Number of ways two knights can be placed such that they don't attack.

What are the number of ways two knights can be placed on a k×k chessboard so that they do not attack each other?

For k from 1 to 8, the answer is given below. How do I find a general formula?
0
6
28
96
252
550
1056
1848


Edit:

Here's my approach after @Peter 's help, I came to a conclusion that number of ways such that they attack is equal to two times the number of possible ways I can put an "L" shape on the board. (2 times because knights can swap positions), am I right? I don't know how do I more forward from here.

I tried finding number of ways to place L by this recursive formula: F[n][n]=4+F[i][i-3]+F[i-2][3]; But it's not working.

• Probably, it is easier to determine the number of ways to place them that they attack each other. This has then only to be subtracted from $\frac{k^2(k^2-1)}{2}$ Jun 18, 2019 at 10:44
• Yes, That helped, So I came to a conclusion that number of ways such that they attack is equal to two times the number of possible ways I can put an "L" shape on the board. (2 times because knights can swap positions), am I right? I don't know how do I more forward from here.
– Het
Jun 18, 2019 at 11:07
• Maybe I can use recursion... @Peter
– Het
Jun 18, 2019 at 11:24

Note that when we have two knights threatening each other, it actually forms either a $$2\times3$$ or $$3 \times 2$$ board. And for each of $$2 \times 3$$ and $$3 \times 2$$ boards, there are $$2$$ ways of placing two knights so that they threaten each other. So, what we should do is to count how many $$2 \times 3$$ and $$3 \times 2$$ squares on $$n\times n$$ board. For general $$n$$, the answer is $$(n-1)(n-2)+(n-2)(n-1) = 2(n-1)(n-2)$$ And for each $$2\times3$$ and $$3\times2$$ board, there are $$2$$ ways of placing the knights so that they threaten each other. Therefore, in total there are $$2\cdot2(n-1)(n-2)=4(n-1)(n-2)$$ ways of placing two knights so that they threaten each other. So what you are looking for is $$\frac{n^2(n^2-1)}{2}-4(n-1)(n-2)$$ It is also worth mentioning that we are not over-counting because whenever we place two knights so that they threaten each other, either a $$2 \times 3$$ or $$3 \times 2$$ board must contain both of the knights.

• How did you count 2x3 and 3x2 boards?
– Het
Jun 18, 2019 at 12:04
• Since we have a square board, once we count number of $2 \times 3$ boards, number of $3 \times 2$ boards have the same number by symmetry. After that, in order to count number of $3 \times 2$ boards, you can basically start from top left and shift $1$ square to right and find how many times you can shift it to the right. And shifting it to bottom and finding how many times you can shift it to bottom and multiplying them will give you the answer since we are actually finding how many rows of $3$ adjacent squares are there and how many columns of $2$ adjacent squares are there. Jun 18, 2019 at 12:11
• Aren't there 4 ways to place 2 knights on 2x3 or 3x2 board? since both knight can swap positions.
– Het
Jun 18, 2019 at 12:23
• I think knights are to be considered identical in this case. So swapping positions will give the same placement. Jun 18, 2019 at 12:28
• Can you come up with a recursive solution to this problem? Dec 20, 2020 at 6:15