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\begin{center}
\vskip 1cm{\LARGE\bf Full Subsets of $\mathbb N$}
\vskip 1cm \large {Lila Naranjani}\\
Islamic Azad University\\ 
Dolatabad Branch\\ 
Isfahan\\
Iran\\
\href{mailto:lnaranjani@yahoo.com}{\tt lnaranjani@yahoo.com}  \\
\ \\
Madjid Mirzavaziri\\
Department of Pure Mathematics\\
Ferdowsi University of Mashhad\\
P. O. Box 1159--91775\\ 
Iran\\
\href{mailto:mirzavaziri@gmail.com}{\tt mirzavaziri@gmail.com} \\
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\begin{abstract}
Let $A$ be a subset of $\mathbb N$. We say that $A$ is $m$-full if
$\sum A=[m]$ for a positive integer $m$, where $\sum A$ is the set of
all positive integers which are a sum of distinct elements of $A$ and
$[m]=\{1,\ldots,m\}$. In this paper, we show that a set
$A=\{a_1,\ldots,a_k\}$ with $a_1<\cdots<a_k$ is full if and only if
$a_1=1$ and $a_i\leq a_1+\cdots+a_{i-1}+1$ for each $i, 2\leq i\leq k$.
We also prove that for each positive integer $m\notin\{2,4,5,8,9\}$
there is an $m$-full set. We determine the numbers
$\alpha(m)=\min\{|A|: \sum A=[m]\}, \beta(m)=\max\{|A|: \sum A=[m]\},
L(m)=\min\{\max A: \sum A=[m]\}$ and $U(m)=\max\{\max A: \sum A=[m]\}$
in terms of $m$. We also give a formula for $F(m)$, the number of
$m$-full sets.
\end{abstract}

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\section{Introduction}
Let $n$ be a positive integer and denote by $D(n)$ and $\sigma(n)$ the set of its positive divisors and the sum of its positive divisors, respectively. A positive integer $n$ is called perfect if $\sigma(n)=2n$. Euclid proved that the formula $2^{p-1}(2^p -1)$ gives an even perfect number whenever $2^p- 1$ is prime. It has been proved for the first time by Euler that if an even positive integer $n$ is perfect then $n=2^{p-1}q$, where $p$ and $q=2^p-1$ are primes. 
%(see \cite{Dik} page 19). 
In this case \[D(n)=\{1,2,2^2,\ldots, 2^{p-1},q,2q,2^2q,\ldots, 2^{p-1}q\}.\]
A simple argument shows that each $1\leq \ell\leq 2n=2^pq$ can be written as a sum of distinct elements of $D(n)$. To see this, note that each $\ell, 1\leq \ell\leq 2^p-1$, can be written as a sum $2^{\ell_1}+\cdots+2^{\ell_r}$, where $r\geq1$ and $0\leq\ell_1<\cdots<\ell_r\leq p-1$. Now if $2^p\leq \ell\leq 2^pq$, then we can write $\ell=\alpha q+\beta$, where $0\leq\beta\leq2^p-1$ and $1\leq\alpha\leq2^p-1$. Thus we can write $\alpha=2^{\alpha_1}+\cdots+2^{\alpha_i}$ and $\beta=2^{\beta_1}+\cdots+2^{\beta_j}$, where $i\geq1, j\geq0, 0\leq\alpha_1<\cdots<\alpha_i\leq p-1$ and $0\leq\beta_1<\cdots<\beta_j\leq p-1$. Hence
\[\ell=2^{\alpha_1}q+\cdots+2^{\alpha_i}q+2^{\beta_1}+\cdots+2^{\beta_j}\]
is a sum of distinct elements of $D(n)$.

These considerations motivate us to find all positive integers $n$ having the property that each $1\leq\ell\leq\sigma(n)$ can be written as a sum of distinct elements of $D(n)$. This leads us to the following problem:

Let $A=\{a_1,\ldots,a_k\}$ be a subset of $\mathbb N$. Define the \textit{sum set} of $A$, denoted by $\sum A$, by
\[\sum A=\{a_{i_1}+\cdots+a_{i_r}: a_{i_1}<\cdots<a_{i_r},1\leq r\leq k\}.\]
For what positive integer $m$ does there exist a set $A$ with $\sum A=[m]$, where $[m]=\{1,\ldots,m\}$?

We show that each positive integer $m\notin\{2,4,5,8,9\}$ has this property and determine the numbers \bea \alpha(m)&=&\min\{|A|: \sum A=[m]\},\\ \beta(m)&=&\max\{|A|: \sum A=[m]\},\\L(m)&=&\min\{\max A: \sum A=[m]\},\\ U(m)&=&\max\{\max A: \sum A=[m]\}.\eea 
\section{The Results}
\begin{definition} Let $m$ be a positive integer. A subset $A$ of $\mathbb N$ is called \textit{$m$-full} if $\sum A=[m]$. $A$ is called full if it is $m$-full for some positive integer $m$.
\end{definition}
\begin{theorem}\label{criterion} A subset $A=\{a_1,\ldots,a_k\}$ of $\mathbb N$ with $a_1<\cdots<a_k$ is full if and only if $a_1=1$ and $a_i\leq a_1+\cdots+a_{i-1}+1$ for each $i, 2\leq i\leq k$.
\end{theorem}
\begin{proof} Let $A$ be full and $\sum A=[m]$ for a positive integer $m$. Clearly $a_1=1$. If $a_j> a_1+\cdots+a_{j-1}+1$ for some $j, 2\leq j\leq k$, then $a_1+\cdots+a_{j-1}+1$ is not a sum of distinct elements of $A$. But $1\leq a_1+\cdots+a_{j-1}+1\leq a_1+\cdots+a_k=m$. This contradicts to the fact that $\sum A=[m]$.

Conversely, suppose that $a_1=1$ and $a_i\leq a_1+\cdots+a_{i-1}+1$ for each $i, 2\leq i\leq k$. We claim that $\sum A=[a_1+\cdots+a_k]$. We prove this by induction on $k$. For $k=1$ the result is obvious. Suppose that the result is true for $k-1$. Then $\sum A\setminus\{a_k\}=[a_1+\cdots+a_{k-1}]$. Now suppose that $a_1+\cdots+a_{k-1}+1\leq\ell\leq a_1+\cdots+a_{k}$ and write $\ell=a_{k}+a$. Then $a$ belongs to $\{0,1,\ldots,a_1+\cdots+a_{k-1}\}$ since $a_k\leq a_1+\cdots+a_{k-1}+1$. If $a=0$ then $\ell=a_k\in\sum A$ and if $a\neq 0$ then $a\in [a_1+\cdots+a_{k-1}]=\sum A\setminus\{a_k\}$ can be written as $a_{i_1}+\cdots+a_{i_r}$. Thus $\ell=a_{i_1}+\cdots+a_{i_r}+a_k\in\sum A$.
\end{proof}
\begin{proposition} Let $n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$, with $p_1<\cdots<p_r$ primes, be a positive integer. Then $D(n)=\{d: d|n\}$ is full if and only if $p_1=2$ and $p_i\leq\sigma(p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}})+1$ for each $i, 2\leq i\leq r$.
\end{proposition}
\begin{proof} If $D(n)$ is $m$-full then $m=\sigma(n)$. Since $p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}}|n$ and $p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}}\neq n$, we have $\sigma(p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}})<\sigma(n)$. Hence $\sigma(p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}})+1$ is a member of $[\sigma(n)]$. Thus if $p_i>\sigma(p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}})+1$ for some $i$, then the number $\sigma(p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}})+1$ is a member of $[\sigma(n)]$ which is not a sum of distinct elements of $D(n)$. On the other hand, if the condition $p_i\leq\sigma(p_1^{\alpha_1}\cdots p_{i-1}^{\alpha_{i-1}})+1$ for each $i, 2\leq i\leq r$, is satisfied, then using an argument similar to the one used in Theorem~\ref{criterion}, we can inductively prove that each element of $[\sigma(n)]$ can be written as a sum of distinct elements of $D(n)$.
\end{proof}
In the next theorem, we characterize all $m$ for which there is an $A$ with $\sum A=[m]$.
\begin{theorem}\label{values} Let $m$ be a positive integer. There is a set $A$ such that $\sum A=[m]$ if and only if $m\notin\{2,4,5,8,9\}$.
\end{theorem}
\begin{proof}
By simple inspection, there is no $A$ with $\sum A=[m]$ for $m=2,4,5,8,9$ . Conversely, for $m=1,3,6,7,10$ note that $A=\{1\},\{1,2\},\{1,2,3\},\{1,2,4\},\{1,2,3,4\}$ are $m$-full. The recursive construction below shows us that for each $m\geq10$ there is an $A$ with $\sum A=[m]$. 

Suppose there is an $A=\{a_1,\ldots,a_k\}$ with $a_1<\cdots<a_k$ such that $\sum A=[m]$, for some $m\geq10$. If
$a_k<a_1+\cdots+a_{k-1}+1$ then $\sum(A\setminus\{a_k\})\cup\{a_k+1\}=[m+1]$, and there remains nothing to prove. If
$a_i<a_1+\cdots+a_{i-1}+1$ for some $i$, $3\leq i<k$, then choose $j$ as the greatest $i$ with this property. In this case we have
\[a_j<a_1+\cdots+a_{j-1}+1,\quad a_{j+1}=a_1+\cdots+a_{j-1}+a_j+1.\]
Notice that $j\geq 3$ since $a_1=1, a_2=a_1+1=2$ and $a_3\in\{3,4\}$. 
Hence $a_j+1\neq a_{j+1}$ and if we omit $a_j$ from $A$ and add $a_j+1$ to it, then the resulting set is still full and its sum set is $[m+1]$. 
Otherwise, there is no $i$ with the property $a_i<a_1+\cdots+a_{i-1}+1$ and so $a_i=a_1+\cdots+a_{i-1}+1$ for each $i, 2\leq i\leq k$. We can therefore deduce that $A=\{1,2,4,\ldots,2^{k-1}\}$. Whence $m=2^k-1$ and we must give a class of sets $B$ with $\sum B=[2^k]$, $k\geq4$, to complete the proof.

The class of sets $B$ is defined by
a recursive construction. For $k=4$ choose the set $B=\{1,2,3,4,6\}$. If $\sum B=[2^k]$ for $B=\{b_1,\ldots,b_s\}$ then the set $B'=\{1,2b_1,\ldots,2b_{s-1},2b_s-1\}$ is clearly full and since $1+2b_1+\cdots+2b_{s-1}+2b_s-1=2(b_1+\cdots+b_s)=2\cdot2^k=2^{k+1}$, we have $\sum B'=[2^{k+1}]$ and the proof is complete.
\end{proof}
A natural question is to determine the number of $m$-full sets for a given positive integer $m$. We denote this number by $F(m)$. As an example, a straightforward argument shows that $F(12)=2$ and the two $12$-full sets are $\{1,2,3,6\}$ and $\{1,2,4,5\}$. We give a formula for $F(m)$, but prior to that we discuss some related problems.
\begin{proposition}\label{N} Let $m\notin\{2,4,5,8,9\}$ be a positive integer and
\bea \alpha(m)&=&\min\{|A|: \sum A=[m]\},\\
\beta(m)&=&\max\{|A|: \sum A=[m]\}.\eea
Then
\bea \alpha(m)&=&\min\{\ell: m\leq 2^\ell-1\}=\lceil\log_2(m+1)\rceil\\
\beta(m)&=&\max\{\ell: \frac{\ell(\ell+1)}2\leq m\}.\eea
\end{proposition}
\begin{proof} Let $A=\{a_1,\ldots,a_k\}$ be an arbitrary $m$-full set. Then, by Theorem \ref{criterion},
\[m=a_1+\cdots+a_k\leq1+2+4+\cdots+2^{k-1}=2^k-1.\]
We claim that if $n=\min\{\ell: m\leq 2^\ell-1\}$ then $k\geq n$. In contrary, suppose that $k<n$. Then $2^k-1<m$. Thus $m=a_1+\cdots+a_k\leq2^k-1<m$ which is a contradiction. Hence $k\geq n$ and, since $A$ was arbitrary, we therefore have $\alpha(m)\geq n$. 

On the other hand, we show that the minimum is attained. At first we show this when $m$ is a power of 2. For $m=16$ we can choose $A=\{1,2,3,4,6\}$. Suppose that for $m=2^{n-1}$ there is an $m$-full set $A=\{a_1,a_2,\ldots,a_n\}$. Then $A'=\{1,2a_1,2a_2,\ldots,2a_{n-1},2a_n-1\}$ is a $2^n$-full set with $n+1$ elements.

We now suppose that $m=2^{n-1}+r$, where $r$ belongs to $\{0,1,2,\ldots,2^{n-1}-1\}$ and we prove the existence of an $m$-full set with $n$ elements by induction on $r$. This is proved for $r=0$. Suppose that this is true for $r-1$; namely there is a $(2^{n-1}+r-1)$-full set $A=\{a_1,a_2,\ldots,a_n\}$. If $m=2^{n-1}+r$ is a power of 2 then there is nothing to prove. We may thus assume that there is an $i$ for which $a_i<a_1+\cdots+a_{i-1}+1$. Now if there is the least $j$ with $a_{i+j}\neq a_i+j$ then the set $\{a_1,\ldots,a_{i+j-2},a_{i+j-1}+1,a_{i+j},\ldots,a_n\}$ is an $m$-full set with $n$ elements. Otherwise, the set $\{a_1,\ldots,a_{n-1},a_{n}+1\}$ is an $m$-full set with $n$ elements.

To prove the other assertion, let $n'=\max\{\ell: \frac{\ell(\ell+1)}2\leq m\}$. We claim that if $A=\{a_1,\ldots,a_k\}$ is an arbitrary $m$-full set, then $k\leq n'$. On the contrary, suppose that $k>n'$. Then $\frac{k(k+1)}2>m$. Hence $m=a_1+a_2+\cdots+a_k\geq1+2+\cdots+k=\frac{k(k+1)}2>m$ which is a contradiction.
Thus $k\leq n'$ and, since $A$ was arbitrary, we therefore have $\beta(m)\leq n'$. 

On the other hand, we show that the maximum is attained. Let $m=\frac{n'(n'+1)}2+r$, where $r$ belongs to $\{0,1,\ldots, n'\}$. Then $A=\{1,2,3,\ldots,n'-1,n'+r\}$ is an $m$-full set with $n'$ elements. 
\end{proof}
\begin{remark}
The direct problem for subset sums
is to find lower bounds for $|\sum A|$ in terms of $|A|$ . The inverse
problem for subset sums is to determine the structure of the extremal sets $A$
of integers for which $|\sum A|$ is minimal. M. B. Nathanson gives a complete solution for the
direct and the inverse problem for subset sums in \cite{Nat} and proves  that if $A$ is a set of positive integers with $|A|\geq2$ then $|\sum A|\geq{|A|+1 \choose 2}$. This immediately implies the last part of Proposition \ref{N}.
\end{remark}
\begin{theorem}\label{L} Let $m\notin\{2,4,5,8,9,14\}$ be a positive integer and denote $\min\{\max A: \sum A=[m]\}$ by $L(m)$. If $m=\frac{n(n+1)}2+r$, where $r=0,1,\ldots,n$ then
\[L(m)=\begin{cases}
n, &\qquad \text{if}~r=0;\\
n+1, & \qquad \text{if}~1\leq r\leq n-2;\\
n+2, & \qquad \text{if}~1\leq r=n-1~\text{or}~n.
\end{cases}\]
\end{theorem}
\begin{proof}  Let $m=\frac{n(n+1)}2+r$, where $r=0,1,\ldots,n$. Then there are 3 cases: 

Case 1: $r=0$. Let $A=\{1,\ldots,n\}$. Since $\sum A=[\frac{n(n+1)}2]$ and $\max A$ has the minimum possible value, $L(m)=n$. 

Case 2: $r=1,\ldots,n-2$. We can consider the set
\[A=\{1,\ldots,n,n+1\}\setminus\{n+1-r\}\]
which is $m$-full, since $n+1-r\geq3$ and $1+\cdots+n+(n+1)-(n+1-r)=\frac{n(n+1)}2+r=m$. Hence $L(m)=n+1$.

Case 3: $r=n-1$ or $n$. In this case we cannot omit $n+1-r$ from the set $\{1,\ldots,n,n+1\}$, because $n+1-r$ is $1$ or $2$ and the resulting set is not full. Thus we have to add $n+2$ and omit some other element. In fact for $r=n-1$ the suitable set with minimum possible value for its maximum is $\{1,\ldots,n,n+2\}\setminus\{3\}$, and for $r=n$ the desired set is $\{1,\ldots,n-1,n+1,n+2\}\setminus\{3\}$ (note that in this case we have $n\neq4$ since $m\neq14$ and we can therefore deduce that the latter set is full). We can therefore deduce that $L(m)=n+2$.
\end{proof}
For the remaining case $m=14$, it can be easily verified that $L(14)=7$. The first few values of $L(m)$, Sloane's OEIS \cite[\seqnum{A188429}]{S}, are
\vspace{.5cm}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline
$m$ & 1&3&6&7&10&11&12&13&14&15&16&17&18&19&20\\ \hline
$L(m)$&1&2&3&4&4&5&5&6&7&5&6&6&6&7&7 \\ \hline
\end{tabular}
\end{center}
\vspace{.5cm}

\begin{theorem}\label{U} Let $m\geq20$ be a positive integer and denote $\max\{\max A: \sum A=[m]\}$ by $U(m)$. Then
\[U(m)=\big\lceil\frac{m}2\big\rceil.\]
\end{theorem}
\begin{proof} Let $A=\{a_1,\ldots,a_k\}$ with $a_1<\cdots<a_k$ be $m$-full. By Theorem \ref{criterion},  $a_k\leq a_1+\cdots+a_{k-1}$, which is equivalent to $a_k\leq m-a_k$ since $A$ is $m$-full. Therefore $a_k\leq\lceil\frac{m}2\rceil$ holds. 

Now put $b=\lceil\frac{m}2\rceil$. Since $m-\lceil\frac{m}2\rceil\geq10$, by Theorem~\ref{values} we can find a $B'$ with $\sum B'=[m-\lceil\frac{m}2\rceil]$. Whence $B=B'\cup\{b\}$ satisfies the properties $\sum B=[m]$ and $\max B=\lceil\frac{m}2\rceil$. \end{proof}
For the remaining cases, it can be easily verified that the values of $U(m)$, Sloane's OEIS \cite[\seqnum{A188430}]{S}, are
\vspace{.5cm}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline
$m$ & 1&3&6&7&10&11&12&13&14&15&16&17&18&19\\ \hline
$U(m)$&1&2&3&4&4&5&6&7&7&8&6&7&8&9 \\ \hline
\end{tabular}
\end{center}
\vspace{.5cm}
We can now determine the value of $F(m)$ for each positive integer $m$. We assume that $L(m)=U(m)=0$ whenever $m\in\{2,4,5,8,9\}$.
\begin{lemma} Let $m$ be a positive integer and $F(m,i)$ denote the number of $m$-full sets $A$ with $\max A=i$, where $L(m)\leq i\leq U(m)$. Then
\[F(m,i)=\sum_{j=L(m-i)}^{\min\{U(m-i),i-1\}}F(m-i,j).\]
\end{lemma}
\begin{proof} Let $A=\{a_1,\ldots,a_k\}$ with $a_1<\cdots<a_k=i$ be an $m$-full set. Then $A'=A\setminus\{a_k\}$ is an $(m-i)$-full set such that $j=\max A'<i$. Thus $L(m-i)\leq j\leq \min\{U(m-i),i-1\}$ and the result follows.
\end{proof}
\begin{theorem} Let $m$ be a positive integer and denote the number of $m$-full sets by $F(m)$. Then $F(m)=\sum_{i=L(m)}^{U(m)} F(m,i)$.
\end{theorem}
\begin{proof} Let $A=\{a_1,\ldots,a_k\}$ with $a_1<\cdots<a_k$ be an $m$-full set. Then $L(m)\leq a_k\leq U(m)$ and the result is now obvious.
\end{proof}
The first few values of $F(m)$, Sloane's OEIS \cite[\seqnum{A188431}]{S}, are
\vspace{.5cm}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline
$m$ & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\ \hline
$F(m)$&1&0&1&0&0&1&1&0&0&1&1&2&2&1&2&1&2&3&4&5 \\ \hline
\end{tabular}
\end{center}
\vspace{.5cm}
\begin{example} We evaluate $F(21)$. Using Theorem~\ref{L} and Theorem~\ref{U}, $L(21)=6$ and $U(21)=12$. Thus 
\[ F(21)=F(21,6)+F(21,7)+F(21,8)+F(21,9)+F(21,10)+F(21,11)+F(21,12).\]
We need to compute $L(m)$ and $U(m)$ for $m=15,14,13,12,11,10,9$. We have
\vspace{.5cm}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}\hline
$m$ &9&10&11&12&13&14&15\\ \hline
$L(m)$&0&4&5&5&6&7&5 \\ \hline
$U(m)$&0&4&5&6&7&7&8 \\ \hline
\end{tabular}
\end{center}
\vspace{.5cm}
Noting the facts that $L(6)=U(6)=3$ and $L(7)=U(7)=4$ we therefore have
\bea F(21)&=&F(15,5)+F(13,6)+F(13,7)+F(12,5)+F(12,6)+F(11,5)+F(10,4)\\
&=&F(10,4)+F(7,4)+F(6,3)+F(7,4)+F(6,3)+F(6,3)+F(6,3)\\
&=&F(6,3)+F(7,4)+F(6,3)+F(7,4)+F(6,3)+F(6,3)+F(6,3)\\
&=&7.\eea
The seven $21$-full sets are
\bea &&\{1,2,3,4,5,6\},
\{1,2,4,6,8\},
\{1,2,3,7,8\},
\{1,2,4,5,9\},\\
&&\{1,2,3,6,9\},
\{1,2,3,5,10\},
\{1,2,3,4,11\}.\eea
\end{example}

\section{Acknowledgement} The second author of this research was supported by a grant from Ferdowsi University of Mashhad; No. MP89177MIZ.
The authors wish to acknowledge the referees for their valuable comments and suggestions. 

\bibliographystyle{amsplain}
\begin{thebibliography}{9}

\bibitem{Nat} M. B. Nathanson, Inverse theorems for subset sums,
\textit{Trans. Amer. Math. Soc.} \textbf{347} (1995), 1409--1418.

\bibitem{S} N. J. A. Sloane,
\href{http://www.research.att.com/~njas/sequences/}{The On-Line
Encylopedia of Integer Sequences}, 2011.

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary
05A17; Secondary 11P81.

\noindent \emph{Keywords: } perfect number, $m$-full set, partition of
a positive integer.

\bigskip
\hrule
\bigskip
\noindent

\noindent (Concerned with sequences \seqnum{A188429}, \seqnum{A188430},
and \seqnum{A188431}.)

\bigskip
\hrule
\bigskip

\noindent Received June 3 2010;
revised versions received December 3 2010; April 20 2011.
Published in {\it Journal of Integer Sequences}, April 22 2011.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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