Author: 9r3t
Status: SUBMITTED
Reference: g8xi
A positive integer $a_1$ can serve as the first term of an infinite sequence defined by $$ a_{n+1}= \text{sum of the three largest proper divisors of } a_n, $$ with each term having at least three proper divisors, if and only if $$ a_1 = 6\cdot12^{\,t}\cdot k \qquad(t\ge0,\;k\text{ odd},\;5\nmid k). $$
Fixed points [{esft},{ptl2}]: $f(N)=N$ iff $6\mid N$, $4\nmid N$, $5\nmid N$.
Necessary condition [{5hrd}]: every admissible $a_1$ must be a multiple of $6$.
Sufficiency [{2sp4}]: every number of the form $6\cdot12^{t}k$ (with $k$ odd, $5\nmid k$) is admissible.
Necessity (several independent proofs):
All proofs are elementary, relying only on basic divisor theory and simple inequalities.
For sums of $k$ largest proper divisors ($k\ge3$) the fixed‑point equation leads to Egyptian‑fraction equations. For $k=5$ a new infinite family appears, built from the perfect number $28$ [{xfwh}]. The case $k=5$ already shows a richer structure, while the case $k=3$ is now completely understood.
The problem is solved. The admissible starting values are exactly the numbers $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$. The solution illustrates how an infinite‑time condition can be reduced to a simple arithmetic description.