
COMMENTS

The task here is to surround a hole of size n with the least number of squares. The hole is another polyomino, so we can obtain a lower bound using A027709: minimal perimeter of polyomino with n square cells. We need an extra 3 (or more) diagonal cells to surround any hole. Hence a(n) >= A027709(n) + 3 = 2*ceiling(2*sqrt(n)) + 3.
For rectangular holes we can obtain an upper bound using A262767: minimum perimeter of a rectangle with area n and integer sides. Hence a(n) <= A262767(n) + 3.
Perhaps a(n) is actually equal to A027709(n)+3?


EXAMPLE

For n=1, we have a single square hole, so a(1)=7.
For n=2, we have a domino hole, so a(2)=9.
For n=3, we can use either an L or V tromino hole, so a(3)=11.
For n=4, we use the square tetromino hole, so a(4)=11.
For n=5, we use the P pentomino hole, so a(5)=13.
For n=6, we use the 2 X 3 rectangle hole, so a(6)=13.
