I hesitate to ask this question here, but since it remained unanswered after a bounty on MSE, I ask it here with some reservation.

Is the maximum bootstrap percolation time for $n$ points on an $n\times n$ grid $\big{|}\left \lceil{(n^2-3)/2}\right \rceil + n - 1 \big{|}$ for $1\leq n\leq 8$, and $n(n+3)/2-7$ for $n\geq 9$?

Below are some possible starting positions for $1\leq n\leq 12$:

and a possible construction method for $n\geq 10$ (based on the starting positions of $n-2$):

$\hspace{2em}$

In *Mathematica*, this might be constructed as follows:

```
aa = Uncompress@"1:eJzVlsEOgjAQRLuAgvyF/+PJT/BAwskD/n/UNhGG7mwpqNFLw8K8zuxaCcfL9dy1zrmheiynfrh1wqv+WXUFCIpI0LtI5W8FugS61GlFGu6/FvrQkxWYVMSE6GdOwY4pJKkYXXaQCqp5KgJp0YI7k8kyWfZuPtseGoJKbYiQNEcIw7SKg72vZGjXZfC91TAVqPhURquaWGUD9Ei9zYGOdqMDSz7poYEhN0lcM0UqDayk4rwrvT7Wl5lQlCvTy/znB+owpbBa2qWy0VJqCyox+rOBybzVX4k1YbuarWeNafYCwU+SNmMjLQjyOehmXmL+X/IL4b+Q3zoNqsw+iAqffkmNyx0cTkRo";
a[9] = aa[[9]]; a[10] = aa[[10]];
a[n_] := If[n < 9, aa[[n]], With[{t = Length@#[[1]] + 2}, Flatten[{ReplacePart[Array[0 &, t], # -> 1] & /@ {1, t + 1, t, 1, t + 1, t - 1, 1}, Drop[Flatten[{Take[#, 2], #}, 1] &@(PadLeft[PadRight[#, t - 1], t] & /@ #), 7]}, 1]
] &@a[n - 2]];
```

or, a non-recurrence solution:

```
a[n_] := If[n < 9, aa[[n]], Partition[ReplacePart[ConstantArray[0, n^2], Thread[# -> 1]] &@
With[{v = Join[{1, 3 #1, 1 + 3 #1, -1 + 6 #1, 1 + 6 #1}, LinearRecurrence[{0, 2, 0, -1}, {-2 + 8 #1, 2 + 8 #1, -3 + 10 #1, 3 + 10 #1}, # - 9], {(-3 - 8*#1 + 4*#1^2 + (-1)^#1*(-5 + 2*#1))/4, ((-1)^#1*(1 + (-1)^#1*(-1 - 6*#1 + 4*#1^2)))/4, ((-1)^#1*(1 + (-1)^#1*(-13 - 2*#1 + 4*#1^2)))/4, ((-1)^#1*(-1 + (-1)^#1*(9 - 2*#1 + 4*#1^2)))/4}] &@n},
If[EvenQ@n, v, ReplacePart[v, (Length@v - 4) -> v[[Length@v - 4]] + 1]]], n]];
```

*eg*

```
Manipulate[With[{b = Most@FixedPointList[
CellularAutomaton[{1018, {2, {{0, 2, 0}, {2, 1, 2}, {0, 2, 0}}}, {1, 1}}, {#, 0}][[1, 2 ;; -2, 2 ;; -2]] &, a[n]]},
ArrayPlot[b[[length]], Mesh -> True]],
{length, 1, If[n < 9, {1, 2, 5, 10, 15, 22, 29, 38}[[n]], n (n + 3)/2 - 7], 1, Appearance -> "Open"},
{{n, 10}, 1, 20, 1, Appearance -> "Open"}]
```

The above is smaller than the lower bound shown in this paper of $13 n^2/18-14 n/9-5/3$, but a quick search for all permutations at $n=5$ shows that the maximum percolation time requires $>n$ initial points.

Does the above construction result in the maximum percolation time for $n$ initial startpoints?

# Sets containing $>n$ initial points

In addition, I am looking through the paper by Fabricio Benevides and Michał Przykucki on maximum bootstrap percolation time and I am having trouble finding an example (or seeing how there *could* be a set of points) that takes a greater time to complete than the one given in their example of a set for a $12\times 12$ grid on page $20$:

the following pattern is valid for every multiple of $12$ and requiring $4n/3-1$ initial points, takes $ n(17 n- 10)/24$ moves to complete:

```
manipu[n_, m_] :=
Manipulate[With[{b = Most@FixedPointList[
CellularAutomaton[{1018, {2, {{0, 2, 0}, {2, 1, 2}, {0, 2, 0}}}, {1, 1}},
{#, 0}][[1, 2 ;; -2, 2 ;; -2]] &, n]},
ArrayPlot[b[[length]], Mesh -> True]], {length, 1, m, 1, Appearance -> "Open"}];
m12[n_] :=
With[{y = Length@#}, manipu[#, y (17 y - 10)/24]] &@ With[{t = 12 n},
Flatten[{Take[Flatten[{PadRight[{1}, t], PadLeft[{1}, t],
Array[0 &, t]} & /@ Range@Ceiling[t/6], 1], t/2],
Reverse@(CenterArray[Join[{0, 0, 1}, Array[0 &, #], {1}], t] & /@
Range[8, t, 4]), {CenterArray[{0, 1, 0, 0, 0, 1}, t]},
{CenterArray[{1, 0, 1}, t]}, CenterArray[Join[{1},
Array[0 &, #], {1}], t] & /@ Range[6, t, 4]}, 1]];
m12[3]
```

This differes from their minimum percolation time: the set following the pattern given in their example takes $ 17 n^2/24 +O(n)$, yet they state the lower bound is $13n^2/18+O(n)$. It is close, $(\lim{n\rightarrow\infty (17 n^2/24)/(13n^2/18)=51/52})$, but I can't see how to construct a set of initial points that meets their lower bound. What am I missing?