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\begin{center}
\vskip 1cm{\LARGE\bf 
Ramanujan and Labos Primes, Their \\
\vskip .1in
Generalizations, and Classifications of Primes
}
\vskip 1cm
\large
Vladimir Shevelev \\
Department of Mathematics \\
Ben-Gurion University of the Negev\\
Beer-Sheva 84105 \\
Israel  \\
\href{mailto:shevelev@bgu.ac.il}{\tt shevelev@bgu.ac.il}
\end{center}

\vskip .2 in

\begin{abstract}
  We study the parallel properties of the Ramanujan primes and a symmetric counterpart, the Labos primes.
  Further, we study all primes with these properties (generalized Ramanujan and Labos primes) and construct two kinds of sieves for them. Finally, we give a further natural generalization of these constructions and pose some conjectures and open problems.
\end{abstract}

\section{Introduction}

 The very well-known Bertrand postulate (1845) states that, for every $x&gt;1,$ there exists a prime in the interval $(x, 2x).$ This postulate quickly became a theorem when, in 1851, it was unexpectedly proved by Chebyshev (for a very elegant version of his proof, see Theorem 9.2 in \cite{8}). In 1919, Ramanujan 
 (\cite{6}; \cite[pp.\ 208--209]{7}) gave a quite new and very simple proof of Bertrand's postulate. Moreover, in his proof of a \emph{generalization} of Bertrand's, the following sequence of primes appeared (\cite{11}, A104272)
\begin{equation}\label{1.1}
2,11,17,29,41,47,59, 67, 71, 97, 101, 107, 127, 149, 151, 167,\ldots
 \end{equation}

\begin{definition}\label{d1}
For $n \geq 1$, the \emph{$n$-th Ramanujan prime} is the largest prime $(R_n)$ for which $\pi(R_n)-\pi(R_n/2)=n.$
\end{definition}

Let us show that, equivalently, $R_n$ is \emph{the smallest positive integer  $g(n)$  with the property that, if $x\geq g(n),$ then} $\pi(x) -\pi(x/2)\geq n.$
\begin{proof} Indeed, evidently, $g(n)$ is prime. Note that, if for $x&gt;R_n,$ we have $\pi(x)-\pi(x/2)\leq n-1,$ then, evidently, there exists a prime $q&gt;x$ for which $\pi(q)-\pi(q/2)=n.$ This contradicts the maximality of $R_n.$ Thus $g(n)\leq R_n.$ On the other hand, if $g(n)&lt;R_n,$ then for all $x\in[g(n), R_n]$ we have $\pi(x) -\pi(x/2)\geq n.$ Let $p\geq g(n)$ be the nearest prime less than $R_n.$ Then, for $k\geq0,$ $\pi(p)-\pi(p/2)=n+k.$ Since $\pi(R_n)-\pi(R_n/2)=n,$ there exists exactly $k+1$ primes $\nu_1&lt;\cdots&lt;\nu_{k+1}$ in interval $[\frac{p+1}{2}, \frac{q-1}{2}].$ For $x=2\nu_{k+1}\in [p+1,q-1],$ we have
$$\pi(x)-\pi(x/2)=\pi(2\nu_{k+1})-\pi(\nu_{k+1})\leq\pi(p)-\pi(p/2)-k-1=n-1.$$
 The contradiction shows that $g(n)=R_n.$
 \end{proof}

In \cite{12}, Sondow obtained some estimates for $R_n$ and, in particular,  proved that $R_n&gt;p_{2n}$ for every $n&gt;1.$  \; Laishram \cite{3} proved that $R_n&lt;p_{3n}$ (a short proof of this result follows from the more general Theorem \ref{t6} in this paper, see Remark \ref{r2}). Further, Sondow proved that $ R_n\sim p_{2n}$ as $n\rightarrow\infty.$ From this, denoting  the counting function of the Ramanujan  primes by $\pi_R$, we have $R_{\pi_R(x)}\sim2\pi_R(x)\ln \pi_R(x).$ Since $R_{\pi_R(x)}\leq x&lt;R_{\pi_{R(x)}+1},$ we have
 $ x\sim p_{2\pi_R(x)}\sim2\pi_R(x)\ln \pi_R(x)$ as $x\rightarrow\infty,$ and may conclude that
\begin{equation}\label{1.2}
\pi_R(x)\sim \frac {x} {2\ln x}\sim \frac {\pi(x)}{2}.
 \end{equation}
 Below we prove several other properties of the Ramanujan primes. \emph{Everywhere below $p_n$ denotes the $n$-th prime}.  An important role is played by the following property.
\begin{theorem}\label{t1}
If $p$ is an odd Ramanujan prime such that $p_m&lt; p/2&lt; p_{m+1},$ then the interval  $(p,\; 2p_{m+1})$ contains a prime.
\end{theorem}
In 2003, Labos introduced the following sequence of primes (cf. \cite{11}, sequence A080359). We call them {\sl Labos primes}, denoting the $n$-th Labos prime by $L_n.$
 \begin{definition}\label{d2} For $n \geq 1$, the \emph{$n$-th Labos  prime} is the smallest positive integer  $(L_n)$  for which $\pi(L_n)-\pi(L_n/2)=n.$
  \end{definition}
The first Labos primes are (see sequence A080359 in \cite{11}):
\begin{equation}\label{1.3}
2, 3, 13, 19, 31, 43, 53, 61, 71, 73, 101, 103, 109, 113, 139, 157, 173,\ldots
 \end{equation}
Note that, since (see \cite{11})
  \begin{equation}\label{1.4}
\pi(R_n)-\pi(R_n/2)=n,
 \end{equation}
 by Definition \ref{d2} we have
\begin{equation}\label{1.5}
L_n\leq R_n.
 \end{equation}
 Note also that, obviously, $L_n\sim p_{2n}\;\; as \;\; n\rightarrow\infty.$

 For the Labos primes we prove a symmetric statement to Theorem \ref{t1}.
\begin{theorem}\label{t2}
 If $p$ is an odd Labos prime, such that $p_m&lt; p/2&lt; p_{m+1},$ then the interval  $(2p_m,\; p)$ contains a prime.
\end{theorem}

It is clear that Theorems \ref{t1}-\ref{t2}\; are connected with some left-right symmetry in the distribution of primes. Unfortunately, we do not lave a \emph{precise} left-right symmetry since the inequalities of type $R_1\leq L_1\leq R_2\leq L_2\leq \cdots$ are broken from the very outset. In Theorem \ref {t3} we show that this deficiency could be removed by the consideration some additional primes with the close properties. Based on this theorem, we give a natural simple classification of the primes.

\section{Proof of Theorems \ref{t1},\ref{t2}}

We start with four conditions on an odd prime $p.$
\begin{condition}\label{co1}
Let $p=p_n,$ with  $n&gt;1.$  All integers $(p+1)/2, (p+3)/2, \ldots , (p_{n+1}-1)/2 $ are composite numbers.
\end{condition}
\begin{condition}\label{co2}
Let $p\geq5$ and $p_m&lt; p/2&lt; p_{m+1}.$ The interval  $(p,\; 2p_{m+1})$ contains a prime.
\end{condition}
\begin{condition}\label{co3}
Let $p=p_n$ with $n\geq3.$ All integers $(p-1)/2, (p-3)/2, \ldots , (p_{n-1}+1)/2 $ are composite numbers.
\end{condition}
\begin{condition}\label{co4}
Let $p_m&lt; p/2&lt; p_{m+1}.$ The interval  $(2p_m,\; p)$ contains a prime.
\end{condition}
\begin{lemma}\label{l1}
 Conditions $\ref{co1}$ and $\ref{co2}$ are equivalent.
 \end{lemma}
  \begin{proof}  If Condition \ref{co1} is valid, and $p_m&lt; p/2&lt; p_{m+1},$ then $p_{m+1}&gt;(p_{n+1}-1)/2,$ i.e., $p_{m+1}\geq(p_{n+1}+1)/2.$ Thus $2p_{m+1}&gt; p_{n+1}&gt;p_n=p,$ and Condition \ref{co2} is valid. Conversely, let Condition \ref{co2} be satisfied and $p=p_n.$ Then from the condition $p_{m+1}&gt;p/2&gt;p_m$ we have $2p_m&lt;p_n&lt;p_{n+1}&lt;2p_{m+1},$ or $p_m&lt;p_n/2&lt;p_{n+1}/2&lt;p_{m+1}.$ Since the interval $(p_m, p_{m+1})$ contains no primes, the interval $(p_n/2,p_{n+1}/2)\subset(p_m, p_{m+1})$ also contains no primes, and Condition \ref{co1} follows.
  \end{proof}

Analogously, we obtain the equivalence of the second pair of conditions.
\begin{lemma}\label{l2}
 Conditions $\ref{co3}$ and $\ref{co4}$ are equivalent.
 \end{lemma}
 Now we are able to prove Theorems \ref{t1}-\ref{t2}.
 \begin{proof} In view of Lemma \ref{l1}, to prove of Theorem 1 it is sufficient to prove that, for Ramanujan primes, Condition \ref{co1} is satisfied. If Condition \ref{co1} is not satisfied, then suppose that $p_m=R_n&lt;p_{m+1}$ and  $k$ is the least positive integer such that $q=(p_m+k)/2$ is a prime not exceeding $(p_{m+1}-1)/2.$ Thus
  \begin{equation}\label{2.1}
  R_n=p_m&lt;2q&lt;p_{m+1}-1.
 \end{equation}
 From Definition \ref{d1} it follows that $R_n-1$ is the maximum integer
 for which the equality
  \begin{equation}\label{2.2}
 \pi(R_n-1)-\pi((R_n-1)/2)=n-1
 \end{equation}
 holds. However, according to (\ref{2.1}), $\pi(2q)= \pi(R_n-1)+1,$ and in view of the minimality of the prime $q,$ in the interval $((R_n-1)/2,q)$ there is no prime. Thus $\pi(q)=\pi((R_n-1)/2)+1$ and
 $$ \pi(2q)-\pi(q)=\pi(R_n-1)-\pi((R_n-1)/2)=n-1.$$
 Since, by (\ref{2.1}), $2q&gt;R_n,$ this contradicts the property of maximality of $R_n$ in (\ref{2.2}). Thus Theorem \ref{t1} follows. Theorem \ref{t2} is proved quite analogously, using Lemma \ref{l2}.
 \end{proof}
 \section{Pseudo-Ramanujan primes, over-Ramanujan primes and their Pseudo-Labos and over-Labos analogues}
 \begin{definition}\label{d3}
 Non-Ramanujan primes satisfying Condition $\ref{co2}\; ($or, equivalently, Condition $\ref{co1} )$ are \emph{pseudo-Ramanujan primes}.
  \end{definition}
Denote the sequence of pseudo-Ramanujan primes by $\{R^*_n\}$ (see sequence $({1.7})).$
 The first pseudo-Ramanujan primes are (see sequence A164288 in \cite{11}):
$$ 109, 137, 191, 197, 283, 521, 617, 683, 907, 991, 1033, 1117, 1319,\ldots$$
\begin{definition}\label{d4}
   An \emph{over-Ramanujan prime} is a prime satisfying Condition $\ref{co2}\; ($or, equivalently, Condition $\ref{co1}).$
  \end{definition}
  Denote the sequence of over-Ramanujan primes by $\{R'_n\}$ (see sequence A164368 in \cite{11}).
  Note that all Ramanujan primes greater than 2 are also over-Ramanujan primes. It is easy to see that $R'_1=11.$ Furthermore, let us prove the following simple criterion.
 \begin{proposition}\label{p1}
$p_n\geq5$ is an over-Ramanujan prime if and only if $\pi(\frac {p_n} {2})=\pi(\frac {p_{n+1}} {2}).$
 \end{proposition}
    \begin{proof}  1) Let $\pi(\frac {p_n} {2})=\pi(\frac {p_{n+1}} {2}).$ From this it follows that, if $p_k&lt; p_n/2&lt;p_{k+1},$ then there are no primes between $p_n/2$ and $p_{n+1}/2.$ Thus $p_{n+1}/2&lt;p_{k+1}$ as well. Therefore, we have $2p_k&lt; p_n&lt;p_{n+1}&lt;2p_{k+1},$ i.e., $p_n$ is an over-Ramanujan prime. Conversely, if $p_n$ is an over-Ramanujan prime, then $2p_k&lt; p_n&lt;p_{n+1}&lt;2p_{k+1},$ and $\pi(\frac{p_n} {2})=\pi(\frac{p_{n+1}} {2}).$
    \end{proof}
     \begin{definition}\label{d5}
     Non-Labos primes satisfying Condition $\ref{co4}\; ($or, equivalently, Condition $\ref{co3})$ are \emph{pseudo-Labos primes}.
  \end{definition}
Denote the sequence of pseudo-Labos primes by $\{L^*_n\}.$
The first pseudo-Labos primes are (A164294 in \cite{11}):
$$ 131, 151, 229, 233, 311, 571, 643, 727, 941, 1013, 1051, 1153, 1373,\ldots$$
\begin{definition}\label{d6}
 An \emph{over-Labos prime} is a prime satisfying Condition $\ref{co4}\; ($or, equivalently, Condition $\ref{co3}).$
  \end{definition}
  Denote the sequence of over-Labos primes by $\{L'_n\}$ (see sequence A194598 in \cite{11}).
   Note that all Labos primes grater than 3 are also over-Labos primes. It is easy to verify that $L'_1=13.$
 Analogously to Proposition \ref{p1} we obtain the following criterion for over-Labos primes.
 \begin{proposition}\label{p2}
$p_n\geq5$ is an over-Labos prime if and only if $\pi(\frac {p_{n-1}} {2})=\pi(\frac {p_{n}} {2}).$
 \end{proposition}

\begin{theorem}\label{t3}
Consider ALL primes $\{R'_n\}$ and $\{L'_n\}$ for which Theorems $\ref{t1}$-$\ref{t2}$ are correspondingly true, then
\begin{equation}\label{3.1}
R'_1\leq L'_1\leq R'_2\leq L'_2\leq\cdots
 \end{equation}
\end{theorem}
 \begin{proof} Note that intervals of the form $(2p_m,\; 2p_{m+1})$ containing not more than one prime, contain neither over-Ramanujan nor over-Labos primes. If an interval contains only two primes, then the first prime is an over-Ramanujan prime $(R'),$ while the second one is an over-Labos prime $(L'), $ and we see that $R'&lt;L'; $ on the other hand, if it contains $k$ primes, then beginning with the second one and up to the $(k-1)$-st we have primes which are simultaneously over-Ramanujan and over-Labos primes. Thus, taking into account that the last prime is only an over-Labos prime we have for this interval
 $$R'_1&lt;L'_1=R'_2&lt;L'_2=R'_3&lt;\cdots&lt;L'_{k-1}=R'_{k-1}&lt;L'_{k}.$$
 The following interval containing at least two primes begins with an over-Ramanujan prime and the process repeats. \end{proof}
 \section{On difference $R_n-L_n$}
 Consider positive records of the difference $R_n-L_n.$ They are (A182366 in \cite{11})
 \begin{equation}\label{4.1}
 8, 10, 24, 36, 60, 64, 84, 114, 124, 144, 202, 226, 228,\ldots
 \end{equation}
 at
  $$n=2,4,10,14,43, 95, 145, 167, 287, 415, 560, 635, 982,\ldots\; . $$
  However, we do not know a proof that $\limsup_{n\rightarrow\infty} (R_n-L_n)=\infty.$ We prove a weaker statement.
\begin{proposition}\label{p3} $\limsup_{n\rightarrow\infty} (R_{n+1}-L_n)=\infty.$
 \end{proposition}
 \begin{proof} If there exists a constant $C,$ such that $R_{n+1}-L_n\leq C,$ then for every $n\geq1,$ there exist not more than $C$ intervals $(2,4),(3,6),\ldots$ containing exactly $n$ primes. Therefore, we have $\leq C$ of the first intervals containing one prime, $\leq 2C$ of the first intervals containing one or two primes,\ldots, $\leq nC$ of the first intervals containing not more than $n$ primes. Hence interval $((n+1)C, 2(n+1)C$ contains more than $n$ primes. However, for large $n$ it is impossible, since the number of primes in the interval $(1,N)$ for $N=2(n+1)C$ is equivalent to $N/\ln N=o(n).$
 \end{proof}

Now consider the following problem. Let us call a prime $p$ \emph{compatible} with another prime $q,$ if the intervals $(p/2,q/2)$ and $(p,q],$ if $q&gt;p,$ (or intervals $(q/2,p/2)$ and $(q,p\;],$ if $q&lt;p)$ contain the same number of primes. It is clear that, if $p$ compatible with $q,$ then $q$ compatible with $p.$ If $p$ is compatible with no other prime, we call it a \emph{peculiar} prime. It is required to describe the peculiar primes. We give a solution of this problem in the following form.
\begin{proposition}\label{p4} A prime $p$ is peculiar if and only if it is simultaneously Ramanujan and Labos prime.
\end{proposition}
 \begin{proof} Let $p=L_n=R_n$ (a case $L_{n-1}=R_n,$ evidently, is impossible). Then, by the Definitions \ref{d1}-\ref{d2}, $p$ is the smallest and the largest prime for which $\pi(p)-\pi(p/2)=n.$ Thus the difference $\pi(x)-\pi(x/2)=n,$ where $x$ is prime, occurs only once. However, if there exists $q\neq p$ for which $\pi(q)-\pi(q/2)=\pi(p)-\pi(p/2)=n,$ then the difference $\pi(x)-\pi(x/2)=n$ with a prime $x$ occurs at least twice. The contradiction shows that $p$ is a peculiar prime. Conversely, if a prime $p$ is peculiar, then,
 for any $q\neq p,$ we have $\pi(p/2)-\pi(q/2)\neq\pi(p)-\pi(q)$ and, consequently, $\pi(q)-\pi(q/2)\neq \pi(p)-\pi(p/2).$ Thus the difference $\pi(x)-\pi(x/2)=\pi(p)-\pi(p/2),$ where $x$ is prime, appears only once. This means that $L_n=R_n=p,$ where $n=\pi(p)-\pi(p/2).$
 \end{proof}
 Thus the peculiar primes are (A164554 in \cite{11})
\begin{equation}\label{4.2}
2, 71, 101, 181, 239, 241, 269, 349, 373, 409, 419, 433, 439, 491, \ldots
\end{equation}
\begin{corollary}\label{C1}
If there exist infinitely many peculiar primes, then $\liminf_{n\rightarrow\infty} (R_n-L_n)=0.$
\end{corollary}
  \section{Prime gaps}
 Note that, as it follows from Lemma \ref{l1} and Theorem \ref{t1}, if we consider a run of consecutive Ramanujan primes $p=R_l,\ldots,q=R_k,$ then the interval $[\frac{1}{2}(p+1),\; \frac{1}{2}(q+1)]$ is free from primes. However, this note is far from a complete characterization of the prime gaps. For example, we have a run $\{2521,2531\}$ of consecutive Ramanujan primes which gives a ``prime gap" $[\frac{2521+1}{2},\; \frac{2531+1}{2}]=[1261,1266].$ However, the real prime gap is much larger: (1259,1277). A better result can be obtained using over-Ramanujan primes. Indeed, the considered property of Ramanujan primes is valid for all over-Ramanujan primes, while runs of consecutive over-Ramanujan primes, generally speaking, are longer.
 For example, instead of the run $\{2521,2531\}$ of Ramanujan primes, we have the run $\{2521,2531,2539,2543,2549\}$ of over-Ramanujan primes. This gives the interval $[\frac{2521+1}{2},\enskip \frac{2549+1}{2}]=[1261,1275],$ which is free from primes and very close to the real gap. In general, since  over-Ramanujan primes satisfy Condition 1, to every run of consecutive over-Ramanujan primes $p=R'_l,\ldots,q=R'_k$ there corresponds the interval $[\frac{1}{2}(p+1),\; \frac{1}{2}(q+1)]$ which contains no primes. Note that the prime $q'$ following $q$ gives an additional improvement of the lower estimate of size $(l)$ of the considered prime gap. Indeed, we know that $q'$ is necessarily an over-Labos prime. Since the over-Labos primes satisfy Condition 3, all numbers $\frac{q'-1}{2}, \; \frac{q'-3}{2},\ldots,\frac {q+1}{2}$ are composite. Hence $l\geq\frac{q'-p}{2}.$
For example, consider the run $\{227,229,233,239,241\}$ of over-Ramanujan primes (all of which are Ramanujan). The following prime is $q'=251.$ Thus, for the gap containing $(227+1)/2=114$ we have $l\geq\frac {251-227}{2}=12$ (the exact value of $l$ here is 14).
   \section{The first sieve for the selection of the over-Ramanujan primes from all primes}

 Recall that Bertrand's sequence $\{b(n)\}$ is defined by $ b(1)=2,$ and, for $n\geq2,\; b(n)$ is the largest prime less than $2b(n-1)$ (see A006992 in \cite{11}):
\begin{equation}\label{6.1}
2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 5003,\ldots
 \end{equation}
Put
\begin{equation}\label{6.2}
B_0=\{b^{(0)}(n)\}=\{{b(n)}\}.
 \end{equation}
Further we build sequences $B_1=\{b^{(1)}(n)\}, B_2=\{b^{(2)}(n)\},\ldots$ according to the following inductive rule: if the sequences $B_0,\ldots,B_{k-1}$ have been defined already, let us consider the minimal prime $p^{(k)}\not \in \bigcup_{i=1}^{k-1}B_i.$ Then the sequence $\{b^{(k)}(n)\}$ is defined by $b^{(k)}(1)=p^{(k)},$ and, for $n\geq2,\; b^{(k)}(n) $ is the largest prime less than $2b^{(k)}(n-1).$ Consequently:
\begin{equation}\label{6.3}
B_1=\{11, 19, 37, 73,\ldots\},
 \end{equation}
 \begin{equation}\label{6.4}
B_2=\{17, 31, 61, 113,\ldots\},
 \end{equation}
 \begin{equation}\label{6.5}
B_3=\{29, 53, 103, 199,\ldots\},
 \end{equation}
 etc., such that, putting $p^{(1)}=11,$ we obtain the sequence
 \begin{equation}\label{6.6}
\{p^{(k)}\}_{k\geq1}=\{11, 17, 29, 41, 47, 59, 67, 71, 97, 101, 107, 109, 127, \ldots\}.
 \end{equation}
 Sequence (\ref{6.6}) coincides with sequence (\ref{1.1}) of Ramanujan primes from the second term up to the $12$-th term, but the $13$-th term of this sequence is $109 $ which is the first term of the pseudo-Ramanujan primes.
 \begin{theorem}\label{t4} For $n\geq1:$
 \begin{equation}\label{6.7}
p^{(n)}=R'_n.
 \end{equation}
 \end{theorem}
 \begin{proof} The least omitted prime in (\ref{6.1}) is $p^{(1)}=11=R'_1$;  the least omitted prime in the union of (\ref{6.2}) and (\ref{6.3}) is $p^{(2)}=17=R'_2.$ We use induction. Suppose we have already defined the primes $$p^{(1)}=11, p^{(2)},\ldots,p^{(n-1)}=R'_{n-1}.$$
 Let $q$ be the least prime which is omitted in the union $\bigcup_{i=1}^{n-1}B_i,$  such that $q/2$ is in the interval $(p_{m}, p_{m+1}).$ According to our algorithm, $q$ which is dropped should not be the largest  prime in the interval $(p_{m+1}, 2p_{m+1}).$ Then there are primes in the interval  $(q, 2p_{m+1})$; let $r$ be one of them. We have $2p_m&lt;q&lt;r&lt;2p_{m+1}.$ This means that $q,$ in view of its minimality among the dropped primes which are more than $R'_{n-1}=p^{(n-1)},$ is the least over-Ramanujan prime larger than $R'_{n-1}$ and  the least prime of the form $p^{(k)}$ larger than $p^{(n-1)}$. Therefore, $q=p^{(n)}=R'_{n}.$
 \end{proof}
  Analogously, using sequence $\{c_n\}$ defined by $c(1)=2,$ and, for $n\geq2,\; c(n)$ is the smallest prime more than $2c(n-1)$ (see A055496 in \cite{11}), one can construct a sieve for over-Labos primes.


 \section{The second sieve for the selection of the over-Ramanujan primes from all primes}
  Theorem \ref{t3} on the precise symmetry between distributions of the sequences $\{R'_n\}$ and $\{L'_n\}$ allows us to construct the second  sieve for over-Ramanujan primes. 

Consider consecutive  intervals of the form $(2p_n,\; 2p_{n+1}),\; n=1,2,\ldots$ Remove all the intervals containing less than two primes. For every remaining interval, we write the primes (in increasing order) except for the last one. Then all remaining primes are over-Ramanujan. Indeed, by the definition, an over-Ramanujan prime cannot be the last prime in every such interval. Let us demonstrate this sieve. For the primes 2,3,5,7,11,\ldots consider the intervals
 \begin{equation}\label{7.1}
  (4,6),(6,10),(10,14),(14,22),(22,26),(26,34), (34,38),\ldots\;.
  \end{equation}
Remove those intervals containing less than two primes. We have the following sequence of intervals:
\begin{equation}\label{7.2}
 (10,14),(14,22),(26,34,),(38,46),(46,58),(58,62),(62,74),\ldots\;.
  \end{equation}
 Now we write all primes from these intervals, excluding the \emph{last} primes. Then we obtain sequence (\ref{6.6}).
   Analogously we obtain the second sieve for over-Labos primes. This sequence can be obtained in a parallel way, since, by definition, an over-Labos prime cannot be the first prime in any interval of the considered form. Therefore, here, for every remaining interval, we write the primes (in increasing order) except of the \emph{first} one. Then all remaining primes are over-Labos (cf. sequence A164333 in \cite{11}):
 \begin{equation}\label{7.3}
13, 19, 31, 43, 53, 61, 71, 73, 101, 103, 109, 113, 131, 139, 151, 157,\ldots\;.
 \end{equation}


\section{A classification of primes}

 In connection with the considered construction, let us consider the following classification of primes.

  1) The first two primes 2,3 form a separate set of primes.
 2) If $p\geq11$ is an over-Ramanujan but not an over-Labos prime, then, in connection with the second sieve, we call $p$ a \emph{right prime} (cf. A166307 in \cite{11}):
$$11, 17, 29, 41, 47, 59, 67, 97, 107, 127, 137, 149, 167, 179, 197, 227,\ldots\;.  $$
3) If $p\geq5$ is an over-Labos but not an over-Ramanujan prime, then we call $p$ a \emph{left prime}
 The first terms of this sequence are (cf. A182365 in \cite{11}):
$$13, 19, 31, 43, 53, 61, 73, 103, 113, 131, 139, 157, 173, 193, 199, 251,\ldots\;.  $$
4) If $p$ is simultaneously an over-Ramanujan and an over-Labos prime, then we  call it a \emph{central prime} (sequence A166252 in \cite{11}):
$$71, 101, 109, 151, 181, 191, 229, 233, 239, 241, 269, 283, 311, 349,\ldots\;.  $$
Note that, by Proposition \ref{p4}, every peculiar prime more than 2 is central. Conversely is not true. The non-peculiar central primes are (cf. A182451 in \cite{11}):
$$109, 151, 191, 229, 233, 283, 311, 571, 643, 683, 727, 941, 991, 1033, \ldots$$
5) Finally, the rest of the primes are naturally called \emph{isolated primes} (sequence A166251 in \cite{11}):
$$5, 7, 23, 37, 79, 83, 89, 163, 211, 223, 257, 277, 317, 331, 337, 359,\ldots\;.  $$
 Note that from the second sieve the following results follow.
 \begin{proposition}\label{p5}
Let $l_n,\; r_n$ denote the $n$-th left prime and the $n$-th right prime, respectively. Then $l_n\sim r_n$ as
$n\rightarrow\infty.$
 \end{proposition}
  \begin{proof} The prime number theorem yields $p_{n+1}-p_n=o(p_n).$ Since, by the construction, $l_n$ and $r_n$ belong to the same interval of the form $(2p_{m(n)},\; 2p_{m(n)+1}),$ we have $r_n-l_n&lt;2(p_{m(n)+1}-p_{m(n)})=o(l_n),$ and the statement follows.
 \end{proof}
   \begin{proposition}\label{p6}
The sequence of lengths of the runs of the consecutive isolated primes is unbounded if and only if there exist arbitrary long sequences of consecutive primes $p_k, p_{k+1},\ldots,p_m,$ such that every interval $(\frac {p_i}{2}, \frac {p_{i+1}}{2}),\; i=k,k+1,\ldots,m-1,$ contains a prime.
 \end{proposition}
 \begin{proof} Since the isolated primes more than 3 are the only primes which are neither over-Labos nor over-Ramanujan, then the statement follows from the condition and Propositions \ref{p1},\ref{p2}.
 \end{proof}


\section{On density of over-Ramanujan and over-Labos primes}
Let $\pi_{R'}(x)$ be the counting function of over-Ramanujan numbers not exceeding $x.$ It follows from (\ref{1.2}), Theorem \ref{t1} and Definition \ref{d4} that, if $\lim_{n\rightarrow\infty}\pi_{R'}(x)/\pi(x)$ exists, then it is more than or equal to $\frac{1}{2}.$ Berend \cite{2} beautifully shown that this fact follows from Definition \ref{d4} only.
\begin{proposition}\label{p7} $($\cite{2}$)$
$$\liminf_{n\rightarrow\infty}\pi_{R'}(x)/\pi(x)\geq \frac{1}{2}.$$
\end{proposition}
 \begin{proof} In the range from 7 up to n there are $\pi(n)-3$ primes. Put
$$ h=h(n)=\pi(n/2)-2.$$
 Then $p_{h+2}\leq n/2$ and interval $(p_{h+2}, n/2]$ is free from primes. Furthermore, consider intervals
$$(2p_2, 2p_3), (2p_3, 2p_4), \ldots , (2p_{h+1}, 2p_{h+2}).$$
Our $\pi(n)-3$ primes are somehow distributed in these $h$
intervals. Suppose $k=k(n)$ of these intervals contain at least one prime and
$h-k$ contain no primes. Then, for exactly $k$ primes, there is no primes
between them and the next $2p_j,$ and for the other $\pi(n)-3-k $ there is. Therefore, since $k(n)\leq h(n)\leq \pi(n/2)-2,$ then,
for $\varepsilon&gt;0$ and $n&gt;n_{\varepsilon},$ we have:
$${\frac {\pi(n)-k(n)} {\pi(n)}}\geq {\frac {\pi(n)- \pi(n/2)} {\pi(n)}}\geq\frac{1}{2}-\varepsilon.$$
\end{proof}
 Unfortunately, the analysis of the sieves obtained in section 7 seems much more difficult than the analysis of the sieve of Eratosthenes for primes. Nevertheless, some very simple probabilistic arguments lead to a very plausible conjecture about the density of over-Ramanujan and over-Labos primes. First of all, let us show that events $R':$ ``a prime is over-Ramanujan" and $L':$ ``a prime is over-Labos" are independent.
  \begin{proof}Indeed, denoting events $r:$ ``a prime is right", $l:$ ``a prime is left" and $Is:$ ``a prime is isolated", we have
 \begin{equation}\label{9.4}
 \mathrm{P}[R' | L']=1-\mathrm{P}[l]-\mathrm{P}[Is]; \;\mathrm{P}[R' |\overline{L'} ]=1-\mathrm{P}[r]-\mathrm{P}[Is].
 \end{equation}
 Hence, in view of $\mathrm{P}[l]=\mathrm{P}[r]$ (cf. Proposition 5), we have
 \begin{equation}\label{9.5}
 \mathrm{P}[R' | L']=\mathrm{P}[R' |\overline{L'} ].
 \end{equation}
\end{proof}
\begin{conjecture}\label{con1}
\begin{equation}\label{9.1}
\pi_{R'}(x)\sim (1-e^{-1})\pi(x)=0.63212\cdots\pi(x).
 \end{equation}
\end{conjecture}
 The following proof is heuristic.
 \begin {proof} Consider asymptotically $\frac{\pi(n)}{2}$ intervals of the form $(2p_m,\; 2p_{m+1})$ covering all $\pi(n)$ primes. It is well known (\cite{5}) that, for large $n,$ an interval between two random consecutive primes on the average has length $\ln p_n.$ Thus a random interval of the considered form has length $2\ln p_n$ and, according to the Cram\'{e}r model, the number of primes in such a random interval has the binomial $(2\ln p_n,\; \frac {1} {\ln p_n})$ distribution which, for large $\ln p_n,$ has a good approximation by a Poisson distribution with parameter $\lambda=2.$ Thus we accept that a random interval contains $k$ primes with probability $\mathrm{P}[X=k]=\frac {2^k}{k!}e^{-2},\; k=0,1,2\ldots\; .$ Since $\mathrm{P}[X=0]=e^{-2},$ then the number of intervals containing at least one prime is about $\frac{\pi(n)}{2}(1-e^{-2}).$
This number corresponds to our condition, since we consider \slshape only \upshape such intervals. Furthermore, since $\mathrm{P}[X=1]=2e^{-2},$ then the probability that such an interval contains an only prime is $2e^{-2}/(1-e^{-2})$ and, consequently, we have about $\frac{\pi(n)}{2}(1-e^{-2})\cdot\frac {2e^{-2}}{(1-e^{-2})}=\pi(n)e^{-2}$ intervals
containing, by our terminology, isolated primes and this number coincides with the number of isolated primes. This means that the probability that a prime is isolated is $e^{-2}.$ On the other hand, this probability equals $\mathrm{P}[\overline{R'} \cdot\overline{L'}]=(\mathrm{P}[\overline{R'}])^2.$ Therefore, $\mathrm{P}[\overline{R'}]=e^{-1}$ and $\mathrm{P}[R']=1-e^{-1}$ which justifies the conjecture.
\end{proof}
  Greg Martin \cite{4} did the corresponding calculations for the first million primes $p$, and found that approximately $61.2\% $ of them have a prime in the interval $(p, 2p_{n+1}).$ Since in this case $\ln p_n$ is small (less than 17), an error of about $2\% $ is quite acceptable.
 \begin{remark}\label{r1}
 It could be done also the following simple explanation of appearance of the constant $1-1/e$ in Conjecture $\ref{con1}.$ Consider a random prime $p.$ Let it be in interval $(2p_n, 2p_{n+1}).$ We accept that $p$ could be to the left or to the right from the midpoint $p_n+p_{n+1}$ with the same frequency. Thus the mathematical expectation of the distance between $p$ and $2p_{n+1}$ is the difference $p_{n+1}-p_n$ which in average is $\ln n.$ Accepting for large $n$ that the frequency of appearance a prime approximately is $\frac{1}{\ln n},$ we see that it is natural to accept that the frequency of the appearance of a prime to the right from $p$ in the considered interval is close to $1-(1-\frac{1}{\ln n})^{\ln n}$ which for large $n$ is close to $1-e^{-1}.$ Thus the frequency that a random $p$ is over-Ramanujan is $1-e^{-1}.$
 \end{remark}

Note that, if Conjecture \ref{con1} is true, then, using (\ref{1.2}), for the counting function $\pi_{R^*}(x)$ of pseudo-Ramanujan primes we have
 \begin{equation}\label{9.2}
\pi_{R^*}(x)\sim (\frac {1}{2}-\frac {1}{e})\pi(x)=0.13212...\pi(x),
 \end{equation}
 so that the proportion of Ramanujan primes among all over-Ramanujan primes is approximately 0.79099. Using Theorem \ref{t3}, we note that, if Conjecture \ref{con1} is true, then, for the counting function $\pi_{L'}(x)$ of over-Labos primes we have
  \begin{equation}\label{9.3}
\pi_{L'}(x)\sim \pi_{R'}(x)\sim (1-e^{-1})\pi(x).
 \end{equation}

  Therefore, if Conjecture \ref{con1} is true, then, for the counting functions $\pi_l(x),\;\pi_r(x),\pi_c(x)$ and $\pi_{is}(x) $ of the left, right, central and isolated primes, respectively, of our classes of primes, we have
  \begin{equation}\label{9.6}
\pi_l(x)\sim\pi_r(x)\sim (1-e^{-1})e^{-1}\pi(x)=0.2325\cdots\pi(x),
 \end{equation}
  \begin{equation}\label{9.7}
\pi_c(x)\sim(1-e^{-1})^2\pi(x)=0.3995\cdots\pi(x),
 \end{equation}
 \begin{equation}\label{9.8}
  \pi_{is}(x)\sim e^{-2}=0.1353\cdots\pi(x),
 \end{equation}
 so that $\pi_r(x)+\pi_l(x)+\pi_c(x)+\pi_{is}(x)=\pi(x).$
\section{A generalization}

Let us consider a natural generalization of Ramanujan primes.
\begin{definition}\label{d7}
 For a real $v&gt;1,$ a \emph{$v$-Ramanujan prime} is the largest prime $(R_v(n))$  for which $\pi(R_v(n))-\pi(R_v(n/v))=n.$
\end{definition}
As in case $v=2,$ equivalently $R_v(n)$  is the  \emph{smallest integer with the property: if  $x \geq R_v(n),$ then} $\pi(x) - \pi(x/v) \geq n.$ {\tt}
 Note that, evidently,
\begin{equation}\label{10.1}
R_v(n)\sim p_{((v/(v-1))n)}
\end{equation}
 as $\rightarrow\infty.$ Let $\pi_R^{(v)}(x)$ be the counting function of $v$-Ramanujan primes. Then (cf. (\ref{1.2}))
 \begin{equation}\label{10.2}
 \pi_R^{(v)}(x)\sim(1-1/v)\pi(x).
\end{equation}
Put
 \begin{equation}\label{10.3}
 \kappa(v)=\begin{cases}0,&amp; 
 \text{if $v$ is not the ratio between two primes}; \\
r,&amp; \text{if $v=\frac{r}{q}$ where $r$ and $q$ are primes}.
\end{cases}
\end{equation}
The following theorem is proved in the same way as Theorem 1.
\begin{theorem}\label{t5}
Let $v&gt;1$ be a given real number. If $p&gt;\max(2v,\; \kappa(v))$ is  $v$-Ramanujan prime such that $p_m&lt; p/v&lt; p_{m+1},$ then the interval  $(p,\; \lceil vp_{m+1}\rceil+\varepsilon)$ contains a prime.
\end{theorem}
\begin{remark}\label{r2}
The condition $p&gt;\max(2v,\; \kappa(v))$ allows us to avoid the cases $p=2v$ and $p=vq$ with a prime $q,$ when the condition $p_n&lt; p/v&lt; p_{n+1}$ is impossible.
\end{remark}
Let us find an upper bound on the $n$-th $v$-Ramanujan prime.
\begin{theorem}\label{t6} If $n\geq\frac{1}{k}\max (6k,\; e^v,\; v^{(0.79677\frac{k-1}{k}v-1)^{-1}}),$ then, for $v\geq1.25507\frac{k}{k-1},$ we have
\begin{equation}\label{10.4}
R_v(n)\leq p_{kn}.
\end{equation}
\end{theorem}
\begin{proof} It is sufficient to show that $\pi(\frac{p_{kn}}{v})\leq(k-1)n.$ Indeed, then we have
$\pi(p_{kn})-\pi(\frac{p_{kn}}{v})\geq kn-(k-1)n=n.$ We use the following known results (\cite{1}, \cite {9}-\cite{10}):
\begin{equation}\label{10.5}
p_n&lt;n\ln n+n\ln\ln n,\; n\geq6;
\end{equation}
\begin{equation}\label{10.6}
p_n&gt;n\ln n;
\end{equation}
\begin{equation}\label{10.7}
\pi(x)&lt;1.25506 \frac {x} {\ln x},\; x&gt;1.
\end{equation}
Note that $\frac {p_{kn}}{v}&gt;\frac {kn}{v}&gt;\frac{kn}{e^v}.$  Hence, by the condition, $\frac{p_{kn}}{v}&gt;1.$
 By (\ref{10.5})-(\ref{10.7}),
$$\pi(\frac{p_{kn}}{v})&lt;1.25506\frac {p_{kn}}{v \ln (\frac {p_{kn}}{v})}&lt;
 1.25506\frac{kn}{v}\cdot\frac {\ln(kn)+\ln(\ln(kn))}{\ln (\frac {kn\ln(kn)}{v})}$$
 $$=1.25506\frac{kn}{v}(1+\frac {\ln v}{\ln (\frac {kn\ln(kn)}{v})}).$$
Taking into account that, by the condition, $\ln (kn)&gt;v,$ we have
$$\pi(\frac{p_{kn}}{v})&lt;1.25506\frac{kn}{v}(1+\frac {\ln v}{\ln (kn)}).$$
Finally, note that, by the condition, $\frac {\ln v}{\ln (kn)}\leq 0.7968\frac{k-1}{k}v-1.$ Therefore,
$$\pi(\frac{p_{kn}}{v})&lt;1.25506\cdot0.79677(k-1)n&lt;(k-1)n. $$
\end{proof}
\begin{corollary}\label{C2}
\begin{equation}\label{10.8}
R_3(n)&lt;p_{2n},\; n\geq1.
\end{equation}
\begin{equation}\label{10.9}
R_{1.8}(n)&lt;p_{4n},\; n\geq1.
\end{equation}
\end{corollary}
\begin{proof} By Theorem \ref{t6}, for $v=3,\; k=2,$  we get the required inequality for $n\geq279.$ Using a computer verification for $n&lt;279,$ we obtain (\ref{10.8}). In case  $v=1.8,\; k=4,$ we get the required inequality for $n\geq2370.$ Using a computer verification for $n&lt;2370,$ we obtain ({10.9}).
\end{proof}
 The first $1.8$-Ramanujan primes are
\begin{equation}\label{10.10}
2,11,17,37,43,59,61,79,97,101,103,137,163,167,191,211,\ldots \;.
\end{equation}
\begin{remark}\label{r3}
In case $v=2,\; k=3,$ by Theorem $6,$ we find that $R_n=R_2(n)&lt;p_{3n},$ for $n\geq22398.$ Computer verification for $n&lt;22398$ leads to Laishram's result \cite{3}.
\end{remark}
\begin{definition}\label{d8}
  A prime $p&gt;\max(2v,\; \kappa(v))$ is a \emph{$v$-over-Ramanujan prime} if, as soon as $p_m&lt; p/v&lt; p_{m+1},$ the interval  $(p,\; vp_{m+1})$ contains a prime.
 \end{definition}
\begin{definition}\label{d9}
 A $v$-over-Ramanujan not $v$-Ramanujan prime is a \upshape $v$-pseudo-Ramanujan prime.
 \end{definition}
\emph{Now $v$-Labos primes, $v$-over-Labos primes and $v$-pseudo-Labos primes are introduced quite symmetrically} (see Section 3). In particular, the following statements are valid.
\begin{theorem}\label{t7}
 Let $v&gt;1$ be a given real number. If $p&gt;\max(2v,\; \kappa(v))$ is $v$-Labos prime, such that $p_m&lt; p/v&lt; p_{m+1},$ then the interval  $(\lfloor vp_m\rfloor-\varepsilon,\; p)$ contains a prime.
\end{theorem}
\begin{theorem}\label{t8}
For the sequences $\{R'_v(n)\}$ and $\{L'_v(n)\}$ of $v$-over-Ramanujan and $v$-over-Labos primes, we have
\begin{equation}\label{10.11}
R'_v(1)\leq L'_v(1)\leq R'_v(2)\leq L'_v(2)\leq\cdots
 \end{equation}
\end{theorem}
A generalization of the first sieve for $v$-over-Ramanujan primes, $v\geq2,$ is based on the Bertrand-like sequence $\{b_v(n)\},$ defined by $ b_v(1)=2,$ and, for $n\geq2,$ as the largest prime less than $\lceil vb_v(n-1)\rceil+\varepsilon.$ A generalization of the second sieve for $v$-over-Ramanujan primes is based on the sequence of intervals
\begin{equation}\label{10.12}
(\lfloor2v\rfloor-\varepsilon,\lceil3v\rceil+\varepsilon),(\lfloor3v\rfloor-\varepsilon,\lceil5v\rceil+\varepsilon),
(\lfloor5v\rfloor-\varepsilon,\lceil7v\rceil+\varepsilon),\ldots
 \end{equation}
 with the removing intervals containing less than two primes (cf. (\ref{7.2})).  For every remaining interval, we write the primes (in increasing order) except for the last one. Then all remaining primes are $v$-over-Ramanujan.
 For example, if $v=3,$ we obtain the following sequence of $3$-over-Ramanujan primes (sequence A164952 in \cite{11}):
 \begin{equation}\label{10.13}
2, 3, 11, 17, 23, 29, 41, 43, 59, 61, 71, 73, 79, 97, 101, 103, 107,\ldots \;.
 \end{equation}
 Furthermore, one can obtain a $v$-classification of primes, including $v$-left, $v$-right, $v$-central and $v$-isolated primes (see Section 8). In particular, if $l_v(n),\; r_v(n)$ denote the $n$-th $v$-left prime and the $n$-th $v$-right prime, respectively, then $l_n\sim r_n$ as $n\rightarrow\infty.$
  Consider now a natural generalization of Proposition \ref{p4} with the similar proof.
 \begin{proposition}\label{p8}
 Let $\pi_{R_v'}(x)$ be the counting function of $v$-over-Ramanujan numbers not exceeding $x.$ Then
$$\liminf_{n\rightarrow\infty}\pi_{R_v'}(x)/\pi(x)\geq 1-\frac {1} {v}.$$
\end{proposition}
 A generalization of Conjecture \ref{con1} (with a similar heuristic proof) is the following.
\begin{conjecture}\label{con2}
\begin{equation}\label{10.14}
\pi_{R_v'}(x)\sim (1-e^{-(v-1)})\pi(x).
 \end{equation}
\end{conjecture}
 Note that, if Conjecture \ref{con2} is true, then, using (\ref{10.2}), for the counting function $\pi_{R_v^*}(x)$ of $v$-pseudo-Ramanujan primes we have
 \begin{equation}\label{10.15}
\pi_{R_v^*}(x)\sim (\frac{1}{v}-e^{-(v-1)})\pi(x),
 \end{equation}
 so that the proportion of $v$-pseudo-Ramanujan primes among all $v$-over-Ramanujan primes is
 $(\frac{1}{v}-e^{-(v-1)})/(1-e^{-(v-1)}).$ This proportion tends to 1 as $v\rightarrow1,$
and decreases to 0 as $v\rightarrow\infty.$
 Using Theorem \ref{t8}, we note that, if Conjecture \ref{con2} is true, then, for the counting function $\pi_{L_v'}(x)$ of $v$-over-Labos primes, we have
  \begin{equation}\label{10.16}
\pi_{L_v'}(x)\sim \pi_{R_v'}(x)\sim (1-e^{-(v-1)})\pi(x).
 \end{equation}
 Furthermore, if Conjecture \ref{con2} is true, then, for the counting functions $\pi_{l_v}(x),\pi_{r_v}(x),\; \pi_{c_v}(x)$ and $\pi_{is_v}(x) $ of the $v$-left, $v$-right, $v$-central and $v$-isolated primes, respectively, of the considered classes of primes, we have
  \begin{equation}\label{10.17}
\pi_{l_v}(x)\sim\pi_{r_v}(x)\sim (1-e^{-(v-1)})e^{-(v-1)}\pi(x),
 \end{equation}
  \begin{equation}\label{10.18}
\pi_{c_v}(x)\sim (1-e^{-(v-1)})^2\pi(x),
 \end{equation}
 \begin{equation}\label{10.19}
   \pi_{is_v}(x)\sim e^{-2(v-1)}\pi(x),
 \end{equation}
so that $\pi_{r_v}(x)+\pi_{l_v}(x)+\pi_{c_v}(x)+\pi_{is_v}(x)=\pi(x).$


\section{Other open problems}

  \begin{conjecture}\label{con3}$(cf. \; Proposition \;\ref{p6}).$ There exist arbitrary long sequences of consecutive primes $p_k, p_{k+1}, \ldots ,p_m,$ such that every interval $(\frac {p_i}{2}, \frac {p_{i+1}}{2}),\; i=k,k+1, \ldots ,m-1,$ contains a prime.
 \end{conjecture}
\begin{conjecture}\label{con4}$(cf. \; Proposition \;\ref{p3}).$
 $\limsup_{n\rightarrow\infty} (R_n-L_n)=\infty.$
 \end{conjecture}
  \begin{conjecture}\label{con5}$(cf. \; Proposition \;\ref{p4}).$ There exist infinitely many peculiar primes.
 \end{conjecture}
  \begin{problem} For $v&gt;1,$ to estimate the smallest pseudo-$v$-Ramanujan prime, the smallest $v$-central prime and the smallest $v$-isolated prime.
   \end{problem}

\section{Acknowledgments}
The author is grateful to Daniel Berend (Ben-Gurion University, Israel) and Greg Martin (University of British Columbia, Canada) for important private communications \cite{2}, \cite{4} and very useful discussions. He is also grateful to Peter J. C. Moses (UK) for improvement the text and to the anonymous referee for very important remarks.

\begin{thebibliography}{16}
\bibitem{1}  E. Bach, and J. Shallit, \emph{Algorithmic Number Theory}, MIT Press,  233 (1996). ISBN 0-262-02405-5.

\bibitem{2} D.  Berend,  Private communication.

\bibitem{3} S.  Laishram,  On a conjecture on Ramanujan primes, \emph{Int. J. Number Theory}  \textbf{6} (2010), 1869--1873.

\bibitem{4} G.  Martin,  Private communication.

\bibitem{5} K.  Prachar, \emph{Primzahlverteilung}, Springer-Verlag, 1957.

\bibitem{6} S.  Ramanujan, A proof of Bertrand's postulate,  \emph{J. Indian Math. Soc.} \textbf{11} (1919), 181--182.

\bibitem{7} S.  Ramanujan, in
G. H. Hardy, S. Aiyar, P. Venkatesvara, and B. M. Wilson, eds.,
\emph{Collected Papers of Srinivasa Ramanujan},
Amer. Math. Soc.,
2000.

\bibitem{8} D.  Redmond,  \emph{Number Theory, An Introduction},  
Marcel Dekker, 1996.

\bibitem{9} J. B.  Rosser, The $n$-th prime is greater than $n\log n,$  \emph{Proc. Lond. Math. Soc.} \textbf{45} (1938), 21--44.

\bibitem{10} J. B.  Rosser and L. Schoenfeld,  Approximate formulas for some functions of prime numbers, \emph{Illinois J. Math.}  \textbf{6} (1962), 64--97.

\bibitem{11}  N. J. A. Sloane,
\emph{The On-Line Encyclopedia of Integer Sequences},
\url{http://oeis.org}.

\bibitem{12} J.  Sondow,  Ramanujan primes and Bertrand's postulate,
\emph{Amer. Math. Monthly}, \textbf{116} (2009), 630--635.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11N05.

\noindent \emph{Keywords: }
Bertrand postulate, Ramanujan prime, Labos prime, over-Ramanujan prime,
over-Labos prime, prime gaps.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A006992},
\seqnum{A055496},
\seqnum{A060715},
\seqnum{A080359},
\seqnum{A104272},
\seqnum{A164288},
\seqnum{A164294},
\seqnum{A164333},
\seqnum{A164368},
\seqnum{A164554},
\seqnum{A164952},
\seqnum{A166251},
\seqnum{A166252},
\seqnum{A166307},
\seqnum{A182365},
\seqnum{A182366},
\seqnum{A182391},
\seqnum{A182392},
\seqnum{A182423},
\seqnum{A182426},
\seqnum{A182451},
\seqnum{A193507},
\seqnum{A193761},
\seqnum{A193880},
\seqnum{A194184},
\seqnum{A194186},
\seqnum{A194217}, and
\seqnum{A194598}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received August 4 2011;
revised versions received  September 7 2011; May 8 2012; May 16 2012.
Published in {\it Journal of Integer Sequences}, May 29 2012.

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


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