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\DeclareMathOperator{\li}{li}
% How far is http://www.trnicely.net/gaps/gaplist.html complete?
\newcommand{\NicelyLimit}{\ensuremath{2\times 10^{17}} }
% and s(p)-p/\log p for the prime with the maximal gap is
\newcommand{\NicelyM}{\ensuremath{0.92064} }
% the bound on M from these limits
\newcommand{\NicelyMaxM}{\ensuremath{199262} }

\begin{document}

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\epsfxsize=4in \leavevmode\epsffile{logo129.eps}
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\begin{center}
\vskip 1cm{\LARGE\bf Determining Mills' Constant and a Note on
Honaker's Problem} \vskip 1cm \large
Chris K.~Caldwell \\
Department of Mathematics and Statistics\\
University of Tennessee at Martin \\
Martin, TN 38238\\
USA \\
\href{mailto:caldwell@utm.edu}{\tt caldwell@utm.edu} \\
\ \\
Yuanyou Cheng\\
Durham, North Carolina\\
USA \\
\href{mailto:cync98@netzero.com}{\tt cync98@netzero.com}\\
\end{center}

\vskip .2 in

\begin{abstract}
In 1947 Mills proved that there exists a constant $A$ such that
$\lfloor A^{3^n} \rfloor$ is a prime for every positive integer $n$.
Determining $A$ requires determining an effective Hoheisel type
result on the primes in short intervals---though most books ignore
this difficulty. Under the Riemann Hypothesis, we show that there
exists at least one prime between every pair of consecutive cubes
and determine (given RH) that the least possible value of Mills'
constant $A$ does begin with $1.3063778838$.  We calculate this
value to $6850$ decimal places by determining the associated primes
to over $6000$ digits and probable primes (PRPs) to over $60000$
digits. We also apply the Cram\'er-Granville Conjecture to Honaker's
problem in a related context.
\end{abstract}

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\section{Introduction}

In 1947 Mills \cite{14} proved that there exists a
constant $A$ such that $\lfloor A\sp{3\sp{n}}\rfloor$ is a~prime
for every positive integer $n$. Mills' proof used Ingham's result
\cite{Ingham1937} that there is always a prime in the interval
$(x,x+k x\sp{5/8})$ for some constant $k$. Since $k$ was not
determined explicitly, Mills could not determine $A$ or even a
range for $A$. Many authors citing Mills' constant either follow
his example and remain silent about $A$'s value (e.g., \cite{9},
\cite{18}, \cite{23}), or explicitly state that it is unknown
(e.g., \cite{1}, \cite{5}). But recently a few authors have begun
to state that $A$ is approximately $1.3063778838$ (e.g., \cite{6},
\cite{8}, \cite{20}, \cite{25} and Sloane \seqnum{A051254}). They
do this despite the fact that it is currently impossible to
compute $A$ without any further unproven assumptions, and
despite the fact that for each $c\ge 2.106$, there are uncountably
many possible values $A$ for which $\lfloor A\sp{c\sp{n}} \rfloor$
is prime for each positive integer $n$.\par

The authors who approximate Mills' constant $A$ also implicitly
define \textit{Mills' constant} to be the least $A$ such that
$\lfloor A\sp{3\sp{n}}\rfloor$ is prime for all positive integers
$n$.  Mills' original article contained no numerics---it only showed that
such an $A$ exists.

In Section~\ref{sec:gaps} we observe that, assuming the Riemann
Hypothesis, it easily follows that there is at least one
prime between $x\sp{3}$ and $(x+1)\sp{3}$ for every $x\ge 0.26$
(Lemma~\ref{lemma:between_cubes}). We then use this to determine
Mills' constant $A$ to over $6850$ decimal places in Section
\ref{sec:Mills}. This requires a substantial computing effort to
calculate the associated \textit{Mills' primes} $\lfloor
A\sp{3\sp{n}}\rfloor$ explicitly for $n=1$, $2\ldots , 10$. In
particular we find prove the following:

\begin{theorem} \label{thm:A250digits}
Assume the Riemann Hypothesis. The minimum Mills' constant (for the
exponent c=$3$) begins with the following $600$ digits:
\begin{center}
\begin{tabular}{rllll}
$1.3063778838$&amp; $6308069046$&amp; $8614492602$&amp; $6057129167$&amp; $8458515671$ \\
  $3644368053$&amp; $7599664340$&amp; $5376682659$&amp; $8821501403$&amp; $7011973957$ \\
  $0729696093$&amp; $8103086882$&amp; $2388614478$&amp; $1635348688$&amp; $7133922146$ \\
  $1943534578$&amp; $7110033188$&amp; $1405093575$&amp; $3558319326$&amp; $4801721383$ \\
  $2361522359$&amp; $0622186016$&amp; $1085667905$&amp; $7215197976$&amp; $0951619929$ \\
  $5279707992$&amp; $5631721527$&amp; $8412371307$&amp; $6584911245$&amp; $6317518426$ \\
  $3310565215$&amp; $3513186684$&amp; $1550790793$&amp; $7238592335$&amp; $2208421842$ \\
  $0405320517$&amp; $6890260257$&amp; $9344300869$&amp; $5290636205$&amp; $6989687262$ \\
  $1227499787$&amp; $6664385157$&amp; $6619143877$&amp; $2844982077$&amp; $5905648255$ \\
  $6091500412$&amp; $3788524793$&amp; $6260880466$&amp; $8815406437$&amp; $4425340131$ \\
  $0736114409$&amp; $4137650364$&amp; $3793012676$&amp; $7211713103$&amp; $0265228386$ \\
  $6154666880$&amp; $4874760951$&amp; $4410790754$&amp; $0698417260$&amp; $3473107746$ \\
\end{tabular}
\end{center}
\end{theorem}

\noindent We also find the two probable-primes which are (most
likely) the next terms in the sequence of Mills' primes.  The last
of these has $61684$ digits.

Concerning other problems that involve prime gaps, let $p\sb{0}=2$,
$p\sb{n}$ be the $n$-th odd prime number and
$g\sb{n}=p\sb{n+1}-p\sb{n}$, the gap between consecutive prime
numbers. The Cram\'er-Granville conjecture \cite{12} states that
$g\sb{n}\le M\log\sp{2}n$ for some constant $M&gt;1$. The last section
is an application of this conjecture on prime gaps to Honaker's
problem.

Honaker's problem \cite{7} is to find all trios of consecutive
prime numbers $p&lt;q&lt;r$, such that $p|(qr+1)$. We call such triples
of consecutive prime numbers \textit{Honaker trios} and conjecture
that there are only three Honaker trios: $(2,3,5)$, $(3,5,7)$, and
$(61,67,71)$. We establish that there are only three Honaker trios
$(p,q,r)$ for $p\le \NicelyLimit$ and prove:

\begin{theorem} \label{thm:3HonakerTrios} The Cram\'er-Granville
conjecture with any constant $M$ implies that there are only
a finite number of Honaker trios.  If also $M\le \NicelyMaxM$, then there
are exactly three trios.\end{theorem}

\section{Gaps Between Prime Numbers}
\label{sec:gaps}

Let $s(q)$ represent the next prime after $q$:
$s(q)=\min\{\mbox{$p$ prime} \mid p &gt; q\}$. Define the sequence of
maximal gap primes as follows:
$$q_j = \left\{ \begin{array}{ll}
  2, &amp; \mbox{if $j = 0$}; \\
  3, &amp; \mbox{if $j = 1$}; \\
  \min\{\mbox{$p$ prime} \mid s(p)-p &gt; s(q_{j-1})-q_{j-1}\}, &amp; \mbox{if $j &gt;
  1$}.
 \end{array} \right.$$
In other words $\{q_k\}$ is the sequence of primes that are
followed by maximal gaps (Sloan \seqnum{A002386}).  This sequences
starts $2$, $3$, $7$, $23$, $89$, $113$, $523$, $887$, $1129$,
$1327$ $\ldots$  and exists for all subscripts $k$ because
$\limsup_{p \rightarrow + \infty} s(p)-p = +\infty$ \cite{11}. The
list of maximal gap primes $q_j$ has been extended through all
primes below \NicelyLimit by Herzog and Silva \cite{HS2005} with
much of their work verified by Nicely and others \cite{16,17}.

With these definitions it is clear that
$$\frac{s(p)-p}{\log^2 p} \leq \frac{s(q_k)-q_k}{\log^2 q_k},$$
for every $q_k \leq p &lt; q_{k+1}$. Now
using Nicely's tables of maximal gaps \cite{16} we
can easily verify the following result:

% reverified 6/2005
\begin{lemma} \label{cor:1} For $11\le p\sb{n}&lt; \NicelyLimit$, we
have $\frac{s(p)-p}{\log\sp{2}p}\leq \NicelyM.$ \end{lemma}

To explicitly determine Mills' constant we must know that there is
always a prime between each pair of successive cubes from some
established point onward.  This follows easily from a result of
Schoenfeld \cite{21}.  Recall
$$\li(x) := \lim_{\epsilon \rightarrow 0^+} \left( \int\sb{0}\sp{1-\epsilon}\frac{dt}{\log t}
+\int\sb{1+\epsilon}\sp{x}\frac{dt}{\log t} \right) .$$

\begin{lemma}[Schoenfeld] \label{lemma:Schoenfeld} Assume the Riemann Hypothesis. For
$x\ge 2657$, we have  $$\li(x)-\frac{\sqrt{x}\log x}{8\pi}&lt;\pi(x) &lt;
\li(x) +\frac{\sqrt{x}\log x}{8\pi}.$$ \end{lemma}

\noindent (Much stronger bounds are now available \cite{RS2003}, but
this is sufficient for our current needs.)

Using this Lemma we see that if $x &gt; 2657^{1/3}$, then
$$\aligned \pi((x+1)\sp{3})-\pi(x\sp{3})
&amp;&gt;\li((x+1)\sp{3})-\li(x\sp{3}) -\frac{3}{4\pi}(x+1)\sp{3/2}\log(x+1)\\
&amp;\ge\int\sb{x\sp{3}}\sp{(x+1)\sp{3}} \frac{dt}{\log
t}-\frac{3}{4\pi}
  (x+1)\sp{3/2}\log(x+1)\\
&amp;\ge\frac{3x\sp{2}+3x+1}{3\log(x)}-\frac{3}{4\pi}(x+1)\sp{3/2}\log(x+1).\\
\endaligned$$
This last lower bound is an increasing function of $x$ and it is
greater than one for $x &gt; 2657^{1/3}$.  After using a computer
program to check the smaller values of $x$ we have shown the
following:

\begin{lemma} \label{lemma:between_cubes} Assume the Riemann Hypothesis. There is at
least one prime between $x\sp{3}$ and $(x+1)\sp{3}$ for every
$x\ge x_0 = 2^{1/3}-1$.\end{lemma}

This lemma states that, under the Riemann Hypothesis, there are
prime numbers between consecutive cubes; we will see that this is
necessary to calculate Mills' constant in the next section.

Without the Riemann Hypothesis it is still possible to use explicit
upper bounds on the zeta function \cite{ChengsThesis,Ford2002} to
get a version of Lemma \ref{lemma:between_cubes} with a far larger
value of $x_0$, roughly $10^{6000000000000000000}$
\cite{Cheng2002,Ramare2002}. In the next section we will show that
calculating Mills' constant involves explicitly finding a prime
larger than the cube-root of this bound.  This bound so dramatically
exceeds current computing abilities, that any current calculation of
Mills' constant must involve an unproven assumption.  This bound
$x_0$ should get much smaller as the bounds on the zeta function are
improved.

\section{Mills' constant}
\label{sec:Mills}

We begin this section with a lemma.

\begin{lemma} \label{lemma:xc} If $x&gt;1$ and $c&gt; 2$, then $1+x\sp{c}+x\sp{c-1}
&lt;(1+x)\sp{c}$.\end{lemma}

\begin{proof} Dividing by $x\sp{c}$ and replacing $x$ with $1/x$ we
arrive at the equivalent inequality $0&lt;(1+x)\sp{c}-(1+x+x\sp{c})$
($0&lt;x&lt;1$). The inequality clearly holds when $c=2$ (because it
reduces to $x&gt;0$) and when $x=0$. Now if $x&gt;0$, differentiate the
right side with respect to $c$ to get $(1+x)\sp{c}\log(1+x)
-x\sp{c}\log(x)$, which is clearly positive, so the inequality
above holds for all $c&gt;2$.\end{proof}

It will be useful to next recall a proof of a simple
generalization of Mills' theorem.\par

\begin{theorem} \label{thm:Mills} Let $S=\{a\sb{n}\}$ be any sequence of integers
satisfying the following property: there exist real numbers
$x\sb{0}$ and $w$ with $0 &lt;w&lt; 1$, for which the open interval
$(x,x+x\sp{w})$ contains an element of $S$ for all real numbers $x
&gt; x\sb{0}$. Then for every real number
$c&gt;\min\displaystyle\left(\frac{1} {1-w} ,2\right)$ there is a
number $A$ for which $\lfloor A\sp{c\sp{n}}\rfloor$ is a subsequence
of $S$.\end{theorem}

\begin{proof} (We follow Ellison \&amp; Ellison \cite{10}.) Define a
subsequence ${b\sb{n}}$ of $S$ recursively by
\begin{quote}
\begin{trivlist}
\item[(a)] $b_1$ is equal to the least member of $S$ for which
$b_1^c$ is greater than $x\sb{0}$. \item[(b)] $b\sb{n+1}$ is the
least member of $S$ satisfying $b\sb{n}\sp{c}&lt; b\sb{n+1} &lt;
b\sb{n}\sp{c} + b\sb{n}\sp{wc}$.
\end{trivlist}
\end{quote}

Because $c&gt;1/(1-w)$ and $c&gt;2$, (b) can be written as:
$$b_n^c &lt; b_{n+1} &lt; 1+b_{n+1} &lt; 1+b_n^c + b_n^{wc}
&lt; 1 + b_n^c + b_n^{c-1} &lt; (1+b_n)^c$$ the last inequality following
from Lemma \ref{lemma:xc}. For all positive integers $n$ we can
raise this to the $c\sp{-(n+1)}$\textit{th} power to get
$$b_n^{c\sp{-n}} &lt; b\sb{n+1}\sp{c\sp{-(n+1)}} &lt; (1+b\sb{n+1})\sp{c\sp{-(n+
1)}} &lt; (1+b\sb{n})\sp{c\sp{-n}}.$$

This shows that the sequence$\{b\sb{n} \sp{c\sp{-n}}\}$
is monotonic and bounded, therefore converges. Call its limit $A$.
Finally,
$$b\sb{n}&lt;A\sp{c\sp{n}}&lt;b\sb{n}+1,$$ so $\lfloor
A\sp{c\sp{n}}\rfloor = b\sb{n}$, yielding the chosen subsequence
of $S$ and completing the proof. \end{proof}

When constructing the sequence $\{b_n\}$ in the previous proof, the
condition (a) can be relaxed in two important ways.  First, by
replacing the words `the least' with `any', we see there are
infinitely many choices for $b_1$, and hence for the resulting value
$A$.  Second, it is not necessary that $b_1^c &gt; x_0$ be satisfied by
$b_1$, as long as the terms satisfying (b) exist to a term $b_n$
which does satisfy $b_n^c &gt; x_0$.  This will someday be important in
removing the assumption of the Riemann Hypothesis from Theorem
\ref{thm:A250digits} by calculating a sequence of primes
$\{b_i\}_{i=1}^n$ that extends to the cube root of the bound $x_0$
discussed at the end of the previous section.

It is proved by Baker \textit{et. al.} \cite{3} that
$p\sb{n+1}-p\sb{n}=O(p\sb{n}\sp{0.525})$. From the above theorem,
one obtains the following proposition.\par

\begin{prop} \label{prop:Infinite_As} For every $c\ge 2.106$, there
exist infinitely many $A$'s such that $\lfloor
A\sp{c\sp{n}}\rfloor$ is a prime for every $n$.\end{prop}

Wright \cite{24} showed that the set of possible values of the
constants $A$ and $c$ in this proposition (and several
generalizations \cite{19}, \cite{20}, \cite{23}) have the same cardinality
as the continuum and are nowhere dense. So authors approximating
``Mills' constant'' must first decide how to choose just one. Mills
specified the exponent $c=3$.  Mills also used the lower bound
$k\sp{8}$ for the first prime in the sequence, where $k$ is the
\textit{integer} constant in Ingham's result \cite{Ingham1937}. So
to follow his proof literally would require that the first prime
be $257$ (and producing a constant $A \approx 6.357861928837$).
% The pair 7, 11 requires k &gt;= 1.1854, so k must be 2 to be an integer.
% 6.3578619288370721819188983391960758868377910599712542300877510989726894517887811
But all authors offering an approximation agree implicitly on
starting with the prime $2$ and then choosing the least possible
prime at each step, so we take this as the definition of 
\textit{Mills' constant}.

The sequence of minimal primes satisfying the criteria of the
proof with $c=3$ (Sloan \seqnum{A051254}) begins with
$$\aligned
b_1 = &amp; 2,\\
b_2 = &amp; 11,\\
b_3 = &amp; 1361,\\
b_4 = &amp; 2521008887,\\
b_5 = &amp; 16022236204009818131831320183,\\
b_6 = &amp; 41131 0114921510 4800030529 5379159531 7048613\backslash \\ &amp; 962 3539759933 1359499948 8277040407 4832568499.\\
\endaligned$$

These first six terms were well known. We have now shown
that assuming the Riemann Hypothesis, prime numbers exist between
consecutive cubes, so (with RH) we know this sequence can be
continued indefinitely.

To make these fast growing primes $b\sb{n}$ easier to present,
define a sequence $a_n$ by $b_{n+1} = b_n^3 + a_n$.  The sequence
$a\sb{n}$ begins with $3, 30, 6, 80, 12, 450, 894, 3636, 70756$
(Sloan \seqnum{A108739}). The primality of the new terms  $b\sb{7}$,
$b\sb{8}$ and $b\sb{9}$ ($2285$ digits) were proven using the
program Titanix \cite{13} which is based on an elliptic curve test
of Atkin (\cite{2}, \cite{15}). The test for $b\sb{9}$, which at
the time was the third largest proven `general' prime, was
completed by Bouk de Water in 2000 using approximately five weeks
of CPU time. The certificate was then verified using Jim
Fougeron's program Cert-Val.\par

Fran\c{c}ois Morain has verified the primality of the next term
$b_{10}$ ($6854$ digits, July 2005) using his current
implementation of fastECPP \cite{FKMW2004, Morain2005}. The
computation was done on a cluster of six Xeon biprocessors at 2.6
GHz. Cumulated CPU time was approximately 68 days (56 for the
DOWNRUN, 12 for the proving part).

In late 2004 Phil Carmody used his self-optimizing sieve generator
to generate an appropriate sieve.  Using it and pfgw he verified
the above results (as PRPs) and found the next two probable-primes
in the sequence $b_n$. They are the values corresponding to
$a_{10} = 97220$ and $a_{11} = 66768$. These yield probable primes
of $20562$ and $61684$ digits respectively, so their primality may
not be proven for some time.

The first ten terms of $\{b_n\}$ are sufficient to determine
Mills' constant $A$ to over $6850$ decimal places because we have
from the proof of Theorem \ref{thm:Mills}:
$$b\sb{n}\sp{c\sp{-n}}&lt;A&lt;(b\sb{n}+1)\sp{c\sp{-n}}.$$
A quick calculation now completes the proof of Theorem \ref{thm:A250digits}.\par

\section{Honaker's Problem}
\label{sec:Honaker}

Let $g(p)$ be the length of the prime gap after $p$:
$g(p) = s(p)-p$.

Cram\'er's original conjecture says that $g(p)=(1+o(1))\log^2 p$.
This conjecture could be too strong to be valid, so Granville
\cite{12} proposed there is a bound $M$ for which $g(p)&lt;M\log^2 p$
(we refer to this conjecture as the Cram\'er-Granville
conjecture).  As we discussed in Section~\ref{sec:gaps},
$M=\NicelyM$ is valid for $11\le p&lt; \NicelyLimit$. However, the
constant involved in this conjecture may be $M$ = $2
e\sp{-\gamma}\approx 1.123$ \cite{12}.

\begin{proof}[Proof of Theorem \ref{thm:3HonakerTrios}] Searching
for Honaker trios with $p &lt; 50$ we find only $(2,3,5)$ and
$(3,5,7)$.  In what follows we will assume $(p,q,r)$ is a Honaker
trio and $p &gt; 50$.

Reformulate the problem by letting $q=p+2k$ and $r=q+2l$. Since
$p|(qr+1)$, we have $p|(4k\sp{2}+4kl+1)$. Thus,
\begin{equation}
\label{eq:1} p\leq 4k\sp{2}+4kl+1.
\end{equation}

If $k\geq l$, then $p \leq 8k^2+1$ and
$g(p)=2k\geq\sqrt{(p-1)/2}$.  Otherwise $k &lt; l$, $p\leq
4(l-1)^2+4(l-1)l+1$ and again $g(q)=2l\geq\sqrt{(p-1)/2}$. These
gaps are bounded by the Cram\'er-Granville conjecture, so either
$p$ or $q$ must satisfy:
\begin{equation}
\label{eq:2} M\geq \sqrt{\frac{p-1}{2}}\frac{1}{\log^2 p}.
\end{equation}
The function on the right is increasing for $p &gt; 50$, and is
unbounded, so for any fixed $M$ the number of Honaker trios is
finite.

Next suppose $p &lt; \NicelyLimit$.  By Lemma \ref{cor:1} we know
$M=\NicelyM$ will suffice in this range, so equation (\ref{eq:1})
gives $p&lt;14142$.  A computer search in this range adds the third
trio $(61,67,71)$.  Thus there are exactly three trios with $p &lt;
\NicelyLimit$.

Finally, since any additional solution must satisfy $p &gt;
\NicelyLimit$, equation \ref{eq:2} shows $M &gt; \NicelyMaxM$. This
completes the proof of Theorem 3.\end{proof}

Clearly this simple proof can be extended to make similar
statements about finding $k$ consecutive primes for which one of
the $k$ divides the product of the other $k-1$ plus or minus a
fixed integer.


\bibliographystyle{amsplain}
\begin{thebibliography}{10}

% \bibliography{c:/viewers/tex/primes}

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\bibitem{3}
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% \bibitem{CG2004}
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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11Y60; Secondary 11Y11, 11A41.

\noindent \emph{Keywords: } Mills' constant, primes in short
intervals, prime gaps, elliptic curve primality testing,
Cram\'er-Granville conjecture, Honaker's problem.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A051021},
\seqnum{A051254} and \seqnum{A108739}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received July 14 2005;
revised version received August 15 2005.
Published in {\it Journal of Integer Sequences}, August 24 2005.

\bigskip
\hrule
\bigskip

\noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.
\vskip .1in

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