[l\_pre\_5\_b\].

    We distinguish two cases.

    -   If $0 \in \tilde{K}$ and $\tilde{j}_0 \not = 0$, then $$\begin{aligned}
        x & =  \tilde{j}_0 \cdot c + \sum_{i \in \tilde{K} \backslash \{ 0 \}} \tilde{j}_i \cdot t_i(a,b,c) \\
        & >   \sum_{i \in \tilde{K} \backslash \{ 0 \}} \tilde{j}_i \cdot t_i(a,b,c) \equiv b l \pmod{c}.\end{aligned}$$

    -   If $0 \not \in \tilde{K}$ or $\tilde{j}_0 = 0$, then $\{ j_i^{(l)} \}_{i=1}^{k_l} \preceq \{ \tilde{j}_i \}_{i \in \tilde{K}}$ and $x_l \leq x$ according to Lemma \[l\_pre\_7\].

        Indeed, suppose by contradiction that $\{ \tilde{j}_i \}_{i \in \tilde{K}} \prec \{ j_i^{(l)} \}_{i=1}^{k_l}$.

        We notice that $$\begin{aligned}
        x & \equiv b \cdot \sum_{i \in \tilde{K}} \tilde{j}_i \cdot s_i(a) \equiv b l \pmod{c},\end{aligned}$$ namely $$\sum_{i \in \tilde{K}} \tilde{j}_i \cdot s_i(a) \equiv l \pmod{c}.$$ This latter is absurd because $$1 \leq \sum_{i \in \tilde{K}} \tilde{j}_i \cdot s_i(a) < \sum_{i = 1}^{k_l} j_i^{(l)} \cdot s_i(a) = l$$ according to Lemma \[l\_pre\_9\].

2.  This assertion follows from (1).

3.  Since $x_i < x_{c-1}$ for any $i \in [0, c-1[$, we have that $F( G_{a,b} (c)) = x_{c-1} - c$.

4.  This assertion follows from Selmer’s formulas.
