# A Note on Generalized Repunit Numerical Semigroups

Feihu Liu1, Guoce Xin2,∗, Suting Ye3 and Jingjing Yin4 1,2,3,4School of
Mathematical Sciences, Capital Normal University, Beijing 100048, PR China
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(Date: June 19, 2023)

###### Abstract.

Let $A=(a_{1},a_{2},...,a_{n})$ be relative prime positive integers with
$a_{i}\geq 2$. The Frobenius number $F(A)$ is the largest integer not
belonging to the numerical semigroup $\langle A\rangle$ generated by $A$. The
genus $g(A)$ is the number of positive integer elements that are not in
$\langle A\rangle$. The Frobenius problem is to find $F(A)$ and $g(A)$ for a
given sequence $A$. In this note, we study the Frobenius problem of
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$
and obtain formulas for $F(A)$ and $g(A)$ when $a\geq k-1$. Our formulas
simplifies further for some special cases, such as repunit, Mersenne and
Thabit numerical semigroups. The idea is similar to that in
[[13],arXiv:2306.03459].

$*$ This work was partially supported by NSFC(12071311).

_Mathematic subject classification_ : Primary 11D07; Secondary 11A67, 11B75,
20M14.

_Keywords_ : Numerical semigroup; Apéry set; Frobenius number; Genus; Pseudo-
Frobenius number.

##  1\. Introduction

In [13], we studied the Frobenius problem for the numerical semigroup $\langle
A\rangle=\langle a,2a+d,2^{2}a+3d,...,2^{k}a+(2^{k}-1)d\rangle$. We show in
this note that the same idea applies to the Frobenius problem for the
numerical semigroup $\langle A\rangle=\left\langle
a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right\rangle$.
Let $b=2$, we can obtain the results in [13]. The material in this note is
better merged with [13], but that paper has been submitted.

Throughout this paper, $\mathbb{Z}$, $\mathbb{N}$ and $\mathbb{P}$ denote the
set of all integers, non-negative integers and positive integers,
respectively. We need to introduce some basic definitions in [25].

A subset $S$ is a _submonoid_ of $\mathbb{N}$ if $S\subseteq\mathbb{N}$, $0\in
S$ and $S$ is closed under the addition in $\mathbb{N}$. If
$\mathbb{N}\setminus S$ is finite, then we say that $S$ is a _numerical
semigroup_. For a sequence (or set) $A=(a_{1},a_{2},...,a_{n})$ with
$a_{i}\in\mathbb{P}$, we denoted by $\langle A\rangle$ the submonoid of
$\mathbb{N}$ generated by $A$, that is,

$\langle A\rangle=\left\\{\sum_{i=1}^{n}x_{i}a_{i}\mid
n\in\mathbb{P},x_{i}\in\mathbb{N},1\leq i\leq n\right\\}.$

If $\gcd(A)=1$, then $\langle A\rangle$ is a numerical semigroup (see [25]).
If $\langle A\rangle$ is a numerical semigroup, then we say that $A$ is a
_system of generators_ of $\langle A\rangle$. If no proper subsequence (or
subset) of $A$ generates $\langle A\rangle$, then we say that $A$ is a
_minimal system of generators_ of $\langle A\rangle$, and the number of $A$ is
the embedding dimension of $\langle A\rangle$, denoted by $e(A)$. Moreover,
the minimal system of generators is finite and unique (see [25]).

If $\gcd(A)=1$, then we have the following definitions:

  1. (1)

_The Frobenius number_ $F(A)$: The greatest integer not belonging to $\langle
A\rangle$.

  2. (2)

_The genus (or Sylvester number)_ $g(A)$: The cardinality of
$\mathbb{N}\backslash\langle A\rangle$.

  3. (3)

_The pseudo-Frobenius numbers_ $u$: If $u\in\mathbb{Z}\backslash\langle
A\rangle$ and $u+s\in\langle A\rangle$ for all $s\in\langle
A\rangle\backslash\\{0\\}$.

  4. (4)

The set $PF(A)$: The set of pseudo-Frobenius numbers of $\langle A\rangle$
(see [21]).

  5. (5)

_The type_ $t(A)$ of $\langle A\rangle$: The cardinality of $PF(A)$.

For more knowledge about numerical semigroup, see [2, 25].

The Frobenius number $F(A)$ has been widely studied. For $A=(a_{1},a_{2})$,
Sylvester [29] obtained $F(A)=a_{1}a_{2}-a_{1}-a_{2}$ in 1882. For $n\geq 3$,
F. Curtis [6] proved that the $F(A)$ can not be given by closed formulas of a
certain type. However, many special cases have been studied such as arithmetic
progression in [3, 18, 26], geometric sequences in [16], triangular and
tetrahedral sequences in [19]. For more special sequences, see [17, 31, 30,
12, 11, 20]. Many special numerical semigroups are also considered, such as
Fibonacci in [15], Mersenne in [24], repunit in [23], squares and cubes in
[10], Thabit in [22] and other numerical semigroups in [8, 27, 28].

The motivation of this paper comes from the repunit numerical semigroup:
$\langle\\{\frac{b^{n+i}-1}{b-1}\mid i\in\mathbb{N}\\}\rangle$. Essentially,
its minimal system of generators is
$S(b,n)=\left(\frac{b^{n}-1}{b-1},\frac{b^{n+1}-1}{b-1},...,\frac{b^{2n-1}-1}{b-1}\right)$
([23]), that is the embedding dimension $e(S(b,n))=n$. In this paper, we study
the following more general case

$A=(a,Ha+dB)=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right),$

where $a,d,b,k\in\mathbb{P},b\geq 2$ and $\gcd(a,d)=1$. Here we have the
embedding dimension $e(A)\leq k+1$. We can observe that

  1. (1)

If $a=2^{n}-1$, $b=2$, $d=1$, $k=n-1$, then $\langle A\rangle$ is the Mersenne
numerical semigroup $S(n)$ in [24].

  2. (2)

If $a=3\cdot 2^{n}-1$, $b=2$, $d=1$, $k=n+1$, then $\langle A\rangle$ is the
Thabit numerical semigroup $T(n)$ in [22].

  3. (3)

If $a=(2^{m}-1)\cdot 2^{n}-1$, $b=2$, $d=1$, $k=n+m-1$, then $\langle
A\rangle$ is a class of numerical semigroups in [7].

  4. (4)

If $a=(2^{m}+1)\cdot 2^{n}-(2^{m}-1)$, $b=2$, $d=2^{m}-1$,
$k\in\\{n+1,n+m-1,n+m\\}$, then $\langle A\rangle$ is a class of numerical
semigroups in [28].

  5. (5)

If $a=m(2^{k}-1)+2^{k-1}-1$, $b=2$, $k\geq 3$, then $\langle A\rangle$ is a
class of numerical semigroups in [13].

  6. (6)

If $a=\frac{b^{n}-1}{b-1}$, $b\geq 2$, $d=1$, $k=n-1$, then $\langle A\rangle$
is the repunit numerical semigroups $S(b,n)$ in [23].

  7. (7)

If $a=b^{n+1}+\frac{b^{n}-1}{b-1}$, $b\geq 2$, $d=1$, $k=n+1$, then $\langle
A\rangle$ is a class of numerical semigroups in [8].

  8. (8)

If $a=(b+1)b^{n}-1$, $b\geq 2$, $d=b-1$, $k=n+1$, then $\langle A\rangle$ is
the Thabit numerical semigroup $T_{b,1}(n)$ of the first kind base $b$ in
[27].

The main purpose of this paper is to solve the Frobenius problem of
$A=A=(a,Ha+dB)=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$.
Using similar discussions as in paper [13], we can obtain formulas of $F(A)$
and $g(A)$ in Theorem 3.5. Specifically, if $k=n-1$ and
$a=\frac{b^{n}-1}{b-1}$, then we can further obtain

  1. (1)

The Frobenius number $F(A)=(b^{n}+d-1)\cdot\frac{b^{n}-1}{b-1}-d$.

  2. (2)

The genus $g(A)=\frac{(b^{n}-b)(b^{n}+d-1)}{2(b-1)}+\frac{b^{n}(n-1)}{2}$.

  3. (3)

The set of Pseudo-Frobenius number $PF(A)=\\{F(A),F(A)-d,...,F(A)-(n-2)d\\}$.

  4. (4)

The type $t(A)=n-1$.

The paper is organized as follows. In Section 2, we provide some necessary
lemmas and related results. In Section 3, we establish Theorem 3.5, which
gives formulas of Frobenius number $F(A)$ and genus $g(A)$ where
$\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$
and $a\geq k-1$. Section 4 focus on application of Theorem 3.5 for special $a$
and $k$. It is related to repunit numerical semigroup. Section 6 is a
concluding remark.

##  2\. Preliminary

It is convenient to use the short hand notation
$A:=(a,B)=(a,b_{1},b_{2},...,b_{k})$. Let $\gcd(A)=1$, $a,b_{i}\in\mathbb{P}$.
The set

$\langle A\rangle=\left\\{ax+\sum_{i=1}^{k}b_{i}x_{i}\ \mid
x,x_{i}\in\mathbb{N}\right\\}$

is a numerical semigroup. Let $w\in\langle A\rangle\backslash\\{0\\}$. The
_Apéry set_ of $w$ in $\langle A\rangle$ is $Ape(A,w)=\\{s\in\langle
A\rangle\mid s-w\notin\langle A\rangle\\}$. In [25], we can obtain

$Ape(A,w)=\\{N_{0},N_{1},N_{2},...,N_{w-1}\\},$

where $N_{r}:=\min\\{a_{0}\mid a_{0}\equiv r\mod w,\ a_{0}\in\langle
A\rangle\\}$, $0\leq r\leq w-1$. We usually take $w:=a$.

A. Brauer and J. E. Shockley [4], E. S. Selmer [26] gave the following results
respectively.

######  Lemma 2.1 ([4], [26]).

Suppsoe $A:=(a,B)=(a,b_{1},b_{2},...,b_{k})$. The _Apéry set_ of $a$ in
$\langle A\rangle$ is $Ape(A,a)=\\{N_{0},N_{1},N_{2},...,N_{a-1}\\}$. Then the
Frobenius number and genus of $A$ are respectively:

$\displaystyle F(A)$
$\displaystyle=F(a,B)=\max_{r\in\\{0,1,...,a-1\\}}N_{r}-a,$ $\displaystyle
g(A)$ $\displaystyle=g(a,B)=\frac{1}{a}\sum_{r=1}^{a-1}N_{r}-\frac{a-1}{2}.$

Now we define the following order relation in $\mathbb{Z}$: $a\preceq_{\langle
A\rangle}b$ if $b-a\in\langle A\rangle$. It is proved that this relation
$\preceq_{\langle A\rangle}$ is an order relation in [25].

######  Lemma 2.2 (Proposition 2.20, [25]).

Let $\langle A\rangle$ be a numerical semigroup. Then

$PF(A)=\left\\{w-a\mid w\in\max\nolimits_{\preceq_{\langle
A\rangle}}Ape(A,a)\right\\},$

where $Ape(A,a)=\\{N_{0},N_{1},...,N_{a-1}\\}$.

Note that $N_{0}=0$ for all $A$. By the definition of $N_{r}$, we can easily
obtain the following result.

######  Proposition 2.3.

Let $A=(a,b_{1},...,b_{k})$, $\gcd(a,d)=1,\ d\in\mathbb{P}$. Then we have

$\\{N_{0},N_{1},N_{2},...,N_{a-1}\\}=\\{N_{d\cdot 0},N_{d\cdot 1},N_{d\cdot
2},...,N_{d\cdot(a-1)}\\}.$ (1)

######  Lemma 2.4 ([13]).

Let $A=(a,h_{1}a+db_{1},...,h_{k}a+db_{k})$, $k,h,d\in\mathbb{P}$ and
$\gcd(A)=1$, $m\in\mathbb{N}$, $\gcd(a,d)=1$. For a given $0\leq r\leq a-1$,
we have

$N_{dr}=\min\left\\{O_{B}^{H}(ma+r)\cdot a+(ma+r)d\mid
m\in\mathbb{N}\right\\},$ (2)

where

$O_{B}^{H}(M):=\min\left\\{\sum_{i=1}^{k}h_{i}x_{i}\mid\sum_{i=1}^{k}b_{i}x_{i}=M,\
x_{i},M\in\mathbb{N},1\leq i\leq k\right\\}.$

By Lemma 2.4, we can define an intermediate function with respect to
$m\in\mathbb{N}$, namely:

$N_{dr}(m):=O_{B}^{H}(ma+r)\cdot a+(ma+r)d.$

We also need the following definition. For a given positive integers sequence
$B=(b_{1},b_{2},...,b_{k})$, $1=b_{1}<b_{2}<\cdots<b_{k}$ and
$M\in\mathbb{N}$, let

$opt_{B}(M):=\min\left\\{\sum_{i=1}^{k}x_{i}\mid\sum_{i=1}^{k}b_{i}x_{i}=M,\ \
M,x_{i}\in\mathbb{N},1\leq i\leq k\right\\}.$ (3)

The problem $opt_{B}(M)$ is called _the change-making problem_ [1]. A strategy
is called greedy, that is to use as many of the maximum as possible, than as
many of the next one as possible, and so on. We denote $grd_{B}(M)$ the number
of elements used in $B$ by the greedy strategy. Then we have $opt_{B}(M)\leq
grd_{B}(M)$. If the greedy solution is always optimal, i.e.,
$opt_{B}(M)=grd_{B}(M)$ for all $M>0$, then we call the sequence $B$ _orderly_
; Otherwise, we call sequence $B$ _non-orderly_.

######  Lemma 2.5 (One-Point Theorem, [5, 9, 14]).

Suppose $B^{\prime}=(1,b_{1},...,b_{k})$ is orderly and $b_{k+1}>b_{k}$. Let
$s=\lceil b_{k+1}/b_{k}\rceil$. Then the sequence
$B=(1,b_{1},...,b_{k},b_{k+1})$ is orderly if and only if
$opt_{B}(sb_{k})=grd_{B}(sb_{k})$.

##  3\. The Frobenius Problem for
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$

Now we consider the case
$A=(a,Ha+dB)=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$,
$b\geq 2$, i.e., $H=(b,b^{2},...,b^{k})$ and
$B=\left(1,\frac{b^{2}-1}{b-1},...,\frac{b^{k}-1}{b-1}\right)$. We first
consider

$opt_{B}(M)=\min\left\\{\sum_{i=1}^{k}x_{i}\mid\sum_{i=1}^{k}\frac{b^{i}-1}{b-1}x_{i}=M,\
\ M,x_{i}\in\mathbb{N},1\leq i\leq k\right\\}.$

By the _One-Point Theorem_ , we can obtain the following result.

######  Lemma 3.1.

Let $k\in\mathbb{P}$ and $b\geq 2$. Then the sequence
$B=\left(1,\frac{b^{2}-1}{b-1},...,\frac{b^{k}-1}{b-1}\right)$ is orderly,
i.e., $opt_{B}(M)=grd_{B}(M)$ for all $M\in\mathbb{P}$.

###### Proof.

By induction, we first know that the sequence $(1)$ and
$\left(1,\frac{b^{2}-1}{b-1}\right)$ are orderly sequences. Suppose
$\left(1,\frac{b^{2}-1}{b-1},...,\frac{b^{k-1}-1}{b-1}\right)$ is orderly. We
have
$s=\left\lceil\frac{b^{k}-1}{b^{k-1}-1}\right\rceil=\left\lceil\frac{b(b^{k-1}-1)+b-1}{b^{k-1}-1}\right\rceil=b+1$.
By
$(b+1)\cdot\frac{b^{k-1}-1}{b-1}=\frac{b^{k}-1}{b-1}+b\cdot\frac{b^{k-2}-1}{b-1}$,
we have
$opt_{B}(s(\frac{b^{k-1}-1}{b-1}))=grd_{B}(s(\frac{b^{k-1}-1}{b-1}))=b+1$. By
Lemma 2.5, the sequence $B$ is orderly. This completes the proof. ∎

For a given $M\in\mathbb{P}$ and
$B=\left(1,\frac{b^{2}-1}{b-1},...,\frac{b^{k}-1}{b-1}\right)$, by
$b\cdot\frac{b^{i}-1}{b-1}+1=\frac{b^{i+1}-1}{b-1}$ and Lemma 3.1, we have the
following properties for $x_{i}$ in $opt_{B}(M)$.

  1. (1)

$x_{k}=\left\lfloor\frac{(b-1)M}{b^{k}-1}\right\rfloor$.

  2. (2)

$x_{i}\in\\{0,1,...,b\\}$ for every $1\leq i\leq k-1$.

  3. (3)

if $2\leq i\leq k-1$ and $x_{i}=b$, then $x_{1}=\cdots=x_{i-1}=0$.

If a solution $X=(x_{1},x_{2},...,x_{k})$ of $opt_{B}(M)$ satisfies the above
conditions (1), (2) and (3), we call $X$ the _greedy presentation_ of $M$. We
define $R(M)$ to be the set of all greedy presentations of $0,1,2,...,M$. We
define a colexicographic order on $R(M)$ as follows:

$\displaystyle(x_{1}^{\prime},x_{2}^{\prime},...,x_{k}^{\prime})\preceq(x_{1},x_{2},...,x_{k})$
$\displaystyle\Longleftrightarrow$ $\displaystyle x_{i}^{\prime}=x_{i}\ \
\text{for any}\ \ i>0\ \ \text{or}$ $\displaystyle
x_{j}^{\prime}<x_{j},x_{i}^{\prime}=x_{i}\ \ \text{for a certain}\ \ j>0\ \
\text{and any}\ \ i>j.$

Obviously, the order relation $\preceq$ is a total order on $R(M)$.

######  Remark 3.2.

Lemma 3.1 essentially provides the construction process of
$X=(x_{1},x_{2},...,x_{k})$ in $R(M)$.

Now we consider the $O_{B}^{H}(M)$ for
$B=\left(1,\frac{b^{2}-1}{b-1},...,\frac{b^{k}-1}{b-1}\right)$,
$H=(b,b^{2},...,b^{k})$ and any $M\in\mathbb{P}$. We have

$\displaystyle O_{B}^{H}(M)$
$\displaystyle=\min\left\\{\sum_{i=1}^{k}b^{i}x_{i}\mid\sum_{i=1}^{k}\frac{b^{i}-1}{b-1}x_{i}=M,\
x_{i}\in\mathbb{N},1\leq i\leq k\right\\}$
$\displaystyle=\min\left\\{(b-1)M+\sum_{i=1}^{k}x_{i}\mid\sum_{i=1}^{k}\frac{b^{i}-1}{b-1}x_{i}=M,\
x_{i}\in\mathbb{N},1\leq i\leq k\right\\}.$

######  Lemma 3.3.

Let
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$,
$a,d,b,k\in\mathbb{P}$, $b\geq 2$ and $\gcd(a,d)=1$. Suppose $a\geq k-1$. Then
$N_{dr}(m)$ is increasing with respect to $m\in\mathbb{N}$. More precisely, if
$X=(x_{1},x_{2},...,x_{k})$ is a greedy presentation of $r$, then

$N_{dr}=\left(\sum_{i=1}^{k}x_{i}\right)a+r((b-1)a+d)=\left(\sum_{i=1}^{k}b^{i}x_{i}\right)a+rd.$
(4)

###### Proof.

Given a $m\in\mathbb{N}$. Assuming
$X^{\prime}=(x_{1}^{\prime},...,x_{k}^{\prime})$ and $X=(x_{1},...,x_{k})$ are
greedy presentations of $(m+1)a+r$ and $ma+r$, respectively. We can easily get
$x_{k}^{\prime}\geq x_{k}$. Furthermore

$N_{dr}(m)=\left((b-1)(ma+r)+\sum_{i=1}^{k}x_{i}\right)a+(ma+r)d.$

Therefore we have

$\displaystyle N_{dr}(m+1)-N_{dr}(m)$
$\displaystyle=(b-1)a^{2}+ad+\left(\sum_{i=1}^{k}x_{i}^{\prime}-\sum_{i=1}^{k}x_{i}\right)a$
$\displaystyle\geq\left((b-1)a+d+\sum_{i=1}^{k-1}x_{i}^{\prime}-\sum_{i=1}^{k-1}x_{i}\right)a$
$\displaystyle\geq((b-1)a+d-b-(k-2)(b-1))a$
$\displaystyle=((b-1)(a-k+1)+d-1)a\geq 0.$

Then $N_{dr}(m)$ is increasing with respect to $m\in\mathbb{N}$. Moreover
$N_{dr}=N_{dr}(0)$. This completes the proof. ∎

In Equation (4), suppose $X=(x_{1},x_{2},...,x_{k})$ is a greedy presentation
of $r$, we call $w(r)=\sum_{i=1}^{k}b^{i}x_{i}$ the _weight_ of $r$. Now we
have the following result.

######  Lemma 3.4.

Suppose $X^{\prime}=(x_{1}^{\prime},...,x_{k}^{\prime})\in R(M)$ and
$X=(x_{1},...,x_{k})\in R(M)$ are greedy presentations of $r_{1}$ and $r_{2}$,
respectively. If
$(x_{1}^{\prime},...,x_{k}^{\prime})\preceq(x_{1},...,x_{k})$, then we have
$w(r_{1})\leq w(r_{2})$.

###### Proof.

For $X=(x_{1},x_{2},...,x_{k})\in R(M)$, it is easy to obtain that the maximum
of $\sum_{i=1}^{k-1}b^{i}x_{i}$ is $b^{k}$. At this point, the possible values
for $(x_{1},...,x_{k-1})$ are

$(b,b-1,b-1,...,b-1),(0,b,b-1,...,b-1),...,(0,0,0,...,b).$

We consider the following three cases.

If $x_{i}^{\prime}=x_{i}$ for any $1\leq i\leq k$, then $w(r_{1})=w(r_{2})$.

If $x_{k}^{\prime}<x_{k}$, then $x_{k}\geq x_{k}^{\prime}+1$ and

$\displaystyle w(r_{1})$
$\displaystyle=\sum_{i=1}^{k}b^{i}x_{i}^{\prime}=\sum_{i=1}^{k-1}b^{i}x_{i}^{\prime}+b^{k}x_{k}^{\prime}\leq
b^{k}+b^{k}x_{k}^{\prime}=b^{k}(1+x_{k}^{\prime})\leq b^{k}x_{k}\leq
w(r_{2}).$

If there is a certain $1\leq u\leq k-1$, so that $x_{u}^{\prime}<x_{u}$ and
$x_{u+1}^{\prime}=x_{u+1},...,x_{k}^{\prime}=x_{k}$. Similarly we have

$\displaystyle w(r_{1})$
$\displaystyle=\sum_{i=1}^{k}b^{i}x_{i}^{\prime}=\sum_{i=1}^{u}b^{i}x_{i}^{\prime}+\sum_{j=u+1}^{k}b^{j}x_{j}^{\prime}$
$\displaystyle\leq
b^{u}(1+x_{u}^{\prime})+\sum_{j=u+1}^{k}b^{j}x_{j}^{\prime}=b^{u}(1+x_{u}^{\prime})+\sum_{j=u+1}^{k}b^{j}x_{j}$
$\displaystyle\leq b^{u}x_{u}+\sum_{j=u+1}^{k}b^{j}x_{j}\leq w(r_{2}).$

This completes the proof. ∎

Now we can obtain the main result of this section.

######  Theorem 3.5.

Let
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$,
$a,d,b,k\in\mathbb{P}$, $\gcd(a,d)=1$, $b\geq 2$ and $a\geq k-1$. We have

$\displaystyle
F(A)=\left((b-1)a-b+d+\left(\sum_{i=1}^{k}x_{i}\right)_{a-1}\right)a-d,$
$\displaystyle
g(A)=\sum_{r=1}^{a-1}\left(\sum_{i=1}^{k}x_{i}\right)_{r}+\frac{(a-1)((b-1)a+d-1)}{2},$

where $(\sum_{i=1}^{k}x_{i})_{r}$ is the sum of elements in the greedy
presentation of $r$. Obviously
$x_{k}=\left\lfloor\frac{(b-1)r}{b^{k}-1}\right\rfloor$.

###### Proof.

By Lemma 3.4 and Equation (4), we have

$\max N_{dr}=N_{d(a-1)}=\left(\sum_{i=1}^{k}x_{i}\right)_{a-1}\cdot
a+(a-1)((b-1)a+d).$

Further from Lemma 2.1, we can obtain the Frobenius number formula $F(A)$.
From Equation (4) and Lemma 2.1 again, we can obtain the genus $g(A)$. This
completes the proof. ∎

##  4\. A Generalization of Repunit Numerical Semigroup

In this section we mainly focus on the generalization of repunit numerical
semigroup. For
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$,
suppose $k:=n-1$, $a:=\frac{b^{n}-1}{b-1}$, $b\geq 2$. Therefore we have

$A=\Bigg{(}\frac{b^{n}-1}{b-1},b\cdot\frac{b^{n}-1}{b-1}+d,b^{2}\cdot\frac{b^{n}-1}{b-1}+\frac{b^{2}-1}{b-1}d,...,b^{n-1}\cdot\frac{b^{n}-1}{b-1}+\frac{b^{n-1}-1}{b-1}d\Bigg{)}.$

We consider the numerical semigroup $\langle A\rangle$ generated by $A$.

By $a-1=\frac{b^{n}-1}{b-1}-1=b\cdot\frac{b^{n-1}-1}{b-1}$, we have

$\displaystyle R(a-1)$ $\displaystyle=\Big{\\{}(x_{1},x_{2},...,x_{n-1})\mid\
0\leq x_{i}\leq b\ \ \text{for}\ \ 1\leq i\leq n-1;$ $\displaystyle\ \ \ \ \
\text{if}\ \ x_{i}=b,\ \ \text{then}\ \ x_{j}=0\ \ \text{for}\ \ j\leq
i-1\Big{\\}}.$ (5)

######  Theorem 4.1.

For
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$,
suppose $k:=n-1$, $a:=\frac{b^{n}-1}{b-1}$, $b,n\geq 2$ and $\gcd(a,d)=1$. We
have

$\displaystyle F(A)=(b^{n}+d-1)\cdot\frac{b^{n}-1}{b-1}-d,$ $\displaystyle
g(A)=\frac{(b^{n}-b)(b^{n}+d-1)}{2(b-1)}+\frac{b^{n}(n-1)}{2}.$

###### Proof.

By Theorem 3.5 and $a-1=b\cdot\frac{b^{n-1}-1}{b-1}$, we have

$\displaystyle F(A)$
$\displaystyle=\left((b-1)\cdot\frac{b^{n}-1}{b-1}-b+d+b\right)\cdot\frac{b^{n}-1}{b-1}-d$
$\displaystyle=(b^{n}+d-1)\cdot\frac{b^{n}-1}{b-1}-d.$

Now we consider the genus $g(A)$. Based on the composition of $R(a-1)$
mentioned above, we have

$\displaystyle\sum_{r=1}^{a-1}\left(\sum_{i=1}^{n-1}x_{i}\right)_{r}=$
$\displaystyle(n-1)b^{n-2}\cdot\sum_{i=0}^{b-1}i+\sum_{j=1}^{n-1}\left(b\cdot
b^{n-1-j}+(n-1-j)\cdot b^{n-2-j}\cdot\sum_{i=0}^{b-1}i\right)$
$\displaystyle=$
$\displaystyle\frac{(n-1)b^{n-1}(b-1)}{2}+\sum_{j=1}^{n-1}\left(b^{n-j}+\frac{b^{n-1-j}(b-1)(n-j-1)}{2}\right)$
$\displaystyle=$ $\displaystyle\frac{b^{n}-b}{2(b-1)}+\frac{b^{n}(n-1)}{2}.$

By Theorem 3.5, we have

$\displaystyle g(A)$
$\displaystyle=\frac{(b^{n}-b)(b^{n}+d-1)}{2(b-1)}+\frac{b^{n}(n-1)}{2}.$

This completes the proof. ∎

Let $d=1$ in Theorem 4.1, the $\langle A\rangle$ becomes the repunit numerical
semigroup ([23]). Clearly, we have the following corollary.

######  Corollary 4.2 (Theorem 20, Theorem 25, [23]).

Let $n,b\geq 2$, $d=1$ in Theorem 4.1. Then $S(b,n)=\langle A\rangle$ is the
repunit numerical semigroup in [23]. Furthermore, we have

$\displaystyle F(S(b,n))=\frac{b^{n}-1}{b-1}b^{n}-1,$ $\displaystyle
g(S(b,n))=\frac{b^{n}}{2}\left(\frac{b^{n}-1}{b-1}+n-1\right).$

######  Theorem 4.3.

For
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$,
suppose $k:=n-1$, $a:=\frac{b^{n}-1}{b-1}$, $b,n\geq 2$ and $\gcd(a,d)=1$.
Then $t(A)=n-1$. Furthermore

$PF(A)=\\{F(A),F(A)-d,...,F(A)-(n-2)d\\}.$

###### Proof.

Let $X=(x_{1},x_{2},...,x_{n-1})$ be a greedy presentation of $r$, $0\leq
r\leq a-1$. By Equation (4), we have

$\displaystyle N_{dr}$
$\displaystyle=\left(\sum_{i=1}^{n-1}b^{i}x_{i}\right)a+\sum_{i=1}^{n-1}\frac{b^{i}-1}{b-1}x_{i}d$
$\displaystyle=\left(\sum_{i=1}^{n-1}b^{i}x_{i}\right)a+\sum_{i=1}^{n-1}\frac{b^{i}x_{i}}{b-1}d-\sum_{i=1}^{n-1}\frac{x_{i}}{b-1}d.$

For a given $0\leq r\leq a-1$, we have
$r=s\cdot\frac{b^{n-1}-1}{b-1}+r^{\prime}$, $0\leq s\leq b$ and $0\leq
r^{\prime}\leq\frac{b^{n-1}-b}{b-1}$. Therefore, under the order relation
$\preceq_{\langle A\rangle}$, the greedy presentations of the maximal elements
in $R(a-1)$ are contained in the following elements:

$(b,b-1,b-1,...,b-1,b-1),\ (0,b,b-1,...,b-1,b-1),\ (0,0,b,...,b-1,b-1),\ ...,\
(0,0,0,...,0,b).$

For the above greedy presentations, we always have

$\sum_{i=1}^{n-1}b^{i}x_{i}=b^{n}.$

For the above maximal elements, let
$X^{\prime}=(x_{1}^{\prime},...,x_{n-1}^{\prime})\in R(a-1)$ and
$X=(x_{1},...,x_{n-1})\in R(a-1)$ be greedy presentations of $r_{1}$ and
$r_{2}$, respectively. We need to consider the following case

$N_{dr_{1}}-N_{dr_{2}}=\sum_{i=1}^{n-1}\frac{x_{i}}{b-1}d-\sum_{i=1}^{n-1}\frac{x_{i}^{\prime}}{b-1}d=td,\
\ t=1,2,...,n-2.$

Now we need to determine whether $td$ belongs to $\langle A\rangle$. If
$td\in\langle A\rangle$, there would exist $(y_{0},y_{1},...,y_{n-1})$ such
that

$td=y_{0}a+y_{1}(ba+d)+\cdots+y_{n-1}\left(b^{n-1}a+\frac{b^{n-1}-1}{b-1}d\right).$

These $y_{0},...,y_{n-1}\in\mathbb{N}$ are not all $0$. Then we have

$(n-2)d\geq\left(t-y_{1}-y_{2}\frac{b^{2}-1}{b-1}-\cdots-
y_{n-1}\frac{b^{n-1}-1}{b-1}\right)d=(y_{0}+by_{1}+\cdots+b^{n-1}y_{n-1})a>0.$

By $\gcd(a,d)=1$, we have $a|\left(t-y_{1}-y_{2}\frac{b^{2}-1}{b-1}-\cdots-
y_{n-1}\frac{b^{n-1}-1}{b-1}\right)$. However $a>n-2\geq t$, we have
$t=y_{1}+y_{2}\frac{b^{2}-1}{b-1}+\cdots+y_{n-1}\frac{b^{n-1}-1}{b-1}$. So
$(y_{0}+by_{1}+\cdots+b^{n-1}y_{n-1})a=0$, a contradiction.

Therefore we have

$\max\nolimits_{\preceq_{\langle A\rangle}}Ape(A,a)=\\{N_{dr}\mid
r=a-\\{1,2,...,n-1\\}\\}.$

By Lemma 2.2, we have

$\displaystyle PF(A)$ $\displaystyle=\\{N_{dr}-a\mid r=a-\\{1,2,...,n-1\\}\\}$
$\displaystyle=\left\\{(b^{n}-1)a+\frac{b^{n}d}{b-1}-\left\\{\frac{bd}{b-1},\frac{(2b-1)d}{b-1},\frac{(3b-2)d}{b-1},...,\frac{((n-1)b-(n-2))d}{b-1}\right\\}\right\\}$
$\displaystyle=\\{F(A),F(A)-d,...,F(A)-(n-2)d\\}.$

This completes the proof. ∎

Similarly, we can obtain the following corollary.

######  Corollary 4.4 (Theorem 23, [23]).

Let $n,b\geq 2$, $d=1$ in Theorem 4.1. Then $S(b,n)=\langle A\rangle$ is the
repunit numerical semigroup in [23]. We have $t(S(b,n))=n-1$ and

$\displaystyle PF(S(b,n))=\\{F(S(n)),F(S(n))-1,...,F(S(n))-(n-2)\\}.$

##  5\. Concluding Remark

Our model
$A=\left(a,ba+d,b^{2}a+\frac{b^{2}-1}{b-1}d,...,b^{k}a+\frac{b^{k}-1}{b-1}d\right)$
can be used to explain the following eight numerical semigroups in the
literature.

1\. If $a=2^{n}-1$, $b=2$, $d=1$, $k=n-1$ and $n\geq 2$, then

$A=(2^{n}-1,2^{n+1}-1,2^{n+2}-1...,2^{2n-1}-1),$

and the $\langle A\rangle$ is the Mersenne numerical semigroups $S(n)$ in
[24].

2\. If $a=3\cdot 2^{n}-1$, $b=2$, $d=1$, $k=n+1$ and $n\geq 1$, then

$A=(3\cdot 2^{n}-1,3\cdot 2^{n+1}-1,3\cdot 2^{n+2}-1,...,3\cdot 2^{2n+1}-1),$

and the $\langle A\rangle$ is the Thabit numerical semigroups $T(n)$ in [22].

3\. If $a=(2^{m}-1)\cdot 2^{n}-1$, $b=2$, $d=1$, $k=n+m-1$, $n\geq 1$ and
$2\leq m\leq 2^{n}$, then

$A=((2^{m}-1)\cdot 2^{n}-1,(2^{m}-1)\cdot 2^{n+1}-1,...,(2^{m}-1)\cdot
2^{2n+m-1}-1),$

and the $\langle A\rangle$ is a class of numerical semigroups $S(m,n)$ in [7].

4\. Let $m,n\in\mathbb{N}$ and $m\geq 2$ and

$\displaystyle\delta=\left\\{\begin{array}[]{lc}1&\ \text{if}\ \ n=0;\\\ m&\
\text{if}\ \ n\neq 0,m\leq n;\\\ m-1&\ \text{if}\ \ n\neq 0,m>n.\\\
\end{array}\right.$

If $a=(2^{m}+1)\cdot 2^{n}-(2^{m}-1)$, $b=2$, $d=2^{m}-1$ and $k=n+\delta$,
then the $\langle A\rangle$ is a class of numerical semigroups $GT(n,m)$ in
[28].

5\. If $a=m(2^{k}-1)+2^{k-1}-1$, $b=2$, $m\geq 1$, $k\geq 3$, then the
$\langle A\rangle$ is a class of numerical semigroups in [13].

6\. If $a=\frac{b^{n}-1}{b-1}$, $b\geq 2$, $d=1$, $k=n-1$ and $n\geq 2$, then

$A=\left(\frac{b^{n}-1}{b-1},\frac{b^{n+1}-1}{b-1},\frac{b^{n+2}-1}{b-1},...,\frac{b^{2n-1}-1}{b-1}\right),$

and the $\langle A\rangle$ is the repunit numerical semigroups $S(b,n)$ in
[23].

7\. If $a=b^{n+1}+\frac{b^{n}-1}{b-1}$, $b\geq 2$, $d=1$, $k=n+1$, then

$A=\left(b^{n+1}+\frac{b^{n}-1}{b-1},b^{n+2}+\frac{b^{n+1}-1}{b-1},...,b^{2n+2}+\frac{b^{2n+1}-1}{b-1}\right),$

and the $\langle A\rangle$ is a class of numerical semigroups $S(b,n)$ in [8].

8\. If $a=(b+1)b^{n}-1$, $b\geq 2$, $d=b-1$, $k=n+1$, then

$A=((b+1)b^{n}-1,(b+1)b^{n+1}-1,...,(b+1)b^{2n+1}-1),$

and the $\langle A\rangle$ is the Thabit numerical semigroup $T_{b,1}(n)$ of
the first kind base $b$ in [27].

Acknowledgements: This work was partially supported by the National Natural
Science Foundation of China [12071311].

## References

  * [1] A. Adamaszek, M. Adamaszek, _Combinatorics of the change-making problem_ , European. J. Comb. 31 (2010), 47–63.
  * [2] A. Assi, M. D’Anna, P. A. García-Sánchez, _Numerical Semigroups and Applications_ , Second Edition. Vol.3, RSMS Springer, Cham (2020).
  * [3] A. Brauer, _On a problem of partitions_ , Amer. J. Math. 64 (1942), 299–312.
  * [4] A. Brauer, J. E. Shockley, _On a problem of Frobenius_ , J. Reine Angew. Math. 211 (1962), 215–220.
  * [5] L. J. Cowen, R. Cowen, A. Steinberg, _Totally greedy coin sets and greedy obstructions_ , Electron. J. Comb. 15 (2008), #R90.
  * [6] F. Curtis, _On formulas for the Frobenius number of a numerical semigroup_ , Math. Scand. 67 (1990), 190–192.
  * [7] Z. Gu, X. Tang, _The Frobenius probelm for a class of numerical semigroups_ , Int. J. Number Theory. 13 (2017), 1335–1347.
  * [8] Z. Gu, _On the numerical semigroup generated by $\\{b^{n+1+i}+\frac{b^{n+i}-1}{b-1}\mid i\in\mathbb{N}\\}$_, Discrete Math. Appl. 30(4) (2020), 257–264.
  * [9] T. C. Hu, M. L. Lenard, _Optimality of a Heuristic solution for a class of Knapsack problems_ , Operations Research. 24(1) (1976), 193–196.
  * [10] M. Lepilov, J. O’Rourke, I. Swanson, _Frobenius numbers of numerical semigroups generated by three consecutive squares or cubes_ , Semigroup Forum. 91 (2015), 238–259.
  * [11] F. Liu, G. Xin, _On Frobenius formulas of power sequences_ , arXiv:2210.02722, (2022).
  * [12] F. Liu, G. Xin, _A combinatorial approach to Frobenius numbers of some special sequences (Complete Version)_ , arXiv:2303.07149. (2023).
  * [13] F. Liu, G. Xin, _A generalization of Mersenne and Thabit numerical semigroups_ , arXiv:2306.03459. (2023).
  * [14] M. J. Magazine, G. L. Nemhauser, L. E. Trotter Jr, _When the greedy solution solves a class of Knapsack problems_ , Operations Research. 23(2) (1975), 207–217.
  * [15] J. M. Marín, J. L. Ramírez Alfonsín, M. P. Revuelta, _On the Frobenius number of Fibonacci numerical semigroups_ , Integers. 7 (2007), A14.
  * [16] D. C. Ong, V. Ponomarenko, _The Frobenius number of geometric sequences_ , Integers. 8 (2008), A33.
  * [17] J. L. Ramírez Alfonsín, _The Diophantine Frobenius Problem_ , Oxford Lecture Series in Mathematics and Its Applications, vol. 30, Oxford University Press. (2005).
  * [18] J. B. Roberts, _Note on linear forms_ , Proc. Amer. Math. Soc. 7 (1956), 465–469.
  * [19] A. M. Robles-Pérez, J. C. Rosales, _The Frobenius number for sequences of triangular and tetrahedral numbers_ , J. Number Theory. 186 (2018), 473–492.
  * [20] Ö. J. Rödseth, _On a linear diophantine problem of Frobenius II_ , J. Reine Angew Math. 307/308 (1979), 431–440.
  * [21] J. C. Rosales, M. B. Branco, _Numerical semigroups that can be expressed as an intersection of symmetric numerical semigroups_ , J. Pure Appl. Algebra. 171 (2002), 303–314.
  * [22] J. C. Rosales, M. B. Branco, D. Torrão, _The Frobenius problem for Thabit numerical semigroups_ , J. Number Theory. 155 (2015), 85–99.
  * [23] J. C. Rosales, M. B. Branco, D. Torrão, _The Frobenius problem for repunit numerical semigroups_ , Ramanujan J. 40 (2016), 323–334.
  * [24] J. C. Rosales, M. B. Branco, D. Torrão, _The Frobenius problem for Mersenne numerical semigroups_ , Math. Z. 286 (2017), 1–9.
  * [25] J. C. Rosales, P. A. García-Sánchez, _Numerical Semigroups_ , Developments in Mathematics. Vol.20, Springer, New York (2009).
  * [26] E. S. Selmer, _On the linear Diophantine problem of Frobenius_ , J. Reine Angew. Math. 293/294 (1977), 1–17.
  * [27] Kyunghwan Song, _The Frobenius problem for numerical semigroups generated by the Thabit numbers of the first, second kind base $b$ and the Cunningham numbers_, Bull. Korean Math. Soc. 57 (2020), 623–647.
  * [28] Kyunghwan Song, _The Frobenius problem for extended Thabit numerical semigroups_ , Integers. 21 (2021), A17.
  * [29] J. J. Sylvester, _On sub-invariants, i.e., semi-invariants to binary quanties of an unlimited order_ , Amer. J. Math. 5 (1882), 119–136.
  * [30] A. Tripathi, _On the Frobenius problem for $\\{a,ha+d,ha+bd,ha+b^{2}d,...,ha+b^{k}d\\}$_, J. Number Theory. 162 (2016), 212–223.
  * [31] A. Tripathi, _Formulate for the Frobenius number in three variables_ , J. Number Theory. 170 (2017), 368–389.