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\begin{center}
\vskip 1cm{\LARGE\bf Some Properties of a Sequence Defined with \\
\vskip .1in the Aid of Prime Numbers} \vskip 1cm \large
Br\u{a}du\c{t} Apostol \\
 ``Spiru Haret'' Pedagogic High School \\
5 Timotei Cipariu St. \\
620004 Foc\c{s}ani \\
Romania \\
\href{maito:apo_brad@yahoo.com}{apo\_brad@yahoo.com}\\ 
\ \\
Lauren\c{t}iu Panaitopol\footnote{Deceased.} \\
\ \\ 
Lucian Petrescu \\
``Henri Coand\u{a}'' Technical College \\
2 Tineretului St. \\
820235 Tulcea \\
Romania \\
\href{mailto:petrescuandreea@yahoo.com}{\tt petrescuandreea@yahoo.com}
\\ \ \\ L\'aszl\'o T\'oth \\
Department of Mathematics \\
University of P\'ecs \\
Ifj\'us\'ag \'utja 6 \\
7624 P\'ecs \\
Hungary  \\
\href{mailto:ltoth@gamma.ttk.pte.hu}{\tt ltoth@gamma.ttk.pte.hu}\\
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\begin{abstract}
For every integer $n\ge 1$ let $a_n$ be the smallest positive integer such that
$n+a_n$ is prime. We investigate the behavior of the sequence
$(a_n)_{n\ge 1}$, and prove asymptotic results for the sums
$\sum_{n\le x} a_n$, $\sum_{n\le x} 1/a_n$, and $\sum_{n\le x} \log
a_n$.
\end{abstract}

\section{Introduction} For every integer $n\ge 1$ let $a_n$
be the smallest positive integer such that $n+a_n$ is prime. Here
$a_1=1$, $a_2=1$, $a_3=2$, $a_4=1$, $a_5=2$, $a_6=1$, $a_7=4$, etc.
This is sequence \seqnum{A013632} in Sloane's Online Encyclopedia of Integer
Sequences \cite{OEIS}. For $n\ge 2$, $a_n$ is the smallest positive
integer such that $\gcd(n!,n+a_n)=1$. In this paper we study the
behavior of the sequence $(a_n)_{n\ge 1}$, and prove asymptotic
results for the sums $\sum_{n\le x} a_n$, $\sum_{n\le x} 1/a_n$ and
$\sum_{n\le x} \log a_n$.

We are going to use the following standard notation:

\pont $\pi(x)$ is the number of primes $\le x$,

\pont $\pi_2(x)$ is the number of twin primes $p,p+2$ such that
$p\le x$,

\pont $p_n$ is the  $n$-th prime,

\pont $d_n=p_{n+1}-p_n$,

\pont $f(x) \ll g(x)$ means that $|f(x)|\le C g(x)$, where $C$ is an
absolute constant,

\pont $g(x) \gg f(x)$ means that $f(x) \ll g(x)$,

\pont $f(x) = F(x)+ O(g(x))$ means that $f(x) - F(x) \ll g(x)$,

\pont $f(x) \asymp g(x)$ means that $c f(x) \le g(x) \le Cf(x)$ for
some positive absolute constants $c$ and $C$,

\pont $f(x) \sim g(x)$ means that $\lim_{x\to \infty} f(x)/g(x)=1$.

We will apply the following known asymptotic results concerning the
distribution of the primes:
\begin{equation*}
\pi(x) \sim \frac{x}{\log x}, \quad p_n\sim n\log n \quad {\text{\rm
(Prime number theorem)}},
\end{equation*}
\begin{equation} \label{H}
\sum_{p_n\le x} d_n^2\ll x^{23/18+\varepsilon} \quad {\text{\rm for
every $\varepsilon &gt;0$ (unconditional result of Heath-Brown
\cite{H})}},
\end{equation}
\begin{equation} \label{S}
\sum_{p_n\le x} d_n^2\ll x (\log x)^3 \quad  {\text{\rm (assuming
the Riemann hypothesis, result of Selberg \cite{S})}},
\end{equation}
\begin{equation} \label{P}
\left(\frac{d_2d_3\cdots d_n}{(\log 2)(\log 3)\cdots (\log
n)}\right)^{1/n}\asymp 1 \quad {\text{\rm (due to Panaitopol
\cite[Prop.\ 3]{P})}}.
\end{equation}

This research was initiated by Lauren\c{t}iu Panaitopol
(1940--2008), former professor at the Faculty of Mathematics,
University of Bucharest, Romania. The present paper is dedicated to
his memory.

\section{Equations and identities}

By the definition of $a_n$, for every $n\ge 1$ we have
$n+a_n=p_{\pi(n)+1}$, that is
\begin{equation} \label{exp_form}
a_n= p_{\pi(n)+1}-n.
\end{equation}

From \eqref{exp_form} we deduce that for every $k\ge 1$,
\begin{equation} \label{values}
a_{p_k}= p_{k+1}-p_k, a_{p_k+1}= p_{k+1}-p_k-1,\ldots,
a_{p_{k+1}-1}=1.
\end{equation}

\begin{proposition} For every integer $a\ge 1$  the equation $a_n=a$ has infinitely many solutions.
\end{proposition}

\begin{proof} Let $A_k=\{1,2,\ldots, p_{k+1}-p_k\}$. Since $\limsup_{k\to \infty}
(p_{k+1}-p_k)=\infty$, it follows from \eqref{values} that for every
integer $a\ge 1$ there exist infinitely many integers $k\ge 1$ such
that $a\in A_k$, whence the equation $a_n=a$ has infinitely many
solutions.
\end{proof}

Now we compute the sum $\displaystyle S_n=\sum_{i=1}^n a_i$.

\begin{proposition} For every prime $n\ge 3$ we have
\begin{equation} \label{sum_prime}
S_n= \frac1{2}\left(2p_{\pi(n)+1}-p_{\pi(n)}+\sum_{k=1}^{\pi(n)-1}
d_k^2 \right),
\end{equation}
and for every composite number $n\ge 4$,
\begin{equation} \label{sum_composite}
S_n= \frac1{2}\left(p_{\pi(n)}^2+ 2(n+1-p_{\pi(n)}) p_{\pi(n)+1}
+\sum_{k=1}^{\pi(n)-1} d_k^2 -n^2-n \right).
\end{equation}
\end{proposition}

\begin{proof} If $n\ge 3$ is a prime, then $n=p_m$ for some $m\ge 2$. By using
\eqref{exp_form},
\begin{align*}
S_n &amp; = \sum_{i=1}^n \left(p_{\pi(i)+1}-i\right) \\
&amp; = 2+3+(5+5)+\cdots+(p_m-p_{m-1})p_m+p_{m+1}-\frac{n(n+1)}{2} \\
&amp; = 2+ \sum_{k=2}^m p_k(p_k-p_{k-1})+p_{m+1}- \frac{n(n+1)}{2}\\
&amp; = \frac1{2}\left(p_1^2+ 2 \sum_{k=2}^m p_k^2 - 2\sum_{k=2}^m p_k
p_{k-1} +2p_{m+1}- n^2-n\right)\\
&amp; = \frac1{2}\left(2p_{m+1}-n+ \sum_{k=1}^{m-1} (p_{k+1}-p_k)^2
\right)
\end{align*}
and \eqref{sum_prime} follows by using that $m=\pi(n)$.

Now let $t\ge 4$ be composite. Let $m\ge 2$ be such that
$p_m&lt;t&lt;p_{m+1}$. By applying \eqref{sum_prime} for $n=p_m$, where
$m=\pi(n)=\pi(t)$, we deduce
\begin{align*}
S_t &amp; = S_n+ \sum_{i=n+1}^t a_i = S_n + \sum_{i=n+1}^t \left(p_{\pi(i)+1}-i \right)\\
&amp; = \frac1{2}\left(2p_{\pi(t)+1}-p_{\pi(t)}+\sum_{k=1}^{\pi(t)-1}
(p_{k+1}-p_k)^2 \right)+ \frac{(2p_{\pi(t)+1}-n-t-1)(t-n)}{2}\\
&amp; = \frac1{2}\left(2p_{\pi(t)+1}-p_{\pi(t)}+\sum_{k=1}^{\pi(t)-1}
(p_{k+1}-p_k)^2 + 2p_{\pi(t)+1}(t-n)-t^2-t+n^2+n\right)\\
&amp; = \frac1{2}\left(p_{\pi(t)}^2+ 2(t+1-p_{\pi(t)}) p_{\pi(t)+1}
+\sum_{k=1}^{\pi(t)-1} (p_{k+1}-p_k)^2-t^2-t \right)
\end{align*}
and \eqref{sum_composite} is proved.
\end{proof}

\begin{remark} If $n$ is prime, then \eqref{sum_composite} reduces to \eqref{sum_prime}. Therefore,
the identity \eqref{sum_composite} holds for every integer $n\ge 3$.
\end{remark}

Next we compute the product $\displaystyle P_n=\prod_{i=1}^n a_i$.

\begin{proposition} For every prime $n\ge 3$ we have
\begin{equation} \label{prod_prime}
P_{n-1} = \prod_{k=1}^{\pi(n)-1} d_k!,
\end{equation}
and for every composite number $n\ge 4$,
\begin{equation}
\label{prod_composite} P_{n-1} = \prod_{k=1}^{\pi(n)-1} d_k!
\prod_{k=1}^{n-p_{\pi(n)}} (p_{\pi(n)+1}-p_{\pi(n)}-k+1).
\end{equation}
\end{proposition}

\begin{proof} Let $n=p_m\ge 3$ be a prime. By using
\eqref{values},
\begin{align*}
P_{n-1} &amp; = \prod_{i=2}^m (p_i-p_{i-1})!= \prod_{i=1}^{m-1}
(p_{i+1}-p_i)!,
\end{align*}
which proves \eqref{prod_prime}.

Now let $t\ge 4$ be composite such that $p_m&lt;t&lt;p_{m+1}$. By applying
\eqref{prod_prime} for $n=p_m$, where $m=\pi(n)=\pi(t)$, we deduce
\begin{align*}
P_{t-1} &amp; = P_{n-1} \prod_{i=n}^{t-1} a_i = P_{n-1} \prod_{i=n}^{t-1} \left(p_{\pi(i)+1}-i \right)\\
&amp; = \prod_{k=1}^{\pi(t)-1} d_k! \prod_{j=1}^{t-p_m} \left(p_{m+1}-p_m-j+1 \right) \\
&amp; = \prod_{k=1}^{\pi(t)-1} d_k! \prod_{k=1}^{t-p_{\pi(t)}}
\left(p_{\pi(t)+1}-p_{\pi(t)}-k+1 \right)
\end{align*}
and \eqref{prod_composite} is proved.
\end{proof}

\begin{remark} If $n$ is prime, then the second product in \eqref{prod_composite} is empty and
\eqref{prod_composite} reduces to \eqref{prod_prime}. Hence the identity \eqref{prod_composite} holds for every integer $n\ge 3$.
\end{remark}


\section{Asymptotic results}

\begin{theorem} For every $\varepsilon &gt;0$,
\begin{equation} \label{asymp_sum}
x\log x \ll \sum_{n\le x} a_n \ll x^{23/18+\varepsilon},
\end{equation}
where $23/18\doteq 1.277$. If the Riemann hypothesis is true, then
the upper bound in \eqref{asymp_sum} is $x(\log x)^3$.
\end{theorem}

\begin{proof} Let $x\ge 2$ and let $p_k\le x&lt; p_{k+1}$. By using \eqref{sum_prime}
for $n=p_{k+1}$,
\begin{align*}
\sum_{n\le x} a_n &amp; \le \sum_{i=1}^{p_{k+1}} a_i =
\frac1{2}\left(2p_{k+2}-p_{k+1} +\sum_{i=1}^k d_i^2
\right)\\
&amp; \ll p_{k+2} + \sum_{p_i\le x} d_i^2.
\end{align*}

Taking into account the estimate \eqref{H} due to Heath-Brown, and
the fact that $p_{k+2}\sim p_k \le x$ we get the unconditional upper
bound in \eqref{asymp_sum}. If the Riemann hypothesis is true, then
by using Selberg's result \eqref{S} we obtain the upper bound
$x(\log x)^3$.

Now, for the lower bound we use the trivial estimate
\begin{equation*}
\sum_{p_n\le x} d_n^2 \gg x\log x,
\end{equation*}
which follows from the inequality between the arithmetic and
quadratic means. We deduce that
\begin{align*}
\sum_{n\le x} a_n &amp; \ge \sum_{i=1}^{p_k} a_i =
\frac1{2}\left(2p_{k+1}-p_k +\sum_{i=1}^{k-1} d_i^2
\right)\\
&amp; \gg  \sum_{p_i\le p_{k-1}} d_i^2 \gg p_{k-1} \log p_{k-1} \sim
x\log x,
\end{align*}
since $p_{k-1}\sim k\log k$ and $k=\pi(x)\sim x/\log x$, $\log k
\sim \log x$.
\end{proof}

To prove our next result we need the following

\begin{lemma} We have
\begin{equation} \label{asympt_lemma}
\sum_{2\le n\le x} \log d_n = x\log \log x + O(x).
\end{equation}
\end{lemma}

\begin{proof} The inequalities \eqref{P} can be written as
\begin{equation*}
c n &lt; \sum_{i=2}^n \log d_i -\sum_{i=2}^n \log \log i &lt; C n
\end{equation*}
for some positive absolute constants $c$ and $C$. Now
\eqref{asympt_lemma} emerges by applying the well known asymptotic
formula
\begin{equation*}
\sum_{2\le n\le x} \log \log n = x\log \log x + O(x).
\end{equation*}
\end{proof}

\begin{theorem} We have
\begin{equation}\label{asymp_harm}
\sum_{n\le x} \frac1{a_n} = \frac{x\log \log x}{\log x} +
O\left(\frac{x}{\log x}\right).
\end{equation}
\end{theorem}

\begin{proof} For $x=p_m-1$ ($m\ge 2$) we have by \eqref{values},
\begin{equation*}
\sum_{n\le p_m-1} \frac1{a_n} = 1+ \sum_{i=2}^m
\left(1+\frac1{2}+\cdots+\frac1{p_i-p_{i-1}}\right).
\end{equation*}

For an arbitrary $x\ge 3$ let $p_k$ ($k\ge 2$) be the prime such
that $p_k\le x&lt; p_{k+1}$. Using the familiar inequalities
\begin{equation*}
\log m &lt; 1+\frac1{2}+\cdots+\frac1{m}\le 1+\log m \quad (m\ge 1)
\end{equation*}
we deduce
\begin{equation*}
\log (p_i-p_{i-1})&lt; 1+\frac1{2}+\cdots+\frac1{p_i-p_{i-1}} \le
1+\log (p_i-p_{i-1}) \quad (i\ge 2)
\end{equation*}
and
\begin{equation*}
1+\sum_{i=2}^k \log (p_i-p_{i-1}) +\frac1{d_k} &lt; \sum_{n\le p_k-1}
\frac1{a_n}+ \frac1{a_{p_k}}
\end{equation*}
\begin{equation*}
\le \sum_{n\le x} \frac1{a_n} \le \sum_{n\le p_{k+1}-1} \frac1{a_n}
\le 1+ k + \sum_{i=2}^{k+1} \log (p_i-p_{i-1}).
\end{equation*}

By \eqref{asympt_lemma} we obtain
\begin{equation*}
\sum_{n\le x} \frac1{a_n} = k\log \log k + O(k),
\end{equation*}

Here $k=\pi(x)\sim x/\log x$, $\log k \sim \log x$ and we deduce
\eqref{asymp_harm}.
\end{proof}

\begin{theorem} One has
\begin{equation*}
x \ll \sum_{n\le x} \log a_n  \ll x\log x.
\end{equation*}
\end{theorem}

\begin{proof} For an arbitrary $x\ge 3$ let $p_k$ ($k\ge 2$) be the prime such
that $p_k\le x&lt; p_{k+1}$. Using the elementary inequalities
\begin{equation*}
m \log m - m + 1 \le \log m! \le m\log m \quad (m\ge 1)
\end{equation*}
we deduce by applying \eqref{prod_prime} that
\begin{equation*}
\sum_{n\le x} \log a_n \le \sum_{n\le p_{k+1}-1} \log a_n =
\sum_{i=1}^k \log d_i! \le \sum_{i=1}^k d_i \log d_i
\end{equation*}
\begin{equation*}
&lt; \sum_{i=1}^k d_i \log p_i &lt; (\log p_k) \sum_{i=1}^k d_i &lt; (\log
p_k)p_{k+1},
\end{equation*}
where we also used that $d_i=p_{i+1}-p_i&lt;p_i$ by Chebyshev's
theorem. Here
\begin{equation} \label{k_x}
p_k\sim k\log k, \quad k=\pi(x)\sim x/\log x, \quad \log k \sim \log
x,
\end{equation}
and we obtain the upper bound $x\log x$.

On the other hand,
\begin{equation*}
\sum_{n\le x} \log a_n &gt; \sum_{n\le p_k-1} \log a_n  =
\sum_{i=1}^{k-1} \log d_i!
\end{equation*}
\begin{equation*}
&gt;  \sum_{i=1}^{k-1} (d_i \log d_i-d_i+1 ) = \sum_{i=2}^{k-1} d_i\log d_i - p_k+
k+1.
\end{equation*}

Here
\begin{align*}
\sum_{i=2}^{k-1} d_i \log d_i &amp; = \sum_{\substack{i=2\\ d_i\ge
3}}^{k-1} d_i\log d_i + 2\log 2 \sum_{\substack{i=2\\ d_i=2}}^{k-1}
1 \\
&amp; \ge (\log 3) \sum_{\substack{i=2\\ d_i\ge 3}}^{k-1} d_i + (2\log
2)\pi_2(k-1)\\
&amp; = (\log 3)\left( \sum_{i=2}^{k-1} d_i - \sum_{\substack{i=2\\
d_i=2}}^{k-1} d_i \right) + (2\log 2)\pi_2(k-1) \\
&amp; = (\log 3)\left(p_k-p_2  - 2\pi_2(k-1) \right) + (2\log
2)\pi_2(k-1)\\
&amp; = (\log 3)p_k -2\log(3/2)\pi_2(k-1)-3\log 3 \\
&amp; &gt; (\log 3)p_k -2\log(3/2)k- 3\log 3,
\end{align*}
where it is sufficient to use the obvious estimate $\pi_2(k-1)&lt; k$.
Note that $\log 3\doteq 1.09$, $2\log(3/2)\doteq 0.81$, $3\log
3\doteq 3.29$.

We deduce that
\begin{equation*}
\sum_{n\le x} \log a_n &gt; 0.09 p_k - 3.
\end{equation*}

Now \eqref{k_x} gives the lower bound $x$.
\end{proof}

\begin{thebibliography}{9}

\bibitem{H} D.~R.~Heath--Brown, The differences between consecutive primes, III,
{\it J. London Math. Soc.} (2) {\bf 20} (1979), 177--178.

\bibitem{P} L.~Panaitopol, Properties of the series of differences of prime numbers,
{\it Publ. Centre de Rech. Math. Pures Neuch\^{a}tel (Serie 1)} {\bf
31} (2000), 21--28.

\bibitem{S} A.~Selberg, On the normal density of primes in small intervals, and the difference between
consecutive primes, {\it Arch. Math. Naturvid.} {\bf 47} (1943), No.~6,
87--105.

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

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
11A41; Secondary 11N05.

\noindent \emph{Keywords:} prime numbers, difference of consecutive
primes, asymptotic behavior.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A000040},
\seqnum{A001223}, \seqnum{A013632}, and \seqnum{A076821}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 3 2015;
revised version received  April 3 2015.
Published in {\it Journal of Integer Sequences}, May 21 2015.

\bigskip
\hrule
\bigskip

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