The site's wording holds trivially for every $n\ge3$. The integer $1$ has no prime factors, so it meets the rule vacuously and every sequence ends at it: for $n=3$ the rule gives $2,1$, and for $n=8$ it gives $7,5,1$. Since $1$ is not prime, no sequence is all prime. The failure is this page's own elementary check. The change replaces the range $[1,a_{k-1})$ by $[2,a_{k-1})$, so that the sequence stops when no integer greater than $1$ qualifies; nothing else changes. The evidence is the site's own commentary: its worked example for $n=8$ stops after $a_1=7$ and $a_2=5$, which holds only when $1$ is excluded, and it keeps the label OPEN and the equivalence with Problem 385 credited to Sarosh Adenwalla, both of which fit only the corrected question. The defect is already in the poser's text: Erdős and Graham [ErGr80, p. 85] index the sequence from the other end, with $a_1=1$ and $a_k$ the least integer exceeding $a_{k-1}$ for which all prime factors of $n-a_k>0$ are greater than $a_k$, and the value $n-a_k=1$ qualifies vacuously in the same way. Their report that Selfridge's preliminary calculations point to a yes answer "but no proof is in sight" fits only the corrected question. Above the excluded value no vacuous case remains, since every integer at least $2$ has a prime factor, and the corrected question is a real one: for $n=8$ the sequence $7,5$ is all prime. No result about the site's wording exists beyond the trivial check recorded here.