(This query was beforehand posted on Math SE, however obtained no solutions.)

3D contaminated cubes puzzle with threshold $4$:
On an $ntimes ntimes n$ dice, some cells are contaminated; if a cell shares a face with $4$ contaminated squares, it turns into contaminated. What’s the minimal variety of initially contaminated cells required to contaminate the entire dice?

The two-dimensional, threshold 2 model is a traditional. The resolution to that puzzle (usually merely known as the “contaminated squares puzzle”) is $n$.

The two-dimensional, threshold 3 model is extra attention-grabbing. When $n$ is of the shape $2^k-1$, the answer is $frac{4^k-1}3=frac13n^2+frac23n$ with an attention-grabbing recursive sample. When $n$ is just not of that type, I imagine that the answer is $lceilfrac13n^2+frac23n+frac13rceil$ for odd $n$ and $lceilfrac13n^2+frac23n+frac43rceil$ for even $n$. (I haven’t got a proof however I feel another person does.) In abstract: $frac13n^3+frac23n^2+O(1)$.

Up a dimension, the three-dimensional, threshold 3 model is easy once more. The reply is $n^2$. In truth, the $d$-dimensional, threshold $d$ model is solved for all $d$: see right here.

The logical subsequent step, then, is the three-dimensional, threshold 4 model. After some pondering, I’ve some conjectural higher bounds:

$n=1$ is $1$, trivially.
$n=2$ is $8$. (In truth, for all $nge2$, the $8$ vertices should begin contaminated, as they solely have three neighbors.)
$n=3$ must be $14$ (nook cells and face cells).
$n=4$ must be $33$.
$n=5$ must be $53$ (on Math SE I initially wrote $52$ however I do not assume that works really).

What extra progress might be made? Are the options I discovered for $nle5$ minimal? Is there a method (even an assymptotic one) for common $n$?

For what it is value, I can handle a decrease certain of $frac14n^3+frac34n^2$. However, given the information above, this does not appear to be an particularly shut certain.

A useful remark: Consider the $(n+1)^3$ factors which can be vertices of a cell. I imagine that this set (“the grid factors”) have to be linked by way of the contaminated cells: that’s, the set of those grid factors union the set of contaminated cells have to be a linked set. (This remark is true of the two-dimensional, threshold 3 model as properly. However, in that case, it was each a needed and ample situation; in our case, that is nonetheless needed however not ample.)