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\begin{center}
\vskip 1cm{\LARGE\bf
The Inverse Problem on Subset Sums, II
}
\vskip 1cm
\large
Jian-Dong Wu\footnote{This work is supported by the National
Natural Science Foundation of China (Grant No.\ 11371195)
and the Program of Natural Science Research of 
Jiangsu Higher Education Institutions of China (Grant No.\ 13KJD110005).}\\
School of Mathematical Sciences and Institute of Mathematics\\
Nanjing Normal University\\
Nanjing  210023\\
P. R. China\\
\href{mailto:wujiandong@njnu.edu.cn}{\tt wujiandong@njnu.edu.cn}
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\begin{abstract}
For a set $T$ of integers, let $P(T)$ be the set
of all finite subset sums of $T$, and let $T(x)$ be the set of all
integers of $T$ not exceeding $x$. Let $B=\{b_{1}&lt;b_{2}&lt;\cdots \}$
be a sequence of integers and $d_1=10$, $d_2=3b_1+4$, and
$d_{n}=3b_{n-1}+2$ $(n\ge 3)$. In this paper, we prove that

(i) if $b_n&gt;d_n$ for all $n\ge 1$, then  there
exists a sequence of positive integers $A=\{a_{1}&lt;a_{2}&lt;\cdots \}$
such that, for all $k\ge 2$, $P(A(b_k))=[0,2b_k]\setminus \{b_u,
2b_k-b_u : 1\le u\le k\}$;

(ii) if $b_m=d_m$ for some $m\ge 1$ and
$b_n&gt;d_n$ for all $n\not= m$, then there is no such sequence
$A$.

We also pose a problem for further research.
\end{abstract}


\section{Introduction}

For a sequence of integers $A=\{a_{1}&lt;a_{2}&lt;\cdots \}$, let
$$P(A)=\left\lbrace \sum \varepsilon_{i}a_{i}:\ a_i\in A, \varepsilon_{i}=0\
\text{or}\ 1;\ \sum \varepsilon_{i}&lt;\infty \right\rbrace.$$
Here $0\in
P(A)$. Burr \cite{B} asked  the following question: which sets $S$
of integers are equal to $P(A)$ for some $A$? Let $B=\{ b_1&lt;
b_2&lt;\cdots \}=\mathbb{N}\setminus S$, where
$\mathbb{N}=\{0,1,2,\ldots\}$ is the set of all natural numbers.
Burr mentioned that if $b_1&gt;b_0$ and $b_{n+1}\geq b_{n}^2$ for all
$n\geq 1$, then there exists an $A$ such that
$P(A)=\mathbb{N}\setminus B$. Hegyv\'{a}ri \cite{H} proved that if
$b_{1}\geq b_0$ and $b_{n+1}\geq 5b_{n}$ for all $n\geq 1$, then
such $A$ exists. The condition $b_{n+1}\geq 5b_{n}$ has been
improved to $b_{n+1}\geq 3b_{n}+5$ by  Chen and Fang \cite{C}.
Recently, Chen and the author \cite{CW} proved that, if
$B=\{b_{1}&lt;b_{2}&lt;\cdots \}$ is a sequence of integers with
$b_{1}\in \{4,7,8\}\cup \{b:\ b\geq 11,\ b\in \mathbb{N}\}$,
$b_{2}\geq 3b_{1}+5$, $b_{3}\geq 3b_{2}+3$ and
$b_{n+1}&gt;3b_{n}-b_{n-2}$ for all $n\geq 3$, then there exists a
sequence of positive integers $A=\{a_{1}&lt;a_{2}&lt;\cdots \}$ such
that $P(A)=\mathbb{N}\setminus B$.

 For any set $T$ of integers and any real number $x$, let
 $T(x)$ be the set of all integers of $T$ not exceeding $x$. Let $[a,b]=\{n:n\in
\mathbb{N},\ a\leq n\leq b\}$, and let $x+T=\{x+a:a\in T\}$.

 In this paper, we prove the following result.

\begin{theorem}\label{thm1}
Let $B=\{b_{1}&lt;b_{2}&lt;\cdots \}$ be a sequence of integers, and let
$d_1=10$, $d_2=3b_1+4$, and $d_{n}=3b_{n-1}+2$ $(n\ge 3)$. Then

(i) if $b_n&gt;d_n$ for all $n\ge 1$, then  there exists a sequence
of positive integers $A=\{a_{1}&lt;a_{2}&lt;\cdots \}$ such that, for
all $k\ge 2$,
$$P(A(b_k))=[0,2b_k]\setminus \{b_u, 2b_k-b_u : 1\le u\le k\};$$

(ii) if $b_m=d_m$ for some $m\ge 1$ and $b_n&gt;d_n$ for all $n\not=
m$, then, for any sequence of positive integers
$A=\{a_{1}&lt;a_{2}&lt;\cdots \}$, there exists an index $k\ge 2$ such
that
$$P(A(b_k))\not= [0,2b_k]\setminus \{b_u, 2b_k-b_u : 1\le u\le k\} .$$
\end{theorem} \vskip 2mm

\begin{remark}
Theorem \ref{thm1} gives a segment version of the original problem. The symmetry of the missing set is related to the structure of a subset sum. The analogous result for $P(A(b_k-b_{k-1}))$ was
given in \cite{CW}.
\end{remark}

We pose a problem here.

\begin{problem} Determine all sequences of integers $B=\{b_{1}&lt;b_{2}&lt;\cdots \}$ for which  there exist two
sequences
of positive integers $A=\{a_{1}&lt;a_{2}&lt;\cdots \}$  and $X=\{
x_1&lt;x_2&lt;\cdots \}$ such that, for all $k\ge 2$,
$$P(A(x_k))=[0,2b_k]\setminus \{b_u, 2b_k-b_u : 1\le u\le k\}.$$
\end{problem}


\section{Proof of Theorem \ref{thm1}}

First, we prove Theorem \ref{thm1} (i). By the proof of
\cite[Theorem 1]{C},  there exists a subset $A_{2}$ of
$[1,b_2-b_1]\subset[1,b_{2}]$ such that
$P(A_2)=[0,2b_{2}]\setminus \{ b_1,b_2, 2b_2-b_1\}$. Suppose that
$k\ge 2$, $A_k\subseteq [1, b_k]$ and
\begin{equation}\label{ab}P(A_k)=[0,2b_k]\setminus\{b_u, 2b_k-b_u : 1\le u\le k\}.\end{equation}


We deal with the case $k+1$. If $b_{k+1}\geq3b_k+5$, then, by the
proof of \cite[Theorem 1]{C}, we can construct the required
$A_{k+1}$. So we consider the case $3b_{k}+3\leq b_{k+1}\leq
3b_k+4$. Similar to the arguments in \cite{C} and \cite{CW}, we
have
$$ P(A_k\cup\{b_k+1\})=[0,3b_k+1]\setminus \{ b_u, 3b_k-b_u+1 : 1\le u\le k\} ,$$
$$ P(A_k\cup\{b_k+1,b_{k+1}-2b_k-1\})=[0,b_{k+1}+b_k]\setminus \{ b_u, b_{k+1}+b_k-b_u : 1\le u\le k\} ,$$
$$P(A_k\cup\{b_k+1,b_{k+1}-2b_k-1,b_{k+1}-b_k\})\\
    =[0,2b_{k+1}]\setminus \{ b_u, 2b_{k+1} -b_u : 1\le u\le k+1\} .$$
Let
$$A_{k+1}=A_k\cup\{b_k+1,b_{k+1}-2b_k-1,b_{k+1}-b_k\}.$$
Thus, we have constructed a sequence of sets $\{
A_k\}_{k=1}^\infty$ such that $A_1\subseteq A_2\subseteq \cdots $,
$A_{k+1}\setminus A_k\subseteq ( b_k, b_{k+1}]\ (k\ge 1)$ and
\eqref{ab} holds for all $k\ge 2$. Let
$A=\bigcup_{k=1}^{\infty}A_k$. Then, for all $k\ge 2$,
$$P(A(b_k))=[0,2b_k]\setminus \{b_u, 2b_k-b_u : 1\le u\le k\}.$$


Now we prove Theorem \ref{thm1} (ii). The proof is similar to that
of \cite[Theorem 2]{CW}.

 Suppose that there exists
a sequence $A=\{a_{1}&lt;a_{2}&lt;\cdots \}$ of positive integers such
that
\begin{equation*}\label{A_t}P(A(b_s))=[0,2b_s]\setminus
\{ b_k, 2b_s-b_k : 1\le k\le s\}\end{equation*} for all $s\geq 2$.
Then $P(A)=\mathbb{N}\setminus B$. By \cite{C}, we may assume that
$m\ge 3$. Thus $b_{m}=3b_{m-1}+2$. Let
$A(b_{m-1})=A\cap[0,b_{m-1}]=\{a_1,\ldots,a_{m'}\}$. Then
$$a_{m'+1}+P(A(b_{m-1})) =[a_{m'+1},a_{m'+1}+2b_{m-1}]\setminus
B_{m,1},$$ where $B_{m,1}=\{ a_{m'+1}+b_k,a_{m'+1}+2b_{m-1}-b_k :
1\le k\le m-1\}$. If $a_{m'+1}&gt;2b_{m-1}-b_{m-2}$, then
$2b_{m-1}-b_{m-2}\notin P(A)$, a contradiction. Hence
$a_{m'+1}\leq 2b_{m-1}-b_{m-2}$. By $a_{m'+1}\notin
A\cap[0,b_{m-1}]$, we have $a_{m'+1}&gt;b_{m-1}$.

\bigskip

{\bf Case 1:} $a_{m'+1}=b_{m-1}+1$. Similar to the arguments in
\cite{C} and \cite{CW}, we have
$$P(A(b_{m-1})\cup \{a_{m'+1}\})=[0,b_{m}-1]\setminus B_{m,2},$$
where $B_{m,2}=\{b_k, b_{m}-1-b_k : 1\le k\le m-1\}$. Thus
\begin{eqnarray*}
a_{m'+2}+P(A(b_{m-1})\cup \{a_{m'+1}\})=[a_{m'+2},
a_{m'+2}+b_{m}-1]\setminus B_{m,3},
\end{eqnarray*}
where $B_{m,3}=\{a_{m'+2}+b_k,a_{m'+2}+b_{m}-1-b_k : 1\le k\le
m-1\}$. If $a_{m'+2}\le b_{m}-1-b_{m-1}$, then
$$b_{m}\in [a_{m'+2}, a_{m'+2}+b_{m}-1],\quad a_{m'+2}+b_{m-1}&lt;b_{m}&lt;a_{m'+2}+b_{m}-1-b_{m-1}.$$
Thus $b_{m}\in a_{m'+2}+P(A(b_{m-1})\cup \{a_{m'+1}\})$,  a
contradiction. If $a_{m'+2}&gt; b_{m}-1-b_{m-1}$, then, by
$b_{m}-1-b_{m-1}\notin P(A(b_{m-1})\cup \{a_{m'+1}\})$, we have
$b_{m}-1-b_{m-1}\notin P(A)$, a contradiction.

\bigskip

{\bf Case 2:} $b_{m-1}+2\leq a_{m'+1}\leq 2b_{m-1}-b_{m-2}.$  By
$b_{m}\in [a_{m'+1},a_{m'+1}+2b_{m-1}]$ and $a_{m'+1}+b_{m-1}\leq
3b_{m-1}-b_{m-2}&lt;b_{m}$, there exist some $u_0(1\leq u_0\leq m-2)$
such that $a_{m'+1}+2b_{m-1}-b_{u_0}=b_{m}=3b_{m-1}+2$. Hence
$a_{m'+1}=b_{m-1}+b_{u_0}+2$.

If there exist $u,\ v$ with $1\leq u,\ v\leq m-2$ such that
$2b_{m-1}-b_u=a_{m'+1}+b_v$, then, by
$a_{m'+1}=b_{m-1}+b_{u_0}+2$, we have
$b_{m-1}=b_u+b_v+b_{u_0}+2\leq 3b_{m-2}+2$. This contradicts the
condition $b_{m-1}&gt; 3b_{m-2}+2$. Hence $2b_{m-1}-b_u\neq
a_{m'+1}+b_v$$(1\leq u,\ v\leq m-2)$ and then
$$
    P(A(b_{m-1})\cup \{a_{m'+1}\})=[0,b_{m}+b_{u_0}]\setminus \{
    b_k, b_{m}+b_{u_0}-b_k : 1\le k\le m-1\}.
$$
Thus, for $i\ge 2$, we have
$$
   a_{m'+i}+P(A(b_{m-1})\cup \{a_{m'+1}\})=[a_{m'+i},a_{m'+i}+b_{m}+b_{u_0}]\setminus
   B_{m,4},
$$
where $B_{m,4}=\{a_{m'+i}+b_k,a_{m'+i}+b_{m}+b_{u_0}-b_k : 1\le
k\le m-1\}$.

If $a_{m'+2}&gt;b_{m}+b_{u_0}-b_{m-1}$, then
$b_{m}+b_{u_0}-b_{m-1}\notin P(A)$, a contradiction. So
$b_{m-1}+b_{u_0}+2=a_{m'+1}&lt;a_{m'+2}\leq b_{m}+b_{u_0}-b_{m-1}$.
Let $b_{m-1}+b_{u_0}+2&lt;a_{m'+i}\leq b_{m}+b_{u_0}-b_{m-1}$. Then
\begin{equation}\label{xy}a_{m'+i}+b_{m-2}&lt;b_{m}&lt;a_{m'+i}+b_{m}+b_{u_0}-b_{m-1}.\end{equation}
Since $b_m\notin P(A)$, it follows that $b_m\notin
a_{m'+i}+P(A(b_{m-1})\cup \{a_{m'+1}\})$. Hence $b_m\in B_{m,4}$.
Thus, by \eqref{xy}, $b_{m}=a_{m'+i}+b_{m-1}$, i.e.,
$a_{m'+i}=b_{m}-b_{m-1}$. So $i=2$ and
$a_{m'+3}&gt;b_{m}+b_{u_0}-b_{m-1}$. Thus
\begin{eqnarray*}&amp;&amp;b_{m}+b_{u_0}-b_{m-1}=a_{m'+i}+b_{u_0}\\
&amp;\notin &amp; (P(A(b_{m-1})\cup \{a_{m'+1}\}))\cup
(a_{m'+i}+P(A(b_{m-1})\cup \{a_{m'+1}\}))\\
&amp;=&amp; P(A(b_{m-1})\cup \{a_{m'+1},a_{m'+2}\}).\end{eqnarray*} By
$a_{m'+3}&gt;b_{m}+b_{u_0}-b_{m-1}$, we have
$b_{m}+b_{u_0}-b_{m-1}\notin P(A)$,  a contradiction. This
completes the proof of  Theorem \ref{thm1} (ii).



\section{Acknowledgments}
I sincerely thank my supervisor Professor Yong-Gao Chen for his
valuable suggestions and useful discussions. I am also grateful to
the referee for his/her valuable comments.

\begin{thebibliography}{10}

\bibitem{B}
S. A. Burr, in: P. Erd\H{o}s, A. R\'{e}nyi, V. T. S\'{o}s, eds.,
\emph{Combinatorial Theory and its Applications III},
 Coll. Math. Soc. J. Bolyai, Vol.~4, North-Holland, 1970, p.\ 1155.

\bibitem{C}
Y. G. Chen and J. H. Fang, On a problem in additive number theory,
\emph{Acta Math. Hungar.} \textbf{134} (2012), 416--430.

\bibitem{CW}
Y. G. Chen and J. D. Wu, The inverse problem on subset sums,
\emph{European J. Combin.} \textbf{34} (2013), 841--845.

\bibitem{H}
N. Hegyv\'{a}ri, On representation problems in the additive number theory,
\emph{Acta Math. Hungar.} \textbf{72} (1996), 35--44.


\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B13.

\noindent \emph{Keywords: }
subset sum, representation problem, Burr's problem, complement.

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received May 7 2013;
revised version received  September 5 2013.
Published in {\it Journal of Integer Sequences}, October 12 2013.

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