We want to determine all possible values of $a_1$ such that the sequence defined by $a_{n+1} = S(a_n)$ is an infinite sequence of positive integers, where $S(N)$ is the sum of the three largest proper divisors of $N$, and each $a_n$ has at least three proper divisors (i.e., the total number of divisors $\tau(a_n) \geq 4$).

Let $R(N) = S(N)/N$. If $\tau(N) \geq 5$, and $1 = d_1 < d_2 < d_3 < d_4$ are the smallest divisors of $N$, then $R(N) = \frac{1}{d_2} + \frac{1}{d_3} + \frac{1}{d_4}$.

\textbf{Step 1: Proving that $a_n$ is even for all $n$.}

Suppose $N$ is odd and $\tau(N) > 4$. The divisors of $N$ are odd. If $\tau(N) \geq 5$, $d_2 \geq 3$, $d_3 \geq 5$, $d_4 \geq 7$. $R(N) \leq \frac{1}{3} + \frac{1}{5} + \frac{1}{7} = \frac{71}{105} < 1$. If $\tau(N) = 4$, $N = p^3$ or $N = pq$ for odd primes $p < q$. The proper divisors are $1, p, p^2$ or $1, p, q$. $S(p^3) = 1 + p + p^2$. For $p \geq 3$, $p^3 - (1 + p + p^2) = p^2(p - 1) - p - 1 \geq 9(2) - 3 - 1 = 14 > 0$. $S(pq) = 1 + p + q$. For $p \geq 3, q \geq 5$, $pq - (1 + p + q) = (p - 1)(q - 1) - 2 \geq 2 \cdot 4 - 2 = 6 > 0$. In all cases, $S(N) < N$. Furthermore, the three largest proper divisors are odd, so their sum $S(N)$ is odd.

If $a_n$ were odd for some $n$. Since $\tau(a_n) \geq 4$ by the problem statement, $a_{n+1} = S(a_n) < a_n$ and $a_{n+1}$ is odd. By induction, $(a_k)_{k>n}$ would be a strictly decreasing infinite sequence of positive integers. This contradicts the Well-Ordering Principle. Thus, $a_n$ is even for all $n$.

\textbf{Step 2: Proving that $a_n$ is divisible by 3 for all $n$.}

Suppose $N$ is even, $\tau(N) \geq 4$, and $3 \nmid N$. $d_2 = 2$. Since $3 \nmid N$, $d_3 \geq 4$. If $\tau(N) \geq 5$, $d_4 \geq 5$. $R(N) \leq \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = \frac{19}{20} < 1$. If $\tau(N) = 4$. $N = 8$ or $N = 2p$ (prime $p \geq 5$). $S(8) = 7 < 8$. $S(2p) = p + 3 < 2p$. In all cases, $S(N) < N$.

We prove a lemma: Lemma: Let $N$ be even, $\tau(N) > 4$, and $3 \nmid N$. If $3|S(N)$, then $S(N)$ is odd. Proof: If $\tau(N) = 4$, $S(8) = 7$, $S(2p) = p + 3$. Since $3 \nmid p$, $3 \nmid p + 3$. So $3 \nmid S(N)$. The implication holds vacuously. If $\tau(N) \geq 5$. $R(N) = \frac{1}{2} + \frac{1}{d_3} + \frac{1}{d_4}$. Since $3 \nmid N$, $3 \nmid d_i$. If $3|S(N)$, since $3 \nmid N$, we must have $v_3(R(N)) > 0$. $R(N) = \frac{d_3 d_4 + 2 d_3 + 2 d_4}{2 d_3 d_4}$. The denominator is not divisible by 3. The numerator $X = d_3 d_4 + 2 d_3 + 2 d_4$ must be divisible by 3. $X \equiv d_3 d_4 - d_3 - d_4 = (d_3 - 1)(d_4 - 1) - 1 \pmod 3$. $X \equiv 0 \Rightarrow (d_3 - 1)(d_4 - 1) \equiv 1 \pmod 3$. This requires $d_3 \equiv 2$ and $d_4 \equiv 2 \pmod 3$. If $4|N$. Since $3 \nmid N$, the divisors start $1, 2, 4$. So $d_3 = 4$. But $4 \equiv 1$ (mod 3). But $4 \equiv 1 \pmod 3$, contradicting $d_3 \equiv 2 \pmod 3$. Thus $v_2(N) = 1$. $N = 2M$ with $M$ odd, $3 \nmid M$. Let $p$ be the smallest prime factor of $M$ ($p \geq 5$). $d_3 = p$. We need $p \equiv 2 \pmod 3$. $d_4$ is the next smallest divisor. $2p \equiv 2(2) = 4 \equiv 1 \pmod 3$. Since $d_4 \equiv 2 \pmod 3$, $d_4 \neq 2p$. So $d_4$ must be the next smallest divisor of $M$, call it $m_3$. $d_4$ is odd. $S(N) = N/2 + N/p + N/d_4 = M + 2M/p + 2M/d_4$. Since $M$ is odd and $p, d_4$ are odd divisors of $M$, $M/p$ and $M/d_4$ are odd integers. $S(N) = \text{Odd} + \text{Even} + \text{Even} = \text{Odd}$. The lemma is proved.

Now, suppose $3 \nmid a_n$ for some $n$. We know $a_n$ is even and $a_{n+1} = S(a_n) < a_n$. If $3 | a_{n+1}$, by the Lemma applied to $a_n$, $a_{n+1}$ must be odd. This contradicts Step 1. Thus $3 \nmid a_{n+1}$. By induction, $(a_k)_{k>n}$ is a strictly decreasing infinite sequence of positive integers. Contradiction. Therefore, $3|a_n$ for all $n$.

Combining Step 1 and Step 2, $6|a_n$ for all $n$. Note that $6|N$ implies $\tau(N) \geq 4$.

\textbf{Step 3: Analyzing the dynamics when $6|N$.}

If $6|N$, the smallest divisors are $1, 2, 3$. The fourth smallest divisor $d_4$ must be $4, 5,$ or $6$. $R(N) = \frac{1}{2} + \frac{1}{3} + \frac{1}{d_4} = \frac{5}{6} + \frac{1}{d_4}$. (This holds even if $\tau(N) = 4$, i.e., $N = 6$, where $S(6) = 6$, $R(6) = 1$, and $d_4$ is formally $N = 6$).

We identify three regimes: Regime A (Growth): $d_4 = 4$. Occurs if $12|N$. $R(N) = 13/12$. Regime B (Boost): $d_4 = 5$. Occurs if $30|N$ and $4 \nmid N$ ($v_2(N) = 1$). $R(N) = 31/30$. Regime C (Fixed Point): $d_4 = 6$. Occurs if $6|N, 4 \nmid N, 5 \nmid N$. $R(N) = 1$.

\textbf{Step 4: Evolution of the sequence and constraints on $a_1$.}

If $a_n \in B$. $v_2(a_n) = 1$. $a_{n+1} = (31/30)a_n$. $v_2(a_{n+1}) = v_2(a_n) + v_2(31/30) = 1 - 1 = 0$. $a_{n+1}$ is odd. This contradicts Step 1. Thus, the sequence must remain in A $\cup$ C.

If $a_n \in A$. $a_{n+1} = (13/12)a_n$. $v_2(a_{n+1}) = v_2(a_n) - 2$. $v_3(a_{n+1}) = v_3(a_n) - 1$. Since $6|a_k$ for all $k$, $v_2(a_k) \geq 1$ and $v_3(a_k) \geq 1$. As the valuations decrease in Regime A, the sequence cannot stay in A indefinitely. It must eventually reach Regime C and stabilize there ($a_{n+1} = a_n$).

In Regimes A ($R = 13/12$) and C ($R = 1$), $v_5(R(N)) = 0$. Thus $v_5(a_n)$ is constant. Let $L$ be the stable value in C. By definition of C, $5 \nmid L$. So $v_5(L) = 0$. Therefore, $v_5(a_1) = 0$.

\textbf{Step 5: Characterization of $a_1$.}

Let $K \geq 0$ be the number of steps the sequence spends in Regime A before reaching Regime C. $a_1, \ldots, a_K \in A$ (if $K \geq 1$) and $a_{K+1} \in C$. Since $5 \nmid a_1$, $5 \nmid a_n$ for all $n$.

Let $A = v_2(a_1)$ and $B = v_3(a_1)$. $a_{K+1} = (13/12)^K a_1$. $v_2(a_{K+1}) = A - 2K$. $v_3(a_{K+1}) = B - K$. Since $a_{K+1} \in C$, we must have $v_2(a_{K+1}) = 1$ (as $6|a_{K+1}$ and $4 \nmid a_{K+1}$) and $v_3(a_{K+1}) \geq 1$. $A - 2K = 1 \Rightarrow A = 2K + 1$. $B - K \geq 1 \Rightarrow B \geq K + 1$.

We verify that these conditions are sufficient. We must ensure $a_i \in A$ for $1 \leq i \leq K$. This means $12|a_i$. For $1 \leq i \leq K$: $v_2(a_i) = A - 2(i - 1) = 2K + 1 - 2i + 2 = 2(K - i) + 3$. Since $i \leq K$, $v_2(a_i) \geq 3$. $v_3(a_i) = B - (i - 1) \geq (K + 1) - (i - 1) = K - i + 2$. Since $i \leq K$, $v_3(a_i) \geq 2$. Thus $2^3 \cdot 3^2 = 72$ divides $a_i$. This implies $12|a_i$, so $a_i \in A$. This also ensures that $a_{i+1} = (13/12)a_i$ is an integer. The sequence is valid.

We express the possible values of $a_1$. $a_1 = 2^{2K+1} 3^B M$, where $K \geq 0$, $B \geq K + 1$, and $M$ is a positive integer such that $\gcd(M, 30) = 1$ (since $v_5(a_1) = 0$). We rewrite this as: $a_1 = (2^{2K+1} 3^{K+1}) \cdot (3^{B-(K+1)} M)$. $2^{2K+1} 3^{K+1} = (2 \cdot 4^K) \cdot (3 \cdot 3^K) = 6 \cdot (12^K)$. Let $J = 3^{B-K-1} M$. $J$ is a positive integer. Since $\gcd(M, 30) = 1$, $J$ is not divisible by 2 or 5. That is, $\gcd(J, 10) = 1$. Conversely, any positive integer $J$ such that $\gcd(J, 10) = 1$ can be represented in this form for a given $K$ (by taking $B = K+1+v_3(J)$ and $M = J/3^{v_3(J)}$).

The set of all possible values of $a_1$ consists of the integers of the form $6J \cdot 12^K$, where $K \geq 0$ is an integer and $J$ is a positive integer such that $\gcd(J, 10) = 1$.
