p^k$. \end{proposition} \begin{proof} We proceed by considering two cases. \medskip \noindent Case I: If all nontrivial divisors of $n$ are divisible by $p$, then $n=p^k$ for some $k \ge 1$. \medskip \noindent Case II: Assume that $n$ has a divisor $d> \sqrt{n}$ that is not divisible by $p$. If $d$ is composite, then one of its prime divisors must be less than or equal to $\sqrt{d}< \sqrt{n}$, and must hence be divisible by $p$. This contradicts our assumption, and $d$ must thus be some prime $q$. \medskip We now prove uniqueness. If there exists another prime $r$ in the interval $(\sqrt{n}, n)$ that divides $n$, then $qr$ must divide $n$ but $qr > \sqrt{n} \cdot \sqrt{n} =n$, another contradiction. Hence, $q$ is the sole divisor of $n$ that is not divisible by $p$, and thus we have $n=p^kq$. \end{proof} \begin{proposition} In a recurrent number $n$, if $\mathcal{S}_n$ satisfies a recurrence $U(1,p,0,b)$ (i.e.\ , $a=0$), then exactly one of the following possibilities holds, where $p,q,r$ are primes and $k \in \mathbb{N}$: \begin{enumerate} \item $n=p^k$. \item $n=p^kq$ for $p^kp$). The former gives rise to $\mathcal{S}_n=\{1,p,p^2, \dots , p^{k}\}$, while the latter implies that $\mathcal{S}_n=\{1,p,q,pq,q^2,\dots, pq^k\}$ or $\{1,p,q,pq,q^2,\dots, q^k\}$, depending on parity. Identical to the proof of Proposition \ref{pkq}, there is at most one prime divisor greater than $\sqrt{n}$ that divides $n$, and hence the possibilities follow directly. \end{proof} \section{The case $\gcd(a,b)=1$} We now turn our attention to the major case where $\gcd(a,b)=1$. We also assume, henceforth, that $n$ has at least two distinct small prime factors---otherwise we are simply in the second and third cases of Theorem \ref{thm}. In particular, $ab \ne 0$ since if one of the numbers is zero, then the other cannot have absolute value $1$, as negative numbers do not appear in $\mathcal{S}_n$ and no number appears twice in the sequence of small divisors. \begin{lemma} \label{bco} For all small divisors $d_i$ of a recurrent number we have $\gcd({b,d_i})=1$. \end{lemma} \begin{proof} Suppose that $q \mid \gcd({b,d_i})$ for some prime $q$ and some index $i$. Notice that this $i \ne 1$. Similarly, if $i=2$, then $p \mid b$ and inductively, all small divisors are divisible by $p$ which contradicts the assumption of having two distinct small prime factors. Suppose $i \ge 3$. Since $d_i=ad_{i-1}+bd_{i-2}$, we have $q \mid ad_{i-1}$. The coprimality of $a$ and $b$ implies that $q \mid d_{i-1}$. We can inductively show that smaller divisors, including $d_1=1$, are divisible by $q$, which is absurd. Hence the coprimality follows. \end{proof} \begin{corollary} \label{coprime} For all $i$ such that $d_i, d_{i+1}$ are small divisors of a recurrent number, we have $\gcd({d_i, d_{i+1}})=1$. \end{corollary} \begin{proof} We proceed by induction. The base case clearly holds. For the inductive step, observe that $$\gcd{(d_i, d_{i+1}})=\gcd{(d_i, ad_{i}+bd_{i-1})} = \gcd{(d_i, bd_{i-1})}.$$ The inductive hypothesis implies that $\gcd{(d_i, d_{i-1})}=1$, while Lemma \ref{bco} implies that $\gcd{(d_i,b})=1$, and hence the coprimality follows. \end{proof} As a direct consequence of this, we conclude that $d_3$ cannot be $p^2$ in a recurrent number, and hence $d_3$ must be prime. Henceforth, we set $d_3=q$. \begin{lemma} All configurations of the first four nontrivial divisors $(d_2,d_3,d_4,d_5)$ except possibly $(p,q,p^2,r), (p,q,r,p^2)$, and $(p,q,r,s)$, where $p,q,r,s$ are distinct primes, are impossible in a recurrent number. \end{lemma} \begin{proof} We have already determined $d_2,d_3$. Now, $d_4$ can be $p^2,q^2,pq$, or $r$ (where $r$ is another prime). By Corollary \ref{coprime}, $d_4$ cannot be $pq$ or $q^2$. Hence, we have either $d_4=p^2$ or $d_4=r$. If $d_4=p^2$, then we can similarly see that the possibilities for the next divisor $d_5$ are $p^3,q^2,pq,r$ for a prime $r$. However, Corollary \ref{coprime} once again implies that $p^3, pq$ are impossible. If $d_5=q^2$, then since $d_5=ap^2+bq$ we would get that $q \mid ap^2$, and so, $q \mid a$. Yet, since $q=ap+b, q \mid a$ would imply that $q \mid b$ and thus contradicts Lemma \ref{bco}. Hence, $d_5$ ought to be another prime, $r$. If $d_4=r$, then we see that our possibilities for the next divisor $d_5$ are $p^2,pq,s$, for a new prime $s$. A reasoning similar to above shows that $q \nmid d_5$, so the only possibilities left are those of $d_5=p^2$ and $d_5=s$. \end{proof} \begin{lemma} \label{bounds} The only pair $(u,x) \in \mathbb{N} \times \mathbb{N}_{\ge 2}$ that satisfies $$-ux^5+(u+3)x^4-\frac{2u+3}{u}x^3+\frac{3u+1}{u^2}x^2-\frac{2}{u^2}x+\frac{1}{u^2} \ge x^2+1$$ is $(1,2)$. \end{lemma} \begin{proof} Define $$P(x):=-ux^5+(u+3)x^4-\frac{2u+3}{u}x^3+\frac{3u+1}{u^2}x^2-\frac{2}{u^2}x+\frac{1}{u^2} - (x^2+1).$$ Notice that \begin{align*} P'(x) &=-5ux^4+4(u+3)x^3-\frac{3(2u+3)}{u}x^2+\frac{2(3u+1)}{u^2}x-\frac{2}{u^2}-2x \\ &= -5ux^4+4(u+3)x^3-\frac{3(2u+3)}{u}x^2+\frac{-2u^2+6u+2}{u^2}x-\frac{2}{u^2}. \end{align*} For $u \ge 4$ and $x \ge 2$ we have $$5ux^4 \ge 10ux^3 \ge 4(u+3)x^3,$$ and in addition, we have $-\frac{3(2u+3)}{u}x^2,\frac{-2u^2+6u+2}{u^2}x,-\frac{2}{u^2}$ are each negative. Notice that $P'(x)$ is negative. Hence, for $u \ge 4$, the function $P(x)$ is strictly decreasing in the interval $[2, \infty)$. Moreover, it is easy to check that $P(2)<0$. Hence, it suffices to simply consider $u \in \{1,2,3\}$. One may easily check that the only possible value is $u=1$. Another simple computation reveals that the inequality only holds for $x=2$. \end{proof} \begin{theorem} The only recurrent number with configuration $(d_2,d_3,d_4,d_5)=(p,q,p^2,r)$ is $60$. \end{theorem} \begin{proof} In this case, we have \begin{equation} \label{kp} p^2=d_4=aq+bp=a(ap+b)+bp. \end{equation} Since $p \mid ab$ and $p \nmid{b}$ by Lemma \ref{bco}, we have $p \mid a$. Let $a=kp$ for some integer $k$. We can rewrite (\ref{kp}) as \[ k^2p^2 +b(k+1)=p, \] which is equivalent to \[ b = \frac{p(1-k^2p)}{k+1}. \] However, since we know that $p \nmid b$, we obtain that $p \mid k+1$. In other words, there exists a positive integer $u$ such that $k+1=up$. From here, we are able to express $a,b$ in terms of $u,p$, and use that to obtain $r$ in terms of $u,p$ only. We end up obtaining $$a=up^2-p,$$ $$b=\frac{1-p}{u}-up^3+2p^2,$$ and $$r=ap^2+bq=-up^5+(u+3)p^4-\frac{2u+3}{u}p^3+\frac{3u+1}{u^2}p^2-\frac{2}{u^2}p+\frac{1}{u^2}.$$ Since $r=d_5\ge d_4+1 = p^2+1$, Lemma \ref{bounds} gives that $(u,p)=(1,2)$ and substituting reveals $a=2, b=-1$. Hence the first five divisors would be $1,2,3,4,5$. This reduces to the case of an arithmetic sequence with common difference $1$, which, using the results of Iannucci \cite[Lemma 2]{Iannucci}, implies that $s(n)=5$ or $s(n)=6$. If $s(n)=5$, we have $n \ge \lcm(1,2,3,4,5)=60$, but $6$ is a small divisor that does not appear in $\mathcal{S}_n$, a contradiction. When $s(n)=6$, we see that $n$ is a multiple of $60$. However, if $n \ge 180$ then $12 \le \sqrt{n}$ does not appear in the list of small divisors, contradicting the recurrency. Clearly, $n=120$ is not recurrent neither as $8 \notin \{1, 2, 3,4,5,6\}$. Hence, the only solution is $\mathcal{S}_n=\{1,2,3,4,5,6\}$ for when $n=60$, which is indeed recurrent. \end{proof} \begin{corollary} \label{three} For an initial configuration $(p,q,r,p^2)$ in a recurrent number, we have that $p \mid d_{j}$ if and only if $j \equiv 2 \pmod{3}$. \end{corollary} \begin{proof} Firstly, the statement holds for $d_2=p$. Since $p^2=d_5=(a^3+2ab)p+b(a^2+b)$, Lemma \ref{bco} implies that $p \mid (a^2+b)$. Hence, $$d_{i+3}=ad_{i+2}+bd_{i+1}=a(ad_{i+1}+bd_i)+bd_{i+1}=(a^2+b)d_{i+1}+abd_{i}.$$ Lemma \ref{bco} once again gives us that $p \mid d_{i+3}$ if and only if $p \mid d_{i}$, which is the inductive step. \end{proof} \begin{theorem} \label{impossible} The configuration of divisors $(p,q,r,p^2)$ cannot occur in a recurrent number. \end{theorem} \begin{proof} Assume otherwise, that is, there exists a recurrent number $n$ of this configuration. We prove, by induction, that $s(n) \ge 3i+2$ for all $i \in \mathbb{N}$, which is absurd. The base case is already assumed. Assume that $s(n) \ge 3i+2$ for some $i \in \mathbb{N}$. By Corollary \ref{three}, $p\nmid d_{3i}d_{3i+1}$ and since $\gcd(d_{3i},d_{3i+1})=1$ by Corollary \ref{coprime}, we have $p^2d_{3i}d_{3i+1}$ divides $n$. By Proposition \ref{div}, $pd_{3i}$ is a small divisor. We consider two cases. \medskip \noindent Case I: $pd_{3i}=d_{3i+2}$. Hence, we obtain that $d_{3i} \mid ad_{3i+1}$, and by Corollary $\ref{coprime}$, we obtain that $d_{3i} \mid a$. Observe that the sequence $(d_i \bmod{a})$ takes values congruent to $1,p,b,bp,b^2, \dots$, and thus, $d_{3i}$ divides an element of the form $b^kp$ where $k \in \mathbb{N}$. However, Corollary \ref{three} and Lemma \ref{bco} yield $$\gcd(p,d_{3i+1})=1=\gcd(b,d_{3i+1}),$$ which is a contradiction. \medskip \noindent Case II: $pd_{3i}>d_{3i+2}$, hence, $pd_{3i} \ge d_{3i+5}$ by Corollary \ref{three}. Therefore, $s(n) \ge 3i+5$, and the induction is complete. \end{proof} \begin{lemma} \label{lastlem} In a recurrent number of configuration $(p,q,r,s)$, the small divisors $d_i,d_{i+2}$ are coprime. \end{lemma} \begin{proof} Assume a prime $t$ divides $\gcd{(d_{i}, d_{i+2})}$. Thus $t \mid ad_{i+1}$. Clearly, $t \nmid d_{i+1}$ by Lemma \ref{coprime}, and therefore, $t \mid a$. As such, the sequence $(d_i \bmod{t})$ takes values congruent to $1,p,b,bp,b^2,\dots$ and hence $t \mid b^kp$ for some $k \ge 0$. Yet Lemma \ref{bco} implies that $\gcd{(b,t)}=1$, hence $t \mid p$ and so $t=p$. However, this would imply that $t \mid r=(a^2+b)p+ab$, which is a contradiction as $pp$. By Lemma \ref{lastlem}, $p \nmid d_{j-2}d_{j-1}$, and we conclude that $d_{j-2}d_{j-1}pv \mid n$, so Proposition \ref{div} implies that $pd_{j-2}\in \mathcal{S}_n$. Moreover, $pd_{j-2} \ne d_{j}$ by Lemma \ref{lastlem}, and so $pd_{j-2}>d_{j}$ is a small divisor greater than $d_j$, a contradiction. \end{proof} \section{Concluding remarks} In conclusion, imposing the condition $s(n) \ge 5$ returns certain infinite families of recurrent integers $n$ with at most $3$ distinct prime divisors, in addition to the sole ‘‘sporadic’’ case of $60$. We now revisit the condition we imposed and relax it to find all recurrent numbers by relating $s(n)$ to $\tau(n)$ using Equation (\ref{tau}). \begin{enumerate}[label=(\roman*)] \item $s(n)=1$. Hence, $\tau(n) \in \{1,2\}$. That is, $n=1$ or $n=p$ for some prime $p$, and in both cases, $n$ is vacuously recurrent. \item $s(n)=2$. Thus, $\tau(n) \in \{3,4\}$, and we have the possibilities that $n=p^2,p^3$, or $pq$ for primes $p r$, then $\mathcal{S}_n=\{1,p,q,r\}$. Setting up a linear system of equations, we see that it is both necessary and sufficient for $\frac{pq-r}{p^2-q}$ to be an integer. \end{enumerate} \end{enumerate} \end{enumerate} We summarize our findings in the following result. \begin{theorem} All recurrent numbers $n$ fall into one of the following categories. \begin{enumerate} \item $n=p^k$, for some prime $p$ and a natural number $k$, hence $\mathcal{S}_n=\{1,p, \dots ,p^{\lfloor \frac{k}{2}\rfloor}\}$. \item $n=p^kq$, for some primes $p,q$ and a natural number $k$, such that $q>p^k$. Hence, $\mathcal{S}_n=\{1,p, \dots ,p^{k}\}$. \item $n=pq^k$, for some primes $ppq^k$. Hence, $\mathcal{S}_n=\{1,p,q,pq, \dots ,pq^k\}$. \item $n=60$, with $\mathcal{S}_n=\{1,2,3,4,5,6\}$. \item $n=p^3q$ for some primes $p,q$ such that $p