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Here is a way to get $26$ for the $C_3(4)$ case, and $2(n^2-n) + 2$ in general.

In the bottom level, add $4n-5$ cubes around the outside, leaving one out adjacent to a corner. For the second level, fill in the interior $(n-2)^2$ cubes, add the two diagonal corners (one of which is next to the missing cube from the base), and then put a cube in the missing cube's column. For the remaining $n-2$ levels, stack on the diagonal.

 1 1 1 1  1 0 0 0  1 0 0 0  1 0 0 0 
 1 0 0 1  0 1 1 0  0 1 0 0  0 1 0 0
 1 0 0 1  0 1 1 0  0 0 1 0  0 0 1 0
 1 1 0 1  0 0 1 1  0 0 0 1  0 0 0 1

In general, that works out to $C_3(n) \leq (4n-5) + (n-2)^2 + 3 + n(n-2) = 2(n^2 - n) + 2$

Edit: My $26$ bound for $C_3(4)$ has since been improved, but here is an optimal $C_3(5) = 37$ arrangement.

1 0 0 0 0
1 1 0 0 0
0 0 1 0 0
0 0 0 1 1
0 0 0 0 1

0 1 0 0 0
0 1 0 0 0
1 1 1 1 1
0 0 0 1 0
0 0 0 1 0

0 0 1 0 0
0 0 1 1 1
0 0 1 0 0
1 1 1 0 0
0 0 1 0 0

0 0 0 1 1
0 0 0 1 0
0 0 1 0 0
0 1 0 0 0
1 1 0 0 0

0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 1 0 0 0
1 0 0 0 0
show/hide this revision's text 5 Rollback to Revision 3

Here is a way to get $25$ 26$ for the $C_3(4)$ case, and $2(n^2-n) + 1$ 2$ in general.

In the bottom level, add $4n-5$ cubes around the outside, leaving one out adjacent to a corner. For the second level, fill in the interior $(n-2)^2$ cubes, minus their diagonal $(n-2)$ cubes, then add the two outer diagonal corners (one of which is next to the missing cube from the base). For the third level, add the two corners, the missing cube from the base level, and then put a cube in the diagonal $(n-2)$, along with their connecting $(n-3)$ cubesmissing cube's column. For the remaining $n-3$ n-2$ levels, stack on the diagonal.

 1 1 1 1  1 0 0 0  1 0 0 0  1 0 0 0 
 1 0 0 1  0 0 1 1 0  0 1 1 0 0  0 1 0 0
 1 0 0 1  0 1 0 1 0  0 0 1 0  0 0 1 0
 1 1 0 1  0 0 0 1 1  0 0 1 0 1  0 0 0 1

In general, that works out to $C_3(n) \leq (4n-5) + (n-2)^2 - (n-2) + 2 + 3 + (n-2) + (n-3) + n(n-3)$$= n(n-2) = 2(n^2 - n) + 1$2$
show/hide this revision's text 4 added 178 characters in body; added 2 characters in body

Here is a way to get $26$ 25$ for the $C_3(4)$ case, and $2(n^2-n) + 2$ 1$ in general.

In the bottom level, add $4n-5$ cubes around the outside, leaving one out adjacent to a corner. For the second level, fill in the interior $(n-2)^2$ cubes, minus their diagonal $(n-2)$ cubes, then add the two outer diagonal corners (one of which is next to the missing cube from the base). For the third level, add the two corners, the missing cube from the base level, and then put a cube in the missing cube's columndiagonal $(n-2)$, along with their connecting $(n-3)$ cubes. For the remaining $n-2$ n-3$ levels, stack on the diagonal.

 1 1 1 1  1 0 0 0  1 0 0 0  1 0 0 0 
 1 0 0 1  0 1 0 1 0  0 1 0 1 0  0 1 0 0
 1 0 0 1  0 1 1 0 0  0 0 1 0  0 0 1 0
 1 1 0 1  0 0 1 1  0 1  0 0 1 1  0 0 0 1

In general, that works out to $C_3(n) \leq (4n-5) + (n-2)^2 - (n-2) + 2 + 3 + n(n-2) = (n-2) + (n-3) + n(n-3)$$= 2(n^2 - n) + 2$1$
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