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\begin{document}

\newcommand{\LOOP}{{\rm LOOP}}

\title{Small NFA's for Cofinite Unary Languages}

\addauthor{William Gasarch}{University of Maryland}
\addauthor{Erik Metz}{University of Maryland}
\addauthor{Zan Xu}{University of Maryland}
\addauthor{Yuang Shen}{University of Maryland}
\addauthor{Sam Zbarsky}{Princeton University}

\jyear{2020}
\authorrunning{Gasarch, Metz, Xu, Shen, \& Zbarsky}
\titlerunning{Small NFA's for Cofinite Unary Languages}


\maketitle

\newcommand{\Ot}{\widetilde{O}}

%\keywords{whatever}

\begin{abstract}

For all $n$ there is a DFA for $\{ a^i \st i\ne n\}$ of size $n+2$; however there is no smaller DFA.
What about NFA's? We show that there is an NFA  for $\{ a^i \st i\ne n\}$ of size $\sqrt{n} + \Ot(1)$. We also
find small NFA's for many other cofinite unary sets. How small can we go? We show that any NFA for $\{ a^i \st i\ne n\}$
must have at least $\sqrt{n}$ states.
\end{abstract}


\section{Introduction}

Consider the language

$$\MN(n)=\{ a^i \st i\ne n \}.$$

($\MN$ stands for {\it Missing Number}.)

It is easy to show that (1) there is a DFA for $\MN(n)$ with $n+2$ states, and
(2) any DFA for $\MN(n)$ has at least $n+2$ states.
What about an NFA for $\MN(n)$? We show that there is an NFA for $\MN(n)$ that has substantially fewer than $n$ states.
We also obtain small NFA's for many other cofinite unary languages.

\begin{notation}
$\N$ is $\{0,1,2,\ldots\}$ (that is, we include 0).
\end{notation}

\begin{definition}
If $A\subseteq \N$ then 
$$\MN(A) = \{ a^i \st i \notin A\}.$$
We will only use this definition when $A$ is finite.
We will write $\MN(a,b,c)$ instead of the formally correct $\MN(\{a,b,c\})$.
\end{definition}

\begin{notation}
If $f$ and $g$ are functions then, informally,  $f\le \Ot(g)$ means that $f$ is less than $g$ if we ignore polylog factors.
Formally it means that
$$(\exists n_0)(\exists c)(\forall n\ge n_0)[f(n) \le c(\log n)^c g(n)].$$
\end{notation}

\begin{enumerate}
\item
In Section~\ref{se:n} we show that (1) there is an NFA for $\MN(100)$ on 29 states, and
(2) for all $n$ there is an NFA for $\MN(n)$ with $\le n^{1/2} + \Ot(1)$ states.
\item
In Section~\ref{se:A} we show that (1) there is an NFA for $\MN(998,999,1000)$ on 104 states, 
(2) for any $A\subseteq \{998,999,1000\}$ there is an NFA for $\MN(A)$ on 104 states,
(3) for all $n$, for all $0<\delta<1$ there is an NFA for $\MN(n-n^\delta,\ldots,n)$ on $5n^{\max\{1/2, \delta\}} + \Ot(1)$ states, and
(4) for any $A\subseteq \{n-n^\delta,\ldots,n\}$ there is an NFA for $\MN(A)$ on $5n^{\max\{1/2, \delta\}} + \Ot(1)$ states.
\item
In Section~\ref{se:alphan} we show that, for all $n$, for all $0<\alpha<1$ such that $\alpha n\in\N$, there is an NFA for $\MN(\alpha n,n)$ on $2n^{1/2}\ln(n) + \Ot(1)$  states.
\item
In Section~\ref{se:gap} we prove a general theorem about unary sets with big gaps. 
We obtain the following corollary:
for all $0<\delta<1$ there is an NFA for $\MN(n^\delta,n)$ on $n^{1/2} + n^\delta + \Ot(1)$ states.
\item
In Section~\ref{se:lower} we show that any NFA for $\MN(n)$ requires at least $n^{1/2}$ states.
\item
In Section~\ref{se:emp} we discuss our empirical results. 
\item
In Section~\ref{se:open} we state open problems.
\end{enumerate}

\begin{definition}
A set $X$ has a {\it small NFA} if there is an NFA that accepts it that is much smaller than
any DFA for it. We do not define the term {\it much smaller} rigorously.
However, all of our results are about small NFA's.
\end{definition}

All of our general results are asymptotic; however, we will present empirical evidence that indicates
the results hold for small $n$ as well.


\section{Needed Lemma}
The following problem is attributed to Frobenius:

\bigskip

\noindent
{\it Given a set of relatively prime positive integers $\{a_1,\ldots,a_m\}$ find
the set \\$\{ \sum_{i=1}^n a_ix_i \st x_1,\ldots,x_m\in\N \}$.}

\bigskip

It is known that this set is always cofinite. The $m=2$ case was solved by
James Joseph Sylvester in 1884:

\begin{lemma}\label{le:syl}
Let $c,d\in\N$ be relatively prime. 
\begin{enumerate}
\item
For all $i\ge cd-c-d+1$ there exists $x,y\in\N$ such that $i=cx+dy$.
\item
There is no $x,y\in\N$ such that $cd-c-d=cx+dy$.
\item
There is no $x,y,C,D\in\N$ such that $cd-c-d-Cc-Dd=cx+dy$.
(If there was then $cd-c-d= (C+x)c + (D+y)d$.)
We use this part in Section~\ref{se:A}.
\end{enumerate}
\end{lemma}

\section{Small NFA's for \texorpdfstring{$\MN(100)$}{MN(100)} and \texorpdfstring{$\MN(n)$}{MN(n)}}\label{se:n}

\subsection{Small NFA for \texorpdfstring{$\MN(100)$}{MN(100)}}

\begin{theorem}\label{th:100}~
\begin{enumerate}
\item
For all $i\ge 96$ there exists $x,y\in \N$ such that $i=13x+9y$.
\item
There does not exist $x,y\in \N$ such that $95=13x+9y$.
\item
For all $i\ge 101$ there exists $x,y\in \N$ such that $i=13x+9y+5$.
\item
There does not exist $x,y\in \N$ such that $100=13x+9y+5$.
\item
There exists an NFA $M$ such that the following are true:
\begin{enumerate}
\item
For all $i\ge 101$, $M$ accepts $a^i$.
\item
$M$ rejects $a^{100}$.
\item
We have no comment on the behavior of $M$ on other $a^i$.
\item
$M$ has 13 states.
\end{enumerate}
\item
There exists an NFA on 29 states that accepts $\MN(100)$.
\end{enumerate}
\end{theorem}


\begin{proof}

\noindent
1,2) These follow from  Lemma~\ref{le:syl}, though they can be proven directly by an easy induction.

\noindent
3,4) These follow from Parts 1 and 2

\noindent
5) The NFA is constructed as follows: (also see Figure~\ref{nfa:good}, the caption will be explained later).

\begin{itemize}
\item
$M$  has states $0,\ldots,12$,  0 is the start state, and 5 is the only final state.
For $0\le j \le 12$, $\delta(j,a)=j+1 \pmod {13}$. 
($\delta$ is not fully defined yet.)
\item
If we go no further then $M$ accepts $\{ a^{13x+5} \st x\in\N\}.$

\item
We put in an $e$-transition from state 5 to state 9. Now $M$ accepts
$$\{ a^{13x+9y+5} \st x,y\in\N\}.$$ 

(The $9y$ is not because the $e$-transition went to state 9. It is because the distance from state 9 back to state 5 is 9.)
\end{itemize}


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\label{nfa:good}
\end{figure}


By Parts 3,4 $M$ satisfies 5a and 5b. $M$ clearly has 13 states, so it satisfies 5d.

\bigskip

\noindent
6) Let $Q=\{3,5,7\}$. Note that $3\times 5\times 7 = 105>100$. 
For each $p\in Q$ let  $M_p$ be the DFA that accepts $\{ a^i \st i\not\equiv 100 \pmod p\}$.

The NFA is constructed as follows: (see also Figure~\ref{nfa:100})

\begin{enumerate}
\item
The NFA $M$ is part of our new NFA. We create a new start state, and then put an e-transition from this new state to $M$'s original start state. 
Note that $M$ 
(a) accepts all $a^i$ with $i\ge 101$ (it also accepts other strings), 
(b) rejects $a^{100}$, and
(c) has 13 states.
\item
For each $p\in Q$ put an $e$-transition from our new start state to the
start state of $M_p$.
Note that $M_p$ 
(a) accepts all $a^i$ with $i\not\equiv 100 \pmod p$, 
(b) rejects $a^{100}$, and
(c) has $p$ states.

\end{enumerate}

Clearly the NFA has $13+3+5+7+1=29$ states and rejects $a^{100}$. We show that it accepts everything else.

Let $a^i$ be rejected by this NFA. 
\begin{itemize}
\item
Since the $M$ part rejects $a^i$, $i\le 100$ (note, hence $i\le 3\times 5\times 7=105$).
\item
Since the $M_3$ part rejects $a^i$, $i\equiv 100\pmod 3$
\item
Since the $M_5$ part rejects $a^i$, $i\equiv 100\pmod 5$
\item
Since the $M_7$ part rejects $a^i$, $i\equiv 100\pmod 7$
\end{itemize}

By the Chinese Remainder Theorem there is a unique number $0\le z\le 3\times 5\times 7=105$
such that, for every $p\in \{3,5,7\}$,  $z\equiv 100 \pmod p$.
Since both $i$ and $100$ satisfies these criteria, $i=n$.
\end{proof}


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\draw [->] (0.8887755161117985, 2.026550540999647) .. controls (0.8689217397391911,1.8948293413836375) .. (0.8296576875507625, 1.7675384105587018);
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}
\caption{NFA for $\MN(100)$}
\label{nfa:100}
\end{figure}




\subsection{Small NFA for \texorpdfstring{$\MN(n)$}{MN(n)}}

We generalize the construction of a small NFA for $\MN(100)$ to get
a small NFA for $\MN(n)$.


\begin{definition}\label{de:loop}
Let $c,d,e\in\N$ be such that $c<d$ and  $c,d$ are relatively prime.
$\LOOP(c,d,e)$ is the NFA defined as follows. There are two cases.

\noindent
{\bf Case 1:} $e\le d-1$.
\begin{enumerate}
\item
The NFA has states  $0,\ldots,d-1$, with 0 as the start state and $e$ as the only final state.
For $0\le j \le d-1$, $\delta(j,a)=j+1 \pmod d$.


\item
So far this NFA accepts $\{ a^{dx+e} \st x\in\N\}.$
\item
We put in an $e$-transition from state $e$ to state $e-c \pmod d$.
Note that the distance from state $e-c \pmod d$ to state $e$ is $c$.
Now the NFA accepts
$$\{ a^{cx+dy+e} \st x,y\in\N\}.$$

\item
This NFA has $d$ states.
\end{enumerate}
 Note that Figure~\ref{nfa:good} is $\LOOP(9,13,5)$ which is an example of a Case 1 LOOP.

\bigskip

\noindent

{\bf Case 2:} $e \ge d$
\begin{enumerate}
    \item 
    The NFA has states $s_0,s_1,\ldots,s_{e-d+2}$ such that $s_0$ is the start state. For $0\le j \le e-d+1$, $\delta(s_j,a)=s_{j+1}$.

    \item
    The NFA has states  $0,\ldots,d-1$, with $d-1$ as the only final state.
For $0\le j \le d-1$, $\delta(j,a)=j+1 \pmod d$.
The state 0 is identical to the state $s_{e-d+2}$.

    \item
    In total there are $(e-d+1) + (d-1) = e$ transitions to get to the final state the first time, after which each loop of length d brings you back to the same state, so the NFA accepts $\{ a^{dx+e} \st x\in\N\}.$
    
    \item
    We put in an $e$-transition from state $d-1$ to state $d-c-1$.
Note that the distance from state $d-c-1$ to state $d-1$ is $c$.
Now the NFA accepts
$$\{ a^{cx+dy+e} \st x,y\in\N\}.$$

\item
This NFA has $e+1$ states.


\end{enumerate}
Figure~\ref{nfa:bad} is $\LOOP(9,13,17)$ which is an example of a Case 2 LOOP\@.
\end{definition}

\bigskip

\begin{figure}[htp]
\centering
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\caption{NFA LOOP(9,13,17) Case 2}
\label{nfa:bad}
\end{figure}


The following is clear:

\begin{lemma}\label{le:loop}
Let $c,d,e\in\N$ be such that $e,c<d$ and $c,d$ are relatively prime.
\begin{enumerate}
\item
$\LOOP(c,d,e$) accepts $\{ a^i \st i\ge cd-c-d+e+1\}$
\item
$\LOOP(c,d,e)$ rejects $\{ a^{cd-c-d+e}\}.$
\item
$\LOOP(c,d,e)$ rejects $\{ a^{cd-Cc-Dd+e} \st C,D\in\N\}$
(since if $a^{cd-Cc-Dd+e}$ can reach the accept state then by adding $C$ $c$'s and $D$ $d$'s
the NFA gets back to the accept state).
We will use this part in Section~\ref{se:A}.
\item
If we used Case 1 then $\LOOP(c,d,e)$ has $d$ states.
\item
%REMINDER: Not needed
If we used Case 2 then $\LOOP(c,d,e)$ has $e+1$ states.

\end{enumerate}
\end{lemma}

\begin{note}
Below, we use o(1) to denote a number that may be positive or negative, but that goes to 0 as our variable of interest 
($N$ in Lemma~\ref{le:primes}), goes to infinity. This may be non-standard.
\end{note}


\begin{lemma}\label{le:primes}
Let $N\in\N$. Let $Q_N$ be the set of the first $N$ primes.
\begin{enumerate}
\item
$\prod_{p\in Q_N} p \sim e^{((1+o(1))N\log N}$. (This is well known.)
\item
$\sum_{p\in Q_N} p \sim O(N^2\log N)=\Ot(N^2)$. (For references and more precise estimates see Axler~\cite{axler}.)
\item
Let $n\in\N$. The product of the first $\Omega(\log n)$ primes is $\ge n$.
The sum of the first $O(\log n)$ primes is $\le O((\log n)^2\log\log n)\le \Ot(1)$.
(This follows from parts 1 and 2.)
\item
$\prod_{p\le_N, \hbox{ $p$ prime} } p \sim e^{(1+o(1))N}$. (This is well known.)
\end{enumerate}
\end{lemma}




\begin{theorem}\label{th:n}
Let $n\in\N$. 
\begin{enumerate}
\item
There exists an NFA $M$ such that the following are true:
\begin{enumerate}
\item
For all $i\ge n+2\ceil{n^{1/2}}$, $M$ accepts $a^i$.
\item
$M$ rejects $a^n$.
\item
We have no comment on the behavior of $M$ on other $a^i$.
\item
$M$ has $\le n^{1/2}+O(1)$ states.
\end{enumerate}
\item
There exists an NFA on $\le n^{1/2}+\Ot(1)$ states that accepts $\MN(n)$.
\end{enumerate}
\end{theorem}

\begin{proof}

\noindent
1) Let $c=\ceil{n^{1/2}}+1$ and $e=n+1 \pmod c$. Note that $e \leq c$. 
Let $M$ be $\LOOP(c,c+1,e)$.
Note that
$$c(c+1) - c - c- 1 + e = c^2 -c -1 + e\le c^2-c-1+c=c^2-1$$
By Lemma~\ref{le:loop} 
$M$ accepts $a^i$ where $i\ge c^2-1+1=c^2\ge n+2\ceil{n^{1/2}}$.

We show that $M$ rejects $a^n$.
Assume, by way of contradiction, that $M$ accepts $a^n$. 
Then there exists $x,y\ge 0$ such that

$$cx+(c+1)y + e = n$$

Take this equation mod $c$. Then

$$0x + 1\times y+ (n+1)\equiv n \pmod c$$

$$y+1\equiv 0 \pmod c$$

$$y\equiv -1 \pmod c.$$

Since $y\ge 0$, $y\ge c-1$. Hence

$$
n=cx + (c+1)y +e \ge (c+1)(c-1) =c^2-1  =  (\ceil{n^{1/2}}+1)^2-1 = n + 2n^{1/2}-1.
$$

This is a contradiction.

Since $e \leq c$, $M$ has $c+1=n^{1/2}+O(1)$ states.

\bigskip

\noindent
2) By Lemma~\ref{le:primes}.3 there is a set of primes $Q$ such that
\begin{itemize}
\item
$\prod_{p\in Q} p \ge n+2\ceil{n^{1/2}}$.
\item
$\sum_{p\in Q} p \le \Ot(1)$.
\end{itemize}
For each $p\in Q$ let $M_p$ be the DFA that accepts $\{a^i \st i\not\equiv n\pmod p\}$. 

The NFA is constructed as follows:

\begin{enumerate}
\item
The NFA $M$ is part of our NFA.  We create a new start state, and then put an e-transition from this new state to $M$'s original start state.
Note that 
(1) $M$ accepts $a^i$ for $i\ge n+2\ceil{n^{1/2}}$ (it also accepts other strings), 
(2) $M$ rejects $a^n$, and
(3) $M$ has $\le n^{1/2}+O(1)$ states.
\item
For each $p\in Q$ there is an $e$-transition from our new start state to the
start state of $M_p$.
Note that (1) $M_p$ accepts $a^i$ if $i\not\equiv n\pmod p$, (2) $M_p$ rejects $a^n$, 
and (3) $M_p$ has $p$ states.
\end{enumerate}

Clearly the NFA has $\le n^{1/2} + \sum_{p\in Q} p \le n^{1/2} + \Ot(1)$ states and rejects $a^n$.
We show that it accepts everything else.

Let $a^i$ be rejected by this NFA.
\begin{itemize}
\item
Since the $M$ part rejects $a^i$, $i\le n+2\ceil{n^{1/2}}$ (note, hence $i\le \prod_{p\in Q} p$).
\item
For each $p\in Q$, since the $M_p$ part rejects $a^i$, $i\equiv n\pmod p$.
\end{itemize}

By the Chinese Remainder Theorem there is a unique number  
$0\le z < \prod_{p\in Q} p\ge n+2\ceil{n^{1/2}}$ such that, 
for every $p\in Q$,  $z\equiv n \pmod p$.
Since both $i$ and $n$ satisfy the criteria, $i=n$.
\end{proof}

\section{Small NFA's for \texorpdfstring{$\MN(998,999,1000)$}{MN(998,999,1000)} and \texorpdfstring{$\MN(A)$}{MN(A)}}\label{se:A}

\subsection{Small NFA for \texorpdfstring{$\MN(998,999,1000)$ and $\MN(998,1000)$}{MN(998,999,1000) and MN(998,1000)}}


\begin{theorem}\label{th:98,99,100}~
\begin{enumerate}
\item
There exists an NFA $M$ such that the following are true:
\begin{enumerate}
\item
For all $i\ge 1067$, $M$ accepts $a^i$.
\item
For all $i\in \{998,999,1000\}$ $M$ rejects $a^i$.
\item
We have no comment on the behavior of $M$ for other $a^i$'s.
\item
$M$ has 34 states.
\end{enumerate}
\item
There exists an NFA with 104 states that accepts $\MN(998,999,1000)$.
\item
There exists an NFA with 104 states that accepts $\MN(A)$ where\\ $A\subseteq\{998,999,1000\}$.
(For $A=\es$ this is trivial.)
\end{enumerate}
\end{theorem}

\begin{proof}

\noindent
1) Let $M$ be $\LOOP(33,34,11)$.
From Lemma~\ref{le:loop} we know the following:
\begin{enumerate}
\item[a)]
For all $i\ge 1067$, $M$ accepts $a^i$.
\item[b)]
For all $C,D\in \N$, $M$ rejects $a^{1066-33C-34D}$
which we write as\\ $a^{1066 -33(C+D) - D}.$
\end{enumerate}

We set $(C,D)$ carefully to obtain, using item b,  strings that $M$ rejects.

\begin{itemize}
\item
If $(C,D)=(2,0)$ then we get $1066-33\times 2 - 0 = 1000$
\item
If $(C,D)=(1,1)$ then we get $1066-33\times 2 - 1 = 999$
\item
If $(C,D)=(0,2)$ then we get $1066-33\times 2 - 2  = 998$
\end{itemize}

Clearly $M$ has 34 states. 

\bigskip


\noindent
2) We will once again use primes and mods. We can't use mod 2 or mod 3 since then one of
$a^{998}$, $a^{999}$, $a^{1000}$ will be accepted.

We can use any mod from 5 up.
We need another trick, as you will see.

Let $Q=\{5,7,11\}$. Note that $3\times 5\times 7\times 11 = 1155>1066$
(That is not a typo. We really do mean to multiply by 3. 
We chose 3 because $|\{998,999,1000\}|=3$. We chose $\{5,7,11\}$ since
 none of them divide 3 and the product $3\times 5\times 7\times 11>1066$.
The fact that 3 is a prime is not important.)

For each $p \in Q$ let $M_{3p}$ be the DFA that accepts

$$\{ a^i \st 
i\not\equiv 998,999,1000 \pmod {3p}
\}
$$

Note that $\mathrm{LCM}(3\times 5, 3\times 7,3\times 11)=3\times 5\times 7\times 11=105>100$.

The NFA is constructed as follows:

\begin{enumerate}
\item
The NFA $M$ is part of our new NFA.
We create a new start state, and then put an e-transition from this new state to $M$'s original start state. 
Note that $M$ 
(1) accepts all $a^i$ with $i\ge 1067$ (it also accepts other strings),
(2) rejects any of $a^i$ with $i\in \{998,999,1000\}$, and 
(3) has 34 states.
\item
For each $p\in Q$, there is an $e$-transition from our new start state to the start state of $M_{3p}$.
Note that $M_{3p}$ (1) accepts $a^i$ when $i\not\equiv 998,999,1000 \pmod {3p}$,
(2) rejects any $a^i$ with $i\in \{998,999,1000\}$, and
(3) has $3p$ states.
\end{enumerate}

This NFA has $34+3(5+7+11) + 1= 104$ states and rejects any $a^i$ with $i\notin \{998,999,1000\}$.
We show that it accepts everything else.

Let $a^i$ be rejected by this NFA.
\begin{itemize}
\item
Since the $M$ part rejects $a^i$, $i\le 1066$ (note, hence $i\le 3\times 5\times 7\times 11=1155$).
\item
For all $p\in Q$, since the $M_{3p}$ part rejects $a^i$, there exists $x\in\{998,999,1000\}$ such that $i\equiv x\pmod {3p}$.
\end{itemize}

We cannot use the Chinese Remainder Theorem (yet) since it is possible that, say $i\equiv 998 \pmod {3\times 7}$ but $i\equiv 1000 \pmod {3\times 11}$.
We need that $i$ is equivalent to the same $x$ with all of those mods.

Let $i\equiv x \pmod 3$ where $x\in \{998,999,1000\}$. Note that $x$ is unique. 
Let $p\in Q$.
Let $y\in \{998,999,1000\}$ be such that $i\equiv y\pmod {3p}$. 
Note that $y$ is unique since $3<3p$. 

We show that $x=y$. 

Since $i\equiv x \pmod 3$ there exists $a\in \Z$ such that 
$$\hbox{Eq 1 } i=x+3a.$$

Since $i\equiv y \pmod {3p}$ there a $b\in \Z$ such that

$$\hbox{Eq 2 } i=y+3pb.$$

By subtracting Eq 2 from Eq 1 we get

$$x-y=3pb-3a\equiv 0 \pmod 3$$

Since $x,y\in \{998,999,1000\}$ and $x\equiv y \pmod 3$, $x=y$.
To recap we now have that there exists $x\in \{998,999,1000\}$ such
that, for all $p\in Q$, $i\equiv x \pmod {3p}$. 

By the Chinese Remainder Theorem there is a unique number 
$0\le z\le \mathrm{LCM}(3\times 5,3\times 7,3\times 11)=1155$ 
such that, for all $p\in Q$,  $z\equiv x \pmod {3p}$. 
Since both $i$ and $x$ satisfy those criteria, $i=x$.

\bigskip

\noindent
3) We look at $\MN(998,1000)$ as an example. The construction is similar to the one
for $\MN(998,1000)$ except that, at the end, use the DFA for
$\{a^i \st i\not\equiv 998,1000\pmod {3p}\}$ instead of 
$\{a^i \st i\not\equiv 998,999,1000\pmod {3p}\}$. The other cases are similar.
\end{proof}


\begin{theorem}\label{th:A}
Let $0 < \delta <1$.
Let $n\in\N$. 
(We will assume $n^\delta\in\N$ and leave it to the reader
to adjust the statement and the proof for when $n^\delta\notin\N$.)
Assume $n=c^2+f$ where $0\le f\le 2c$.
\begin{enumerate}
\item
There exists an NFA $M$ such that the following are true:
\begin{enumerate}
\item
For all $i\ge n+ n^{1/2+\delta} + n^{2\delta} +1$,
$M$ accepts $a^i$.
\item
For all $i\in \{ n-n^\delta, n-n^\delta+1,\ldots,n\}$, $M$ rejects $a^i$.
\item
We have no comment on the behavior of $M$ for other $a^i$'s. 
\item
$M$ has $\le 5n^{\max\{1/2,\delta\}}+O(1)$ states.
\end{enumerate}
\item
There exists an NFA on 
$$\le 5n^{\max\{1/2,\delta\}} + \Ot(1)\hbox{ states }$$
that accepts 
$\MN(n-n^\delta,n-\delta+1,\ldots,n)$.
\item
Let $A\subseteq \{ n-n^\delta,\ldots,n\}$. 
There exists an NFA on 
$$5n^{\max\{1/2,\delta\}}  + \Ot(1)\hbox{ states }$$
that accepts $\MN(A)$.
(For $A=\es$ this is trivial.)
\end{enumerate}
\end{theorem}

\begin{proof}

1) 
Let $M$ be the NFA 
$\LOOP(c+k,c+k+1,f+1+x(k))$ where we determine $k$ and $x(k)$ later.

\noindent
{\bf Claim:}
\begin{enumerate}
    \item 
If $M$ rejects $a^{n + n^\delta(c+k)}$ then, for $i=n-n^\delta,\ldots,n$, $M$ rejects $a^i$.  
\item
If $x(k) = n^\delta(c+k) - k^2 - 2c k + c + k$ then $M$ rejects $a^{n + n^\delta(c+k)}$.
\item
If $x(k) = n^\delta(c+k) - k^2 - 2c k + c + k$ then, for $i=n-n^\delta,\ldots,n$, $M$ rejects $a^i$
(this follows from parts 1 and 2).
\end{enumerate}

\noindent
{\bf Proof of Claim:}

\noindent
1) 
Assume $M$ rejects $n + n^\delta(c+k)$. Then it also rejects 
everything of the form 

$$n + n^\delta(c+k) - (c+k)C - (c+k+1)D = n + n^\delta(c+k) -(C+D)(c+k) - D$$

(since otherwise $M$ would accept $n+n^\delta(c+k) - (c+k)C - (c+k+1)D + (c+k)C + (c+k+1)D= n+n^\delta(c+k)$).

We  set $(C,D)$ as follows:

\begin{itemize}
\item
If $(C,D)=(n^\delta,0)$ then we get $n+n^\delta (c+k) - n^\delta (c+k) -0 = n$.
\item
If $(C,D)=(n^\delta-1,1)$ then we get $n+n^\delta (c+k) - n^\delta (c+k) -1 = n-1$.
\item
$\qquad \vdots$
\item
If $(C,D)=(0,n^\delta)$ then we get $n+n^\delta (c+k) - n^\delta (c+k) -n^\delta = n-n^\delta$.
\end{itemize}


\noindent
2) 
For $k\in\N$ we need $x(k)\in\N$ such that 
$\LOOP(c+k,c+k+1,f+1+x(k))$ rejects $n + n^\delta(c+k)$. Note that this NFA rejects
\[(c+k)(c+k+1) - (c+k) - (c+k+1) + f + 1 + x(k)\] 
\[= c^2 + k^2 + 2ck + c + k - 2c - 2k - 1 + f + 1 + x(k) \]
\[= n + k^2 + 2ck - c - k + x(k)\]
Hence we find $x(k)$ via:
\[n + k^2 + 2ck - c - k + x(k) = n + n^\delta(c+k)\] or equivalently
\[x(k) = n^\delta(c+k) - k^2 - 2c k + c + k\]

\noindent
{\bf End of Proof of Claim}

We choose $k$ such that the max of $\{c+k+1,f+1+x(k)\}$ is small. We look at what
happens to $x(k)$ for $k\in \{0,\ldots,n^\delta\}$. We consider only when $n \ge 9$, with smaller $n$ being expressed within the $O(1)$ term. Note that 

\begin{itemize}
    \item
 $x(0)=n^\delta c + c > 0$.
 \item
$x(n^\delta)=n^{2\delta }+ n^\delta c -n^{2\delta} - 2c n^\delta + c + n^\delta =c+n^\delta - cn^\delta<0$ (since $n\ge 9$).
\item
there exists $k_o$ such that $x(k_o)\ge 0$ and $x(k_o+1)\le 0$ (this follows from the first two points).
\end{itemize}

Note that 

$$x(k_o) \le x(k_o) - x(k_o+1) \le |-2c +n^\delta - 2k_o| \le 2c+n^\delta.$$

Let $M=\LOOP(c+k_o,c+k_o,f+1+x(k_o))$. Since $c\le n^{1/2}$, $f\le 2n^{1/2}$, $k_o \le n^\delta$, and 
$x(k_o)\le 2c + n^\delta \le 3n^{\max\{1/2,\delta\}}$. $e = x(k_o) + f + 1$ has $\le 5n^{\max\{1/2,\delta\}} + O(1)$ states, while $c+k_o+1$ has $\le 3n^{\max\{1/2,\delta\}} + O(1)$ states, so $M$ must have $\le 5n^{\max\{1/2,\delta\}} + O(1)$ states overall.

$M$ satisfies conditions of what to reject and how many states it has. We now consider
what it accepts. Note that $x(k)$ was chosen so that the largest number $M$ (with $k=k_o$) rejects is 
$a^{n+n^\delta(c+k_o)}$. We need to estimate this.

$$n+n^\delta(c+k_o)\le n+n^\delta(n^{1/2}+n^\delta) \le n+ n^{1/2+\delta} + n^{2\delta}.$$

By Lemma~\ref{le:loop} $M$ accepts what it should.


\noindent
2) To simplify the algebra we just use that the NFA in Part 1 accepts
$\{ a^i \st i\ge n^2\}$.

By Lemma~\ref{le:primes} there is a set of primes $Q'$ such that
(1) $\prod_{p\in Q'} p \ge n^2$,
(2) $\sum_{p\in Q'} p \le \Ot(1)$.
We form $Q$ as follows:
(1) remove from $Q'$ all of the primes that divide $n^\delta$,
(2) add in the smallest primes possible that do not divide $n^\delta$ so that $n^\delta\prod_{p\in Q}p\ge n^2$.

One can show that $\sum_{p\in Q} p\le  O(\sum_{p\in Q'} p) \le \Ot(1)$.
Hence we have a set $Q$ such that
(1) $n^\delta \prod_{p\in Q} p \ge n^2$
(2) $\sum_{p\in Q} p \le \Ot(1)$, and
(3) for all $p\in Q$, $p$ does not divide $n^\delta$.
For each $p\in Q$ let $M_{n^\delta p}$ be the DFA that accepts $\{a^i \st i\not\equiv n\pmod {n^\delta p}\}$. 

Note that the $\mathrm{LCM}\{ n^\delta p \st p\in Q\} = n^\delta\prod_{p\in Q} p \ge n$.


The NFA is constructed as follows:

\begin{enumerate}
\item
The NFA $M$ is part of our new NFA .  We create a new start state, and then put an e-transition from this new state to $M$'s original start state. 
Note that $M$ 
(1) accepts all $a^i$ with $i\ge n^2$ (it also accepts other strings),
(2) rejects any of $a^i$ with $i\in \{n-n^\delta,\ldots,n\}$, and 
(3) has $\le 5n^{\max\{1/2,\delta\}}+O(1)$ states.
\item
For each $p\in Q$ put an $e$-transition from our new start state to the
start state of $M_{n^\delta p}$.
Note that $M_{n^\delta p}$ (1) accepts $a^i$ with $i\not\equiv n-n^\delta,\ldots,n \pmod {n^\delta p}$,
(2) rejects any of $a^i$ with $i\in \{n-n^\delta,\ldots,n\}$, and 
(3) has $n^\delta p$ states.
\end{enumerate}

This NFA has $5n^{\max\{1/2,\delta\}} + \Ot(1)$ states and rejects any $a^i$ with
$i\in \{n-n^\delta,\ldots,n\}$.
We show that it accepts everything else.

Let $a^i$ be rejected by this NFA.
\begin{itemize}
\item
Since the $M$ part rejects $a^i$, $i\le n^2$ (note, hence $i\le n^\delta\prod_{p\in Q} p$).
\item
For each $p\in Q$, since the $M_{n^\delta p}$ 
part rejects $a^i$, there exists $x\in \{n-n^\delta,\ldots,n\}$ such that $i\equiv x\pmod {n^\delta p}$.
\end{itemize}

We cannot use the Chinese Remainder Theorem (yet) since it is possible that, 
say $i\equiv 95 \pmod {n^\delta\times 7}$ but $i\equiv 92 \pmod {n^\delta\times 11}$.
We need that $a^i$ is equivalent to the same $n^\delta p$ with all those mods.

Let $i\equiv x \pmod {n^\delta}$ where $x\in \{n-n^\delta,\ldots,n\}$.
Note that $x$ is unique. 
Let $n^\delta p\in n^\delta Q$.
Let $y\in \{n-n^\delta,\ldots,n\}$ be such that $i\equiv y \pmod n^\delta p$.
Note that $y$ is unique since $n^\delta < n^\delta p$.
We show that $x=y$. 

Since $i\equiv x  \pmod {n^\delta}$ there exists $a\in\Z$ such that:

$$\hbox{Eq 1 } i=x+n^\delta a.$$

Since  $i\equiv y\pmod {n^\delta}$ there exists a $b\in \Z$ such that

$$\hbox{Eq 2 } i=y+n^\delta p b.$$

By subtracting Eq 2 from Eq 1 we get

$$x-y=n^\delta pb-n^\delta a\equiv 0 \pmod {n^\delta}$$

Since $x,y\in \{n-\delta,\ldots,n\}$ and $x\equiv y \pmod {n^\delta}$, $x=y$.
To recap we now have that there exists $x\in \{n-\delta,\ldots,n\}$ such
that, for all $n^\delta p\in Q$, $i\equiv x \pmod {n^\delta p}$. 

By the Chinese Remainder Theorem there is a unique number
$0\le z\le \mathrm{LCM}\{ n^\delta p \st p\in Q\}\ge n$ such that, for all $p\in Q$, 
$z\equiv x \pmod {n^\delta p}$.
Since both $i$ and $x$ satisfy those criteria, $i=x$.

\bigskip

\noindent
3) This is an easy modification of Part 2 which we leave to the reader.
\end{proof}


\section{Small NFA's for \texorpdfstring{$\MN(\alpha n,n)$}{MN(alpha n, n)}}\label{se:alphan}

\begin{lemma}\label{le:zero}
Let $x,x',y,y',c\in \N$ with $c\ge 1$ be such that the following hold.
\begin{enumerate}
\item
$c(x-x') + (c+1)(y-y')=0$.
\item
$|x-x'|\le c$.
\end{enumerate}
Then $x=x'$ and $y=y'$.
\end{lemma}

\begin{proof}
Since $c+1$ divides $c(x-x')$ and $c+1$ is rel prime to $c$ we have that $c+1$ divides $x-x'$.
Since $|x-x'|\le c$, $x=x'$. Hence $y=y'$.
\end{proof}

\begin{theorem}\label{th:alphan}
Let $n\in\N$ and $0<\alpha<1$ be such that $\alpha n\in \N$.
\begin{enumerate}
\item
There exists an NFA $M$ such that the following are true:
\begin{enumerate}
\item
For all $i\ge 2n\ln n$, $M$ accepts $a^i$.
\item
For all $i\in \{\alpha n, n\}$, $M$ rejects $a^i$.
\item
We have no comment on the behavior of $M$ for any other $a^i$'s.
\item
$M$ has $\le 2n^{1/2}\ln n +\Ot(1)$ states. 
\end{enumerate}
\item
There exists an NFA on $\le 2n^{1/2}\ln n+\Ot(1)$ 
states that accepts $\MN(\alpha n, n)$.
\end{enumerate}
\end{theorem}

\begin{proof}
Let $c=\ceil{n^{1/2}}+1$ and $e=n+1 \pmod c$. Note that $e\le c$. 

\noindent
1) Let $M'$ be $\LOOP(c,c+1,e)$.
By the proof of Theorem~\ref{th:n} we have:

\begin{enumerate}
\item
%%%%%%%%% this used to say sqrt(n) %%%%%%%%%%%%%%%%%
For all $i\ge n+2\ceil{n^{1/2}}$, $M'$ accepts $a^i$. Note that for all $i\ge 2n\ln n$, $M'$ accepts $a^i$.
\item
$M'$ rejects $a^n$.
\item
We have no comment on the behavior of $M'$ on other $a^i$.
\item
$M'$ has $\le n^{1/2}+O(1)$ states.
\end{enumerate}

\noindent
{\bf Case 1:} $M'$ rejects $a^{\alpha n}$. Then take $M$ to be $M'$.

\noindent
{\bf Case 2:} $M'$ accepts $a^{\alpha n}$. We use the very acceptance of 
$a^{\alpha n}$ to find an NFA $M$ that satisfies the theorem.

\noindent
{\bf Claim 1:}  There exists a unique $x,y$,  such that $cx + (c+1)y+e=\alpha n$.
Both $x,y$ are $\le c-1\le n^{1/2}$.

\noindent
{\bf Proof of Claim:} Since $M'$ accepts $a^{\alpha n}$ there is at least one such
$x,y$ such that:

$$\hbox{Eq 1 } cx + (c+1)y+e=\alpha n$$

$x\le c-1$ since otherwise Eq 1 implies:

$$\alpha n = cx + (c+1)y+e\ge c^2 + (c+1)y+e \ge c^2  = n.$$

\noindent
$y\le c-1$ by a similar argument.

Assume that $x',y'$ also works.

$$\hbox{Eq 2 } cx' + (c+1)y'+e=\alpha n$$

By the same reasoning that $x\le c-1$, we have $x'\le c-1$, so $|x-x'|\le c-1$.

Subtract the second equation from the first to obtain:

$$c(x-x') + (c+1)(y-y')=0$$

By Lemma~\ref{le:zero}, $x=x'$ and $y=y'$.


\noindent
{\bf End of Proof of Claim 1}

Let $p$ be the least prime that does not divide $yc$ (hence does not divide $y$ or $c$).
Since $y\le c^{1/2}$ and $c=\ceil{n^{1/2}}+1$, $yc\le n+n^{1/2}+O(1)$. By Lemma~\ref{le:primes}.3, 
$p \le (1+o(1))\ln(n+n^{1/2}+O(1)) \le 2\ln(n) + O(1)$.
Let $M$ be $\LOOP(c,p(c+1),e)$.
Note that $M$ has $\le 2n^{1/2}\ln n + \Ot(1)$ states.
We need to show that (1) for all $i\ge n\ln n$, $M$ accepts $a^i$,
and (2) $M$ rejects $a^{\alpha n}$ and $a^n$.
Note that

$$cp(c+1) - c - cp - p + e = c^2p + cp - c - cp - p + e = c^2p  - c - p + e \le 2n\ln n-1$$
By Lemma~\ref{le:loop}

\begin{itemize}
\item
For all $i\ge 2n\ln n$, $M$ accepts $a^i$.
\item
For all $C,D$, $M$ rejects $a^i$ where $i=c^2p - c - p + e -Cc  - D(p(c+1))$.
(We will not be using this.)
\end{itemize}

\noindent
{\bf Claim 2:} $M$ rejects $a^{\alpha n}$.

\noindent
{\bf Proof of Claim 2:} 

Assume, by way of contradiction, that $M$ accepts $\alpha n$. Then there exists $x',y'$ such that

$$\hbox{Eq 1 } \alpha n= cx'+ (c+1)y' + e$$

Recall that from Claim 1 there exists unique $x,y$ such that

$$\hbox{Eq 2 } \alpha n =  cx + (c+1)y+e.$$

Hence $x=x'$ and $y=y'p$. This contradicts that $p$ does not divide $y$.

\noindent
{\bf End of Proof of Claim 2}


\noindent
{\bf Claim 3:} $M$ rejects $a^{n}$.

\noindent
{\bf Proof of Claim 3:} 

Assume, by way of contradiction, that $M$ accepts $n$. Then there exists $x',y'$ such that

$$n= cx' + p(c+1)y' + e$$

Then $a^n$ is accepted by $\LOOP(c,c+1,e)$, which is a contradiction.

\noindent
{\bf End of Proof of Claim 3}

\bigskip

\noindent
2) This proof is similar to that of Theorem~\ref{th:A}.2.

\end{proof}

\section{Small NFA's for \texorpdfstring{$\MN(A)$ where $A$}{MN(A) where A} has large gaps}\label{se:gap}

\begin{theorem}\label{th:gap}
Let $A\subseteq \N$ with maximum element $n'$.
Let $n'<n$. Then there is an NFA for $\MN(A\cup \{n\})$ of
size $n^{1/2} + n' + \Ot(1)$.
\end{theorem}

\begin{proof}
By Theorem~\ref{th:n} there exists an NFA $M'$ for $\MN(n-n')$ of size
$(n-n')^{1/2} + \Ot(1)\le n^{1/2}+\Ot(1)$. We form $M$ as follows:

\begin{enumerate}
\item
Add states $0,1,\ldots,n'$ where state $n'$ is the start state of $M'$.
\item
0 is the start state of $M$.
\item
For $0\le i\le n'-1$ we have transitions $\delta(i,a)=i+1$.
\item
For all $0\le i\le n'$, make $i$ an accept state iff $i\in A$.
\end{enumerate}

Clearly $M''$ accepts $A\cup \{n\}$ and has $n^{1/2} + n' + \Ot(1)$ states.
\end{proof}

\begin{corollary}~
\begin{enumerate}
\item
For all $n$, for all $0<\delta<1$, there is an NFA for $MN(n^\delta,n)$ of size $n^{1/2} + n^{\delta} + \Ot(1)$. 
\item
For all $n$, for all $\beta\in\R^+$, there is an NFA for $MN((\log n)^\beta ,n)$ of size $n^{1/2} + \Ot(1)$. 
\end{enumerate}
\end{corollary}

\section{Every NFA for \texorpdfstring{$MN(n)$ has $\ge n^{1/2}$}{MN(n) has at least sqrt(n)} States}\label{se:lower}

This section is due to Jeff Shallit who shared it with us.


Chroback~\cite{unaryreg} proved the following.

\begin{theorem}\label{th:normal}
Let $L$ be a cofinite unary regular language.
If there is an NFA for $L$ with $n$ states then there is an NFA for $L$ of
the following form:
\begin{itemize}
\item
There is a sequence of $\le n^2$ states from the start state to a state we
will call $X$. Note that there is no nondeterminism involved yet.
\item
From $X$ there are $e$-transitions to $X_1,\ldots,X_m$.
(This is nondeterministic.)
\item
Each $X_i$ is part of a cycle $C_i$. All of the $C_i$ are disjoint.
\end{itemize}
\end{theorem}

\begin{theorem}\label{th:lower}~
\begin{enumerate}
\item
Let $L$ be a cofinite unary language where the shortest string that is not in $L$ is
of length $n$.
Then any NFA for $L$ requires $n^{1/2}$ states
\item
Any NFA for $MN(n)$ has $n^{1/2}$ states
(this follows from part 1).
\end{enumerate}
\end{theorem}

\begin{proof}

Assume there was an NFA with $< n^{1/2}$ states for $L$.
Then by Theorem~\ref{th:normal} there would be an NFA for $L$ with
a path from the start state to a state $X$ of length $<n$ and then from $X$ a branch to many cycles.
Let $X_i$ and cycle's $C_i$ as described in Theorem~\ref{th:normal}.

Run $a^n$ through the NFA and try out all paths. For each $i$ there will be
a point in $C_i$ that you end up at. Let $n_i$ be the length of $C_i$.
For every $i$ there is a state on $C_i$ that rejects. Hence the strings
$a^{n+Kn_1n_2\cdots n_m}$ are all rejected. This is an infinite number of strings.
This is a contradiction.
\end{proof}

\section{Empirical Results}\label{se:emp}

We have written a program that, given $n$, tries to find the smallest NFA for $\MN(n)$.
We first set $c=\ceil{\sqrt{n}}$, $d=c+1$, and $e$ such that $\LOOP(c,d,e)$ 
(1) rejects $a^n$ and (2) for all $i\ge n+1$, accepts $a^i$.
We then looked at sets of prime powers (these work as well as primes)
so that the usual $M_{p^b}$ machines will accept all $a^i$ such that  $i\le n-1$.
We took the smallest NFA among all of these choices. We ran this program for
$1\le n\le 10^{27}$. 
Here is what we discovered:
\begin{enumerate}
    \item 
    The smallest NFA for $\MN(n)$ was around $n^{1/2} + (\ln n)^g$ where $g$ had the following values:
    \begin{enumerate}
        \item 
        For $1\le n\le 10^6$ $g$  decreases from 2 to around 1.55.
        \item
        For $10^6 \le n\le 10^{27}$ $g$ fluctuates  around 1.55 but slowly increases.
    \end{enumerate}
\end{enumerate}

The log-term is actually the inverse of Landau's function. As such, it is known
that $g$ is bounded by 2. 

We also wrote a second  program that tries to find smaller NFAs than the first program. How? Note that, in the first program, we found a set of $M_{p^b}$ machines to accept all $a^i$ such that $i \leq n - 1$. However, $\LOOP(c,d,e)$ already accepts some of those strings. Our second program finds a set of $M_{p^b}$ machines that accepts all $a^i$ that $\LOOP(c,d,e)$ did not accept. We ran this program for $1\le n\le 1700$
(this program took much longer to run then the first one).
Here is what we discovered:
\begin{enumerate}
    \item 
For slightly more than half of the $n$, we found a smaller NFA this way.
\item
The most common improvement was 1. Then 2. $\ldots$ Then 6.
There were no improvements  bigger than 6. There was a slight tendency of getting
bigger improvements for bigger $n$. 
\end{enumerate}

We conjecture that, for all $L$, there exists $n$, so that the second program will produce an  NFA
that has at least $L$ states fewer than the first program.

\section{Open Questions}\label{se:open}

We conjecture that every cofinite unary language has a small NFA; however,
this is hard to state rigorously. 

The NFA for $\MN(n)$ is optimal up to $\Ot(1)$ terms. We would like to know if the other NFA's we
have presented are optimal up to $\Ot(1)$ terms. 

\section{Acknowledgments}

We would like to thank Jeff Shallit for his slides~\cite{shallitslidesfrob} that got us started on this
subject, his sharing the proof of Theorem~\ref{th:lower} with us,  and for his encouragement to pursue this work.

\begin{thebibliography}{999}

\bibitem{axler}
Christian Axler.
\newblock On the sum of the first $n$ prime numbers, 2014.
\newblock \url{https://arxiv.org/pdf/1409.1777.pdf}.

\bibitem{unaryreg}
M.~Chrobak.
\newblock Finite automata and unary languages.
\newblock {\em TCS}, 47:149--158, 1986.
\newblock Erratum for paper is at
  \url{https://dl.acm.org/citation.cfm?id=860232}.

\bibitem{shallitslidesfrob}
J.~Shallit.
\newblock The {F}robenius problem and its generalization.
\newblock Slides: \url{https://cs.uwaterloo.ca/~shallit/Talks/frob14.pdf}.

\end{thebibliography}

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