Problem: An infinite sequence $a_1$, $a_2$, \dots\ consists of positive integers
has each of which has at least three proper divisors.
Suppose that for each $n\geq 1$,
$a_{n+1}$ is the sum of the three largest proper divisors of $a_n$.
Determine all possible values of $a_1$.

Solution: The answer is $a_1 = 12^e \cdot 6 \cdot \ell$
for any $e, \ell \ge 0$ with $\gcd(\ell, 10) = 1$.

Let $\mathbf{S}$ denote the set of positive integers with at least three divisors.
For $x \in \mathbf{S}$, let $\psi(x)$ denote the sum of the three largest ones,
so that $\psi(a_n) = a_{n+1}$.

\paragraph{Proof that all such $a_1$ work.}
Let $x = 12^e \cdot 6 \cdot \ell \in \mathbf{S}$ with $\gcd(\ell, 10) = 1$.
As $\frac12+\frac13+\frac14 = \frac{13}{12}$, we get
\[ \psi(x) = \begin{cases}
    x & e = 0 \\
    \frac{13}{12} x & e > 0
  \end{cases} \]
so by induction on the value of $e$ we see that $\psi(x) \in \mathbf{S}$
(the base $e=0$ coming from $\psi$ fixing $x$).

\paragraph{Proof that all $a_1$ are of this form.}
In what follows $x$ is always an element of $\mathbf{S}$,
not necessarily an element of the sequence.

\begin{claim*}
  Let $x \in \mathbf{S}$. If $2 \mid \psi(x)$ then $2 \mid x$.
\end{claim*}
\begin{proof}
  If $x$ is odd then every divisor of $x$ is odd,
  so $\psi(x)$ is the sum of three odd numbers.
\end{proof}

\begin{claim*}
  Let $x \in \mathbf{S}$.  If $6 \mid \psi(x)$ then $6 \mid x$.
\end{claim*}
\begin{proof}
  We consider only $x$ even because of the previous claim.
  We prove the contrapositive that $3 \nmid x \implies 6 \nmid \psi(x)$ (for even $x$).
  \begin{itemize}
    \ii If $4 \mid x$, then letting $d$ be the third largest proper divisor of $x$,
    \[ \psi(x) = \frac x2 + \frac x4 + d = \frac 34 x + d \equiv d \not\equiv 0 \pmod 3. \]
    \ii Otherwise, let $p \mid x$ be the smallest prime dividing $x$, with $p > 3$.
    If the third-smallest nontrivial divisor of $x$ is $2p$, then
    \[ \psi(x) = \frac x2 + \frac xp + \frac{x}{2p} = \frac{3}{2p} x + \frac x2
      \equiv \frac x2 \not\equiv 0 \pmod 3. \]
    If the third-smallest nontrivial divisor of $x$ is instead an odd prime $q$, then
    \[ \psi(x) = \frac x2 + \frac xp + \frac{x}{q} \equiv 1+0+0 \equiv 1 \pmod 2. \qedhere \]
  \end{itemize}
\end{proof}

To tie these two claims into the problem, we assert:
\begin{claim*}
  Every $a_i$ must be divisible by $6$.
\end{claim*}
\begin{proof}
  The idea is to combine the previous two claims
  (which have no dependence on the sequence) with a size argument.
  \begin{itemize}
    \ii For odd $x \in \mathbf{S}$ note that $\psi(x) < \left( \frac13+\frac15+\frac17 \right) x < x$
    and $\psi(x)$ is still odd.
    So if any $a_i$ is odd the sequence is strictly decreasing and that's impossible.
    Hence, we may assume $a_1$, $a_2$, \dots\ are all even.

    \ii If $x \in \mathbf{S}$ is even but $3 \nmid x$ then $\psi(x) < \left( \half+\frac14+\frac15 \right) x < x$
    and $\psi(x)$ is still not a multiple of $3$.
    So if any $a_i$ is not divisible by $3$ the sequence is again strictly decreasing.
    \qedhere
  \end{itemize}
\end{proof}

On the other hand, if $x$ is a multiple of $6$, we have the following formula for $\psi(x)$:
\[ \psi(x) = \begin{cases}
  \frac{13}{12}x & 4 \mid x \\
  \frac{31}{30}x & 4 \nmid x \text{ but } 5 \mid x \\
  x & 4 \nmid x \text{ and } 5 \nmid x.
  \end{cases} \]
Looking back on our sequence of $a_i$ (which are all multiples of $6$),
the center case cannot happen with our $a_i$, because $\frac{31}{30}x$
is odd when $x \equiv 2 \pmod 4$.
Hence in actuality
\[ a_{n+1} = \frac{13}{12} a_n \quad\text{or}\quad a_{n+1} = a_n \]
for every $n$.

Let $T$ be the smallest index such that $a_T = a_{T+1} = a_{T+2} = \dotsb$
(it must exist because we cannot multiply by $\frac{13}{12}$ forever).
Then we can exactly describe the sequence by
\[ a_n = a_1 \cdot \left( \frac{13}{12} \right)^{\min(n,T)-1}. \]
Hence $a_1 = \left( \frac{12}{13} \right)^{T-1} a_T$,
and since $a_T$ is a multiple of $6$ not divisible by $4$ or $5$,
it follows $a_1$ has the required form.