\documentclass[12pt,reqno]{article}
\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}
\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}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}
\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}}
\DeclareMathOperator{\lcm}{lcm}
\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\Q{{\mathbb Q}}
\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}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\begin{center}
\vskip 1cm{\LARGE\bf
Consecutive Integers Divisible by a Power \\
\vskip .1in
of their Largest Prime Factor
}
\vskip 1cm
\large
Jean-Marie De Koninck\\
D\'ep. de math\'ematiques et de statistique \\
Universit\'e Laval\\
Qu\'ebec, G1V 0A6\\
Canada\\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca} \\
\ \\
Matthieu Moineau\\
\'Ecole des mines de Nancy \\
Universit\'e de Lorraine \\
54042 Nancy Cedex\\
France\\
\href{mailto:matthieu.moineau@mines-nancy.org}{\tt matthieu.moineau@mines-nancy.org}\\
\end{center}
\vskip .2in
\begin{abstract}
We construct families of consecutive polynomials with integer coefficients
that allow for the discovery of consecutive integers divisible by a
power of their largest prime factor.
\end{abstract}
\section{Introduction}\label{sect:1}
Let $P(n)$ stand for the largest prime factor of an integer $n\ge 2$ and set $P(1)=1$.
Given an arbitrary positive integer $\ell$ and $k$ distinct primes $p_0,p_1,\ldots,p_{k-1}$, the Chinese remainder theorem guarantees the existence of infinitely many integers $n$ such that $p_i^\ell \mid n+i$ for $i=0,1,\ldots,k-1$.
However, this theorem does not guarantee that such integers $n$ will also have the property that $P(n+i)=p_i$ for $i=0,1,\ldots,k-1$, although such is the case in some particular instances. For example when $\ell=2$, $k=3$ and $n=1\,294\,298$, we indeed have
\begin{eqnarray*}
1\,294\,298 & = & 2 \cdot 61 \cdot 103^2, \\
1\,294\,299 & = & 3^4 \cdot 19\cdot 29^2, \\
1\,294\,300 & = & 2^2 \cdot 5^2 \cdot 7 \cdot 43^2.
\end{eqnarray*}
In fact, one can show that the above number $n$ is the smallest positive integer with that property.
This motivates the following definitions. Given fixed integers $k\ge 2$ and $\ell\ge 2$,
set
\begin{eqnarray*}
E_{k,\ell} &:= & \{n\in \N: P(n+i)^\ell\mid n+i \ \mbox{ for each } i=0,1,\ldots,k-1\}\\
E_{k,\ell}(x) &:=& \#\{ n\le x: n\in E_{k,\ell} \}.
\end{eqnarray*}
Many elements of $E_{2,2}$, $E_{2,3}$, $E_{2,4}$, $E_{2,5}$ and $E_{3,2}$ are given in the 2009 book of the first author \cite{kn:jm-book}, whereas no elements of the sets $E_{3,3}$, $E_{2,6}$ and $E_{4,2}$ were known at that time.
But, in 2014, Burcsi and G\'evay (private communication), found the 77-digit number $n_0$
which satisfies
\begin{eqnarray*}
n_0-1 & =& 2^7 \cdot 53 \cdot 4253 \cdot 27631 \cdot 27953 \cdot 1546327 \cdot 2535271 \\
& & \qquad \qquad \cdot 17603683 \cdot 1472289739 \cdot 16476952799^3,\\
n_0 & = & 3^6 \cdot 19 \cdot 37 \cdot 787 \cdot 711163 \cdot 2181919 \cdot 137861107 \\
& & \qquad \qquad \cdot 318818473 \cdot 937617607\cdot 7323090133^3, \\
n_0+1 & = & 2 \cdot 12899 \cdot 133451 \cdot 421607 \cdot 2198029 \cdot 8046041 \\
& & \qquad \qquad \cdot 19854409 \cdot 555329197\cdot 32953905599^3,
\end{eqnarray*}
thereby establishing that $n_0-1\in E_{3,3}$. Perhaps, this number is the smallest element of $E_{3,3}$, but this has not been shown.
Even though no elements of $E_{k,\ell}$ for $k\ge 4$ and $\ell\ge 2$ are known, it seems reasonable to conjecture that, given any fixed integers $k\ge 2$ and $\ell\ge 2$, the corresponding set $E_{k,\ell}$ is infinite.
The fact that
$\#E_{k,\ell} = \infty$ is certainly true in the particular case $k=\ell=2$, as it is an immediate consequence of the fact that the Fermat-Pell equation $a^2-2b^2=1$ has infinitely many integer solutions $(a,b)$, thereby also ensuring that $E_{2,2}(x)\gg \log x$. However, $E_{2,2}(x)$ can be proved to be much larger. Indeed,
De Koninck, Doyon, and Luca \cite{kn:jm-nic-fl} focused their attention on the size of
$E_{2,2}(x)$
and proved that
$$x^{1/4}/\log x \ll E_{2,2}(x)\ll x\exp\{ -c \sqrt{2\log x \log \log x} \},$$
where $c=25/24\approx 1.042$. Note that de la Bret\`eche and Drappeau \cite{kn:drappeau} have recently showed that one can choose $c=4/\sqrt 5 \approx 1.789$, whereas, as we will see in Section \ref{sect:9}, one can expect that the true order of $E_{2,2}(x)$ is $x\exp\{ - (1+o(1)) 2\sqrt{2\log x \log \log x} \}$ as $x\to \infty$.
\vskip 5pt
At this point, we introduce additional notation. Given $k$ integers $\ell_0,\ell_1,\ldots,\ell_{k-1}$, each $\ge 2$, consider the set
$$F(\ell_0,\ell_1,\ldots,\ell_{k-1}):=\{n\in \N: P(n+i)^{\ell_i}\mid n+i \mbox{ for }i=0,1,\ldots,k-1\},$$
so that in particular $E_{k,\ell} = F(\underbrace{\ell,\ldots,\ell}_{k})$.
Also, for each integer $\ell\ge 2$, we set $G_\ell:=\{n\in \N: P(n)^\ell \mid n\}$ and $G_\ell(x):=\#\{n\le x: n\in G_\ell\}$.
Most likely, each set $F(\ell_0,\ell_1,\ldots,\ell_{k-1})$ is infinite, but besides the set $F(2,2)$, no such statement has been proved.
Here, we first show that if we assume that there exist infinitely many primes of the form $9k^2+6k+2$ (respectively $4k^2+2k+1$), then the set $F(3,2)$ (respectively $F(4,2)$) is infinite. We then explore some identities involving consecutive polynomials whose algebraic structure provides the potential for revealing infinitely many members of $E_{k,\ell}$ for any given pair of integers $k\ge 2$, $\ell\ge 2$ and of
$F(\ell_0,\ell_1,\ldots,\ell_{k-1})$ for any given $k$-tuple of integers $\ell_0\ge 2,\ell_1\ge 2,\ldots,\ell_{k-1}\ge 2$.
\section{Preliminary results and conjectures}
\subsection{Friable numbers and the Dickman function}
For $2\le y \le x$, the function $\Psi(x,y):=\#\{n\le x: P(n)\le y\}$, which counts the number of ``$y$-friable'' or ``$y$-smooth'' numbers not exceeding $x$, has been studied extensively. In particular, it is known (see for instance Hildebrand and Tenenbaum \cite{kn:hildebrand}), that, given $\varepsilon>0$ and setting $u=\log x/\log y$,
$$\Psi(x,y)=x\rho(u) \left(1+ O_\varepsilon \left( \frac{\log(u+1)}{\log y} \right)
\right)$$
uniformly for $x\ge 3$, $\displaystyle{\exp\{(\log \log
x)^{\frac 53 + \varepsilon}\}\le y \le x }$,
where $\rho(u)$ stands for the Dickman function defined for $0\le u \le 1$ by $\rho(u)=1$ and for $u>1$ by the differential equation $u\rho'(u)=-\rho(u-1)$.
It can also be shown (see for instance Corollary 9.18 in the book of De Koninck and Luca \cite{kn:DL}) that
$$
\rho(u) = \exp\{-u(\log u + \log \log u -1 +o(1)) \} \qquad (u\to \infty),
$$
indicating that $\rho(u)$ decreases very rapidly as $u\to \infty$. In fact the following table provides the approximate values of $\rho(u)$ for $u=1,2,\ldots, 7$.
\vskip 5pt
\begin{center}
\begin{tabular}{|c||c|c|c|c|c|c|c|}\hline
$u$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
$\rho(u)$ & 1.0 & 0.3068 & 0.0486 & 0.00491 & 0.000354 & 0.0000196 & 0.00000087 \\ \hline
\end{tabular}
\vskip 3pt
\centerline{\sc Table 1}
\end{center}
\vskip 5pt
Related to the above is the difficult problem of estimating the number of friable (or smooth) values of polynomials. To do so, given a polynomial $f\in \Z[x]$ with positive leading coefficient, we set
$$\Psi(f;x,y):=\#\{n\le x: P(f(n))\le y\}.$$
Now, given $k$ irreducible polynomials $f_1,\ldots, f_k\in \Z[x]$, consider the polyonimal
$f(t):=f_1(t)\cdots f_k(t)$. Then, as stated by Martin \cite{kn:martin},
if we assume that the multiplicative
properties of the various $f_i(n)$ are independent of one another, we are led to the probabilistic
prediction that
\begin{equation} \label{eq:approx-f}
\Psi(f;x,y) \sim x\,\prod_{i=1}^k \rho\left( \frac{\log f_i(x)}{\log y} \right) \qquad (x\to \infty).
\end{equation}
As we will later see in Section \ref{sect:9}, this probabilistic relation will prove useful in estimating the expected size of the smallest elements of the various sets $E_{k,\ell}$.
\subsection{Estimates for the size of $G_\ell(x)$}
It was established by Ivi\'c and Pomerance \cite{kn:ivic-pomerance}
that
\begin{equation} \label{eq:ivic-pom}
G_2(x)=x\,\exp\{-(1+o(1))\sqrt{2\log x \log \log x}\} \qquad (x\to \infty).
\end{equation}
Observe that a more explicit expression for the right hand side of (\ref{eq:ivic-pom}) was later obtained by Ivi\'c \cite{kn:ivic}.
Now, the technique used in \cite{kn:ivic-pomerance} can be used to establish a more general result, namely that, for any fixed integer $\ell\ge 2$,
\begin{equation} \label{eq:ivic-pom-ell}
G_\ell(x) = x\,\exp\{-(1+o(1))\sqrt{2(\ell-1)\log x \log \log x}\} \qquad (x\to \infty).
\end{equation}
\subsection{The Bunyakovsky conjecture}
As we will see in the next section, there is an unexpected connection between the size of the sets $F(3,2)$ and $F(4,2)$ and a particular case of an old conjecture of Bunyakovsky \cite{kn:bun}, which essentially says that any irreducible polynomial with no fixed prime divisor contains infinitely many prime values.
\vskip 5pt
\noindent
{\bf Conjecture A (Bunyakovsky).} {\sl Let $f\in \Z[x]$ be an irreducible polynomial of positive degree and with positive leading coefficient such that the greatest common divisor of $f(1), f(2), f(3),\ldots$ is 1. Then there exist infinitely many values of $n$ for which $f(n)$ is prime.}
\vskip 10pt
In trying to prove that $F(4,2)$ is infinite and as will be seen in Section \ref{sect:3}, it would be helpful if we could say that there are infinitely many primes $p$ such that $P(p^2+1)< p$. However, no such claim has been proved, so far. Interestingly, although one can easily show that the sequence $(n^2+1)_{n\ge 1}$ is such that $P(n^2+1)< n$ for infinitely many integers $n$ (simply consider the subsequence $n=2m^2$, $m=1,2,\ldots$, for which $n^2+1=4m^4+1=(2m^2+2m+1)(2m^2-2m+1)$, and observe that $5\mid 2m^2+2m+1$ provided $m\equiv 1,3 \pmod 5$, in which case $P(2m^2+2m+1)0$ for $i=1,\ldots,k-1$, in such a way that the $k$ linear polynomials $x,c_1x+d_1,\ldots,c_{k-1}x+d_{k-1}$ are pairwise linearly independent over $\Q$.
We shall first construct $k$ consecutive polynomials $Q_i(x)$ with rational coefficients, namely
\begin{equation} \label{eq:poly}
Q_0(x) = x^2 \sum_{r=0}^{2k-1} a_r x^r, \qquad Q_i(x) = (c_i x+d_i)^2 \sum_{r=0}^{2k-1} b_{i,r} x^r \qquad (i=1,\ldots,k-1),
\end{equation}
where each $Q_i(x)\in \Q[x]$ and $Q_i(x)=Q_0(x)+i$ for $i=1,\ldots,k-1$.
In order for the equation $Q_1(x)=Q_0(x)+1$ to hold for all $x$, that is for the equation
\begin{equation} \label{eq:star-1}
(c_1 x +d_1 )^2 (b_{1,0} + b_{1,1} x + \cdots + b_{1,{2k-1}} x^{2k-1} ) = 1+ a_0 x^2 + a_1 x^3 + \cdots + a_{2k-1} x^{2k+1}
\end{equation}
to hold,
we need to equate the respective coefficients of $x^r$, $r=0,1,\ldots,2k+1$, on both sides of the above identity.
Equating the coefficients of $x^0$ and of $x^1$, we find
$$d_1^2 b_{1,0} = 1 \qquad \mbox{ and } \qquad d_1^2 b_{1,1} + 2 c_1 d_1 b_{1,0} =0,$$
which allows us to express
the values of $b_{1,0}$ and $b_{1,1}$ in terms of $c_1$ and $d_1$. Equating the coefficients of $x^2,x^3,\ldots,x^{2k+1}$ on both sides of (\ref{eq:star-1}), we find that
\begin{eqnarray*}
a_0 & = & d_1^2 b_{1,2} + 2 c_1 d_1 b_{1,1} +c_1^2 b_{1,0} ,\\
a_1 & = & d_1^2 b_{1,3} + 2 c_1 d_1 b_{1,2} +c_1^2 b_{1,1} ,\\
& \vdots & \\
a_{2k-3} & = & d_1^2 b_{1,2k-1} + 2 c_1 d_1 b_{1,2k-2} +c_1^2 b_{1,2k-3} , \\
a_{2k-2} & = & \qquad \qquad \quad 2 c_1 d_1 b_{1,2k-1} +c_1^2 b_{1,2k-2}, \\
a_{2k-1} & = & \qquad \qquad \qquad \qquad \qquad \quad c_1^2 b_{1,2k-1}.
\end{eqnarray*}
We then move to equation $Q_2(x)=Q_0(x)+2$ and again equate coefficients.
Equating the coefficients of $x^0$ and of $x^1$, we find
$$d_2^2 b_{2,0} = 2 \qquad \mbox{ and } \qquad d_2^2 b_{2,1} + 2 c_2 d_2 b_{2,0} =0,$$
which allows us to express $b_{2,0}$ and $b_{2,1}$ in terms of $c_2$ and $d_2$.
Equating the coefficients of $x^2,x^3,\ldots,x^{2k+1}$, we find that
\begin{eqnarray*}
a_0 & = & d_2^2 b_{2,2} + 2 c_2 d_2 b_{2,1} +c_2^2 b_{2,0} ,\\
a_1 & = & d_2^2 b_{2,3} + 2 c_2 d_2 b_{2,2} +c_2^2 b_{2,1} ,\\
& \vdots & \\
a_{2k-3} & = & d_2^2 b_{1,2k-1} + 2 c_2 d_2 b_{2,2k-2} +c_2^2 b_{2,2k-3} , \\
a_{2k-2} & = & \qquad \qquad \quad 2 c_2 d_2 b_{2,2k-1} +c_2^2 b_{2,2k-2}, \\
a_{2k-1} & = & \qquad \qquad \qquad \qquad \qquad \quad c_2^2 b_{2,2k-1}.
\end{eqnarray*}
Similarly, we construct $k-3$ other systems each with $2k$ equations. Then, from these $k-1$ systems, we see that
$$a_0 = d_1^2 b_{1,2} + 2 c_1 d_1 b_{1,1} +c_1^2 b_{1,0} =
d_2^2 b_{2,2} + 2 c_2 d_2 b_{2,1} +c_2^2 b_{2,0} = \cdots = d_2^2 b_{k-1,2} + 2 c_2 d_2 b_{k-1,1} +c_2^2 b_{k-1,0}$$
and obtain analogous identities for $a_1,a_2,\ldots,a_{2k-1}$. Hence, recalling that $c_1,\ldots,c_{k-1}$ and $d_1,\ldots,d_{k-1}$ are given, in all we obtain $2k(k-2)$ equations involving a total of $2(k-1)^2$ unknowns, namely $b_{i,j}$ where $1\le i \le k-1$ and $2\le j\le 2k-1$. To summarize, we have constructed a system of $2k^2-4k$ linear equations involving $2k^2-4k+2$ unknowns. This means that if we fix any two of these unknowns, we will obtain a unique solution for the set of $b_{i,j}$'s.
Finally, having obtained the values of $b_{i,j}$ for $1\le i \le k-1$ and $2\le j\le 2k-1$, we can use any of the $k-1$ systems of equations to determine the unique values of $a_0,a_1,\ldots,a_{2k-1}$.
Moreover, since all of the above equations are linear and involve rational coefficients, the coefficients $a_r$ are also rational. Let us then write each $a_r$ as
$$a_r = \frac{p_r}{q_r} \mbox{ where }p_r,q_r\in \Z,\ (p_r,q_r)=1,\ q_r>0 \quad \mbox{ and set } D:= \lcm[q_1,\ldots,q_r]$$
and consider the polynomials
$$L_i(x): = Q_i(D\,x) \qquad (i=0,\ldots,k-1).$$
We then have
$$L_0(x)=Q_0(Dx) = D^2\, x^2 \, \sum_{r=0}^{2k-1} a_r D^r \cdot x^r
=D\cdot x^2 \sum_{r=0}^{2k-1} D \cdot a_r \cdot (D x)^r.$$
By the nature of $D$, the coefficients of $L_0(x)/x^2$ are therefore all integers.
This implies that each of the polynomials $L_0(x),\ldots,L_{k-1}(x)$ has integer coefficients. Moreover, these polynomials are clearly consecutive.
We have thus created an infinite family of consecutive polynomials each with a squared factor, as required.
\end{proof}
\noindent
{\bf Example.} In the case $(k,\ell)=(4,2)$, the above construction yields the four consecutive polynomials
\begin{eqnarray*}
L_0(x) & = & x^2 \Big(17220 + a - 5 b + (815440 + 52 a - 248 b) x + (14339520 +
1108 a - 4916 b) x^2\\
& & + (120865536 + 12384 a - 48624 b) x^3 + (494346240 + 76608 a - 234432 b )x^4\\
& & + (789626880 + 248832 a - 324864 b) x^5 + (331776 a +
1327104 b )x^6 + 3981312 b x^7\Big)
\end{eqnarray*}
and $L_1(x)$, $L_2(x)$ and $L_3(x)$ whose squared factors are $(6x+1)^2$, $(8x+1)^2$ and $(12x+1)^2$, respectively.
We have thus constructed infinitely many such quadruples with parameters $a$ and $b$. Choosing $a=b=0$ allows for the more simple quadruple
\begin{small}
\begin{eqnarray*}
L_0(x) & = & x^2 \Big(17220 + 815440 x + 14339520 x^2 + 120865536 x^3 + 494346240 x^4 + 789626880 x^5 \Big)\\
L_1(x) & = & L_0(x)+1 = (1+ 6x)^2 \Big( 1 - 12 x + 17328 x^2 + 607936 x^3 + 6420480 x^4 + 21934080 x^5 \Big) \\
L_2(x) & = & L_0(x)+2 = 2 (1 + 8 x)^2 \Big(1 - 16 x + 8802 x^2 + 267912 x^3 + 2319840 x^4 +
6168960 x^5\Big) \\
L_3(x) & = & L_0(x)+3 = (1 + 12 x)^2 \Big(3 - 72 x + 18516 x^2 + 381424 x^3 + 2519040 x^4 +
5483520 x^5 \Big)
\end{eqnarray*}
\end{small}
Using the method of proof of Theorem \ref{thm:101}, the following result also holds.
\begin{theorem} \label{thm:102}
Given arbitrary integers $\ell_i\ge 2$, $i=0,1,\ldots,k-1$, there exist $k$ consecutive polynomials $L_0(x),L_1(x),\ldots,L_{k-1}(x)\in \Z[x]$ such that each $L_i(x)$, $i=0,1,\ldots,k-1$, is divisible by the $\ell_i$-th power of some linear polynomial.
\end{theorem}
\begin{proof}
We only provide a sketch of the proof. The idea is to let
$s:=\ell_0+\cdots+\ell_{k-1}$, to set
$$Q_0(x) := x^{\ell_0} \sum_{r=0}^{s-\ell_0-1} a_r x^r\quad \mbox{ and } \quad
Q_i(x) := (c_i x + d_i)^{\ell_i} \sum_{r=0}^{s-\ell_i-1} b_{i,r}x^r \mbox{ for } i=1,\ldots,k-1
$$
and then to search for the coefficients $a_r\in \Q$ and $b_{i,r}\in \Q$ in the same manner as we did in the proof of Theorem \ref{thm:101}. This allows us to obtain
$k$ consecutive polynomials $Q_0(x)$, $Q_1(x)$, $\ldots$, $Q_{k-1}(x)\in \Q[x]$ with the property that $Q_0(x)$ is divisible by $x^{\ell_0}$ whereas each $Q_i(x)$, for $i=1,\ldots,k-1$, is divisible by the $\ell_i$-th power of a linear polynomial $c_ix+d_i$ with $c_i,d_i\in \Q,\, c_i>0$. Using these $k$ polynomials with rational coefficients, we proceed as in the proof of Theorem \ref{thm:101} and obtain $k$ consecutive polynomials $L_0(x),L_1(x),\ldots,L_{k-1}(x)\in \Z[x]$ with the same properties.
\end{proof}
\begin{remark} \label{rem:8.1}
It follows from Theorems \ref{thm:101} and \ref{thm:102} that if Martin's probabilistic prediction (\ref{eq:approx-f}) is true, then each one of the sets $E_{k,\ell}$ and $F(\ell_0,\ell_1,\ldots,\ell_{k-1})$ is infinite.
\end{remark}
\section{Final remarks and heuristics}\label{sect:9}
Let us now examine the expected size of the smallest elements of $E_{k,\ell}$.
A consequence of estimate (\ref{eq:ivic-pom-ell}) is that the probability that a given large integer $n$ is such that $P(n)^\ell$ divides $n$ is
approximately $1/e^{\sqrt{2(\ell-1)\log n \log \log n}}$.
On the other hand, given an arbitrary integer $k \ge 2$, it is reasonable to assume that $P(n),P(n+1),\ldots,P(n+k-1)$ are independent events and therefore to conclude that the probability that $P(n+i)^\ell\mid n+i$ for $i=0,1,\ldots,k-1$ is around $1/e^{k\sqrt{2(\ell-1)\log n \log \log n}}$.
Using this approach, one can expect the smallest element of $E_{3,3}$ to have around 82 digits (that is, roughly the size of the numbers $n_0$ and $n_1$ obtained in Sections \ref{sect:1} and \ref{sect:6}, respectively) and that the smallest element of $E_{4,2}$ to have around 71 digits.
On the other hand, in line with our algebraic approach,
the following system (similar to the one displayed in Section \ref{sect:8}, but with smaller coefficients for each of the four degree 5 polynomials) clearly has the potential of generating infinitely many elements of $E_{4,2}$:
\begin{eqnarray*}
f(x) & = & 4x^2 (184896 x^5+ 292320 x^4 + 172500 x^3 + 46500 x^2 +5501 x + 195), \\
f(x)+1 & = & (2x+1)^2 (184896 x^5+ 107424 x^4 + 18852 x^3 + 792 x^2 -4x +1),\\
f(x)+2 & = & 2(4x+1)^2 (23112 x^5 + 24984 x^4 +7626 x^3 + 438 x^2 -8x +1),\\
f(x)+3 & = & (6x+1)^2 (20544 x^5 +25632 x^4 + 10052 x^3 + 1104 x^2 -36 x +3).
\end{eqnarray*}
Unfortunately, in order to find a number $n_1\in E_{4,2}$ using the above four polynomials, one would need much computer time since, in light of the conjectured estimate (\ref{eq:approx-f}) and of Table 1,
one can expect, as $x$ runs through the positive integers, that the probability that each of the above degree 5 co-factors has its largest prime factor smaller than the largest prime factor of their respective squared factors is smaller than $\rho(5)\approx 0.000354$, implying that the smallest integer $x$ meeting these four requirements would be larger than $1/\rho(5)^4 > 10^{14}$ and therefore that
$$n_1=f(x)> 4\cdot x^2 \cdot 184896 \cdot x^5 > 739584 \cdot (10^{14})^7 > 10^{103}.$$
Of course, one could perhaps come up with a smaller element of $E_{4,2}$ using a totally new approach.
Finally, if Martin's probabilistic estimate (\ref{eq:approx-f}) could be proved, not only would each set $E_{k,\ell}$ be infinite (as already mentioned in Remark \ref{rem:8.1}), but one could hope to find the approximate size of
$E_{k,\ell}(x)$.
\section{Acknowledgment}
The authors would like to thank the referee for pointing out a previous result of Hildebrand (see Remark \ref{rem:9}). The first author was supported in part by a grant from NSERC.
\begin{thebibliography}{99}
\bibitem{kn:bun} V. Bunyakovsky, Nouveaux th\'eor\`emes relatifs \`a la distinction des nombres premiers et \`a la d\'ecomposition des entiers en facteurs, {\it M\'em. Acad. Sc. St. P\'etersbourg} {\bf 6} (1857), 305--329.
\bibitem{kn:jm-book} J.-M. De Koninck, {\it Those Fascinating Numbers},
American Mathematical Society, 2009.
\bibitem{kn:jm-nic-fl} J.-M. De Koninck, N. Doyon, and F. Luca,
Consecutive integers divisible by the square of their largest prime
factors, {\it J. Comb. Number Theory} {\bf 5}
(2013), 81--93.
\bibitem{kn:DL} J.-M. De Koninck and F. Luca, {\it Analytic Number Theory: Exploring
the Anatomy of Integers}, Graduate Studies in Mathematics, Vol.~134, American Mathematical Society,
2012.
\bibitem{kn:drappeau} R. de la Bret\`eche et S. Drappeau, Majoration du nombre de valeurs friables d'un polyn\^ome, preprint.
\bibitem{kn:hildebrand-1986} A. Hildebrand,
On consecutive values of the Liouville function, {\it Enseign. Math.} (2) {\bf 32} (1986), 219--226.
\bibitem{kn:hildebrand} A. Hildebrand and G. Tenenbaum, Integers without large prime factors, {\it Journal de Th\'eorie des Nombres de Bordeaux} {\bf 5} (1993), 411--484.
\bibitem{kn:ivic} A. Ivi\'c, On certain large additive functions, in
G\'abor Hal\'asz, L\'aszl\'o Lov\'asz, Mikl\'os Simonovits and Vera T. S\'os,
eds.,
{\it Paul Erd\H{o}s and his Mathematics I},
Bolyai Society Mathematical Studies, Vol.~11,
Springer-Verlag, 2002, pp.~ 319--331.
\bibitem{kn:ivic-pomerance} A. Ivi\'c and C. Pomerance,
Estimates for certain sums involving the largest prime factor of
an integer, in {\it Topics in Classical Number Theory}, Vol.~I, II
(Budapest, 1981), {\it Colloq. Math. Soc. J\'anos Bolyai}
{\bf 34}, North-Holland, 1984, pp.~769--789.
\bibitem{kn:martin} G. Martin, An asymptotic formula for the number of smooth values of a polynomial, {\it J. Number Theory} {\bf 93} (2002), 108--182.
\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}: Primary 11N25; Secondary 11N32. \\
\noindent \emph{Keywords: } largest prime factor.
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received June 6 2018;
revised version received October 15 2018.
Published in {\it Journal of Integer Sequences}, December 6 2018.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in
\end{document}
\end{document}