Three-colour solutions from four-colour cubes

For every $n\ge2$, any $16n-3$ cubes, each using all four colours of a fixed palette, contain an $n\times n\times n$ frame whose exterior faces are monochromatic in exactly three colours; this bound is sharp. For a fixed palette, exactly two collections of $16n-4$ cubes fail, and reflection exchanges them. For $n\ge3$, every failure of size at least $14n$ lies in one ordinary orientation class with four bounded colour buckets. This gives $2\binom{d+15}{15}$ failures at size $16n-4-d$, for $0\le d\le2n-4$, of which $2\binom{d+11}{11}$ are inclusion-maximal. The threshold for a complete cube is $\max\{n^3,16n-3\}$; in particular, any $n^3$ source cubes suffice when $n\ge4$. The case $n=2$ answers the threshold question in Problem 2 of Berkove, Eickemeyer and Schulman (2018). The proof combines their corner-containment reduction with orientation constraints and double counting.