The conjecture: For those $k$ that have a saturated sublist $a_{j}$, the first occurrence is: $$j \geq k+3.$$ A proof will imply Oppermann and will be a start to a pattern-based attack on the quadruplet-primes conjecture.
Edit This post contains counter-examples to this conjecture.
Construct a list for an arbitrary $k\geq12$: $$A=\left\lbrace \text{lpf}(n,k):1\leq n\leq\text{lcm}(1,2,...,k)\right\rbrace,$$
where$$\text{lpf}(n,k):=\left\{\begin{array}{cc} 1\hfill&\text{lpf}(n)>k\hfill\\ \text{lpf}(n)&\text{otherwise}\\ \end{array} \right.,$$
is the least prime factor $\leq k$, or a placeholder (which marks the position of a totient).
Partition $A$ into sublists of size $k$: $(a_1,a_2,\dots,a_{\text{lcm}(1,2,...,k)/k}).$
For most $k,$ the construction results in a few saturated sublists (i.e., sublists that only contain least prime factors---no placeholders exist). For $k=15,$ no saturated sublists exist.
Q1: Is $k=15$ the only exception?
Q2: Is there a more precise lower limit?
Edit First occurrences of saturated sublists: $$ \begin{array}{ccc} \textit{k} & \textit{j} & \textit{a$_j$}\hfill \\ 12 & 183 & (5,2,3,2,11,2,7,2,3,2,5,2)\hfill \\ 13 & 169 & (5,2,3,2,11,2,7,2,3,2,5,2,13)\hfill \\ 14 & 157 & (5,2,3,2,11,2,7,2,3,2,5,2,13,2)\hfill \\ 15 & - & \text{None}\hfill \\ 16 & 591 & (3,2,7,2,5,2,3,2,11,2,13,2,3,2,5,2)\hfill \\ 17 & 1211 & (3,2,7,2,5,2,3,2,13,2,11,2,3,2,5,2,7)\hfill \\ 18 & 1547 & (17, 2, 3, 2, 13, 2, 5, 2, 3, 2, 7, 2, 11, 2, 3, 2, 5, 2)\hfill \\ 19 & 3659 & (7, 2, 5, 2, 3, 2, 11, 2, 13, 2, 3, 2, 5, 2, 7, 2, 3, 2, 19)\hfill \\ 20 & 239 & (3, 2, 11, 2, 5, 2, 3, 2, 19, 2, 13, 2, 3, 2, 5, 2, 17, 2, 3, 2)\hfill \\ 21 & 696 & (2, 11, 2, 13, 2, 3, 2, 17, 2, 5, 2, 3, 2, 7, 2, 19, 2, 3, 2, 5, 2)\hfill \\ 22 & 5950 & (7, 2, 3, 2, 17, 2, 5, 2, 3, 2, 11, 2, 19, 2, 3, 2, 5, 2, 13, 2, 3, 2)\hfill \\ \end{array} $$