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\title[Generalized H-fold sumset and Subsequence sum]{Generalized H-fold sumset and Subsequence sum}

\author{\lastname{Mohan}}
\address{Department of Mathematics, Indian Institute of Technology Roorkee, Uttarakhand 247667, India}
\email{mohan@ma.iitr.ac.in}
\email{mohan98math@gmail.com}

\author{\firstname{Ram} \middlename{Krishna} \lastname{Pandey}\IsCorresp}
\address[1]{Department of Mathematics, Indian Institute of Technology Roorkee, Uttarakhand, 247667, India}
\email{ram.pandey@ma.iitr.ac.in}

\thanks{The first author was funded by the Council of Scientific and Industrial Research (CSIR), India (Grant No. 09/143(0925)/2018-EMR-I). The second author was funded by the Department of Atomic Energy and National Board for Higher Mathematics, India (Grant No. 02011/15/2021-NBHM (R.P)/R\&D II/8161).}
\CDRGrant[Council of Scientific and Industrial Research (CSIR)]{09/143(0925)/2018-EMR-I}
\CDRGrant[Department of Atomic Energy and National Board for Higher Mathematics]{02011/15/2021-NBHM (R.P)/R\&D II/8161}

\subjclass{11P70, 11B75, 11B13}
\keywords{\kwd{sumset}
\kwd{subset sum}
\kwd{subsequence sum}}

\begin{abstract}
Let $A$ and $H$ be nonempty finite sets of integers and positive integers, respectively. The \emph{generalized $H$-fold sumset}, denoted by $H^{(r)}A$, is the union of the sumsets $h^{(r)}A$ for $h\in H$ where, the sumset $h^{(r)}A$ is the set of all integers that can be represented as a sum of $h$ elements from $A$ with no summand in the representation appearing more than $r$ times. In this paper, we find the optimal lower bound for the cardinality of $H^{(r)}A$, i.e., for $|H^{(r)}A|$ and the structure of the underlying sets $A$ and $H$ when $|H^{(r)}A|$ is equal to the optimal lower bound in the cases $A$ contains only positive integers and $A$ contains only nonnegative integers. This generalizes recent results of Bhanja. Furthermore, with a particular set $H$, since $H^{(r)}A$ generalizes \emph{subsequence sum} and hence \emph{subset sum}, we get several results of subsequence sums and subset sums as special cases.
\end{abstract}

\dateposted{2024-02-02}
\begin{document}
\maketitle

\section{Introduction}

Let $\mathbb{N}$ be the set of positive integers. Let $A=\lbrace a_{1}, \dots, a_{k}\rbrace$ be a nonempty finite set of integers and $h$ be a positive integer. The \emph{h-fold sumset}, denoted by $hA$, and the \emph{restricted h-fold sumset}, denoted by $h^{\wedge}A$ of $A$, are defined, respectively, by
\begin{align*}
hA&:=\left\lbrace \sum_{i=1}^{k} \lambda_{i} a_{i}: \lambda_{i} \in \mathbb{N} \cup \left\lbrace 0\right\rbrace \ \text{for} \ i= 1, \dots, k \ \text{with} \ \sum_{i=1}^{k} \lambda_{i}=h\right\rbrace,
\\
h^{\wedge}A&:=\left\lbrace \sum_{i = 1}^{k} \lambda_{i} a_{i}: \lambda_{i} \in \left\lbrace 0, 1\right\rbrace \ \text{for} \ i= 1, \dots, k \ \text{with} \ \sum_{i=1}^{k} \lambda_{i}=h\right\rbrace.
\end{align*}

Mistri and Pandey~\cite{MISTRIPANDEY2014} generalized $hA$ and $h^{\wedge}A$, into the generalized $h$-fold sumset, denoted by $h^{(r)}A$, as follows:

Let $r$ be a positive integer such that $1 \leq r \leq h$. The \emph{generalized h-fold sumset} $h^{(r) }A$, is defined~by
\[
h^{(r)}A:=\left\lbrace \sum_{i=1}^{k} \lambda_{i} a_{i} : 0 \leq \lambda_{i} \leq r \ \text{for} \ i=1, \dots, k \ \text{with} \sum_{i=1}^{k} \lambda_{i}=h\right\rbrace.
\]
So, the generalized $h$-fold sumset $h^{(r)}A$ is the set of all sums of $h$ elements of $A$, in which every summand can repeat at most $r$ times. Therefore, $hA$ and $h^{\wedge}A$ are particular cases of $h^{(r)}A$ for $r=h$ and $r=1$, respectively.

For a finite set $H$ of positive integers, Bajnok~\cite{BAJNOK2018} introduced the sumset
\[
HA := \bigcup_{h \in H} hA,
\]
and the restricted sumset
\[
H^{\wedge}A := \bigcup_{h \in H} h^{\wedge}A.
\]
In a recent article, Bhanja and Pandey~\cite{JBHANJA2021} considered a generalization of $HA$ and $H^{\wedge}A$, the \emph{generalized H-fold sumset}, denoted by $H^{(r)}A$, defined by
\[
H^{(r)}A:=\bigcup_{h \in H}h^{(r)}A.
\]
Observed that, if $r\geq\max(H)$, then $H^{(r)}A=HA$ and if $r=1$, then $H^{(r)}A=H^{\wedge}A$. The sumset $H^{(r)}A$ becomes more important as it also generalizes \emph{subset sums} and \emph{subsequence sums}.


\subsection{Subset sum and Subsequence sum}\label{sec1.1}

Let $A$ be a finite set of integers. The sum of all the elements of a given subset $B$ of $A$ is called \emph{subset sum} and it is denoted by $s(B)$. That is,
\[
s(B)=\sum_{b \in B}b.
\]
The set of all nonempty subset sum of $A$, denoted by $\sum(A)$, that is
\[
\sum(A)=\Big\{s(B): \emptyset \not= B\subseteq A\Big\}.
\]
Also we define, for $1 \leq \alpha \leq k$
\[
\sum_{\alpha}(A)=\Big\{s(B): \emptyset \not= B\subseteq A \ \text{and} \ \left|B\right| \geq \alpha\Big\}.
\]
Similarly, we define subsequence sum of a given sequence of integers. Let $A=\{a_{1}, a_{2}, \dots,a_{k}\}$ be a set of $k$ integers and $r$ be a positive integer, with $a_{1} < a_{2} < \cdots < a_{k}$. Then we define a sequence associated with $A$ as
\[
\mathbb{A}=(\underbrace{a_{1},\dots,a_{1}}_{r-\text{times}},\underbrace{a_{2},\dots,a_{2}}_{r-\text{times}},\dots,\underbrace{a_{k},\dots,a_{k}}_{r-\text{times}})=(a_{1},a_{2},\dots,a_{k})_{r} (\text{say}).
\]
Let $\mathbb{B}$ be a subsequence of $\mathbb{A}$. Then
\[
\mathbb{B}=(\underbrace{a_{1},\dots,a_{1}}_{r_{1}-\text{times}},\underbrace{a_{2},\dots,a_{2}}_{r_{2}-\text{times}},\dots,\underbrace{a_{k},\dots,a_{k}}_{r_{k}-\text{times}}) \ \text{with} \ 0\leq r_{i}\leq r.
\]
Given any subsequence $\mathbb{B}$ of $\mathbb{A}$, the sum of all terms of the subsequence $\mathbb{B}$ is called the \emph{subsequence sum}, is denoted by $s(\mathbb{B})$ and we write
\[
s(\mathbb{B})=\sum_{b \in \mathbb{B}}b.
\]
The set of all subsequence sums of a given sequence $\mathbb{A}$ is the set
\[
\sum(\mathbb{A}) = \left\lbrace s(\mathbb{B}): \text{$\mathbb{B}$ is subsequence of $\mathbb{A}$ of length $\geq 1$}\right\rbrace.
\]
For $1\leq \alpha\leq kr$, define
\[
\sum_{\alpha}(\mathbb{A}) = \left\lbrace s(\mathbb{B}) : \text{$\mathbb{B}$ is subsequence of $\mathbb{A}$ of length $\geq $} \ \alpha\right\rbrace.
\]

Note that, we can write
\[
h^{(r)}A=\left\lbrace s(\mathbb{B}): \text{$\mathbb{B}$ is subsequence $\mathbb{A}$ of length $h$}\right\rbrace.
\]
With suitable sets $H$, we can express $\sum(A), \sum_{\alpha}(A),\sum(\mathbb{A})$ and $\sum_{\alpha}(\mathbb{A})$ in terms of $H^{\wedge}A$ and $H^{(r)}A$, as~follows:
\begin{itemize}
\item If $H=\{1,2,\dots,k\}$, then $H^{\wedge}A = \bigcup_{h=1}^{k} h^{\wedge}A=\sum(A)$.
\item If $H= \{\alpha,\alpha+1,\dots,k\}$, then $H^{\wedge}A = \bigcup_{h=\alpha}^{k} h^{\wedge}A=\sum_{\alpha}(A)$.
\item If $H=\{1,2,\dots,kr\}$, then $H^{(r)}A=\bigcup_{h=1}^{kr} h^{(r)}A=\sum(\mathbb{A})$.
\item If $H=\{\alpha,\alpha+1,\dots,kr\}$, then $H^{(r)}A=\bigcup_{h=\alpha}^{kr} h^{(r)}A=\sum_{\alpha}(\mathbb{A})$.
\end{itemize}

Let $A=\{a_{1},a_{2},\dots,a_{k}\}$ be a nonempty set of integers with $a_{1}<a_{2}<\cdots<a_{k}$. For an integer $c$, we write $c \ast A = \{ ca : a \in A \}$ and for integers $ a$ and $b$ with $a < b$, we write $[a,b] = \{ a, a+1, \dots, b \}$. For a nonempty set $S = \{ s_{1}, s_{2}, \dots, s_{n-1}, s_{n} \} $, we let $\max(S), \min(S), \max\nolimits_{-}(S), \min_{+}(S)$ be the largest, smallest, second largest and second smallest elements of $S$, respectively. For a given real number $x$, $\lfloor x \rfloor$ and $\lceil x \rceil$ denote, floor function and ceiling function of $x$, respectively. We assume $\sum_{i=1}^{t} f(i) = 0$ if $t < 1$.

Two standard problems associated with a sumset in additive number theory are to find best possible lower bound for the cardinality of sumset when the set $A$ is known (called the direct problem) and to find the structure of the underlying set $A$ when the size of the sumset attains its lower bound (called the inverse problem). These two types of problems have been solved for the sumsets in various types of groups. We have several classical results on sumsets for the case when $A$ is a subset of group of integers, (see~\cite{BAJNOK2018, JAGBHANJA2021, MISTRIPANDEY2014, MONO2015, NATHAN1995, NATHAN1996, YCHEN2015}), and for subsequence sums and subset one may refer to~\cite{BHANJA2021, BHANJA2020, JBHANJA2021, MISTRIPANDEYPRAKASH2015}. We mention now, some of these results that are applied in this paper.

\begin{theo}[{\cite[Theorem~1.3, Theorem~1.6]{NATHAN1996}}]\label{Nathanson Theorem I}
Let $h\geq 1$, and let $A$ be a nonempty finite set of integers. Then
\begin{center}
$\left| hA\right| \geq h\left| A\right| - h + 1$.
\end{center} This lower bound is best possible. Furthermore, if $\left|hA\right|$ attains this lower bound with $h \geq 2 $, then $A$ is an arithmetic progression.
\end{theo}

\begin{theo}[{\cite[Theorem~1, Theorem~2]{NATHAN1995, NATHAN1996}}]\label{Nathanson Theorem II}
Let A be a nonempty finite set of integers, and let $1 \leq h \leq \left| A\right| $. Then
\begin{center}
$\left| h^{\wedge}A \right| \geq h \left| A\right| - h^{2} + 1$.
\end{center} This lower bound is best possible. Furthermore, if $\lvert h^{\wedge}A\rvert$ attains this lower bound with $\left|A \right| \geq 5$ and $2 \leq h \leq |A| - 2$, then $A$ is an arithmetic progression.
\end{theo}

Mistri and Pandey~\cite{MISTRIPANDEY2014} generalized above results as follows:

\begin{theo}[{\cite[Theorem~2.1]{MISTRIPANDEY2014}}]\label{MistriPandey Theorem I}
Let A be a nonempty finite set of k integers. Let r and h be integers such that $1 \leq r \leq h \leq kr$. Set $m= \left \lfloor{h/r}\right \rfloor$. Then
\begin{center}
$\lvert h^{(r)}A \rvert \geq mr\left(k-m\right) + \left(h-mr\right) \left(k-2m-1\right) +1$.
\end{center} This lower bound is best possible.
\end{theo}

\begin{theo}[{\cite[Theorem~3.1, Theorem~3.2]{MISTRIPANDEY2014}}]\label{MistriPandey Theorem II}
Let $k \geq 3$. Let $r$ and $h \geq 2$ be integers such that $1 \leq r \leq h \leq kr -2$ and $\left(k, h, r\right) \neq \left(4, 2, 1\right)$. Set $m= \left \lfloor{h/r}\right \rfloor$. If A is a finite set of k integers such that
\begin{center}
$\lvert h^{(r)}A \rvert = mr\left(k-m\right) + \left(h-mr\right) \left(k-2m-1\right) +1$,
\end{center} then $A$ is an arithmetic progression.
\end{theo}

Further generalization of $h^{(r)}A$ was considered in~\cite{MISTRIPANDEY2014} for which the direct and inverse results were proved by Yang and Chen~\cite{YCHEN2015}. Direct results for $h^{(r)}A$ when $A$ is a subset of the group of residual classes modulo a prime and $A$ is a subset of a finite cyclic group were given, respectively, by Monopoli~\cite{MONO2015} and Bhanja~\cite{JAGBHANJA2021}.

The direct and inverse theorems for the sumsets $HA$ and $H^{\wedge}A$ proved by Bhanja~\cite{BHANJA2021} are the following:

\begin{theo}[{\cite[Theorem~3]{BHANJA2021}}]\label{Bhanja I}
Let A be a set of k positive integers. Let H be a set of t positive integers with $\max(H) = h_{t}$. Then
\[
\left|HA\right| \geq h_{t}(k-1) + t.
\]
This lower bound is optimal.
\end{theo}

\begin{theo}[{\cite[Theorem~5]{BHANJA2021}}]\label{Bhanja II}
Let A be a set of $k \geq 2 $ positive integers and H be a set of $t \geq 2 $ positive integers with $\max(H) = h_{t}$. If
\[
\left|HA\right| = h_{t}(k - 1)+ t,
\]
then H is an arithmetic progression with common difference d and A is an arithmetic progression with common difference $d \ast \min(A).$
\end{theo}

\begin{theo}[{\cite[Theorem~6, Corollary~7]{BHANJA2021}}]\label{Bhanja III}
Let A be a set of $k$ nonnegative integers and $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace $ be a set ofpositive integers with $h_{1} < h_{2} < \cdots < h_{t}$. Set $h_{0}=0$. If\/ $0\notin A$ and $h_{t} \leq k$, then
\begin{equation*}
\bigl\lvert H^{\wedge}A\bigr\rvert \geq \sum_{i=1}^{t} (h_{i} - h_{i-1}) (k - h_{i}) + t.
\end{equation*}
If $0\in A$ and $ h_{t} \leq k-1$, then
\begin{equation*}
\bigl\lvert H^{\wedge}A\bigr\rvert \geq h_{1}+ \sum_{i=1}^{t} (h_{i} - h_{i-1}) (k - h_{i}-1) + t.
\end{equation*}
The lower bounds are optimal.
\end{theo}

\begin{theo}[{\cite[Theorem~9, Corollary~10]{BHANJA2021}}]\label{Bhanja IV}
Let A be a set of k nonnegative integers. Let $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace $ be a set ofpositive integers with $h_{1} < h_{2} < \cdots < h_{t}$. Set $h_{0}=0$. If\/ $0 \notin A$, $k\geq 6$, $h_{t} \leq k-1$, and
\[
\bigl\lvert H^{\wedge}A\bigr\rvert = \sum_{i=1}^{t} (h_{i} - h_{i-1}) (k - h_{i}) + t,
\]
then $H = h_{1} + [0,t-1]$ and $ A = \min(A) \ast [1,k]$.

If $0 \in A$, $k\geq 7$, $h_{t} \leq k-2$, and
\[
\bigl\lvert H^{\wedge}A\bigr\rvert = h_{1}+ \sum_{i=1}^{t} (h_{i} - h_{i-1}) (k - h_{i}-1) + t,
\]
then $H = h_{1} + [0,t-1]$ and $ A = \min(A\setminus\{0\}) \ast [0,k-1]$.
\end{theo}

In this paper, we prove similar direct and inverse results for the sumset $H^{(r)}A$ when $A$ is a finite nonempty set of positive integers. In Sections~\ref{sec2} and \ref{sec3}, we prove our main theorems, Theorem~\ref{Direct Theorem} and Theorem~\ref{Inverse Theorem}, the direct and inverse theorems for sumset $H^{(r)}A$, when $A$ is a finite set of positive integers. Consequentaly we prove direct and inverse theorems when $A$ contains nonnegative integers with $0 \in A$.

To state our main results we need some notation that are used throughout the paper. Let $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace$ be a set of positive integers with $0 = h_{0}< h_{1} < h_{2} < \cdots < h_{t}$ and $r$ be a positive integer. If $t=1$, then $H^{(r)}A = h_{1}^{(r)}A$. So, we assume $t \geq 2$. If $r>h_{t}$, then $h^{(r)}_{i}A=h_{i}A$ for $1 \leq i \leq t$, giving $H^{(r)}A =HA$. So we assume that $r \leq h_{t}$. There always exists a unique positive integer $l$ such that $h_{l-1} < r \leq h_{l}$, where $1 \leq l \leq t$. For $i=1,2, \dots, t$, let $ h_{i}=m_{i}r+\epsilon_{i}$, where $0 \leq \epsilon_{i} \leq r-1$. For given set $H$ of positive integers and set of integers $A$ with $|H|=t$ and $|A|=k$, let
\begin{multline}
\mathcal{L}(H^{(r)}A) = h_{l-1}(k-1)+(l-1) + \sum_{i=l}^{t} r(m_{i}-m_{i-1})(k-m_{i}) \\
+ \sum_{i=l}^{t}\Big((\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\max\{\epsilon_{i},\epsilon_{i-1}\}(m_{i}-m_{i-1})+1 \Big).
\end{multline}
Note that, if $0\leq i \leq l-1$, then $m_{i}=0$ and $\epsilon_{i}=h_{i}$. So, we can also write
\[
\mathcal{L}(H^{(r)}A) = \sum_{i=1}^{t} \Big(r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\max\{\epsilon_{i},\epsilon_{i-1}\}(m_{i}-m_{i-1})+1 \Big).
\]
For $i=1, \dots,t$, define
\[
M_{i}=\left\lfloor \frac{h_{i}-h_{i-1}}{r} \right \rfloor
\]
and for $j=0, \dots,t-1$, define
\[
N_{j}=
\begin{cases}
\left\lceil \frac{h_{j}}{r} \right\rceil & \text{if } l-1 \leq j \leq t-1\\
\:0 & \text{otherwise}.
\end{cases}
\]
Also, let $\lbrace 0 \rbrace ^{(r)}A=\lbrace 0 \rbrace$.


\section{Direct Theorems}\label{sec2}

\begin{theo}\label{Direct Theorem}
Let $A$ be a nonempty finite set of $k \geq 3$ positive integers. Let $r$ be a positive integer and $H$ be a set of $t \geq 2$ positive integers with $1 \leq r \leq \max(H) \leq (k-1)r-1$. Then
\begin{equation}\label{Direct Theorem Eq- 1}
\bigl\lvert H^{(r)}A\bigr\rvert \geq \mathcal{L}(H^{(r)}A).
\end{equation}
This lower bound is best possible.
\end{theo}


\begin{proof}
Let $A=\lbrace a_{1}, a_{2}, \dots, a_{k} \rbrace$ and $H=\lbrace h_{1},h_{2}, \dots, h_{t}\rbrace$ be such that
\[
0<a_{1}<a_{2}< \cdots< a_{k} \quad \text{and} \quad 0=h_{0}<h_{1}< h_{2}<\cdots<h_{t}.
\]
For $i=0,1, \dots, t $, write $h_{i}=m_{i}r+ \epsilon_{i}, \ \text{where} \ 0\leq \epsilon_{i} \leq r-1$. Then, we have
\[
0 = m_{0}\leq m_{1} \leq m_{2} \leq \cdots \leq m_{t} \leq k-2.
\]
Since $l$ is a positive integer satisfying $h_{l-1} < r \leq h_{l}$, we have $m_{i}=0$ and $\epsilon_{i}=h_{i}$ for $i=0, \dots, l-1$. Set $S_{0}=\emptyset$. Define
\begin{equation*}\label{generalized sumset eq 2}
S_{i}=(h_{i}-h_{i-1})^{(r)}A_{i}+\max\bigl\lbrace h^{(r)}_{i-1}A \bigr\rbrace \quad\text{for } i=1, 2, \dots, t,
\end{equation*}
where
\[
A_{i}= \bigl\lbrace a_{1}, \dots, a_{k-N_{i-1}} \bigr\rbrace\quad \text{for } i=1, 2, \dots, t.
\]
Note that $S_{i} \subseteq h^{(r)}_{i}A \subseteq H^{(r)}A$ and $\max(S_{i})<\min(S_{i+1})$ for all $i\in[1,t-1]$. We shall define sets $T_{i} \subseteq h_{i}^{(r)}A$ that satistfy $T_{i}\cap S_{i}=\emptyset$ for $i\in[0,t]$. Let $R_{i} = S_{i}\cup T_{i} \subseteq h_{i}^{(r)}A, \ \text{for} \ i= 0,1,\dots,t$. If $i\in[0,l-1]$, then define $T_{i}=\emptyset$. So, $|R_{0}| = 0$ for $l\geq 1$, and by Theorem~\ref{Nathanson Theorem I}, we have $\lvert R_{i}\rvert=\left|S_{i}\right| \geq (h_{i}-h_{i-1})(k-1)+1$ for $l \geq 2$ and $i\in[1,l-1]$. If $i\in [l,t]$, then we define $T_{i}$ for every possible values of $\epsilon_{i-1}$ and $\epsilon_{i}$, and consequently find $|R_{i}|$.

\mcs
Let $i\in [l,t]$ be such that $\epsilon_{i-1}=0$ and $\epsilon_{i} \geq 0$. Then $M_{i}=m_{i}-m_{i-1}$ and $N_{i-1} = m_{i-1}$. Let $T_{i}=\emptyset$ in this case. Then by Theorem~\ref{MistriPandey Theorem I}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\
&\geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1 \\
& = r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $\epsilon_{i} > \epsilon_{i-1} >0$ and $m_{i}= m_{i-1}$. Then $M_{i}=m_{i}-m_{i-1}=0$ and $N_{i-1} = m_{i-1}+1$. For $j=0,1, \dots, \epsilon_{i}-\epsilon_{i-1}$, define
\[
T^{0}_{i,j}=(\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i}-1} + (\epsilon_{i-1}+j)a_{k-m_{i}} + \sum_{p=1}^{m_{i}}ra_{k-m_{i}+p}.
\]
Then, we have $\max(S_{i})=T^{0}_{i,0}< T^{0}_{i,1}< \cdots<T^{0}_{i,\epsilon_{i}-\epsilon_{i-1}}=\max(h_{i}^{(r)}A)<\min(S_{i+1})$. Let
\begin{equation}\label{DT-eq-1}
T_{i} = \Big\{T^{0}_{i,j} : j=1, \dots, \epsilon_{i}-\epsilon_{i-1}\Big\}.
\end{equation}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-1}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\
&\geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1 + (\epsilon_{i}-\epsilon_{i-1})\\
& = r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $\epsilon_{i} > \epsilon_{i-1} > 0$ and $m_{i}= m_{i-1} + 1$. Then $M_{i}=m_{i}-m_{i-1}=1$ and $N_{i-1}=m_{i-1}+1=m_{i}$. For $j=0, \dots, \epsilon_{i}-\epsilon_{i-1}-1$, define
\begin{align*}
T^{0}_{i,j}&=(\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i-1}-2}+ ra_{k-m_{i-1}-1} + (\epsilon_{i-1}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p},
\\
T^{1}_{i,j}&=(\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i-1}-2}+ (r-1)a_{k-m_{i-1}-1} + (\epsilon_{i-1}+j+1)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}
\end{align*}
and for $j=0, 1, \dots, r-\epsilon_{i}$,
\[
U^{0}_{i,j}= (r-j)a_{k-m_{i-1}-1} + (\epsilon_{i}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\]
Then, we have $\max(S_{i})=T^{0}_{i,0}<T^{1}_{i,0}<T^{0}_{i,1}<T^{1}_{i,1}< \cdots < T^{0}_{i,\epsilon_{i}-\epsilon_{i-1}-1} < T^{1}_{i,\epsilon_{i}-\epsilon_{i-1}-1} < U^{0}_{i,0}<U^{0}_{i,1} < \cdots < U^{0}_{i,r-\epsilon_{i}}=\max(h_{i}^{(r)}A) <\min(S_{i+1})$. Assume $\bigl\{T^{0}_{i,j}: j=1,\dots,\epsilon_{i}-\epsilon_{i-1}-1 \bigr\} = \emptyset,$ if $\epsilon_{i}-\epsilon_{i-1}=1$. Let
\begin{equation}\label{DT-eq-2}
T_{i}= \bigl\{T^{0}_{i,j}: j=1,\dots, \epsilon_{i}-\epsilon_{i-1}-1\bigr\} \cup \bigl\{T^{1}_{i,j}: j=0,\dots, \epsilon_{i}-\epsilon_{i-1}-1\bigr\} \cup \bigl\{U^{0}_{i,j}: j=0,\dots, r-\epsilon_{i}\bigr\}.
\end{equation}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-2}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\
& \geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1 + 2(\epsilon_{i}-\epsilon_{i-1})+(r-\epsilon_{i})\\
& =r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $\epsilon_{i} > \epsilon_{i-1} > 0$ and $m_{i}> m_{i-1} + 1$. Then $M_{i}=m_{i}-m_{i-1} \geq 2$ and $N_{i-1}=m_{i-1}+1$. For $j=0, \dots,\epsilon_{i}-\epsilon_{i-1}-1$ and $q=1, \dots, m_{i}-m_{i-1}$, define
\[
T^{0}_{i,j}=(\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i}-1} + \left(\sum_{p=1}^{m_{i}-m_{i-1}}ra_{k-m_{i}-1+p} \right) + (\epsilon_{i-1}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p},
\]
\begin{multline*}
T^{q}_{i,j} = (\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i}-1} + \left(\sum_{p=1, \ p \neq m_{i}-m_{i-1}+1-q }^{m_{i}-m_{i-1}}ra_{k-m_{i}-1+p} \right) + (r-1)a_{k-m_{i-1}-q} \\
+ (\epsilon_{i-1}+j+1)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\end{multline*}
For $j=0,\dots,r-\epsilon_{i}-1$ and $q=1, \dots, m_{i}-m_{i-1}-1$, define
\[
U^{0}_{i,j}=(r-j)a_{k-m_{i}} + \sum_{p=1}^{m_{i}-m_{i-1}-1}ra_{k-m_{i}+p} + (\epsilon_{i}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p},
\]
\begin{multline*}
U^{q}_{i,j}=(r-j)a_{k-m_{i}}+ \left(\sum_{p=1, \ p \neq m_{i}-m_{i-1}-q}^{m_{i}-m_{i-1}-1} ra_{k-m_{i}+p}\right) + (r-1)a_{k-m_{i-1}-q}+(\epsilon_{i}+j+1)a_{k-m_{i-1}}\\
+ \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\end{multline*}
Furthermore, define
\[
U^{0}_{i,r-\epsilon_{i}}=\epsilon_{i}a_{k-m_{i}}+ \sum_{p = k-m_{i}+1}^{k} ra_{p}.
\]
Then $U^{0}_{i,r-\epsilon_{i}} = \max(h_{i}^{(r)}A) <\min(S_{i+1})$ and
\begin{equation*}
\begin{array}{cccccccccccc}
\max(S_{i})=T^{0}_{i,0} & < &T^{1}_{i,0} & < &\cdots & < &T^{m_{i}-m_{i-1}}_{i,0} & < & \\[5pt]
T^{0}_{i,1} & < &T^{1}_{i,1} & < &\cdots & < &T^{m_{i}-m_{i-1}}_{i,1} &<&\\[7pt]
\vdots&\vdots&\vdots&\vdots& & \vdots & \vdots & \vdots\\[7pt]
T^{0}_{i,\epsilon_{i}-\epsilon_{i-1}-1} & <& T^{1}_{i,\epsilon_{i}-\epsilon_{i-1}-1}&<&\cdots&<&T^{m_{i}-m_{i-1}}_{i,\epsilon_{i}-\epsilon_{i-1}-1}&< &\\[7pt]
U^{0}_{i,0}&<&U^{1}_{i,0}&<& \cdots&<&U^{m_{i}-m_{i-1}-1}_{i,0} &<&\\[7pt]
U^{0}_{i,1}&<&U^{1}_{i, 1}&<& \cdots &<& U^{m_{i}-m_{i-1}-1}_{i, 1} &<&\\[7pt]
\vdots&\vdots&\vdots&\vdots& & \vdots & \vdots & \vdots \\[7pt]
U^{0}_{i, r-\epsilon_{i}-1}&<&U^{1}_{i, r-\epsilon_{i}-1}&<&\cdots&<&U^{m_{i}-m_{i-1}-1}_{i, r-\epsilon_{i}-1} & < &U^{0}_{i,r-\epsilon_{i}}.
\end{array}
\end{equation*}
Assume $\{T^{0}_{i,j}: j=1,\dots,\epsilon_{i}-\epsilon_{i-1}-1 \} = \emptyset,$ if $\epsilon_{i}-\epsilon_{i-1}=1$. Let
\begin{multline}\label{DT-eq-3}
T_{i} =\bigl\{T^{0}_{i,j}: j=1,\dots,\epsilon_{i}-\epsilon_{i-1}-1\bigr\} \cup \bigl\{T^{q}_{i,j}: j=0,\dots,\epsilon_{i}-\epsilon_{i-1}-1; q=1,\dots,m_{i}-m_{i-1}\bigr\}\\
\cup \bigl\{U^{0}_{i,j}: j=0,\dots,r-\epsilon_{i} \bigr\} \cup \bigl\{U^{q}_{i,j}: j=0,\dots,r-\epsilon_{i}-1; q=1,\dots,m_{i}-m_{i-1}-1\bigr\}.
\end{multline}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-3}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\
&\geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1\\
& \qquad+ (\epsilon_{i}-\epsilon_{i-1})(m_{i}-m_{i-1}+1)+(r-\epsilon_{i})(m_{i}-m_{i-1})\\
& =r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $\epsilon_{i} = \epsilon_{i-1} > 0$ and $m_{i}= m_{i-1} + 1$. Then $M_{i}=m_{i}-m_{i-1}=1$ and $N_{i-1}=m_{i-1}+1$. For $j=0, \dots, r-\epsilon_{i}$, define
\[
U^{0}_{i,j}= (r-j)a_{k-m_{i-1}-1} + (\epsilon_{i}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\]
Then $\max(S_{i})= U^{0}_{i,0}<U^{0}_{i,1} < \cdots < U^{0}_{i,r-\epsilon_{i}} = \max(h_{i}^{(r)}A) < \min(S_{i+1})$. Let
\begin{equation}\label{DT-eq-4}
T_{i}= \bigl\{ U^{0}_{i,j} : j=1,\dots,r-\epsilon_{i}\bigr\}.
\end{equation}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-4}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\ & \geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1 +(r-\epsilon_{i})\\
& = r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $\epsilon_{i} = \epsilon_{i-1} > 0$ and $m_{i}> m_{i-1} + 1$. Then $M_{i}=m_{i}-m_{i-1} \geq 2$ and $N_{i-1}=m_{i-1}+1$. For $j=0,\dots,r-\epsilon_{i}-1$ and $q=1, \dots, m_{i}-m_{i-1}-1$, define
\[
U^{0}_{i,j}=(r-j)a_{k-m_{i}}+\left(\sum_{p=1}^{m_{i}-m_{i-1}-1} ra_{k-m_{i}+p}\right)+(\epsilon_{i}+j)a_{k-m_{i-1}}+ \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}
\]
and
\begin{multline*}
U^{q}_{i,j}=(r-j)a_{k-m_{i}}+ \left(\sum_{p=1, \ p \neq m_{i}-m_{i-1}-q}^{m_{i}-m_{i-1}-1} ra_{k-m_{i}+p}\right) + (r-1)a_{k-m_{i-1}-q}+(\epsilon_{i}+j+1)a_{k-m_{i-1}} \\
+ \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\end{multline*}
Furthermore, define
\[
U^{0}_{i,r-\epsilon_{i}}=\epsilon_{i}a_{k-m_{i}}+ \sum_{p=1}^{m_{i}} ra_{k-m_{i}+p}.
\]
It is easy to see that $U^{0}_{i,r-\epsilon_{i}} = \max(h_{i}^{(r)}A) <\min(S_{i+1})$ and
\begin{equation*}
\begin{array}{cccccccccc}
\max(S_{i})=U^{0}_{i,0} & < & U^{1}_{i,0} & < & \cdots & < & U^{m_{i}-m_{i-1}-1}_{i,0} & < \\[7pt]
U^{0}_{i,1} & < & U^{1}_{i, 1} & < & \cdots & < & U^{m_{i}-m_{i-1}-1}_{i, 1} & < \\[7pt]
\vdots &\vdots&\vdots&\vdots&&\vdots&\vdots&\vdots & \\[7pt]
U^{0}_{i, r-\epsilon_{i}-1} & < & U^{1}_{i, r-\epsilon_{i}-1} & < & \cdots &< & U^{m_{i}-m_{i-1}-1}_{i, r-\epsilon_{i}-1} & < & U^{0}_{i,r-\epsilon_{i}}.
\end{array}
\end{equation*}
Let
\begin{equation}\label{DT-eq-5}
T_{i} = \bigl\{ U^{0}_{i,j} : j=1,\dots, r-\epsilon_{i} \bigr\} \cup\bigl\{ U^{q}_{i,j} : j=0, \dots,r-\epsilon_{i}-1 \ \text{and} \ q=1,\dots,m_{i}-m_{i-1}-1\bigr\}.
\end{equation}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-5}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\
& \geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1\\
&\quad+(r-\epsilon_{i})(m_{i}-m_{i-1})\\
& = r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $ \epsilon_{i} < \epsilon_{i-1} $ and $m_{i}= m_{i-1} + 1$. Then $M_{i}=m_{i}-m_{i-1}-1 = 0$ and $N_{i-1}=m_{i-1}+1=m_{i}$. For $j=0, \dots, r-\epsilon_{i-1}$, define
\[
T^{0}_{i,j}=(r+\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i-1}-1} + (\epsilon_{i-1}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\]
Then $\max(S_{i}) = T^{0}_{i,0}<T^{0}_{i,1}<\cdots<T^{0}_{i,r-\epsilon_{i-1}} = \max(h_{i}^{(r)}A) < \min(S_{i+1})$. Let
\begin{equation}\label{DT-eq-6}
T_{i}=\bigl\{T^{0}_{i,j} : j=1, \dots, r-\epsilon_{i-1}\bigr\}.
\end{equation}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-6}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\ & \geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1 +(r-\epsilon_{i-1})\\
& = r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i-1}(m_{i}-m_{i-1})+1.
\end{align*}

\mcs
Let $i\in [l,t]$ be such that $ \epsilon_{i} < \epsilon_{i-1} $ and $m_{i}> m_{i-1} + 1$. Then $M_{i}=m_{i}-m_{i-1}-1 \geq 1$ and $N_{i-1}=m_{i-1}+1$. For $j=0, \dots,r-\epsilon_{i-1}-1$ and $q=1, \dots, m_{i}-m_{i-1}-1$, define
\[
T^{0}_{i,j}=(r+\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i}} + \left(\sum_{p=1}^{m_{i}-m_{i-1}-1}ra_{k-m_{i}+p} \right) + (\epsilon_{i-1}+j)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p},
\]
\begin{multline*}
T^{q}_{i,j}=(r+\epsilon_{i}-\epsilon_{i-1}-j)a_{k-m_{i}} + \left(\sum_{p=1, \ p \neq m_{i}-m_{i-1}-q }^{m_{i}-m_{i-1}-1}ra_{k-m_{i}+p} \right) + (r-1)a_{k-m_{i-1}-q} \\+ (\epsilon_{i-1}+j+1)a_{k-m_{i-1}} + \sum_{p=1}^{m_{i-1}}ra_{k-m_{i-1}+p}.
\end{multline*}
Define also
\[
T^{0}_{i,r-\epsilon_{i-1}}=\epsilon_{i}a_{k-m_{i}}+ \sum_{p=1}^{m_{i}} ra_{k-m_{i}+p}.
\]
It is easy to see that $T^{0}_{i,r-\epsilon_{i-1}} = \max(h_{i}^{(r)}A) <\min(S_{i+1})$ and
\begin{equation*}
\begin{array}{ccccccccccc}
\max(S_{i})=T^{0}_{i,0} & < & T^{1}_{i,0} & < & \cdots & < & T^{m_{i}-m_{i-1}-1}_{i,0} & < \\[7pt]
T^{0}_{i,1} & < & T^{1}_{i,1} &< & \cdots & < & T^{m_{i}-m_{i-1}-1}_{i,1} & < \\[7pt]
\vdots& \vdots & \vdots& \vdots & & \vdots & \vdots& \vdots \\[7pt]
T^{0}_{i, r-\epsilon_{i-1}-1} & < & T^{1}_{i, r-\epsilon_{i-1}-1} & < & \cdots & < & T^{m_{i}-m_{i-1}-1}_{i, r-\epsilon_{i-1}-1} &< &T^{0}_{i,r-\epsilon_{i-1}} <\min(S_{i+1}).
\end{array}
\end{equation*}
Let
\begin{equation}\label{DT-eq-7}
T_{i} =\bigl\{T_{i,j}^{0}: j=1, \dots, r-\epsilon_{i-1}\bigr\} \cup \bigl\{T_{i,j}^{q}:j=0,\dots, r-\epsilon_{i-1}-1 \ \text{and} \ q=1, \dots, m_{i}-m_{i-1}-1\bigr\}.
\end{equation}
Then by Theorem~\ref{MistriPandey Theorem I} and~\eqref{DT-eq-7}, we have
\begin{align*}
\lvert R_{i}\rvert &=\left|S_{i}\right| + \left|T_{i}\right|\\
& \geq M_{i}r(k-N_{i-1}-M_{i})+(h_{i}-h_{i-1}-M_{i}r)(k-N_{i-1}-2M_{i}-1)+1\\
&\quad+(r-\epsilon_{i-1})(m_{i}-m_{i-1})\\
& > r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\epsilon_{i-1}(m_{i}-m_{i-1})+1.
\end{align*}
Hence
\begin{align*}
\lvert H^{(r)}A\rvert &\geq \sum_{i=0}^{t}\lvert R_{i}\rvert = \sum_{i=0}^{l-1}\left|S_{i}\right|+\sum_{i=l}^{t}\left|S_{i}\cup T_{i}\right| \\
& \geq \sum_{i=1}^{l-1} (h_{i}-h_{i-1})(k-1)+1 \nonumber \\
& \qquad + \sum_{i=l}^{t} r(m_{i}-m_{i-1})(k-m_{i}) + (\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\max\{\epsilon_{i}, \epsilon_{i-1}\}(m_{i}-m_{i-1}) + 1\nonumber\\
&= h_{l-1}(k-1)+(l-1) \nonumber \\
& \qquad + \sum_{i=l}^{t} r(m_{i}-m_{i-1})(k-m_{i)}+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\max\{\epsilon_{i}, \epsilon_{i-1}\}(m_{i}-m_{i-1}) + 1 \\
& =\mathcal{L}(H^{(r)}A).
\end{align*}
This proves~\eqref{Direct Theorem Eq- 1}. Next, we show that this bound is best possible. Let $H = [1,(k-1)r-1]$, $A= \{ 1, 2, \dots, k \}$. Then $H^{(r)}A \subseteq [1,2(r-1)+3r+ \cdots + kr]$. So $\lvert H^{(r)}A\rvert \leq \frac{rk(k+1)}{2} - r-2$. On the other hand, we have by~\eqref{Direct Theorem Eq- 1}, $\lvert H^{(r)}A\rvert \geq \frac{rk(k+1)}{2} - r-2$. This completes the proof of Theorem~\ref{Direct Theorem}.
\end{proof}

\goodbreak
\begin{rema}\label{Remark 2.2}
Following the notation from Theorem~\ref{Direct Theorem}.
\begin{enumerate}\alphenumi
\item \label{10a} If $0=h_{0}<h_{1}<\cdots<h_{t_{0}-1}<(k-1)r \leq h_{t_{0}}<\cdots<h_{t} \leq kr$ with $t_{0} \geq 2$, then we have
\[
\max(h^{(r)}_{t_{0}-1}A)<\max\nolimits_{-}(h^{(r)}_{t_{0}}A)<\max (h^{(r)}_{t_{0}}A)<\max(h^{(r)}_{t_{0}+1}A)<\cdots<\max(h^{(r)}_{t}A).
\]
So $\lvert H^{(r)}A\rvert \geq \lvert H^{(r)}_{t_{0}-1}A\rvert + t-t_{0}+2 \geq \mathcal{L}(H_{t_{0}-1}^{(r)}A)+t-t_{0}+2$, where $H_{t_{0}-1}=\{h_{1}, \dots,h_{t_{0}-1}\}$, $t_{0} \geq 2 $. This lower bound is best possible, as that can be verified with $ A=[1,k]$ and $H=[1,rk]$. Clearly, we have $\lvert H^{(r)}A\rvert=\frac{rk(k+1)}{2}$.

\item \label{10b} If $0=h_{0}<(k-1)r\leq h_{1}<\cdots<h_{t}\leq kr$, then
\[
H^{(r)}A \supseteq h_{1}^{(r)}A \cup \bigl\{\max(h_{i}^{(r)}A) : i= 2, \dots,t\bigr\}.
\]
Therefore
\[
\bigl\lvert H^{(r)}A\bigr\rvert \geq \bigl| h_{1}^{(r)}A \bigr| + t-1 \geq m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-2m_{1}-1)+t,
\]
where $m_{1}=\lfloor h_{1}/r \rfloor$. To check, this bound is best possible, we take $A=[1,k]$ and $H=[(k-1)r,kr]$. Then $H^{(r)}A=[r+2r+\cdots+(k-1)r,r+2r+\cdots+kr]$ and hence $\lvert H^{(r)}A\rvert=kr+1$.
\end{enumerate}
\end{rema}

\begin{coro}\label{Corollary 2.1}
Let $A$ be a nonempty finite set of $k\geq 4$ nonnegative integers with $0 \in A$. Let $r$ be a positive integer and $H$ be a set of $t \geq 2$ positive integers with $1 \leq r \leq \max (H) \leq (k-2)r-1 $. Let $m = \lceil \min(H)/r \rceil$ and $m_{1}=\lfloor \min(H)/r \rfloor$. Then
\begin{equation}\label{Eq- coro-2.1}
\bigl\lvert H^{(r)}A\bigr\rvert \geq m_{1}r(m-m_{1}+1) + (\min(H)-m_{1}r)(m-2m_{1}) + \mathcal{L}\big(H^{(r)}(A\setminus\{0\}) \big).
\end{equation}
This lower bound is best possible.
\end{coro}

\begin{proof}
Let $A=\{0,a_{1},\dots,a_{k-1}\}$ be a set of nonnegative integers with $0<a_{1}< \cdots <a_{k-1}$ and $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace$ be a set of positive integers with $0 = h_{0}< h_{1}=\min(H) < h_{2} < \cdots < h_{t}=\max(H)$. Consider $A^{'} = A \setminus \lbrace 0 \rbrace$. Then $H^{(r)}A^{'} \subseteq H^{(r)}A$.

Let $m = \lceil h_{1}/r \rceil$, $h_{1}=m_{1}r+\epsilon_{1}$, where $0 \leq \epsilon_{1} \leq r-1$ and $B=\{0,a_{1}, \dots, a_{m}\}$. Then
\[
h^{(r)}_{1}B \subseteq H^{(r)}A
\]
and $h^{(r)}_{1}B \cap H^{(r)}A^{'} = \max(h^{(r)}_{1}B) = \min (H^{(r)}A^{\prime})= ra_{1} + \cdots + ra_{m_{1}} + \epsilon_{1}a_{m_{1}+1} $. Hence by Theorem~\ref{MistriPandey Theorem I} and Theorem~\ref{Direct Theorem}, we have
\begin{align*}
\bigl\lvert H^{(r)}A\bigr\rvert &\geq \bigl| h^{(r)}_{1}B \bigr| + \bigl|H^{(r)}A^{\prime}\bigr| -1 \\
&\geq m_{1}r(m-m_{1}+1) + (\min(H)-m_{1}r)(m-2m_{1}) + \mathcal{L}\big(H^{(r)}(A\setminus\{0\}) \big).
\end{align*}
This proves the Corollary. To check optimallity of the bound, take $A=[0,k-1]$ and $H=\linebreak{} [1,(k-2)r-1] $. Then $H^{(r)}A \subseteq [0,2(r-1)+3r+ \cdots+(k-1)r]$ and $\lvert H^{(r)}A\rvert \leq \frac{rk(k-1)}{2}-r-1$. From~\eqref{Eq- coro-2.1}, we have $\lvert H^{(r)}A\rvert \geq \frac{rk(k-1)}{2}-r-1$.
\end{proof}

\begin{rema}\label{Remark 2.3}
Following the notation from Corollary~\ref{Corollary 2.1}.
\begin{enumerate}\alphenumi
\item \label{12a} If $0=h_{0}<h_{1}<\cdots<h_{t_{0}-1}<(k-2)r \leq h_{t_{0}}<\cdots<h_{t} \leq (k-1)r$ with $t_{0} \geq 2$, then we have
\[
\ \ \ \ \ \ \max\bigl(h^{(r)}_{t_{0}-1}A\bigr)<\max\nolimits_{-}\bigl(h^{(r)}_{t_{0}}A\bigr)<\max \bigl(h^{(r)}_{t_{0}}A\bigr)<\max\bigl(h^{(r)}_{t_{0}+1}A\bigr)<\cdots<\max\bigl(h^{(r)}_{t}A\bigr).
\]
So $\lvert H^{(r)}A\rvert \geq m_{1}r(m-m_{1}+1) + (\min(H)-m_{1}r)(m-2m_{1}) + \mathcal{L}\big(H^{(r)}(A\setminus\{0\}) \big) + t-t_{0}+2$. This lower bound is best possible, as that can be verified with $ A=[0,k-1]$ and $H=[1,(k-1)r]$. Clearly, we have $\lvert H^{(r)}A\rvert=\frac{rk(k-1)}{2}+1$. Also, if we take $H = [1,(k-1)r] \cup X$, where $X \subseteq [(k-1)r + 1, kr]$, then again $\lvert H^{(r)}A\rvert=\frac{rk(k-1)}{2}+1$.

\item \label{12b} If $0=h_{0}<(k-2)r \leq h_{1}<\cdots<h_{t}\leq (k-1)r$, then
\[
H^{(r)}A \supseteq h_{1}^{(r)}A \cup \bigl\{\max(h_{i}^{(r)}A) : i= 2, \dots,t\bigr\}.
\]
Therefore
\[
\lvert H^{(r)}A\rvert \geq | h_{1}^{(r)}A | + t-1 \geq m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-2m_{1}-1)+t,
\]
where $m_{1}=\lfloor h_{1}/r \rfloor$. To check, this bound is best possible, we take $A=[0,k-1]$ and $H=[(k-2)r,(k-1)r]$. Then $H^{(r)}A=[r+2r+\cdots+(k-3)r,r+2r+\cdots+(k-1)r]$ and hence $\lvert H^{(r)}A\rvert=(2k-3)r+1$.
\end{enumerate}
\end{rema}

\begin{rema}\ 
\begin{enumerate}\alphenumi
\item \label{13a} For $r = \max(H) = h_{t}$, Theorem~\ref{Bhanja I} follows from Theorem~\ref{Direct Theorem} as a consequence.
\item \label{13b} For $r=1$, Theorem~\ref{Bhanja III} follows from Remark~\ref{Remark 2.2} and Remark~\ref{Remark 2.3} as a consequence.
\end{enumerate}
\end{rema}


\section{Inverse problem}\label{sec3}

This section deals with the inverse theorems associated with the sumset $H^{(r)}A$. In this section, we charaterize the sets $A$ and $H$, when the cardinality of $H^{(r)}A$ is equal to its optimal lower bound. There are some cases in which either $A$ or $H$ or both may not be arithmetic progression but size of $H^{(r)}A$ is equal to the optimal lower bound (called extremal set). See some extremal sets in~\cite[Section~3]{MISTRIPANDEY2014} and~\cite[Section~2.2]{BHANJA2020}. Here we give some more example of extremal sets.
\begin{enumerate}
\item Let $A$ be a set of $k\left(\geq 3\right)$ integers and $r$ be a positive integer. If $H=\{1,rk\}$ or $H= \{rk-1,rk\}$, then $|H^{(r)}A|=k+1$.
\item Let $A=\{a_{1},a_{2},a_{1}+a_{2}\}$ with $0<a_{1}<a_{2}$ and $H\subseteq\{1,2,3\}$ with $r=1$; or $A=\{0,a_{1},a_{2},\linebreak a_{1}+a_{2},\}$ with $H \subseteq \{1,2,3\}$ and $r=1$. Then the sets $A$ are extremal sets.
\end{enumerate}
We now present the main inverse results associated with $H^{(r)}A$.

\begin{theo}\label{Inverse Theorem}
Let $r \geq 1$ be a positive integer, $A$ be a nonempty finite set of $k \geq 6$ positive integers and $H$ be a set of $t \geq 2$ positive integers with $1\leq r \leq \max(H) \leq (k-1)r-1 $. If
\begin{equation*}\label{Inverse Theorem -Eq-1}
\bigl\lvert H^{(r)}A\bigr\rvert = \mathcal{L}\bigl(H^{(r)}A\bigr),
\end{equation*}
then $H$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$.
\end{theo}

\begin{proof}
Let $A=\lbrace a_{1}, a_{2}, \dots, a_{k} \rbrace$ and $H=\lbrace h_{1},h_{2}, \dots, h_{t}\rbrace$ be such that
\[
0<a_{1}<a_{2}< \cdots< a_{k} \text{ and } 0=h_{0}<h_{1}< h_{2}<\cdots<h_{t}.
\]
For $i=1, \dots, t$, let $ h_{i}=m_{i}r+\epsilon_{i}$, where $0 \leq \epsilon_{i} \leq r-1$. Let $l$ be a positive integer such that $h_{l-1} < r \leq h_{l}$, where $1 \leq l \leq t$. Since $|H^{(r)}A|$ is equal to its lower bound given in~\eqref{Direct Theorem Eq- 1}, we have, from the proof of Theorem~\ref{Direct Theorem} that, $|H^{(r)}A|=\sum_{i=1}^{t} \lvert R_{i}\rvert$. This implies that
\[
\lvert R_{1}\rvert = \bigl| h^{(r)}_{1}A \bigr|=m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-2m_{1}-1)+1
\]
and $\lvert R_{i}\rvert = \left|S_{i}\right|+\left|T_{i}\right|= r(m_{i}-m_{i-1})(k-m_{i})+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-1)-\max\{\epsilon_{i},\epsilon_{i-1}\}(m_{i}-m_{i-1})+1$, for $i =2, \dots, t$. If $h_{1}>1$, then by Theorem~\ref{MistriPandey Theorem II}, the set $A$ is an arithmetic progression. Let $h_{1}=1$ and $h_{2}>2$. Then we have
\[
R_{1}=A \text{ and } R_{2} = S_{2} \cup T_{2}.
\]
Therefore $|S_{2}|$ is minimum and hence $A_{2}=\{a_{1},a_{2},\dots,a_{k-1}\}$ is an arithmetic progression. Now we show that $a_{k-1}-a_{k-2}=a_{k}-a_{k-1}$. Let $m_{2} \leq k-3$. We have
\begin{align*}
a_{m_{2}+1}&<\min\bigl((h_{2}-1)^{(r)}A_{2}\bigr)+a_{m_{2}+1} \\
&<\min\bigl((h_{2}-1)^{(r)}A_{2}\bigr)+a_{m_{2}+2}\\[-0.1em]
&\phantom{i}\vdots\\
&<\min\bigl((h_{2}-1)^{(r)}A_{2}\bigr)+a_{k-1}\\
&<\min\bigl((h_{2}-1)^{(r)}A_{2}\bigr)+a_{k}=\min(R_{2}).
\end{align*}
We also have $R_{1}=A$ and $a_{1}<a_{2}<\cdots<a_{m_{2}+1}<a_{m_{2}+2}<\cdots<a_{k}<\min((h_{2}-1)^{(r)}A_{2})+a_{k}=\min(R_{2})$. So $\min((h_{2}-1)^{(r)}A_{2})+a_{m_{2}+i}=a_{m_{2}+i+1}$ for $i=1,2, \dots, k-m_{2}-1$. This gives $a_{k-1}-a_{k-2}=a_{k}-a_{k-1}$.

Let $m_{2}=k-2$ and $\epsilon_{2}=0$. Then
\begin{align*}
a_{k-2}&<ra_{1}+\dots+ra_{k-2}\\
&<ra_{1}+\dots+ra_{k-3}+(r-1)a_{k-2}+a_{k-1}\\
&<ra_{1}+\dots+ra_{k-3}+(r-1)a_{k-2}+a_{k}=\min(R_{2}).
\end{align*}
This implies that $ra_{1}+\dots+ra_{k-3}+(r-1)a_{k-2}=a_{k-1}-a_{k-2}=a_{k}-a_{k-1}$. Let $m_{2}=k-2$ and $\epsilon_{2}\geq 1$. Then $r \geq 2$, $m_{1}=0$ and $a_{k-1}<\min(h_{2}^{(r)}A))< \min(R_{2})$. Note that
\[
\left|R_{2}\right|=2r(k-2)-\epsilon_{2}(k-3)
\]
and by Theorem~\ref{MistriPandey Theorem I}
\[
\bigl|h_{2}^{(r)}A\bigr|\geq 2r(k-2)-\epsilon_{2}(k-3)+1.
\]
Let $y$ be an element of $ h_{2}^{(r)}A$, which is different from $\min(h_{2}^{(r)}A)$. If $y\notin R_{2}$, then
\[
H^{(r)}A \supseteq \bigl\{a_{1}, a_{2}, \dots,a_{k-1}\bigr\} \cup \bigl\{\min(h_{2}^{(r)}A),y\bigr\} \cup \Bigg(\bigcup_{i=2}^{t}R_{i}\Biggr).
\]
This gives $\lvert H^{(r)}A\rvert > \sum_{i=1}^{t} \lvert R_{i}\rvert$, which is not possible. Therefore $y \in R_{2}$. This gives that $ h_{2}^{(r)}A=R_{2} \cup \bigl\{\min(h_{2}^{(r)}A)\bigr\}$ and
\[
\bigl|h_{2}^{(r)}A\bigr|= 2r(k-2)-\epsilon_{2}(k-3)+1
\]
and so by Theorem~\ref{MistriPandey Theorem II}, A is an arithmetic progression.

Let $h_{1}=1$ and $h_{2}=2$. Then $R_{1}=A$. Consider $R_{1}^{\prime}=\{a_{1}+a_{i}: i=2, \dots,k-1\}$, a subset of $h_{2}^{(r)}A$. Then $\max(R_{1}^{\prime})<\min(R_{2})=a_{1}+a_{k}$. Therefore $R_{1}^{\prime}\subseteq R_{1}=A$. This gives that $a_{1}+a_{i}=a_{i+1}$ for $i=2,\dots,k-1$. Also $a_{k}=a_{1}+a_{k-1}<a_{2}+a_{k-1}<a_{2}+a_{k}=\min_{+}(R_{2})$ and $a_{k}<\min(R_{2})<\min_{+}(R_{2})$ give $a_{2}+a_{k-1}=a_{1}+a_{k}$. Hence $A$ is an arithmetic progression.

Let $A= a_{1}+d_{1}\cdot[0,k-1]$, where $d_{1}$ is the common difference of $A$. We show that $H$ is an arithmetic progression with common difference $d$ and $d_{1}=da_{1}$. Note that, for all $i\in[1,t-1]$, we have
\[
\max\nolimits_{-}(R_{i})< \min \bigl\lbrace (h_{i+1}-h_{i})^{(r)}A_{i+1} \bigr\rbrace+ \max\nolimits_{-}(R_{i}) < \min(R_{i+1}).
\]
But we already know that
\[
\max\nolimits_{-}(R_{i})<\max(R_{i})<\min(R_{i+1}).
\]
So
\[
\min \bigl\lbrace (h_{i+1}-h_{i})^{(r)}A_{i+1} \bigr\rbrace+ \max\nolimits_{-}(R_{i})=\max(R_{i}).
\]
This implies that
\begin{equation}\label{inverse thm eq I}
\min \bigl\{ (h_{i+1}-h_{i})^{(r)}A_{i+1} \bigr\}=\max(R_{i})-\max\nolimits_{-}(R_{i}) = a_{s+1}- a_{s}=a_{2}-a_{1} \ \text{for some $s$}.
\end{equation}
Consider the following cases:
\begin{enumerate}\alphenumi
\item Let $i \in [1,t-1]$ be such that $ \epsilon_{i} = \epsilon_{i+1}$. Then $m_{i+1} > m_{i}$. If $m_{i+1}-m_{i} \geq 2$, then\linebreak $ \min\{ (h_{i+1}-h_{i})^{(r)}A_{i+1} \} = ra_{1}+ \cdots +ra_{m_{i+1}-m_{i}} > a_{2} >a_{2}-a_{1}$, which contradicts~\eqref{inverse thm eq I}. Hence $ m_{i+1}-m_{i} = 1$ and $ra_{1}=(h_{i+1}-h_{i})a_{1}=a_{2}-a_{1}$.

\item Let $i \in [1,t-1]$ be such that $ \epsilon_{i} < \epsilon_{i+1}$. Then $m_{i+1} \geq m_{i}$. If $m_{i+1}-m_{i} \geq 1$, then\linebreak $ \min\{ (h_{i+1}-h_{i})^{(r)}A_{i+1} \} = ra_{1}+ \cdots +ra_{m_{i+1}-m_{i}} + (\epsilon_{i+1}-\epsilon_{i})a_{m_{i+1}-m_{i}+1} > a_{2} >a_{2}-a_{1}$, which contradicts~\eqref{inverse thm eq I}. Hence $ m_{i+1}=m_{i} $ and $(\epsilon_{i+1}-\epsilon_{i})a_{1}=(h_{i+1}-h_{i})a_{1}=a_{2}-a_{1}$.

\item Let $i \in [1,t-1]$ be such that $\epsilon_{i}>\epsilon_{i+1} $. Then $m_{i+1} > m_{i}$. If $m_{i+1}>m_{i}+1$, then\linebreak $ \min\lbrace (h_{i+1}-h_{i})^{(r)}A_{i+1} \rbrace=ra_{1}+ \cdots +ra_{m_{i+1}-m_{i}-1} + (r+\epsilon_{i+1}- \epsilon_{i})a_{m_{i+1}-m_{i}}> a_{2} > a_{2} - a_{1}$, which contradicts~\eqref{inverse thm eq I}. Hence $m_{i+1}=m_{i}+1 \text{ and } (r+\epsilon_{i+1}- \epsilon_{i})a_{1}=(h_{i+1}-h_{i})a_{1}= a_{2} - a_{1}$.
\end{enumerate}
Hence, $(h_{i+1}-h_{i})a_{1}=a_{2}-a_{1}=d_{1}$ for each $i=1,\dots,t-1$. So $H$ is an arithmetic progresion with common difference $d\leq r$ and $d_{1}=da_{1}$. This completes the proof.
\end{proof}

\begin{coro}\label{Corollary 3.1}
Let $r \geq 1$ and $t > t_{0} \geq 2$ be integers. Let $A$ be a nonempty finite set of $k \geq 6$ positive integers and $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace $ be a set of $t$ positive integers with $h_{1} < h_{2} < \cdots < h_{t_{0}-1} \leq (k-1)r-1<h_{t_{0}}<\cdots<h_{t} <kr $. If $(t_{0},h_{1})\not=(2,1)$ and $\lvert H^{(r)}A\rvert = \mathcal{L}(H_{t_{0}-1}^{(r)}A) +t-t_{0}+2,$ then $H$ is an arithmetic progression with common difference $d\leq r$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$, where $H_{t_{0}-1}=\{h_{1},h_{2},\dots,h_{t_{0}-1}\}$.
\end{coro}

\begin{proof}
Note that
\[
\max\bigl(H_{t_{0}-1}^{(r)}A\bigr) = \max\bigl(h_{t_{0}-1}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr)<\max\bigl(h_{t_{0}}^{(r)}A\bigr)<\max\bigl(h_{t_{0}+1}^{(r)}A\bigr)<\cdots<\max\bigl(h_{t}^{(r)}A\bigr)
\]
and
\[
H^{(r)}A \supseteq H_{t_{0}-1}^{(r)}A \cup \bigl\{\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr)\bigr\}\cup \bigl\{\max \bigl(h_{i}^{(r)}A\bigr) : i=t_{0}, \dots, t\bigr\}.
\]
Therefore $\bigl|H_{t_{0}-1}^{(r)}A\bigr|=\mathcal{L}(H_{t_{0}-1}^{(r)}A)$. If $t_{0} \geq 3$, then by Theorem~\ref{Inverse Theorem}, $H_{t_{0}-1}$ is an arithmetic progression with common difference $d$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$. Since $(t_{0},h_{1}) \ne (2,1)$, so if $t_{0}=2$, then $h_{1}>1$. So by Theorem~\ref{MistriPandey Theorem II}, $A$ is an arithmetic progression.

\begin{enonce*}[remark]{Claim}
If $t_{0} \geq 2$, $t\geq t_{0}+1 $, and $A$ is an arithmetic progression with common difference $d_{1}$, then
\begin{enumerate}
\item \label{c1} $\epsilon_{t_{0}} < \epsilon_{t_{0}-1}$,
\item \label{c2} $m_{t_{0}-1}=k-2$,
\item \label{c3} $h_{i}-h_{i-1}=d$ for $i=t_{0}, \dots, t$ and the common difference of $A$ is $d_{1}=da_{1}$.
\end{enumerate}
\end{enonce*}

\begin{proof}[Proof of the claim]
Note that $m_{t_{0}-1}r + \epsilon_{t_{0}-1} = h_{t_{0}-1} \leq (k-1)r -1 = (k-2)r + r-1$. Hence $m_{t_{0}-1} \leq k-2$. Also $h_{t_{0}} \geq (k-1)r$ and $h_{t_{0}} < h_{t} \leq kr-1$, i.e., $h_{t_{0}} \leq kr-2 = (k-1)r + r-2$. Thus $(k-1)r \leq h_{t_{0}} \leq (k-1)r + r-2$. Hence $m_{t_{0}} = k-1$ and $0 \leq \epsilon_{t_{0}} \leq r-2$. Note also that
\begin{align*}
\max\bigl(h_{t_{0}}^{(r)}A\bigr) &= \epsilon_{t_{0}}a_{1} + ra_{2} + \cdots + ra_{k},
\\
\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr) &= (\epsilon_{t_{0}} + 1)a_{1} + (r-1)a_{2} + \cdots + ra_{k}.
\end{align*}

\begin{proof}[\meqref{c1}]
If $\epsilon_{t_{0}} \geq \epsilon_{t_{0}-1}$, then
\[
\max\bigl(h_{t_{0}-1}^{(r)}A\bigr)< y = ra_{1}+\cdots + ra_{k-m_{t_{0}-1}-1}+\epsilon_{t_{0}}a_{k-m_{t_{0}-1}}+ra_{k-(m_{t_{0}-1}-1)}+\cdots+ra_{k}<\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr),
\]
and $y\in h_{t_{0}}^{(r)}A $, which is a contradiction. Hence $\epsilon_{t_{0}} < \epsilon_{t_{0}-1}$.
\let\qed\relax
\end{proof}

\begin{proof}[\meqref{c2}]
If $m_{t_{0}-1}\leq k-3$, then
\begin{align*}
&\max\bigl(h_{t_{0}-1}^{(r)}A\bigr)\\
&<ra_{1}+\cdots + ra_{k-m_{t_{0}-1}-2}+(r-(\epsilon_{t_{0}-1}-\epsilon_{t_{0}}))a_{k-m_{t_{0}-1}-1}+\epsilon_{t_{0}-1}a_{k-m_{t_{0}-1}}+ra_{k-m_{t_{0}-1}+1}+\cdots+ra_{k}\\
&<\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr),
\end{align*}
which is a contradiction. Hence $\epsilon_{t_{0}} < \epsilon_{t_{0}-1}$ and $m_{t_{0}-1}=k-2$. Consequently, we can write
\[
\max\bigl(h_{t_{0}-1}^{(r)}A\bigr)<(r-(\epsilon_{t_{0}-1}-\epsilon_{t_{0}}))a_{1}+\epsilon_{t_{0}-1}a_{2}+ra_{3}+\cdots+ra_{k}<\max\bigl(h_{t_{0}}^{(r)}A\bigr).
\]
But we already know that
\[
\max\bigl(h_{t_{0}-1}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr)<\max\bigl(h_{t_{0}}^{(r)}A\bigr).
\]
This implies that
\[
(r-(\epsilon_{t_{0}-1}-\epsilon_{t_{0}}))a_{1}+\epsilon_{t_{0}-1}a_{2}+ra_{3}+\cdots+ra_{k}=\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr),
\]
which gives $\epsilon_{t_{0}-1}=r-1$. Therefore $h_{t_{0}}-h_{t_{0}-1}=(k-1)r+\epsilon_{t_{0}}-(k-2)r-(r-1)=\epsilon_{t_{0}}+1$. Now we have
\[
\max\nolimits_{-}\bigl(h_{t_{0}-1}^{(r)}A\bigr)<(\epsilon_{t_{0}}+1)a_{1}+ ra_{2}+(r-1)a_{3}+ra_{4}+\cdots+ra_{k}<\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr).
\]
We also have
\[
\max\nolimits_{-}\bigl(h_{t_{0}-1}^{(r)}A\bigr)<\max\bigl(h_{t_{0}-1}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr).
\]
Therefore
\[
(\epsilon_{t_{0}}+1)a_{1}+ ra_{2}+(r-1)a_{3}+ra_{4}+\cdots+ra_{k}=\max\bigl(h_{t_{0}-1}^{(r)}A\bigr).
\]
This gives
\[
(\epsilon_{t_{0}}+1)a_{1}=a_{3}-a_{2}=d_{1}.
\]
This implies that $a_{1}$ divides $d_{1}$, so $d_{1}=da_{1}$ where $d=\epsilon_{t_{0}}+1$. Hence $ h_{t_{0}}-h_{t_{0}-1}=d$.
\let\qed\relax
\end{proof}

\begin{proof}[\meqref{c3}]
Now we show that $h_{i}-h_{i-1}=d$ for $i=t_{0}+1, \dots, t$.

Note that
\[
\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{t_{0}+1}^{(r)}A\bigr)< \cdots <\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr)<\max\bigl(h_{t}^{(r)}A\bigr).
\]
We already have
\[
\max\nolimits_{-}\bigl(h_{t_{0}}^{(r)}A\bigr)<\max\bigl(h_{t_{0}}^{(r)}A\bigr)<\max\bigl(h_{t_{0}+1}^{(r)}A\bigr) \cdots <\max\bigl(h_{t}^{(r)}A\bigr).
\]
Therefore
\[
\max\bigl(h_{i}^{(r)}A\bigr)=\max\nolimits_{-}\bigl(h_{i+1}^{(r)}A\bigr),
\]
which gives $(\epsilon_{i+1}-\epsilon_{i})a_{1}=a_{2}-a_{1}=d_{1}$ for $i=t_{0},t_{0}+1,\dots,t-1$. Hence, $H$ is an arithmetic\linebreak progression with common difference $d$ and $A$ is an arithmetic progression with common difference $da_{1}$.
\end{proof}
\let\qed\relax
\end{proof}
\let\qed\relax
\end{proof}

Now we discuss the case when $t=t_{0}$.

\begin{coro}\label{Corollary 3.2}
Let $r \geq 1$ and $t \geq 2$ be positive integers. Let $A$ be a nonempty finite set of $k \geq 6$ positive integers and $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace $ be a set of $t$ positive integers with $h_{1} < \cdots < h_{t-1} \leq (k-1)r-1<h_{t} <kr $. If $(t,h_{1}) \neq (2,1)$ and $|H^{(r)}A| = \mathcal{L}((H\setminus \{h_{t}\})^{(r)}A) +2,$ then $H$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$.
\end{coro}

\begin{proof}
Note that
\[
\max\bigl((H\setminus \{h_{t}\})^{(r)}A\bigr) = \max\bigl(h_{t-1}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr)<\max\bigl(h_{t}^{(r)}A\bigr)
\]
and
\[
H^{(r)}A \supseteq (H\setminus \{h_{t}\})^{(r)}A \cup \bigl\{\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr), \max \bigl(h_{t}^{(r)}A\bigr) \bigr\}.
\]
Therefore $\bigl|H_{t-1}^{(r)}A\bigr|=\mathcal{L}((H\setminus \{h_{t}\})^{(r)}A)$. Also, if $t=2$ and $h_{1}>1$, then by Theorem~\ref{MistriPandey Theorem II}, $A$ is an arithmetic progression.

\begin{enonce*}[remark]{Claim}
If $t \geq 2$, then
\begin{enumerate}
\item \label{c'1} $h_{t-1} > r$,
\item \label{c'2} $\epsilon_{t} \leq \epsilon_{t-1}$,
\item \label{c'3} $m_{t-1}=k-2$.
\end{enumerate}
\end{enonce*}

\begin{proof}[Proof of the claim]\ 
\begin{proof}[\meqref{c'1}]
If $h_{t-1} \leq r$, then $\max(h^{(r)}_{t-1}A) = h_{t-1}a_{k}$. Note that
\[
h_{t-1}a_{k} < (\epsilon_{t}+1)a_{1} + ra_{2} + (r-1)a_{3} + \cdots + ra_{k} < (\epsilon_{t}+1)a_{1} + (r-1)a_{2} + \cdots + ra_{k} = \max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr),
\]
which is a contradiction. Hence $h_{t-1} > r$ and so $m_{t-1} \geq 1$.

Note that $m_{t-1}r + \epsilon_{t-1} = h_{t-1} \leq (k-1)r -1 = (k-2)r + r-1$. Hence $m_{t-1} \leq k-2$. Also $(k-1)r \leq h_{t} \leq kr -1$. Hence $m_{t} = k-1$. Note also that
\begin{align*}
\max\bigl(h_{t-1}^{(r)}A\bigr) &= \epsilon_{t-1}a_{k-m_{t-1}} + ra_{k-m_{t-1}+1} + \cdots + ra_{k},
\\
\max\bigl(h_{t}^{(r)}A\bigr) &= \epsilon_{t}a_{1} + ra_{2} + \cdots + ra_{k},
\\
\max\nolimits_{-}h_{t}^{(r)}A &= (\epsilon_{t}+1)a_{1} + (r-1)a_{2} + \cdots + ra_{k}.
\end{align*}
\let\qed\relax
\end{proof}

\begin{proof}[\meqref{c'2}]
Let $\epsilon_{t} > \epsilon_{t-1}$. Then
\begin{align*}
\max\bigl(h_{t-1}^{(r)}A\bigr) & <x = ra_{1}+\cdots + ra_{k-m_{t-1}-1}+\epsilon_{t}a_{k-m_{t-1}}+ra_{k-(m_{t-1}-1)}+\cdots+ra_{k} \\
& \leq y = ra_{1} + \epsilon_{t}a_{2} + ra_{3} + \cdots + ra_{k}\\
& \leq (\epsilon_{t}+1)a_{1} + (r-1)a_{2} + \cdots + ra_{k} = \max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr),
\end{align*}
and $x, y\in h_{t}^{(r)}A $. If $x<y$ or $y< \max\nolimits_{-}(h_{t}^{(r)}A) $, then we get a contradiction. So we assume that $x = y = \max\nolimits_{-}(h_{t}^{(r)}A)$. This implies that $\epsilon_{t} = r-1$ and $m_{t-1} = k-2$. Since $\epsilon_{t} > \epsilon_{t-1} $, we have $\epsilon_{t-1} \leq r-2$. Now consider $z = ra_{1} + ra_{2} + (r-1)a_{3} + ra_{4} + \cdots + ra_{k} \in h_{t}^{(r)}A $. Then we have $\max(h_{t-1}^{(r)}A) < z < \max\nolimits_{-}(h_{t}^{(r)}A) $, which is again a contradiction. Hence $\epsilon_{t} \leq \epsilon_{t-1}$.
\let\qed\relax
\end{proof}

\begin{proof}[\meqref{c'3}]
If $m_{t-1}\leq k-3$, then
\begin{align*}
&\max\bigl(h_{t-1}^{(r)}A\bigr)\\
&<ra_{1}+\cdots + ra_{k-m_{t-1}-2}+(r-(\epsilon_{t-1}-\epsilon_{t}))a_{k-m_{t-1}-1}+\epsilon_{t-1}a_{k-m_{t-1}}+ra_{k-m_{t-1}+1}+\cdots+ra_{k}\\
&<\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr),
\end{align*}
which is a contradiction. Hence $\epsilon_{t} \leq \epsilon_{t-1}$ and $m_{t-1}=k-2$. Consequently, we can write
\[
\max\bigl(h_{t-1}^{(r)}A\bigr)<(r-(\epsilon_{t-1}-\epsilon_{t}))a_{1}+\epsilon_{t-1}a_{2}+ra_{3}+\cdots+ra_{k}<\max\bigl(h_{t}^{(r)}A\bigr).
\]
But we already know
\[
\max\bigl(h_{t-1}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr)<\max\bigl(h_{t}^{(r)}A\bigr).
\]
This implies
\[
(r-(\epsilon_{t-1}-\epsilon_{t}))a_{1}+\epsilon_{t-1}a_{2}+ra_{3}+\cdots+ra_{k}=\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr).
\]
This gives $\epsilon_{t-1}=r-1$. Therefore $h_{t}-h_{t-1}=(k-1)r+\epsilon_{t}-(k-2)r-(r-1)=\epsilon_{t}+1$. We have
\[
\max\nolimits_{-}\bigl(h_{t-1}^{(r)}A\bigr)<(\epsilon_{t}+1)a_{1}+ ra_{2}+(r-1)a_{3}+ra_{4}+\cdots+ra_{k}<\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr).
\]
We also have
\[
\max\nolimits_{-}\bigl(h_{t-1}^{(r)}A\bigr)<\max(h_{t-1}^{(r)}A)<\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr).
\]
Therefore
\[
(\epsilon_{t}+1)a_{1}+ ra_{2}+(r-1)a_{3}+ra_{4}+\cdots+ra_{k}=\max\bigl(h_{t-1}^{(r)}A\bigr).
\]
This gives
\begin{equation}\label{Eq- x}
(\epsilon_{t}+1)a_{1}=a_{3}-a_{2}.
\end{equation}
If $t \geq 3$, by Theorem~\ref{Inverse Theorem}, $H\setminus \{h_{t}\}$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$. Therefore
\[
(\epsilon_{t}+1)a_{1}=a_{3}-a_{2} = da_{1},
\]
which implies $h_{t} - h_{t-1} = \epsilon_{2} +1 = d$. Hence, if $t\geq 3$, $H$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$. If $t=2$, then $H = \{h_{1}, h_{2}\}$ is an arithmetic progression with common difference $d= h_{2}-h_{1}=\epsilon_{t} + 1 \leq r$. Since $h_{1}>1 $ and $\bigl|H_{1}^{(r)}A\bigr|=\mathcal{L}((H\setminus \{h_{2}\})^{(r)}A) = |h_{1}^{(r)}A| = m_{1}r(k-m_{1}) + \epsilon_{1}(k-2m_{1}-1) + 1 $, we have from Theorem~\ref{MistriPandey Theorem II} that $A$ is an arithmetic progression with common difference $da_{1}$ from~\eqref{Eq- x}.
\end{proof}
\let\qed\relax
\end{proof}
\let\qed\relax
\end{proof}

\begin{coro}\label{Corollary 3.3}
Let $r \geq 2$ be a positive integer and $A$ be a nonempty finite set of $k \geq 6$ positive integers and $H$ be a set of $t\geq 2$ positive integers with $ (k-1)r-1<\min(H)<\max(H) <kr $. Let $m_{1}=\lfloor \min(H)/r \rfloor$. If
\[
\bigl|H^{(r)}A\bigr| = m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-2m_{1}-1)+t,
\]
then $H$ is an arithmetic progression with common difference $d \leq r-1$ and $A$ is an arithmetic progression with common difference $d\ast \min(A)$.
\end{coro}

\begin{proof}
Note that
\[
\max\bigl(h_{2}^{(r)}A\bigr)<\max\bigl(h_{3}^{(r)}A\bigr)<\cdots<\max\bigl(h_{t}^{(r)}A\bigr)
\]
and
\[
H^{(r)}A \supseteq h_{1}^{(r)}A \cup \bigl\{\max\bigl(h_{i}^{(r)}A\bigr) : 2\leq i \leq t\bigr\}.
\]
Therefore $\bigl|h_{1}^{(r)}A\bigr|=m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-2m_{1}-1)+1$ and by Theorem~\ref{MistriPandey Theorem II}, $A$ is an arithmetic progression. Assume $d_{1}$ is the common difference of $A$. Note that
\[
\max\nolimits_{-}\bigl(h_{1}^{(r)}A\bigr)<\max\nolimits_{-}\bigl(h_{2}^{(r)}A\bigr)< \cdots <\max\nolimits_{-}\bigl(h_{t}^{(r)}A\bigr)<\max\bigl(h_{t}^{(r)}A\bigr).
\]
We already have
\[
\max\nolimits_{-}\bigl(h_{1}^{(r)}A\bigr)<\max\bigl(h_{1}^{(r)}A\bigr)<\max\bigl(h_{2}^{(r)}A\bigr) \cdots <\max\bigl(h_{t}^{(r)}A\bigr).
\]
Therefore
\[
\max\bigl(h_{i}^{(r)}A\bigr)=\max\nolimits_{-}\bigl(h_{i+1}^{(r)}A\bigr),
\]
which gives $(\epsilon_{i+1}-\epsilon_{i})a_{1}=a_{2}-a_{1}=d_{1}$ for $i=1,2,\dots,t-1$. Hence, set $H$ is an arithmetic progression with common difference $d \leq r-1$ and set $A$ is an arithmetic progression with common difference $d\ast \min(A)$.
\end{proof}

\begin{coro}\label{Corollary 3.4}
Let $r$ be a positive integer, $A$ be a finite set of $k\geq 7$ nonnegative integers with $0 \in A$, and $H$ be a set of $t \geq 2$ positive integers with $1 \leq r \leq \max(H) \leq (k-2)r-1 $. Let $ m = \lceil \min(H)/r \rceil$ and $ m_{1}=\lfloor \min(H)/r \rfloor$. If
\begin{equation}\label{Cor 3.4-Eq-1}
\bigl|H^{(r)}A\bigr| = m_{1}r(m-m_{1}+1) + (\min(H)-m_{1}r)(m-2m_{1}) + \mathcal{L}\bigl(H^{(r)}(A\setminus\{0\}) \bigr),
\end{equation}
then $H$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast\min(A\setminus\{0\})$. Moreover, if $\min(H)>1$, then $d=1$.
\end{coro}

\begin{proof}
Let $A=\{0,a_{1},\dots,a_{k-1}\}$ be a set of nonnegative integers with $0<a_{1}< \cdots< a_{k-1}$ and $H = \{h_{1}, h_{2}, \dots, h_{t}\}$ be set of positive integer such that $h_{1} < h_{2} < \cdots < h_{t}$. From~\eqref{Cor 3.4-Eq-1} and Corollary~\ref{Corollary 2.1}, we have
\[
\bigl|h^{(r)}_{1}B\bigr|= m_{1}r(m-m_{1}+1) + (h_{1}-m_{1}r)(m-2m_{1})+1
\]
and
\[
\bigl|H^{(r)}A^{\prime} \bigr| = \mathcal{L}\bigl(H^{(r)}(A^{\prime}) \bigr),
\]
where $A^{\prime}=\{a_{1},a_{2},\dots,a_{k-1}\}$ and $ B=\{0,a_{1}, \dots, a_{m}\}$ with $m=\lceil h_{1}/r \rceil$. Then by Theorem~\ref{Inverse Theorem}, $H$ is an arithmetic progression with common difference $d \leq r$ and $A^{\prime}$ is an arithmetic progression with common difference $d\ast \min(A^{\prime})$. Now, we show that $d=1$, if $h_{1}>1$. To show that $d=1$, it is sufficient to prove that the common difference of arithmetic progression $A$ is $a_{1}$. If $r=1$, then $d=1$. Assume $r\geq 2$. Now, define $R_{i} =S_{i} \cup T_{i}$ for the set $A^{\prime}$ as it was defined in Theorem~\ref{Direct Theorem}. So
\begin{equation*}
R_{1}=S_{1}=h^{(r)}_{1}A^{\prime} \subseteq h^{(r)}_{1}A.
\end{equation*}
Now $\max(h^{(r)}_{1}A^{\prime})=\max(h^{(r)}_{1}A)$ implies that $h^{(r)}_{1}A \cap R_{2}=\emptyset$. We write
\begin{multline}\label{bound}
\bigl|H^{(r)}A\bigr| = m_{1}r(m-m_{1}+1) + (h_{1}-m_{1}r)(m-2m_{1})+ \bigl|h_{1}^{(r)}A^{\prime}\bigr| \\
+ \sum_{i=2}^{t} \Big(r(m_{i}-m_{i-1})(k-m_{i}-1)+(\epsilon_{i}-\epsilon_{i-1})(k-m_{i}-2) \\
-(\max\{\epsilon_{i},\epsilon_{i-1}\})(m_{i}-m_{i-1})+1\Big).
\end{multline}

%\begin{itemize}
%\item
\mcs
If $h_{1} = m_{1}r + \epsilon_{1}$ with $m_{1} \geq 1$ and $\epsilon_{1} \geq 1$, then $m = m_{1} + 1$ and so $|B| \geq 3$. Since $\bigl|h^{(r)}_{1}B\bigr|=\linebreak m_{1}r(m-m_{1}+1) + (h_{1}-m_{1}r)(m-2m_{1})+1$, Theorem~\ref{MistriPandey Theorem II} implies that $B$ is an arithmetic progression with common difference $a_{1}$. Furthermore, as $A^{\prime}$ is also an arithmetic progression, we have $A = B \cup A^{\prime}$ is an arithmetic progression with common difference $a_{1}$.

%\item
\mcs
If $h_{1} = m_{1}r + \epsilon_{1}$ with $m_{1} = 0$ or $\epsilon_{1} = 0$, then $m = 1$ or $m_{1} = m$. Since
\[
h^{(r)}_{1}B \cup h_{1}^{(r)}A^{\prime} \subseteq h_{1}^{(r)}A \ \text{and} \ h^{(r)}_{1}A \cap R_{2}=\emptyset,
\]
we have from~\eqref{bound} that
\begin{align*}
\bigl|h_{1}^{(r)}A \bigr| &= \bigl|h^{(r)}_{1}B \bigr| + \bigl| h_{1}^{(r)}A^{\prime} \bigr|-1\\
&=m_{1}r(m-m_{1}+1) + (h_{1}-m_{1}r)(m-2m_{1})+m_{1}r(k-m_{1}-1)+(h_{1}-m_{1}r)(k-2m_{1}-2)+1\\
&\leq m_{1}r+(h_{1}-m_{1}r)+m_{1}r(k-m_{1}-1)+(h_{1}-m_{1}r)(k-2m_{1}-2)+1\\
&=m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-2m_{1}-1)+1.
\end{align*}
This gives
\[
\bigl|h_{1}^{(r)}A\bigr|= m_{1}r(k-m_{1})+(h_{1}-m_{1}r)(k-m_{1}-1)+1.
\]
So, by Theorem~\ref{MistriPandey Theorem II}, $A$ is an arithmetic progression with common difference $a_{1}$. This implies that $d=1$. Hence,
\[
H=h_{1}+[0,t-1]\quad\text{and}\quad A= \min(A\setminus \{0\})\ast [0,k-1].\qedhere
\]
%\end{itemize}
\end{proof}

As a consequence of Corollary~\ref{Corollary 3.1}, Corollary~\ref{Corollary 3.2}, Corollary~\ref{Corollary 3.3} and Corollary~\ref{Corollary 3.4}, we have the following Corollaries.

\begin{coro}\label{Corollary 3.5}
Let $r \geq 1$ and $t > t_{0} \geq 2$ be integers. Let $A$ be a nonempty finite set of $k \geq 7$ nonnegative integers with $0 \in A$ and $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace $ be a set of $t$ positive integers with $h_{1} < h_{2} < \cdots < h_{t_{0}-1} \leq (k-2)r-1<h_{t_{0}}<\cdots<h_{t} <(k-1)r $. If $(t_{0},h_{1})\not=(2,1)$ and $\lvert H^{(r)}A\rvert \geq m_{1}r(m-m_{1}+1) + (\min(H)-m_{1}r)(m-2m_{1}) + \mathcal{L}\bigl(H^{(r)}(A\setminus\{0\}) \bigr) + t-t_{0}+2$, then $H$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast\min(A\setminus\{0\})$. Moreover, if $\min(H)>1$, then $d=1$.
\end{coro}

\begin{coro}\label{Corollary 3.6}
Let $r \geq 1$ and $t \geq 2$ be integers. Let $A$ be a nonempty finite set of $k \geq 7$ nonnegative integers with $0 \in A$ and $H = \lbrace h_{1}, h_{2}, \dots, h_{t} \rbrace $ be a set of $t$ positive integers with $h_{1} < h_{2} < \cdots < h_{t-1} \leq (k-2)r-1<h_{t} <(k-1)r $. If $(t,h_{1}) \neq (2,1)$ and $\lvert H^{(r)}A\rvert \geq m_{1}r(m-m_{1}+1) + (\min(H)-m_{1}r)(m-2m_{1}) + \mathcal{L}\bigl(H^{(r)}(A\setminus\{0\}) \bigr) +2$, then $H$ is an arithmetic progression with common difference $d \leq r$ and $A$ is an arithmetic progression with common difference $d\ast\min(A\setminus\{0\})$. Moreover, if $\min(H)>1$, then $d=1$.
\end{coro}

\begin{coro}\label{Corollary 3.7}
Let $r \geq 2$ be a positive integer and $A$ be a nonempty finite set of $k \geq 7$ nonnegative integers with $0 \in A$ and $H$ be a set of $t\geq 2$ positive integers with $ (k-2)r-1<\min(H)<\max(H) <(k-1)r $. Let $m_{1}=\lfloor \min(H)/r\rfloor$. If
\[
\bigl|H^{(r)}A\bigr| = m_{1}r(k-m_{1})+(\min(H)-m_{1}r)(k-2m_{1}-1)+t,
\]
then $H$ is an arithmetic progression with common difference $d \leq r-1$ and $A$ is an arithmetic progression with common difference $d\ast\min(A\setminus\{0\})$.
\end{coro}

\section{Conclusions} In Section~\ref{sec1.1}, we have already discussed the relation between generalized $H$-fold sumset and subsequence sum. Choosing a particular $H$ we get some results of subsequence sum.

\begin{coro}[{\cite[Theorem~2.1]{MISTRIPANDEYPRAKASH2015}}]\label{MistriPandeyOM Theorem I}
Let $k$ and $r$ be positive integers. Let $\mathbb{A}$ be a finite sequence of nonnegative integers with $k$ distinct terms each with repetitions $r$.

If\/ $0 \notin \mathbb{A}$ and $k \geq 3$, then
\begin{equation*}
\sum(\mathbb{A}) \geq \frac{rk(k+1)}{2}.
\end{equation*}

If\/ $0\in \mathbb{A}$ and $k \geq 4$, then
\begin{equation*}
\sum(\mathbb{A}) \geq 1 + \frac{rk(k-1)}{2}.
\end{equation*}
The above lower bounds are best possible.
\end{coro}

\begin{proof}
If $0 \notin \mathbb{A}$, then taking $H = [1,kr]$ in Remark~\ref{Remark 2.2}$\MK$\eqref{10b}, we get $\sum(\mathbb{A}) \geq \frac{rk(k+1)}{2}.$ If $0 \in \mathbb{A}$, then taking $H=[1,kr]$ in Remark~\ref{Remark 2.3}$\MK$\eqref{12a}, we get $\sum(\mathbb{A}) \geq 1 + \frac{rk(k-1)}{2}.$
\end{proof}

\begin{coro}[{\cite[Theorem~2.3]{MISTRIPANDEYPRAKASH2015}}]\label{MistriPandeyOM Theorem II}
Let $k$ and $r$ be positive integers. If $\mathbb{A}$ is a finite sequence of nonnegative integers with $k$ distinct terms each with repetitions $r$.

If\/ $0 \notin \mathbb{A}$, $k \geq 6$ and
\begin{equation*}\label{MPO Eqn 3}
\sum(\mathbb{A}) = \frac{rk(k+1)}{2},
\end{equation*}
then $\mathbb{A} = d \ast [1,k]_{r}$ for some positive integer $d$.

If\/ $0\in \mathbb{A}$, $k \geq 7$ and
\begin{equation*}\label{MPO Eqn 4}
\sum(\mathbb{A}) = 1 + \frac{rk(k-1)}{2},
\end{equation*}
then $\mathbb{A} = d \ast [0,k-1]_{r}$ for some positive integer $d$.
\end{coro}

\begin{proof}
If $0 \notin \mathbb{A}$, then taking $H = [1,kr]$ in Corollary~\ref{Corollary 3.1}, we get $\mathbb{A} = d \ast [1,k]_{r}$ for some positive integer $d$. If $0 \in \mathbb{A}$, then taking $H=[1,kr]$ in Corollary~\ref{Corollary 3.5}, we get $\mathbb{A} = d \ast [0,k-1]_{r}$ for some positive integer $d$.
\end{proof}

Taking $H = [\alpha,kr]$ in Theorem~\ref{Direct Theorem} and Remark~\ref{Remark 2.2} and taking $H = [\alpha,(k-1)r]$ in Corollary~\ref{Corollary 2.1} and Remark~\ref{Remark 2.3}, we get the following result.

\begin{coro}[{\cite[Corollary~3.2]{BHANJA2020}}]
Let $k\geq 4$, $r \geq 1$ and $\alpha$ be integers with $1 \leq \alpha < kr$. Let $m \in [1, k]$ be the integer such that $(m - 1)r \leq \alpha < mr$. Let $\mathbb{A}$ be a finite sequence of nonnegative integers with $k$ distinct terms each with repetitions $r$.

If\/ $0 \notin \mathbb{A}$, then
\begin{equation*}
\sum_{\alpha}(\mathbb{A}) \geq \frac{rk(k+1)}{2} - \frac{rm(m+1)}{2} + m(mr-\alpha) + 1.
\end{equation*}

If\/ $0\in \mathbb{A}$, then
\begin{equation*}
\sum_{\alpha}(\mathbb{A}) \geq \frac{rk(k-1)}{2} - \frac{rm(m-1)}{2} + (m-1)(mr-\alpha) + 1.
\end{equation*}
The above lower bounds are best possible.
\end{coro}

Taking $H = [\alpha,kr-2]$, Theorem~\ref{Inverse Theorem} and Corollaries~\ref{Corollary 3.1}--\ref{Corollary 3.7} give the following result.

\begin{coro}[{\cite[Corollary~3.5]{BHANJA2020}}]
Let $k\geq 7$, $r \geq 1$ and $\alpha$ be integers with $1 \leq \alpha \leq kr-2$. Let $m \in [1, k]$ be the integer such that $(m - 1)r \leq \alpha < mr$. Let $\mathbb{A}$ be a finite sequence of nonnegative integers with $k$ distinct terms each with repetitions $r$.

If\/ $0 \notin \mathbb{A}$ and
\begin{equation*}
\sum_{\alpha}(\mathbb{A}) = \frac{rk(k+1)}{2} - \frac{rm(m+1)}{2} + m(mr-\alpha) + 1,
\end{equation*}
then $\mathbb{A} = d \ast [1,k]_{r}$ for some positive integer $d$.

If\/ $0\in \mathbb{A}$ and
\begin{equation*}
\sum_{\alpha}(\mathbb{A}) = \frac{rk(k-1)}{2} - \frac{rm(m-1)}{2} + (m-1)(mr-\alpha) + 1,
\end{equation*}
then $\mathbb{A} = d \ast [0,k-1]_{r}$ for some positive integer $d$.
\end{coro}



\subsection*{Acknowledgment}

The authors thank to the anonymous referee for giving suggestions to improve the paper.
%The first author would like to thank to the Council of Scientific and Industrial Research (CSIR), India for providing the grant to carry out the research with Grant No. 09/143(0925)/2018-EMR-I. The second author would like to thank to the Department of Atomic Energy and National Board for Higher Mathematics, India for providing the grant with Grant No. 02011/15/2021-NBHM (R.P)/R\&D II/8161.


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