Sharedwww / primes.texOpen in CoCalc
Author: William A. Stein
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\documentclass[11pt]{article}
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%\textwidth=1.1\textwidth
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%\hoffset=-.3in
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%\textheight=1.1\textheight
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%\voffset=-.5in
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%
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% The pari code I used:
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%
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% prog(p, a, i=0) = while(isprime(p+a*i),i++);return(i);
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%
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% f(a, l, j=200000, i=0) = for(i=1,j,\
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% if(prog(prime(i),a)>=l,print1(prime(i)," ");i+=l-2));
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\include{macros}
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\title{Arithmetic Progressions of Primes}
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\author{W. Stein\footnote{UC Berkeley, Department of Mathematics, {\tt was@math.berkeley.edu}}}
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\begin{document}
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\maketitle
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\begin{abstract}
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Let $2n$ be an even integer.
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What is the longest arithmetic progression
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$p,\,p+2n,\,\ldots,\,p+2na$
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with $p$ prime? In this paper we guess some answers.
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\end{abstract}
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{\bf Warning:} The following is just for fun! There could be
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a well-known conjecture which supercedes all of this.
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\section{Introduction}
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It is easy to show (see Proposition~\ref{prop}) that if
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$$p,\, p+2,\, p+4$$
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are all prime, then $p=3$.
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Similarly, if
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$$p,\, p+6,\, p+12,\, p+18,\, p+24$$
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are all prime, then $p=5$ and the progression is thus
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$5, \,11,\, 17, \, 23,\, 29.$
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There are probably infinitely many examples where
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$$p,\, p+6,\, p+12, \, p+18,$$
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are all prime, but one would not dare think of proving this
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here as it is probably more difficult than the twin primes conjecture.
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For each even integer $2n$, let $M(2n)$ to be
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the maximal length of an arithmetic progression
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of primes with gap $2n$. Thus $M(2)=3$, $M(6)=5$.
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Let $M_{\infty}(2n)$ be the maximum length of arithmetic
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progressions of gap $2n$ which occur
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infinitely often. Thus one conjectures that
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$M_{\infty}(2)=2$, $M_{\infty}(6)=4$.
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Is there an example in which
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$M_{\infty}(2n)< M(2n)-1?$
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Is there an upper bound on the possible
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values $M(2n)$?
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\section{Table}
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The $M_?$ column was computed by searching the first
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100000 or so primes for an arithmetic progression of
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primes, having gap $2n$.
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The $M_{\infty?}$ column was computed by doing the same search,
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but requiring that at least {\em two} distinct progressions were found
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(usually there were many more). By Proposition~\ref{prop},
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the computed values of $M_?$ are the actual value of
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$M$, except possibly for the last four rows in the table.
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\begin{center}
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\begin{tabular}{|c|c|c|c|l|}\hline
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$2n$& $2n$ & $M_?$ & $M_{\infty?}$& Examples\\\hline\hline
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$2$& $2$ & 3& 2 & 3\\
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$4$& $2^2$ & 3& 2 & 3\\
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$6$& $2\cdot 3$ & 5& 4 & 5\\
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$8$& $2^3$ & 3& 2 & 3\\
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10& $2\cdot 5$ & 3& 2 & 3\\\hline
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12& $2^2\cdot 3$ & 5& 4 & 5\\
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14& $2\cdot 7$ & 3& 2 & 3\\
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16& $2^4$ & 2& 2 & 3\\
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18& $2\cdot 3^2$ & 4& 4 & 5\\
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20& $2^2\cdot 5$ & 3& 2 & 3\\\hline
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22& $2\cdot 11$ & 2& 2 & 7\\
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24& $2^3\cdot 3$ & 4& 4 & 59\\
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26& $2\cdot 13$ & 2& 2 & 3\\
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28& $2^2\cdot 7$ & 2& 2 & 3\\
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30 &$2\cdot 3\cdot 5$ & 6& 6 & 7, 107 \\\hline
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32 &$2^5$ & 2& 2 & 5\\
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34 &$2\cdot 17$& 3& 2 & 3\\
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36 &$2^2\cdot 3^2$& 4& 4 & 31\\
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38 &$2\cdot 19$& 3& 2 & 3\\
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40 &$2^3\cdot 5$ & 3& 2 & 3\\\hline
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42 &$2\cdot 3\cdot 7$& 5& 4 & 5\\
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44 &$2^2\cdot 11$& 2& 2 & 3\\
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46 & $2\cdot 23$ & 2& 2 & 7\\
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48&$2^4\cdot 3$& 5& 4 & 5\\
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50&$2\cdot 5^2$& 3& 2 & 3\\\hline
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210 & $2\cdot 3\cdot 5\cdot 7$ & 10& 9 & 199\\
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2310 & $2\cdot 3\cdot 5\cdot 7\cdot 11$& 9 & 9 & 3823\\
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30030 & $2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13$& 12 & 12 & 23143,\,1498141\\
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510510 & $2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$& 13 & ? & 766439\\
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\hline
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\end{tabular}
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\end{center}
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\section{Conclusions}
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\begin{proposition}\label{prop}
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Let $q$ be a prime.
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If $q\nmid 2n$ then
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$M(2n)\leq q$ and $M_{\infty}(2n)\leq q-1$.
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Furthermore, $M(2n)=q$ iff $q$ is involved in arithmetic
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progression of length $q$ with gap $2n$.
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\end{proposition}
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\begin{proof}
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Reduce the arithmetic progression
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$$p,\,p+2n,\,\ldots,\,p+2na$$
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modulo $q$. Since $2n$ is a unit
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modulo $q$, if the length of the arithmetic progression
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is $\geq q$ then one of
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the terms in the arithmetic progression is divisible
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by $q$, hence equal to $q$.
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\end{proof}
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In the table above, there is not a single example in which $q\nmid 2n$,
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$M(2n)=q$ and $q$ is not the first prime in the arithmetic.
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This suggests
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\begin{question}
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Let $q\nmid 2n$ be a prime and suppose
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$M(2n)=q$. Then must
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$$q,\, q+2n,\,\ldots, \, q+2n(q-1)$$
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all be prime?
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\end{question}
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The data also suggests the following
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\begin{conjecture}
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Both $M(2n)$ and $M_{\infty}(2n)$ are $\geq 2$ for all $n$.
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\end{conjecture}
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The conjecture that $M(2n)\geq 2$ is similar to Golbach's
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conjecture. It asserts that any even integer is
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the difference of two primes; the conjecture that
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$M_{\infty}(2n)\geq 2$ asserts that this can be done in
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infinitely many ways.
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It seems conceivable that $M(2n!)$ is unbounded as
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a function of $n$.
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\end{document}