\documentclass[12pt]{article} \textwidth= 6.25in \textheight= 9.0in \topmargin = -10pt \evensidemargin=10pt \oddsidemargin=10pt \headsep=25pt \parskip=10pt \font\smallit=cmti10 \font\smalltt=cmtt10 \font\smallrm=cmr9 \usepackage{amssymb,amsmath,latexsym,theorem} \def\B{\hfill $\Box$} \begin{document} \vspace*{-60pt} \centerline{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4 (2004), \#A11} \vskip 50pt \begin{center} {\bf \uppercase{A Variation on Perfect Numbers}} \vskip 20pt {\bf Roger Woodford\footnote{Supported by an NSERC undergraduate student research award.}}\\ {\smallit Department of Mathematics, University of Manitoba, Winnipeg, MB R3M 2N2, Canada}\\ {\tt rogerw@math.ubc.ca}\\\vskip 10pt \end{center} \vskip 30pt \centerline{\smallit Received: 7/28/03, Revised: 3/4/04, Accepted: 7/6/04, Published: 7/9/04 } \vskip 30pt \centerline{\bf Abstract} \noindent For $k\in \mathbb{N}$ we define a new divisor function $s_{k}$ called the $k^{th}$ prime symmetric function. By analogy with the sum of divisors function $\sigma$, we use the functions $s_{k}$ to consider variations on perfect numbers, namely $k$-symmetric-perfect numbers as well as $k$-cycles. We find all $k$-symmetric-perfect numbers for $k = 1, 2, 3$. We also consider the problem of whether a natural number $n$ can be expressed in the form $s_{k}(n)$, and show that for $n$ large enough, it always can be for $k = 1, 2$. \pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4 (2004), \#A11\hfill} \thispagestyle{empty} \baselineskip=15pt \vskip 30pt \section*{\normalsize 1. Introduction} \noindent {\bf Definition 1:} Let $k$ be a nonnegative integer. We define $s_{k}: \mathbb{N} \rightarrow \mathbb{N}\cup\{0\}$ as follows: If $k = 0$, $s_{k}(n) \equiv 1$. If $k > 0$, and $n = p_{1}\cdots p_{r}$, where $r = \Omega(n)$ is the number of prime factors (with multiplicity) of $n$, then \begin{equation*} s_{k}(n) = \sum p_{i_{1}}\cdots p_{i_{k}}, \end{equation*} where the sum is taken over all products of $k$ prime factors from the set $\{p_{1}, \ldots , p_{r}\}$. We say $s_{k}$ is the $k^{th}$ prime symmetric function. Note that if $\Omega(n) < k$, we have $s_{k}(n) = 0$. There is an alternate way of defining the functions $s_{k}$. Given $n = p_{1}\cdots p_{r} \in \mathbb{N}$, set \begin{equation*}S_{n}(x) = \prod_{i=1}^{r}(x + p_{i}). \end{equation*} Then $s_{k}(n)$ is the coefficient of $x^{r-k}$ in $S_{n}(x)$. The empty product is taken to be 1. \noindent {\bf Example 1:} $s_{0}(12) = 1$, $s_{1}(12) = 2 + 2 + 3 = 7$, $s_{2}(12) = 2\cdot2 + 2\cdot3 + 2\cdot3 = 16$, $s_{3}(12) = 12$, and $s_{4}(12) = 0$. Several good texts detailing the basic theory of perfect numbers exist, see for instance ~\cite{dB76},~\cite{wD99},~\cite{gL82}, and ~\cite{jT99}. In addition, many variations on perfect numbers have been defined and studied. For examples, see the remaining references. We now define a new variation of perfect, defective, and excessive numbers using the divisor functions $s_{k}$. \noindent {\bf Definition 2:} Let $n \in \mathbb{N}$. If $s_{k}(n) < n$, we say $n$ is $k$-symmetric-defective. If $s_{k}(n) > n$, we say $n$ is $k$-symmetric-excessive. If $s_{k}(n) = n$, and $\Omega(n) = k$, we say $n$ is trivially $k$-symmetric-perfect. If $s_{k}(n) = n$, and $\Omega(n) > k$, we say $n$ is $k$-symmetric-perfect. If $n$ is $k$-symmetric-perfect or $k$-symmetric-excessive, we say $n$ is $k$-symmetric-special. \noindent {\bf Notation:} For the sake of brevity we write $k$-SD for $k$-symmetric-defective, $k$-SP for $k$-symmetric-perfect, $k$-SE for $k$-symmetric-excessive, and $k$-SS for $k$-symmetric-special. \noindent {\bf Example 2:} If $p$ is prime, then $p^{p}$ is a $(p-1)$-SP number, since \begin{equation*} s_{p-1}(p^{p}) = \binom{p}{p-1}p^{p-1} = p^{p}. \end{equation*} In fact, this example has a form of converse: \noindent {\bf Theorem 1:} The prime power $p^{\alpha}$ is $k$-SP if and only if $\alpha=p$ and $k=p-1$. \noindent {\it Proof.} We have seen that this is sufficient, now suppose $k < \alpha$, and $s_{k}(p^{\alpha}) = p^{\alpha}$. Then \begin{equation}\binom{\alpha}{k} = p^{\alpha - k}. \end{equation} For $1 < k < \alpha - 1$, $\binom{\alpha}{k}$ is divisible by two distinct prime factors, hence we must have $k = 1$ or $\alpha - 1$. Now $4 = 2^{2}$, is the only $1$-SP number, and corresponds to the case where $k = 1 = \alpha - 1$. Hence we may assume $k = \alpha - 1$, which from (1) implies that $\alpha = p$ and $k = p - 1$. This proves the theorem. \B \noindent {\bf Definition 3:} A finite sequence $\{n_{0}, \ldots , n_{\ell}\}$ is called a $k$-cycle if the following conditions are satisfied: \begin{enumerate} \item $\ell > 1$, \item $n_0, \dots, n_{\ell-1}$ are distinct and $n_{\ell}=n_0$, and \item $s_k(n_i)=n_{i+1}$, for $i=0,1,\dots,\ell-1$. \end{enumerate} \vskip 30pt \section*{\normalsize 2. Basic Properties of Prime Symmetric Functions} The following proposition is an immediate consequence of the definition. \noindent {\bf Proposition 2:} If $n = p_{1}^{\alpha_{1}}\cdots p_{r}^{\alpha_{r}}$, then \begin{equation*}s_{k}(n) = \sum_{\begin{array}{c} i_1+\cdots+i_r=k \\ i_1,\,\dots,i_r\ge0 \end{array} }\binom{\alpha_{1}}{i_{1}}\cdots\binom{\alpha_{r}}{i_{r}}p_{1}^{i_1}\cdots p_{r}^{i_r}. \end{equation*} \noindent {\bf Proposition 3:} \begin{equation*}s_{k}(mn) = \sum_{i=0}^{k}s_{k-i}(m)s_{i}(n). \end{equation*} \noindent {\it Proof.} If $m = 1$, or $n = 1$, the result is immediate, as it is if $k = 0$. If $k > 0$, $m = p_{1}\cdots p_{r}$, and $n = q_{1}\cdots q_{s}$, let \begin{equation*}S = \{p_{1}, \ldots, p_{r}, q_{1}, \ldots, q_{s}\}. \end{equation*} Then \begin{equation*}s_{k}(mn) = \sum_{\{r_{1}, \ldots, r_{k}\}\subset S}r_{1} \cdots r_{k}. \end{equation*} We collect the terms of this sum having $k-i$ factors from $m$, and $i$ factors from $n$. The sum of these is equal to $s_{k-i}(m) s_{i}(n)$. Summing as $i$ ranges from $0$ to $k$ gives the desired result. \B \noindent {\bf Corollary 4:} Let $n$, $k \in \mathbb{N}$, and let $p$ and $q$ be primes, with $p < q$, and suppose $\Omega(pn) > k$. If $pn - s_{k}(pn) > 0$, then $pn - s_{k}(pn) < qn - s_{k}(qn)$. \noindent {\it Proof.} The following inequality \begin{align*}qn - s_{k}(qn) &= qn - qs_{k-1}(n) - s_{k}(n)\\ &> pn - ps_{k-1}(n) - s_{k}(n)\\ &= pn - s_{k}(pn) \end{align*} is true if $n > s_{k-1}(n)$. But \begin{equation*}pn - s_{k}(pn) = pn - ps_{k-1}(n) - s_{k}(n) > 0 \end{equation*} by assumption, so \begin{equation*}n > s_{k-1}(n) + s_{k}(n)/p > s_{k-1}(n). \end{equation*} \B In searching for $k$-cycles and $k$-SP numbers, it is essential to know when $s_{k}(n) \geq n$. We search by fixing $\Omega(n)$, and systematically checking all products of $\Omega(n)$ primes. The corollary tells us that if $s_{k}(pn) < pn$, then for any $q > p$, $qn$ is also $k$-SD. \noindent {\bf Lemma 5:} Let $k$, $n \in \mathbb{N}$. Then there exists an $r > k$ such that if $\Omega(n) \geq r$, then $n$ is $k$-SD. Let $r(k)$ denote the least such $r$. Then \begin{equation*} r(k)=\min\{\,r\,:\,\binom{r}{k}<2^{r-k}\,\}\,. \end{equation*} \noindent {\it Proof.} There is an $r > k$ such that the function \begin{equation*}f(t) = \binom{t}{k} \end{equation*} satisfies $f(t) < g(t)$ for all $t \geq r$, where \begin{equation*}g(t) = 2^{t-k}, \end{equation*} since $f$ is a polynomial, and $g$ is an exponential function. Now suppose $t \geq r$, and let $p_{1}, \ldots, p_{t}$ be $t$ primes. Then \begin{align*}\binom{t}{k} = \binom{t}{t-k} &< 2^{t-k}, \text{ which implies }\,\, \sum\frac{1}{p_{i_{1}}\cdots p_{i_{t-k}}} \leq \binom{t}{t-k}\frac{1}{2^{t-k}} < 1, \\ \end{align*} where the sum is taken over all $i_1$, \dots, $i_{t-k}$ such that $1\le i_1<\cdots< i_{t-k}\le t$. This implies \begin{equation*} \sum p_{i_{1}}\cdots p_{i_{k}} < p_1\cdots p_t, \end{equation*} where the sum is taken over all $i_1, \dots, i_k$ such that $1\le i_1<\cdots< i_k\le t$. This in turn implies that \begin{equation*} s_{k}(p_1\cdots p_t) < p_1\cdots p_t. \end{equation*} Now we prove the second statement. The inequality \begin{equation*} \binom{2k}{k}\geq2^k \end{equation*} holds for all $k \geq 1$, and so $r(k)>2k$. This in mind, let $r(k)$ be as claimed in the statement of the theorem. We argue inductively. Let $t>r$, and suppose that \begin{equation*} \binom{t-1}{k}<2^{t-1-k}. \end{equation*} Then \begin{equation*} 2\binom{t-1}{k}<2^{t-k}. \end{equation*} Since $t>2k$, we have $t<2(t-k)$, and so \begin{equation*} \binom{t}{k}=\frac{t(t-1)\cdots(t-k+1)}{k!}<\frac{2(t-1)(t-2)\cdots(t-k)}{k!}=2\binom{t-1}{k}. \end{equation*} Hence \begin{equation*} \binom{t}{k}<2\binom{t-1}{k}<2^{t-k}, \end{equation*} and the proof is complete by induction. \B The first few values of $r(k)$ are given in the following table: \begin{center} \begin{tabular}{|c|c|} \hline $k$ & $r(k)$ \\ \hline 1 & 3 \\ \hline 2 & 6 \\ \hline 3 & 10 \\ \hline 4 & 14 \\ \hline 5 & 19 \\ \hline 6 & 23 \\ \hline 7 & 27 \\ \hline 8 & 31 \\ \hline 9 & 36 \\ \hline 10 & 40 \\ \hline \end{tabular} \end{center} The properties of 1-symmetric-perfection etc. corresponding to the first prime symmetric function $s_{1}$ are easily characterized. The primes are the trivial 1-SP numbers, 4 is the only 1-SP number, and all other numbers are 1-SD. Clearly there are no 1-cycles. We now investigate these properties in the second prime symmetric function. \vskip 30pt \section*{\normalsize 3. The Second Prime Symmetric Function} Let $n$ be an integer greater than 1. By a family $E_{k}(n, r)$ of $k$-SS numbers, we mean a set \begin{equation*} E_{k}(n,r) = \{np_{1}\cdots p_{r} |p_{1}, \ldots , p_{r} \text{ are primes} \} \end{equation*} such that if $m \in E_{k}(n, r)$, then $m$ is $k$-SS. The family $E_{2}(4,1)$ of numbers of the form $4p$, where $p$ is prime, is one such set, since the elements satisfy $s_{2}(4p) = 4p + 4 > 4p$. $E_{k}(n, 0)$ merely denotes the singleton set of a $k$-SS number. To find all 2-SP numbers and all 2-cycles we need to find all numbers $n$ such that $2 < \Omega(n) < 6$, with $s_{2}(n) \geq n$, since $r(2) = 6$. To do this we use the algorithm mentioned after Corollary 4. \subsection*{\normalsize 3.1. $\Omega(n) = 3$} \begin{align*}s_{2}(2\cdot2\cdot p) &= 4p + 4 > 4p,\\ s_{2}(2\cdot3\cdot p) &= 5p + 6 < 6p, \text{ when $p > 6$.} \end{align*} This shows that there are no other infinite families of 2-SS numbers satisfying $\Omega(n) = 3$. Below we find all 2-SS numbers not belonging to this family. \begin{align*}s_{2}(2\cdot3\cdot 3) &= 21 > 18,\\ s_{2}(2\cdot3\cdot 5) &= 31 > 30,\\ s_{2}(2\cdot3\cdot 7) &= 41 < 42,\\ s_{2}(3\cdot3\cdot 3) &= 27,\\ s_{2}(3\cdot3\cdot 5) &= 39 < 45. \end{align*} Thus 27 is the only 2-SP number satisfying $\Omega(n) = 3$. Iterating on the above 2-SE numbers shows none belong to 2-cycles. For example \begin{equation*}18 \overset{s_{2}}{\longrightarrow} 21 \overset{s_{2}}{\longrightarrow} 10 \overset{s_{2}}{\longrightarrow} 10 \overset{s_{2}}{\longrightarrow} \cdots. \end{equation*} \subsection*{\normalsize 3.2. $\Omega(n) = 4$} \begin{equation*}s_{2}(2\cdot2\cdot2\cdot p) = 6p + 12 < 8p, \text{ when $p > 6$.} \end{equation*} Thus there are no infinite families of 2-SS numbers with $\Omega(n) = 4$. Iterating on $8p$ for $p = 2$, 3, 5, shows that none belong to a 2-cycle. Checking other cases: \begin{align*}s_{2}(2\cdot2\cdot3\cdot3) &= 37 > 36,\\ s_{2}(2\cdot2\cdot3\cdot5) &= 51 < 60,\\ s_{2}(2\cdot3\cdot3\cdot3) &= 45 < 54. \end{align*} Hence there are no 2-SP numbers satisfying $\Omega(n) = 4$. Iterating on the above 2-SE numbers shows that none belong to a 2-cycle. \subsection*{\normalsize 3.3. $\Omega(n) = 5$} \begin{equation*}s_{2}(2\cdot2\cdot2\cdot2\cdot p) = 8p + 24 < 16p, \text{ when $p > 3$.} \end{equation*} Thus there are no infinite families of 2-SS numbers with $\Omega(n) = 5$. Iterating on $16p$ for $p = 2$, 3, shows that 48 is in fact 2-SP, and 32, which is 2-SE, does not belong to a 2-cycle. Checking other cases: \begin{equation*}s_{2}(2\cdot2\cdot2\cdot3\cdot3) = 57 < 72. \end{equation*} Hence 48 is the only 2-SP number satisfying $\Omega(n) = 5$. We have proved the following theorem. \noindent {\bf Theorem 6:} 27 and 48 are the only 2-SP numbers. \noindent {\bf Theorem 7:} There are no 2-cycles. \noindent {\it Proof.} A 2-cycle must have a least element that is 2-SE. We have shown that any such element must belong to the family of numbers of the form $4p$. We will show that in all but a few trivial cases $s_{2}(s_{2}(4p)) < 4p$, giving a contradiction. Now, $s_{2}(4p) = 8((p + 1)/2)$. We may assume that $p$ is odd, and set $m = (p + 1)/2$. Thus we will have a contradiction if the following holds: \begin{equation*}s_{2}(8m) < 8m - 4. \end{equation*} This is equivalent to \begin{equation}12 + 6s_{1}(m) + s_{2}(m) < 8m - 4, \end{equation} which is equivalent to \begin{equation*} \frac{16}{p_{1}\cdots p_{s}} + 6\sum_{i=1}^{s}\frac{1}{p_{1}\cdots \hat{p_{i}}\cdots p_{s}} + \sum_{1\leq i 8$. It is easily verified for $p = 2$, 3, 5 and 7, that $8p$ does not belong to a 2-cycle. For $s = 2$, we can write $m = pq$. The only values of $m$ for which (2) fails are determined by the prime pairs $(p, q) = (2, 2)$, $(2, 3)$. In both cases, $8m$ does not belong to a 2-cycle. Finally for $s = 3$, if $m = pqr$, only for the triple $(p, q, r) = (2, 2, 2)$ does $m$ fail to satisfy (2). Again, in this case, $8m$ does not belong to a 2-cycle. \B \noindent {\bf Definition 4:} A sequence $\{n_{i}\}$ (finite or infinite) is called a $k$-ascending sequence if $n_{i} < s_{k}(n_{i}) = n_{i+1}$. If $\{n_{i}\} = \{n_{i}\}_{i=0}^{t}$, then $\{n_{i}\}$ is said to have length $t$. \noindent {\bf Remark:} The longest 2-ascending sequence is \begin{equation*}8 \overset{s_{2}}{\longrightarrow} 12 \overset{s_{2}}{\longrightarrow} 16 \overset{s_{2}}{\longrightarrow} 24 \overset{s_{2}}{\longrightarrow} 30 \overset{s_{2}}{\longrightarrow} 31. \end{equation*} \noindent {\bf Definition 5:} Let $k \in \mathbb{N}\cup\{0\}$. We define $r_{k}:\mathbb{N} \rightarrow \mathbb{N}\cup\{0\}$ by \begin{equation*}r_{k}(n) = |\{s_{k}^{-1}[\{n\}]| \end{equation*} \noindent {\bf Example 3:} $r_{1}(1) = 0$, but for all $n \geq 2$, $r_{1}(n) \geq 1$. In fact, $\lim_{n \to \infty} r_{1}(n) = \infty$. To see this, simply set \begin{equation*}n = s_{1}(2^{a}3^{b}) = 2a + 3b, \end{equation*} and observe that the number of pairs $(a, b)$ satisfying this equation can be made arbitrarily large for all $n$ sufficiently large. We prove a weaker result for $r_{2}$. \noindent {\bf Theorem 8:} There exists an $N \in \mathbb{N}$ such that for all $m \geq N$, $r_{2}(m) \geq 1$. \noindent {\it Proof.} It suffices to show that for $m$ sufficiently large, $m = s_{2}(2^{a}3^{b}5^{c}7^{d})$, for some $a$, $b$, $c$, and $d \geq 0$. In general, \begin{align*}s_{2}(2^{a}3^{b}5^{c}7^{d}) &= 4\binom{a}{2} + 9\binom{b}{2} + 25\binom{c}{2} + 49\binom{d}{2}\\\notag &+ 6\binom{a}{1}\binom{b}{1} + 10\binom{a}{1}\binom{c}{1} + 14\binom{a}{1}\binom{d}{1}\\\notag &+ 15\binom{b}{1}\binom{c}{1} + 21\binom{b}{1}\binom{d}{1} + 35\binom{c}{1}\binom{d}{1}\\ &= \frac{1}{2}\left[(2a + 3b + 5c + 7d)^{2} - (4a + 9b + 25c + 49d)\right] \end{align*} So, given $m$, we need only find solutions to the equations: \begin{align*}2a + 3b + 5c + 7d &= R,\\ 4a + 9b + 25c + 49d &= R^{2} - 2m, \end{align*} with nonnegative integers $a, b, c, d$, and $R \in \mathbb{N}$. These equations are equivalent to: \begin{alignat}{2}2a - 10&c - 28&d &= 3R - R^{2} + 2m,\\ 3b + 15&c + 35&d &= R^{2} - 2R - 2m, \end{alignat} Since $a$ and $b$ must be nonnegative integers, we have the following necessary and sufficient conditions for a solution to (3) and (4): \begin{enumerate}\item $2m \equiv R^{2} + R + d$ (mod 3), \item $R^{2} - 3R - 10c -28d \leq 2m$, \item $2m \leq R^{2} - 2R - 15c - 35d$. \end{enumerate} Note that equation (3) is always satisfied modulo 2. Condition 1 results from taking equation (4) modulo 3, and conditions 2 and 3 are derived from the fact that $a, b \geq 0$. Consider the interval \begin{equation*} I_{R}(c, d) = [R^2 - 3R - 10c -28d, R^{2} - 2R - 15c - 35d]. \end{equation*} For fixed $d$, let $c_{R}(d)$ be the least $c$ such that $\ell(I_{R}(c, d)) < 15$, where $\ell(I)$ denotes the length of an interval $I$. We use the notation $L(I)$ and $R(I)$ to denote the left and right endpoints of an interval $I$, respectively. Since $R(I_{R}(c, d)) = R(I_{R}(c + 1, d)) + 15$, when they exist, we have that \begin{equation*} \bigcup_{c=0}^{c_{R}(d)}I_{R}(c, d) = [R^{2} - 3R - 10c_R(d) - 28d, R^{2} - 2R - 35d]. \end{equation*} Denote the above interval by $\mathcal{I}_R(d)$. By definition of $c_R(d)$, \begin{align*} \ell(I_R(c_R(d), d)) &= R - 5c_R(d) - 7d < 15, \text{ so }\,\, -10c_R(d) < -2R + 30 + 14d, \end{align*} and $c_R(d)$ is the least such $c$. Consider the interval $\bigcap_{d=0}^{2}\mathcal{I}_R(d)$. Clearly $R(\bigcap_{d=0}^{2}\mathcal{I}_R(d)) = R^2-2R-70$. We now wish to find an upper bound for $L(\bigcap_{d=0}^{2}\mathcal{I}_R(d))$. From the above inequality, we have that \begin{equation*} L(\mathcal{I}_R(d))=R^2 - 3R - 10c_R(d) - 28d k$, we have $s_{k+1}(n) > 0$. So \begin{align*}s_{k+1}(pn) &= ps_{k}(n) + s_{k+1}(n)\\ &\geq pn + s_{k+1}(n)\\ &> pn. \end{align*} (2) If $s_{k+1}(pn) = ps_{k}(n) + s_{k+1}(n) \geq pn$, for every prime $p$, then $s_{k}(n) \geq n - s_{k+1}(n)/p$. Letting $p \rightarrow \infty$, we have $s_{k}(n) \geq n$. \B \noindent {\bf Corollary 10:} For $k \in \mathbb{N}$, there are only finitely many $k$-SP numbers. \noindent {\it Proof.} By the previous theorem, any family $E_{k+1}(n, r+1)$ is of the form $pE_{k}(n, r)$, where $p$ ranges over the primes. Furthermore, this family contains only $(k+1)$-SE numbers. There are only finitely many $(k+1)$-SS numbers not belonging to any such family. \B Thus the infinite families of $3$-SE numbers are: $E_{3}(4, 2)$, $E_{3}(16, 1)$, $E_{3}(18, 1)$, $E_{3}(24, 1)$, $E_{3}(27, 1)$, $E_{3}(30, 1)$, $E_{3}(32, 1)$, $E_{3}(36, 1)$, $E_{3}(40, 1)$, $E_{3}(48, 1)$. By exhaustive search (as was done with $k = 2$), all other 3-SS numbers can be found. They constitute the following set: \begin{align*} &\{42p | p = 7, 11, \ldots , 41\} \cup \{56p | p = 7, 11, \ldots , 43\} \cup \{64p | p = 2, 3, \ldots , 37\} \cup\\ &\{726, 858, 250, 350, 225, 315, 968, 1144, 300, 420, 162, 270, 378, 243, 400, 560,\\ & 216, 360, 504, 324, 288, 480, 672, 432, 256, 384, 640, 576, 512, 768\}. \end{align*} None of the elements in the above sets are $3$-SP, hence there are no $3$-SP numbers. The diversity of possible $3$-ascending sequences makes it difficult to rule out the existence of $3$-cycles as we did $2$-cycles. This is illustrated in the following example. \noindent {\bf Example 4:} If $p_{1}$, $q_{1}$ are odd primes, then $s_{3}(4p_{1}q_{1}) = 4(p_{1}q_{1} + p_{1} + q_{1})$. It is possible that $p_{1}q_{1} + p_{1} + q_{1} = p_{2}q_{2}$, where $p_{2}$, $q_{2}$ are again odd primes, and so on. Several such sequences exist the longest one with $p_{1}q_{1} < 50000$, and $p_{i}, q_{i} > 3$ is: \begin{alignat*}{3}184892 &= 4\cdot17\cdot2719 \overset{s_{3}}{\longrightarrow} 195836 &= 4\cdot173\cdot283 \overset{s_{3}}{\longrightarrow} 197660 &= 4\cdot5\cdot9883\\ \overset{s_{3}}{\longrightarrow} 237212 &= 4\cdot31\cdot1913 \overset{s_{3}}{\longrightarrow} 244988 &= 4\cdot73\cdot839 \overset{s_{3}}{\longrightarrow} 248636 &= 4\cdot61\cdot1019\\ \overset{s_{3}}{\longrightarrow} 252956 &= 4\cdot11\cdot5749 \overset{s_{3}}{\longrightarrow} 275996 &= 4\cdot7\cdot9857 \overset{s_{3}}{\longrightarrow} 315452 &= 4\cdot17\cdot4639\\ \overset{s_{3}}{\longrightarrow} 334076 &= 4\cdot47\cdot1777 \overset{s_{3}}{\longrightarrow} 341372 &= 4\cdot31\cdot2753 \overset{s_{3}}{\longrightarrow} 352508 &= 4\cdot13\cdot6779\\ \overset{s_{3}}{\longrightarrow} 379676 &= 4\cdot11\cdot8629 \overset{s_{3}}{\longrightarrow} 414236 &= 4\cdot29\cdot3571 \overset{s_{3}}{\longrightarrow} 428636 &= 4\cdot13\cdot8243\\ \overset{s_{3}}{\longrightarrow} 461660 &= 4\cdot5\cdot41\cdot563. \end{alignat*} It seems highly unlikely, however, that a 3-ascending sequence be infinite. 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