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\begin{center}
\vskip 1cm{\LARGE\bf
Two Remarks on the Largest
\vskip .1in
Prime Factors of $n$ and $n+1$
}
\vskip 1cm
\large
Sungjin Kim \\
Department of Mathematics\\
Santa Monica College \\
California State University, Northridge\\
18111 Nordhoff Street\\
Northridge, CA 91330\\
USA\\
\href{mailto:sungjin.kim@csun.edu}{\tt sungjin.kim@csun.edu} \\
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\begin{abstract} 
Let $P(n)$ be the largest prime factor of $n$. We give an
alternative proof of the existence of infinitely many $n$
such that $P(n)>P(n+1)>P(n+2)$. Further, we prove that the
set $\{P(n+1)/P(n)\}_{n\in\N}$ has infinitely many limit points
$\{0,x_n,1,y_n\}_{n\in\N}$ with $0<x_n<1<y_n$ and $\lim x_n = \lim
y_n =1$.
\end{abstract}

\section{Introduction}
Let $n\geq 2$ be a positive integer. Let $P(n)$ denote denote the largest prime factor of $n$. Erd\H{o}s and Pomerance~\cite{EP} proved that the number of $n\leq x$ such that $P(n)<P(n+1)$ is at least $0.0099x$, and the same holds for $P(n)>P(n+1)$. This lower density $0.0099$ was subsequently improved by several authors ($0.05544$ by de la Bret\`{e}che, Pomerance and Tenenbaum~\cite{BPT}, $0.1063$ by Z. Wang~\cite{W}). The current record holders are L\"{u} and Wang~\cite{LW}, who proved that the lower density is at least $0.2017$. 

Erd\H{o}s and Pomerance~\cite{EP} also noted that the three patterns $P(n)<P(n+1)>P(n+2)$, $P(n)>P(n+1)<P(n+2)$, and $P(n)<P(n+1)<P(n+2)$ occur infinitely often. They presented a simple proof for the infinitude of the third pattern. Namely, they took
$$n=p^{2^m}-1, \   n+1=p^{2^m}, \ \textrm{and} \   n+2=p^{2^m}+1,$$
where $p$ is prime and $m=\inf\{k| P(p^{2^k}+1)>p\}$. They left the infinitude of the fourth pattern $P(n)>P(n+1)>P(n+2)$ as an open problem. 

This problem was later solved by Balog~\cite{B}, who showed that the number of occurrence of this pattern for $n\leq x$ is $\gg \sqrt x$. Building on earlier results by Matom\"{a}ki, Radziwi\l{}\l{}, and Tao~\cite{MRT}, and Ter\"{a}v\"{a}inen~\cite{Te}, Tao and Ter\"{a}v\"{a}inen~\cite{TT} proved that the following sets have positive lower density:
$$
\{n\in\N \ | \ P(n)<P(n+1)<P(n+2)>P(n+3)\}\ \textrm{and}$$
$$\{n\in\N \ | \ P(n)>P(n+1)>P(n+2)<P(n+3) \}.$$
Using the Maynard-Tao theorem~\cite{Pb}, in this paper we provide a simple alternative proof of the infinitude of the patterns $P(n)<P(n+1)<P(n+2)$ and $P(n)>P(n+1)>P(n+2)$. We prove that both patterns occur for $\gg x/(\log x)^{50}$ values of $n\leq x$. The result is weaker than Tao and Ter\"{a}v\"{a}inen's,
and stronger than Balog's.
\begin{theorem}
For sufficiently large $x$, we have
$$
\#\{n\leq x \ | \ P(n)<P(n+1)<P(n+2)\} \gg \frac x{(\log x)^{50}} \ \mathrm{and} $$
$$ \#\{n\leq x \ | \ P(n)>P(n+1)>P(n+2)\} \gg \frac x{(\log x)^{50}}.
$$
\end{theorem}
Erd\H{o}s and Pomerance~\cite[Theorem 1]{EP} proved that for any $\epsilon>0$, there is $\dt>0$ such that the number of $n\leq x$ with
$$
x^{-\dt}<\frac{P(n+1)}{P(n)}<x^{\dt}
$$
is less than $\epsilon x$. They remarked that this means $P(n)$ and $P(n+1)$ are usually not close. In the opposite direction, we prove that this ratio can approach arbitrarily close to $1$ from both sides.
\begin{theorem}
For any $\epsilon>0$, we have
$$\#\left\{n\leq x \ \Bigg\vert 1\leq \frac{P(n+1)}{P(n)}<1+\epsilon\right\}\gg_{\epsilon} \frac x{(\log x)^{50}} \ \mathrm{and} $$
$$\#\left\{n\leq x \ \Bigg\vert 1-\epsilon< \frac{P(n+1)}{P(n)}\le 1\right\}\gg_{\epsilon} \frac x{(\log x)^{50}}.$$
\end{theorem}
Let $R:=\{\frac{P(n+1)}{P(n)} \ \vert \ n\in\N\}$.
As a direct consequence of Theorem 2, we obtain that $1\in \overline{R}$. From the proof of Theorem 2, we obtain a finite set $\{a_1,\ldots, a_{50}\}\subseteq \N$ with $1\leq a_i<a_j$ for each $1\leq i<j\leq 50$ such that $a_{j_1}/a_{i_1} \in \overline{R}\cap (1,\infty)$ for some $1\leq i_1<j_1\leq 50$, and $a_{i_2}/a_{j_2}\in\overline{R}\cap (0,1)$ for some $1\leq i_2<j_2\leq 50$. Changing $a_i$ with $\epsilon$, we obtain a sequence rational numbers with $0<x_n<1<y_n$ such that $\lim x_n=\lim y_n=1$ and $\{x_n,y_n\}_{n\in\N}\subseteq\overline{R}$.

In fact, there is an elementary proof of $1\in \overline{R}$. This elementary proof is based on a solution ($\epsilon=1$ in the following argument) that appeared in Mathematics Stack Exchange~\cite{C} by Barry Cipra. For any $0<\epsilon\leq 1$, take primes $p$ and $q$ satisfying $p<q<(1+\epsilon)p$ so that
$$
\frac1{1+\epsilon}<\frac pq <1 < \frac qp <1+\epsilon.
$$
By the extended Euclidean algorithm, there exist
integers $u$ and $v$ with $0<u<q$, $0<v<p$ and $pu-qv=1$. Let $U=q-u$, $V=p-v$. Then $qV-pU=1$. The integers $qu$ and $qV$ have $q$ as largest prime factor. Since
$u+U=q<(1+\epsilon)p$, at least one of $u\leq p$ or $U\leq p$ holds. If $u\leq p$, then $p$ is the largest prime factor of $pu$. If $U\leq p$, then 
$p$ is also the largest prime factor of $pU$. Thus,
either one of the following holds:
$$
n=qv, \ n+1= pu, \ \frac{P(n+1)}{P(n)}=\frac pq,
$$
or
$$
n=pU, \ n+1= qV, \ \frac{P(n+1)}{P(n)}=\frac qp.
$$
Therefore, $1\in\overline{R}$ follows. From this argument the number of $n\leq x$ with $\frac1{1+\epsilon}<\frac{P(n+1)}{P(n)}<1+\epsilon$ is $\gg_{\epsilon} x / (\log x)^2$. Slightly modifying this argument, we have for any $x\in[1,2]$, either $x$ or $1/x$ is in $\overline{R}$. However, this argument does not determine whether a limit point is in $[0,1)$ or $(1,\infty)$.

By Dirichlet's theorem on primes in arithmetic progressions, it is easy to see that $0$ is also a limit point of $R$. For if we take a prime $n=ar-1$, with $a$ large, then $P(n)=ar-1$ and $P(n+1)\leq \max(a, r)$. Assuming the prime $k$-tuples conjecture (Conjecture 4), we prove that all nonnegative real numbers are limit points of $R$.
\begin{theorem}[Conditional]
Assuming the prime $k$-tuples conjecture (Conjecture 4), we have $\overline{R}=[0,\infty)$.
\end{theorem}



\section{Estimates on the numbers of prime $k$-tuples}
A set of $k$-tuple of linear forms $\{a_1x+b_1, \ldots a_kx+b_k\}$ is said to be \it admissible \rm if for any prime $p$ there is $x_p\in\mathbb{Z}$ such that $p\nmid \prod_{i=1}^k (a_i x_p + b_i)$. We consider the tuples with
$$
\prod_i a_i \neq 0 \ \ \mathrm{and} \ \ \prod_{i<j} (a_ib_j-a_jb_i)\neq 0.
$$

The following is a special case of Bateman-Horn conjecture (a quantitative estimate on Dickson's prime $k$-tuples conjecture).
\begin{conjecture}[Bateman-Horn]
Let $k\geq 2$ and $A_k=\{a_1x+ b_1,\ldots, a_kx+b_k\}$ be an admissible set of linear forms. Then for sufficiently large $x$, the number $R_k(x)$ of $r\leq x$ such that $a_i x + b_i$, $1\leq i\leq k$ are all prime satisfies
$$
R_k(x)\gg_{A_k} \frac x{(\log x)^k}.
$$
\end{conjecture}

Substantial progress toward this conjecture begin with Zhang's result~\cite{Z} on bounded gaps in primes. Subsequently, Maynard~\cite{M} and Polymath8b (\cite{Pb} led by Tao) improved upon Zhang's result. We state a quantitative form of the Maynard-Tao theorem for admissible sets of linear forms. The proof requires slight modifications of~\cite{Pb} and the stated lower bound can be found in~\cite[Remark 32]{Pb}. Note that the following is unconditional.
\begin{lemma}[Maynard-Tao-Polymath8b]
Let $A=\{a_1r+b_1, \ldots, a_{50}r+b_{50}\}$ be an admissible set of linear forms. Then for sufficiently large $x$, the number $R(A,x)$ of $r\leq (x-\max_i b_i)/\max_i a_i$ such that at least two of the linear forms are primes satisfies
$$
R(A,x)\gg_{A} \frac x{(\log x)^{50}}.
$$
\label{lemma5}
\end{lemma}
We will apply the above lemma in the following two special cases:

\medskip

\noindent {\bf Case 1.}  $0<a_1<\cdots <a_k$ and $b_i=1$ for all $i=1,\ldots, k$.


\medskip
\noindent{\bf Case 2.}    $0<a_1<\cdots <a_k$ and $b_i=-1$ for all $i=1,\ldots, k$. 


\medskip
The set of linear forms in these cases is always admissible.
\section{The main lemma}
We construct a special sequence $\{a_i\}$ by the following inductive process.
\begin{lemma}[The main lemma]
Let $k\geq 2$ and $e_k=1$. For each $0\leq j\leq k-2$, assume that $\{e_{k-j}, \ldots, e_k\}$ satisfies
$$
\sum_{s< i\leq t} e_i  \ \bigg\vert \ \sum_{k-j\leq i\leq s} e_i \ \textrm{ for any }\ k-j\leq s<t\leq k.
$$
Let $e_{k-j-1}$ be a multiple of
$$
\lcm \left\{\sum_{s\leq i\leq t}e_i \ \bigg\vert k-j\leq s<t\leq k\right\}.
$$
Then $a_i=\sum_{m\leq i} e_m$ satisfies $0<a_j-a_i \ | \ a_i$ for each $1\leq i<j\leq k$.
\label{lemma6}
\end{lemma}
\begin{proof}
The proof is clear from the inductive construction.
\end{proof}
We exhibit some sequences $\{a_i\}$ that can be produced by the main lemma. \\

\begin{example}
If $k=2$, then let $\{e_1,e_2\}=\{1,1\}$ and $\{a_1,a_2\}=\{1,2\}$. \\

If $k=3$, then let $\{e_1,e_2,e_3\}=\{2,1,1\}$ and $\{a_1,a_2,a_3\}=\{2,3,4\}$. \\

If $k=4$, then let $\{e_1,e_2,e_3,e_4\}=\{12,2,1,1\}$ and $\{a_1,a_2,a_3,a_4\}=\{12,14,15,16\}$. \\

If $k=5$, then let $\{e_1,e_2,e_3,e_4,e_5\}=\{1680, 12, 2, 1,1\}$ and \\

 $\{a_1,a_2,a_3,a_4,a_5\}=\{1680, 1692, 1694, 1695, 1696\}$. \\
\end{example}

Note that $e_1$ can be made arbitrarily large in the final inductive step. We will use the sequence $\{a_i\}_{1\leq i\leq 50}$.

\begin{lemma}
Let $\{a_i\}_{1\leq i\leq 50}$ be a sequence produced in the main lemma. That is, $0<a_j-a_i \ | \ a_i$ for each $1\leq i<j\leq 50$. Suppose that $a_ir+1$ and $a_jr+1$ are primes. Then by taking
\begin{equation}
n=\frac{a_i}{a_j-a_i}(a_jr+1), \ n+1=\frac{a_j}{a_j-a_i}(a_ir+1),
\end{equation}
we have for sufficiently large $r$,
$$
P(n)=a_jr+1, \ P(n+1)=a_ir+1.
$$
Now suppose that $a_ir-1$ and $a_jr-1$ are primes. Then by taking
\begin{equation}
n=\frac{a_j}{a_j-a_i}(a_ir-1), \ n+1=\frac{a_i}{a_j-a_i}(a_jr-1),
\end{equation}
we have for sufficiently large $r$,
$$
P(n)=a_ir-1, \ P(n+1)=a_jr-1.
$$
\label{lemma8}
\end{lemma}

\begin{proof}
We take large enough $r$ so that $a_1r-1$ exceeds the largest prime factor of $\prod a_i$.
\end{proof}
\begin{remark}
The author recently learned that a sequence $\{a_i\}$ with the property $0<a_j-a_i \ | \ a_i$ for each $1\leq i< j$ was obtained earlier by Heath-Brown~\cite[Lemma 1]{Hb}, and such a sequence is used in an unpublished work of Maynard and Ford~\cite[Theorem 7.18]{F}. Using such a sequence and Lemma~\ref{lemma8} (1), Maynard and Ford proved that there is a constant $B>0$ so that 
$P(n)\geq n/B$ and $P(n+1)\geq (n+1)/B$
for infinitely many $n$.
\end{remark}

\section{Proof of theorems}
\subsection{Proof of Theorem 1}

\begin{proof}
Let $\{a_i\}_{1\leq i\leq 50}$ be a sequence produced in the main lemma. We apply (1) of Lemma~\ref{lemma8}. By letting $n+2$ divisible by $\frac{a_j}{a_j-a_i}+1$, we obtain $P(n)>P(n+1)>P(n+2)$ for $n$ in (1). For this idea to work, we need to require $r$ to be divisible by $\frac{a_j}{a_j-a_i}+1$ for any choice of $1\leq i<j\leq 50$. To see this, we let
$$
M=\lcm\left\{\frac{a_j}{a_j-a_i}+1 \ \bigg\vert \ 1\leq i<j\leq 50\right\}.
$$
Then we work with the admissible set of linear forms $\{a_iMr+1\}_{1\leq i\leq 50}$. By Lemma~\ref{lemma5} and the pigeonhole principle, there is a pair $(i,j)$, $1\leq i<j\leq 50$ depending on $x$ such that $a_iMr+1$ and $a_jMr+1$ are primes for $\gg x/(\log x)^{50}$ values of $r\leq (x-a_{50})/(a_{50}^2M)$. For such $r\geq r_0$, we have $n=\frac{a_i}{a_j-a_i}(a_jMr+1)\leq x$, $P(n)=a_jMr+1$, $P(n+1)=a_iMr+1$, and $P(n+2)\leq (n+2)/(\frac{a_j}{a_j-a_i}+1)$. Thus, $P(n)>P(n+1)>P(n+2)$ is satisfied for such $r$.

To obtain an analogous result on $P(n)<P(n+1)<P(n+2)$, we apply Part (2) of 
Lemma~\ref{lemma8}. By letting $n-1$ divisible by $\frac{a_j}{a_j-a_i}+1$, we obtain $P(n-1)<P(n)<P(n+1)$ for $n$ in (2). Then we work with the admissible set of linear forms $\{a_iMr-1\}_{1\leq i\leq 50}$. The rest of the argument is similar to the previous case. 
\end{proof}

\subsection{Proof of Theorem 2}
\begin{proof}
Let $\epsilon>0$ be arbitrary. We show that the number of $n\leq x$ with $1-\epsilon< \frac{P(n+1)}{P(n)}\leq 1$ is $\gg_{\epsilon} \frac x{(\log x)^{50}}$. In the inductive process in Lemma~\ref{lemma6},
we let $e_1$ be large enough to have
$$
1-\epsilon< \frac{a_i}{a_j}\leq 1 \ \textrm{ for each }1\leq i<j\leq 50.
$$
Then we apply Lemma~\ref{lemma8} (1) to conclude the existence of a pair $(i,j)$, $1\leq i<j\leq 50$ depending on $x$ such that $a_ir+1$ and $a_jr+1$ are primes for $\gg_{\epsilon} x/(\log x)^{50}$ values of $r\leq (x-a_{50})/a_{50}^2$. It is clear that $\frac{a_ir+1}{a_jr+1}=\frac{a_i}{a_j}+\frac{a_j-a_i}{a_j(a_jr+1)}$. Since $P(n)=a_jr+1$ and $P(n+1)=a_ir+1$ for such $r$ by Lemma~\ref{lemma8} (1), we have
$$
1\geq \frac{P(n+1)}{P(n)}=\frac{a_ir+1}{a_jr+1}> \frac{a_i}{a_j}>1-\epsilon.
$$
The result now follows.
\end{proof}

To obtain an analogous result for $1\leq  \frac{P(n+1)}{P(n)}<1+\epsilon$, we apply Lemma~\ref{lemma8} (2).

\subsection{Proof of Theorem 3}

\begin{proof}
Let $a_1$ be an even positive integer, and $a_2$ be a positive integer with $(a_1,a_2)=1$. By Bezout's identity, we can find positive integers $b_1$ and $b_2$ such that $a_1b_2-a_2b_1=(a_1,a_2)=1$. The sets of linear forms $\{a_1r+b_1, a_2r+b_2\}$ and $\{a_1r-b_1, a_2r-b_2\}$ are admissible. By Conjecture 4, there are infinitely many $r$ such that both of these forms are primes. We take
$$
n=a_2(a_1r+b_1), \ n+1=a_1(a_2r+b_2)
$$
or
$$
n=a_1(a_2r-b_2), \ n+1=a_2(a_1r-b_1).
$$
If we select $r$ to exceed any prime factor of $a_1a_2$, then we see in both cases
$$
\left\{\frac{a_1}{a_2}, \frac{a_2}{a_1}\right\}\subseteq \overline{R}.
$$
Hence, it follows that any positive rational numbers with numerator and denominator of different parity are limit points of $R$, and consequently, $\overline{R}=[0,\infty)$.
\end{proof}

\section{Acknowledgments}
The author thanks Kevin Ford for bringing~\cite{Hb},~\cite{MRT}, and~\cite{Te} to his attention. The author also thanks an anonymous referee and Paul Pollack for careful reading of the paper and helpful comments.

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\end{thebibliography}


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

\noindent \emph{Keywords: } largest prime factor, consecutive prime.

\bigskip
\hrule
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\noindent (Concerned with sequence \seqnum{A006530}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received  June 15 2020;
revised versions received  September 7 2020; October 16 2020.
Published in {\it Journal of Integer Sequences}, October 16 2020.

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\noindent
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