\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Ramanujan Primes: \\ \vskip .1in Bounds, Runs, Twins, and Gaps } \vskip 1cm \large Jonathan~Sondow\\ 209~West~97th~Street\\ New York,~NY~~10025\\ USA\\ \href{mailto:jsondow@alumni.princeton.edu}{\tt jsondow@alumni.princeton.edu} \\ \ \\ John~W.~Nicholson\\ P.~O.~Box~2423\\ Arlington,~TX~~76004\\ USA\\ \href{mailto:reddwarf2956@yahoo.com}{\tt reddwarf2956@yahoo.com} \\ \ \\ Tony~D.~Noe\\ 14025~NW~Harvest~Lane\\ Portland,~OR~~97229\\ USA\\ \href{mailto:noe@sspectra.com}{\tt noe@sspectra.com} \\ \end{center} \vskip .2in \begin{abstract} The $n$th Ramanujan prime is the smallest positive integer $R_n$ such that if $x \ge R_n$, then the interval $\left(\frac12x,x\right]$ contains at least $n$ primes. We sharpen Laishram's theorem that $R_n < p_{3n}$ by proving that the maximum of $R_n/p_{3n}$ is $R_5/p_{15} = 41/47$. We give statistics on the length of the longest run of Ramanujan primes among all primes $p<10^n$, for $n\le9$. We prove that if an upper twin prime is Ramanujan, then so is the lower; a table gives the number of twin primes below $10^n$ of three types. Finally, we relate runs of Ramanujan primes to prime gaps. Along the way we state several conjectures and open problems. An appendix explains Noe's fast algorithm for computing $R_1,R_2,\dotsc,R_n$.\\ \end{abstract} %%% DOCUMENT %Section \section {Introduction} \label{SEC: introd} For $n\ge1$, the $n$th \emph{Ramanujan prime} is defined as the smallest positive integer $R_n$ with the property that for any $x \ge R_n$, there are at least $n$ primes $p$ with $\frac12x1). \label{EQ: ineqs} \end{equation} Moreover, for every $\epsilon > 0$, there exists $N_0(\epsilon)>0$ such that \begin{equation} R_n < (2+\epsilon)n \log n \qquad (n\ge N_0(\epsilon)). \label{EQ: 1+e} \end{equation} In particular, $R_n \sim p_{2n}$ as $n\to\infty$. \end{theorem} \begin{proof} Inequalities of Rosser and Schoenfeld for $\pi(x)$, together with \emph{Rosser's theorem} \cite{rosser} that $p_n>n\log n$, lead to \eqref{EQ: ineqs}. The bound \eqref{EQ: 1+e} follows from the Prime Number Theorem. For details, see Sondow \cite{sondow}. \end{proof} A prediction~\cite[Conjecture~$1$]{sondow} that \eqref{EQ: ineqs} can be improved to \mbox{$p_{2n}0$. For details, see Laishram \cite{laishram}. \end{proof} Using one of those values and a fast algorithm for computing Ramanujan primes (see the Appendix), we sharpen Theorem~\ref{THM: laishram} by giving an optimal upper bound on $R_n/p_{3n}$, namely, its maximum. (Notice that the rational numbers $R_n/p_{3n}$ are all distinct, because the $p_{3n}$ are distinct primes and $0 0.8666\dotso =13/15 $, it suffices to show $R_n/p_{3n} < 13/15$ for $n\neq5$. Set $\epsilon = 3/5$ and substitute $2+\epsilon = 13/5$ into \eqref{EQ: 1+e}. Using Rosser's theorem with $3n$ in place of $n$, we can write the result as \begin{equation*} R_n < \frac{13}{15}\,3n\log n < \frac{13}{15}\,p_{3n} \qquad(n\ge N_0(3/5)). \end{equation*} According to Laishram \cite[Theorem~1]{laishram}, if $0<\epsilon\le1.08$, then $N_0(\epsilon) = (2/\epsilon)^{c/\epsilon}$ in \eqref{EQ: 1+e}, where $c=c(\epsilon)=6$ at $\epsilon= 0.6$. Hence $N_0(3/5) = (10/3)^{10} = 169350.87\dotso$, and so \begin{equation*} \frac{R_n}{p_{3n}} < \frac{13}{15} \qquad (n>169350). \end{equation*} To complete the proof, we compute the first $169350$ Ramanujan primes and then check that $R_n/p_{3n}<13/15$ when $5\neq n\le169350$. \end{proof} Similarly, one can show that \begin{align*} \max_{n\neq5}\frac{R_n}{p_{3n}} &= \frac{R_{10}}{p_{30}} = \frac{97}{113} = 0.8584\dotso,\\ \max_{n\neq5\,\text{\rm or}\,10}\frac{R_n}{p_{3n}} &= \frac{R_{2}}{p_{6}} = \frac{11}{13} = 0.8461\dotso, \end{align*} \noindent and so on down towards $$\lim_{n\to\infty} \frac{R_n}{p_{3n}} = \frac23 = 0.666\dotso.$$ We conclude this section with a related prediction. \begin{conjecture} \label{CONJ: R vs p} For $m=1,2,3,\dotsc,$ let $N(m)$ be given by the following table. \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline $m$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7,8,\dotsc,19$ & $20,21,\dotso$\\ \hline $N(m)$ & $1$ & $1245$ & $189$ & $ 189$ & $85$ & $85$ & $10$ & $2$\\ \hline \end{tabular} \end{center} Then we have $$\pi(R_{mn}) \le m\pi(R_n) \qquad (n \ge N(m)).$$ Equivalently, if we define the function $\rho$ by $\rho(n):=\pi(R_n)$, so that $R_n=p_{\rho(n)}$, then $$\rho(mn) \le m\rho(n) \qquad (n \ge N(m)).$$ \end{conjecture} In the cases $m=2,3,\dotsc,20$, the statement has been verified for all $n$ with $R_{mn}<10^9$. The first few values of $\rho(n)$, for $n=1,2,3,\dotsc,$ are \cite[Sequence \seqnum {A179196}]{oeis} $$\rho(n)=1, 5, 7, 10, 13, 15, 17, 19, 20, 25, 26, 28, 31, 35, 36, 39, 41, 42, 49, 50, 51, 52, 53, \dotsc.$$ Note that Theorems \ref{THM: sondow} and \ref{THM: laishram} imply $2n<\rho(n)<3n$ for all $n>1$, and $\rho(n)\sim2n$ as $n\to\infty$. The latter yields $\rho(mn)\sim2mn\sim m\rho(n)$ as $n\to\infty$, for any fixed $m\ge1$. %Section \section{Runs} \label{SEC: runs} Since $p_{2n} < R_n \sim p_{2n}$ as $n\to\infty$, the probability of a randomly chosen prime being Ramanujan is slightly less than~$1/2$, roughly speaking. More precisely, column~$2$ in Table~\ref{TABLE: runs} gives the probability $P_n$ (rounded to $3$ decimal places) that a prime $p<10^n$ is a Ramanujan prime, for $n=1,2,\dotsc,9$. Let us consider a coin-tossing model. Suppose that a biased coin has probability $P$ of heads. According to Schilling~\cite{schilling}, the expected length E$L_N={\rm E}L_N(P)$ of the longest run of heads in a sequence of $N$ coin tosses is approximately equal to \begin{equation*} {\rm E}L_N \approx \frac{\log N}{\log(1/P)} - \left( \frac12 - \frac{\log (1-P)+\gamma}{\log(1/P)}\right), \end{equation*} where $\gamma = 0.5772\dotso$ is the Euler-Mascheroni constant. The variance Var$L_N =$ Var$L_N(P)$ is close \cite{schilling} to \begin{equation*} {\rm Var}L_N \approx \frac{\pi^2}{6\log(1/P)^2} + \frac{1}{12} \end{equation*} ``and is quite remarkable for the property that it is essentially constant with respect to'' $N$. For example, with a fair coin, \begin{equation} {\rm E}L_N \approx \frac{\log N}{\log2} - \left( \frac32 - \frac{\gamma}{\log2}\right) = \frac{\log N}{\log2} - 0.667\dotso \qquad\left(P=\frac12\right) \label{EQ: fairexp} \end{equation} and \begin{equation} {\rm Var}L_N \approx \frac{\pi^2}{6(\log2)^2} + \frac{1}{12} = 3.507\dotso \qquad\left(P=\frac12\right). \label{EQ: fairvar} \end{equation} Schilling points out that by \eqref{EQ: fairvar} ``the standard deviation of the longest run is approximately $({\rm Var}L_N)^{1/2} \approx 1.873$, an amazingly small value. This implies that the length of the longest run is quite predictable indeed; normally it is within about two of its expectation.'' This is nearly true of the longest run of Ramanujan primes in the sequence of prime numbers below $10^n$ (where $P=P_n\lessapprox1/2$), at least for $n\le9$. But for non-Ramanujan primes (where $P=1-P_n\gtrapprox1/2$), the actual length of the longest run exceeds the expected length by much more than two, at least for $n=6,7,8,9$. (See Table~\ref{TABLE: runs}, in which the two columns marked ``Actual'' are \cite[Sequences \seqnum {A189993} and \seqnum {A189994}]{oeis}.) \begin{table}[H] \begin{center} \begin{tabular}{|c|c|c|c|c|c|} \hline & Probability $P_n$ of a prime& \multicolumn{4}{|c|}{Length of the longest run below $10^n$ of} \\ & $p<10^n$ & \multicolumn{2}{|c|}{Ramanujan primes} & \multicolumn{2}{|c|}{non-Ramanujan primes} \\ \cline{3-6} $n$ & being Ramanujan & Expected & Actual & Expected & Actual \\ \hline $1$ & $.250$ & $1$ & $1$ & $5$ & $3$ \\ $2$ & $.400$ & $5$ & $2$ & $8$ & $4$ \\ $3$ & $.429$ & $8$ & $5$ & $11$ & $7$ \\ $4$ & $.455$ & $11$ & $13$ & $14$ & $13$\\ $5$ & $.465$ & $14$ & $13$ & $18$ & $20$ \\ $6$ & $.471$ & $18$ & $20$ & $21$ & $36$ \\ $7$ & $.476$ & $21$ & $21$ & $24$ & $47$ \\ $8$ & $.479$ & $24$ & $26$ & $28$ & $47$ \\ $9$ & $.482$ & $28$ & $31$ & $31$ & $65$ \\ \hline \end{tabular} \caption{Length of the longest run of (non-)Ramanujan primes below $10^n$.} \label{TABLE: runs} \end{center} \end{table} \begin{problem} Explain the unexpectedly long runs of non-Ramanujan primes among primes $p<10^n$, for $n\ge6$. \end{problem} Formula \eqref{EQ: fairexp} suggests the following predictions supported by Table~\ref{TABLE: runs}. They strengthen an earlier prediction \cite[Conjecture~2]{sondow} that arbitrarily long runs of both types exist. \begin{conjecture} \label{CONJ: runs} We have $$\limsup_{N\to\, \infty} \frac{\text{length of the longest run of Ramanujan primes among primes}\le p_N}{\log N/\log2}\ge1$$ and the same holds true if ``Ramanujan'' is replaced with ``non-Ramanujan''. \end{conjecture} For $n=1,2,\dotsc,$ the first run of $n$ Ramanujan primes begins at $$2, 67, 227, 227, 227, 2657, 2657, 2657, 2657, 2657, 2657, 2657, 2657, 562871,793487, \dotsc,$$ and the first run of $n$ non-Ramanujan primes at $$3, 3, 3, 73, 191, 191, 509, 2539, 2539, 5279, 9901, 9901, 9901, 11593, 11593, 55343, 55343,\dotsc,$$ respectively \cite[Sequences \seqnum {A174602} and \seqnum {A174641}]{oeis}. %Section \section{Twins} \label{SEC: twins} If $p_n + 2 = p_{n+1}$, then $p_n$ and $p_{n+1}$ are \emph{twin primes}; the smallest are $3$ and~$5$. If $R_n + 2 = R_{n+1}$, then $R_n$ and $R_{n+1}$ are \emph{twin Ramanujan primes}; the smallest are $149$ and $151$. Given primes $p$ and $q>p$, a necessary condition for them to be twin Ramanujan primes is evidently that \begin{equation} \pi(p) - \pi\!\left(\frac12p\right) + 1 = \pi(q) - \pi\!\left(\frac12q\right). \label{EQ: twin} \end{equation} \noindent To see that the condition is not sufficient, even when $p$ and $q$ are \emph{consecutive primes} $p_k$ and $p_{k+1}$, verify \eqref{EQ: twin} for any one of the pairs \begin{equation} (p,q)=(p_k,p_{k+1})=(\boldsymbol{11},13),\ (\boldsymbol{47},53),\ (\boldsymbol{67},\boldsymbol{71}),\ (109, 113),\ (137,139), \label{EQ: not suff} \end{equation} where Ramanujan primes are in {\bf bold}. It is less evident that \eqref{EQ: twin}~is a necessary condition for $p$ and $q$ even to be (ordinary) twin primes, but that is not hard to prove \cite[Proposition~1]{sondow}. \begin{proposition} \label{PROP: twin} If $p$ and $q=p+2$ are twin primes with $p>5$, then~\eqref{EQ: twin} holds. \end{proposition} The converse is false, even when $p$ and $q$ are consecutive primes both of which are Ramanujan, as the example $(p_{19},p_{20})=(\boldsymbol{67},\boldsymbol{71})=(R_8,R_9)$ shows. As mentioned, each pair in \eqref{EQ: not suff} consists of consecutive primes $p5$ it suffices, by Proposition~\ref{PROP: twin}, to prove part (ii). Since $q=R_{n+1}$, we have $\pi(x) - \pi\!\left(\frac12x\right)\ge n+1$ when $x\ge q$, and \eqref{EQ: twin} implies that $\pi(p) - \pi\!\left(\frac12p\right)=n$. To prove that $p=R_n$, we have to show that \mbox{$\pi(p-1) - \pi\!\left(\frac12(p-1)\right)2/5$, for $x\ge10^5$. Corollary~\ref{COR: upper twin} implies that whether a twin prime pair is counted in $\pi_{2,1}(x)$ or $\pi_{2,2}(x)$ depends on only one of the two primes being Ramanujan. This suggests that the ratios $\pi_{2,1}(x)/\pi_2(x)$ and $\pi_{2,2}(x)/\pi_2(x)$ should approach $1/2$ as $x$ tends to infinity. We conclude this section with these and other conjectures based on our results and on Table~\ref{TABLE: twins}, as well as with two more open problems. \begin{conjecture} \label{CONJ: 1/4} For all $x \ge 10^5$, we have \begin{align*} \label{EQ: ratios} \frac{\pi_{2,1}(x)}{\pi_2(x)} &< \frac45, \qquad \frac{\pi_{2,2}(x)}{\pi_2(x)} > \frac25, \qquad \frac{\pi_{2,2}(x)}{\pi_{2,1}(x)} > \frac{2/5}{4/5}= \frac12. \end{align*} \end{conjecture} \begin{conjecture} \label{CONJ: lim} If $\pi_2(x)\to\infty$ as $x\to\infty$, then $\pi_{2,1}(x) \sim \pi_{2,2}(x) \sim \frac12\pi_2(x)$. \end{conjecture} Recall Brun's famous theorem \cite{brun} that the series of reciprocals of the twin primes converges or is finite (unlike the series of reciprocals of all the primes, which Euler showed diverges). Its sum \cite[Sequence \seqnum {A065421}]{oeis} is \emph{Brun's constant} $B_2$, $$B_2:=\left(\frac13 + \frac{1}{5}\right)+\left(\frac15 + \frac{1}{7}\right) + \left(\frac{1}{11}+ \frac{1}{13}\right)+ \left(\frac{1}{17}+ \frac{1}{19}\right)+\dotsb \stackrel{?}{=} 1.9021605\dotso.$$ Here $\stackrel{?}{=}$ means that the value of $B_2$ is conditional ``on heuristic considerations about the distribution of twin primes'' (Ribenboim \cite[p.~201]{rib89}). \begin{problem} Compute the analogous constant $B_{2,1}$ for twin primes at least one of which is Ramanujan,$$B_{2,1}:= \left(\frac{1}{11}+ \frac{1}{13}\right)+ \left(\frac{1}{17}+ \frac{1}{19}\right)+ \left(\frac{1}{29}+ \frac{1}{31}\right)+ \left(\frac{1}{41}+ \frac{1}{43}\right)+\dotsb.$$ \end{problem} The numbers $11, 17, 29,$ $41,\dotso$ \cite[Sequence \seqnum {A178128}]{oeis} are the lesser of twin primes if at least one is Ramanujan. By Corollary~\ref{COR: upper twin}, that is the same as the lesser of twin primes if it is Ramanujan. \begin{problem} Compute the analogous constant $B_{2,2}$ for twin Ramanujan primes,$$B_{2,2}:= \left(\frac{1}{149}+ \frac{1}{151}\right)+ \left(\frac{1}{179}+ \frac{1}{181}\right)+ \left(\frac{1}{227}+ \frac{1}{229}\right)+ \left(\frac{1}{239}+ \frac{1}{241}\right)+\cdots.$$ \end{problem} The numbers $149, 179, 227,239,\dotso$ \cite[Sequence \seqnum {A178127}]{oeis} are the lesser of twin Ramanujan primes. %Section \section{Prime gaps} \label{SEC: gaps} Let us say that there is a \emph{prime gap from $a$ to~$b\ge a$} if none of the numbers $a,a+1,a+2,\dotsc, b$ is prime. Given a run of $r$ odd Ramanujan primes starting at $p$, we can associate to it a prime gap of length at least $r$ starting at $\frac12(p+1)$. \begin{proposition} \label{PROP: gaps} {\rm(i).} If $p=R_n$ is odd, then the integer $\frac12(p + 1)$ is not prime. \noindent{\rm(ii).} More generally, given a run of $r\ge1$ odd Ramanujan primes from $p=R_n=p_k$ to $q=R_{n+r-1}=p_{k+r-1}$, there is a prime gap from $\frac12(p + 1)$ to $\frac12(q + 1)$. \noindent{\rm(iii).} Parts {\rm (i)} and {\rm (ii)} are sharp in the sense that, for \emph{certain} runs of Ramanujan primes $p$ to $q$, both $\frac12(p + 1)-1$ and $\frac12(q + 1)+1$ are prime numbers. \noindent{\rm(iv).} But in the case $r=2$, if $p$ and $q$ are \emph{twin} Ramanujan primes, then the prime gap from $\frac12(p + 1)$ to $\frac12(q+1)$ always lies in a longer prime gap of length $5$ or more. \end{proposition} \begin{proof}(i). Since $p=R_n$ is odd, $\pi(p)=\pi(p+1)$, and the quantity $\pi(x) - \pi\!\left(\frac12x\right)$ does not decrease from $x=p$ to $x=p+1$. Hence $\pi\!\left(\frac12p\right)\ge\pi\!\left(\frac12(p+1)\right)$, and so $\frac12(p + 1)$ is not prime. \noindent(ii). By (i), the case $r=1$ holds. Taking $r=2$, let $p=R_n=p_k$ and $q=R_{n+1}=p_{k+1}$ be odd. By (i), neither $\frac12(p + 1)$ nor $\frac12(q + 1)$ is prime. If an integer $i$ lies strictly between them, then the oddness of $p$ and $q$ implies $p+1< j:=2i-1< q-1$. Since $p=p_k$ and $q=p_{k+1}$, we have $\pi(p)=k=\pi(j+1)$. As $p=R_n$ and $q=R_{n+1}$, it follows that $\pi(x) - \pi\!\left(\frac12x\right)$ does not decrease from $x=p$ to $x=j+1$. Hence $\pi\!\left(\frac12p\right)\ge\pi\left(\frac12(j+1)\right) = \pi(i)$, and so $i$ is also not prime. This proves (ii) for runs of length $2$. The general case follows easily by induction on $r$. Namely, given a run of length $r>2$ from $R_n=p_k$ to $R_{n+r-1}=p_{k+r-1}$, break it into a run of length $2$ from $R_n=p_k$ to $R_{n+1}=p_{k+1}$, concatenated with a run of length $r-1$ from $R_{n+1}=p_{k+1}$ to $R_{n+r-1}=p_{k+r-1}$. \noindent(iii). For $r=1$, the composite number $\frac12(R_2 + 1)=\frac12(11+1)=6$ lies between the primes $5$ and~$7$. For an example with $r>1$, take the run $(R_{293}, R_{294})=(4919,4931)=(p_{657}, p_{658})$ of length $r=2$. It is associated to the prime gap from $\frac12(R_{293} + 1) = 2460$ to $\frac12(R_{294} + 1)=2466$, which is bounded by the primes $2459$ and $2467$. \noindent(iv). Since $p>3$ and $q$ are twin primes, $(p,q)=(6k-1,6k+1)$ for some $k$. If $k=2i$ is even, then $\left(\frac12(p + 1),\frac12(q+1)\right)=(6i,6i+1)$ lies in the prime gap from $6i$ to $6i+4$. Now assume that $k=2i+1$ is odd. Then $\left(\frac12(p + 1),\frac12(q+1)\right)=(6i+3,6i+4)$ will lie in a prime gap from $6i+2$ to $6i+6$, unless $6i+5$ is prime. But if $6i+5=\frac12(q+3)$ were prime, then, since $q+2=6k+3$ is not prime, $\pi(x) - \pi\!\left(\frac12x\right)$ would decrease from $x=q$ to $x=q+3$, contradicting the fact that $q$ is a Ramanujan prime. This completes the proof. \end{proof} For part (iii), the first ``sharp'' example of a run of length $r=1,2,\dotsc,11$ begins at the Ramanujan prime $$11, 4919, 1439, 7187, 37547, 210143, 3376943, 663563, 4429739, 17939627, 12034427,$$ respectively \cite[Sequence \seqnum {A177804}]{oeis}. An example of part (iv) is the prime gap associated to the twin Ramanujan primes $R_{14}=149$ and $R_{15}=151$, which lies in the larger prime gap $$74,\frac12(R_{14} + 1) = 75,\frac12(R_{15} + 1)=76,77,78.$$ %Appendix \section{Appendix on the algorithm} To compute a range of Ramanujan primes $R_i$ for $1 \le i \le n$, we perform simple calculations in each interval $\left(k/2,k\right]$ for $k=1,2,\ldots,p_{3n}-1$. To facilitate the calculation, we use a counter $s$ and a list $L$ with $n$ elements $L_i$. Initially, $s$ and all $L_i$ are set to zero. They are updated as each interval is processed. After processing an interval, $s$ will be equal to the number of primes in that interval, and each $L_i$ will be equal either to the greatest index of the intervals so far processed that contain exactly $i$ primes, or to zero if no interval having exactly $i$ primes has yet been processed. Having processed interval $k-1$, to find the number of primes in interval $k$ we perform two operations: add 1 to $s$ if $k$ is prime, and subtract 1 from $s$ if $k/2$ is prime. We then update the $s$th element of the list to $L_s=k$, because now $k$ is the largest index of all intervals processed that contain exactly $s$ primes. After all intervals have been processed, the list $R$ of Ramanujan primes is obtained by adding 1 to each element of the list $L$. These ideas are captured in the following \textit{Mathematica} program for finding the first 169350 Ramanujan primes. \begin{center} \begin{tabular}{cccclccc} $$ & & $$ & $$ & $\texttt{nn = 169350; }$ & $$ & \\ $$ & & $$ & $$ & $\texttt{L = Table[0, \{nn\}]; }$ & $$ & \\ $$ & & $$ & $$ & $\indent\texttt{s = 0; }$ & $$ & \\ $$ & & $$ & $$ & $\indent\texttt{Do[ }$ & $$ & \\ $$ & & $$ & $$ & $\hspace{.6cm}\texttt{If[PrimeQ[k], s++]; }$ & $$ & \\ $$ & & $$ & $$ & $\hspace{.6cm}\texttt{If[PrimeQ[k/2], s--]; }$ & $$ & \\ $$ & & $$ & $$ & $\hspace{.6cm}\texttt{If[s < nn, L[[s+1]] = k], }$ & $$ & \\ $$ & & $$ & $$ & $\hspace{.6cm}\texttt{\{k, Prime[3*nn]-1\}]; }$ & $$ & \\ $$ & & $$ & $$ & $\texttt{R = L + 1 }$ & $$ & $$ & \end{tabular} \end{center} Although it is adequate for computing a modest number of them, to compute many more requires a speedup of several orders of magnitude. That can be achieved by using a lower-level programming language and generating prime numbers via a sieve. With this speedup we computed all Ramanujan primes below $10^9$ in less than three minutes on a 2.8 GHz Pentium 4 computer. \section{Acknowledgment} The authors are grateful to Steven Finch for pointing out Schilling's paper \cite{schilling}. \begin{thebibliography}{99} \bibitem{brun}V.~Brun, La s\'erie $ \frac 15 + \frac 17 + \frac {1}{11} + \frac {1}{13} + \frac {1}{17} + \frac {1}{19} + \frac {1}{29} + \frac {1}{31} + \frac {1}{41} + \frac {1}{43} + \frac {1}{59} + \frac {1}{61} +\dotsb$ o\`u les d\'enominateurs sont ``nombres premiers jumeaux'' est convergente ou finie, \textit{Bull. Soc. Math. France} \textbf{43} (1919), 100--104, 124--128. \bibitem{dusart}P.~Dusart, In\'egalit\'es explicites pour $\psi(X)$, $\theta(X)$, $\pi(X)$ et les nombres premiers, \textit{C. R. Math. Acad. Sci. Soc. R. Can.} \textbf{21} (1999), 53--59. \bibitem{laishram}S.~Laishram, On a conjecture on Ramanujan primes, \textit{Int. J. Number Theory} \textbf{6} (2010), 1869--1873; also available at \\ \url{http://www.isid.ac.in/~shanta/PAPERS/RamanujanPrimes.pdf}. \bibitem{ramanujan}S.~Ramanujan, A proof of Bertrand's postulate, \textit{J. Indian Math. Soc.} \textbf{11} (1919), 181--182; also available at \\ \url{http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper24/}. \bibitem{rib89} P.~Ribenboim, \emph{The Book of Prime Number Records}, 2nd. ed., Springer-Verlag, New York, 1989. \bibitem{rosser}J.~B.~Rosser, The $n$th prime is greater than $n \ln n$, \textit{Proc. London Math. Soc.} \textbf{45} (1938), 21--44. \bibitem{schilling}M.~F.~Schilling, The longest run of heads, \textit{College Math. J.} \textbf{21} (1990), 196--207; also available at \\ \url{http://users.eecs.northwestern.edu/~nickle/310/2010/headRuns.pdf}. \bibitem{shapiro} H.~N.~Shapiro, \emph{Introduction to the Theory of Numbers}, Wiley, New York, 1983; reprinted by Dover, Mineola, NY, 2008. \bibitem {oeis}N.~J.~A.~Sloane, {\em \htmladdnormallink {The On-Line Encyclopedia of Integer Sequences}{http://oeis.org}}, published electronically at \url{http://oeis.org}, 2011. \bibitem{sondow}J.~Sondow, Ramanujan primes and Bertrand's postulate, \textit{Amer. Math. Monthly} \textbf{116} (2009), 630--635; also available at \url{http://arxiv.org/abs/0907.5232}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A41. \noindent \emph{Keywords:} prime gap, Ramanujan prime, twin prime. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum {A007508}, \seqnum {A065421}, \seqnum {A104272}, \seqnum {A173081}, \seqnum {A174602}, \seqnum {A174641}, \seqnum {A177804}, \seqnum {A178127}, \seqnum {A178128}, \seqnum {A179196}, \seqnum {A181678}, \seqnum {A189993}, and \seqnum {A189994}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received December 14 2010; revised version received May 11 2011. Published in {\it Journal of Integer Sequences}, May 17 2011. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}