I checked my chess queens application for larger chess boards, and found some interesting results: It appears that when the minimal distance between queens is more than 1, the minimal chess board size with solutions (not including one queen on the trivial 1×1 board) is at least the square of the minimal distance between queens. Here are a few examples:

- When the minimal distance between queens is 2, the minimal chess board size with solutions is 4×4 (2 solutions).
- When the minimal distance between queens is 3, the minimal chess board size with solutions is 10×10 (4 solutions).
- When the minimal distance between queens is 4, the minimal chess board size with solutions is 16×16 (2 solutions).
- When the minimal distance between queens is 5, the minimal chess board size with solutions is 28×28 (10 solutions).
- When the minimal distance between queens is 6, the minimal chess board size with solutions is 36×36 (2 solutions).

When the minimal distance between queens is 7, the minimal chess board size with solutions must be more than 50×50, because there are no solutions in any board size up to 50×50. When I created my chess queens application I limited the maximal board size to be 50×50, because I didn’t know that larger board sizes can return results in reasonable time. But I checked and found out recently that even boards with size 36×36 return results (when the minimal distance between queens is 6) and even the 50×50 board returns “no solutions” when the minimal distance between queens is 7. It would be interesting to increase the limit of my chess queens application to boards larger than 50×50 and find out the minimal chess board size with solutions when the minimal distance between queens is 7, 8 and more.

From my results above I would guess that with any even minimal distance between queens n, the minimal chess board size with solutions would probably be n^{2}×n^{2} with 2 solutions, but this needs to be proved. It would also be nice if I can find out a formula for the minimal chess board size with solutions as an expression of the minimal distance between queens, and also a formula for the number of solutions. But it looks like it can be a big challenge to find such a formula and prove it.

The numbers 10 and 28 above also appear in Pascal’s triangle, and it would be interesting to see if the minimal chess board size with solutions when the minimal distance between queens is higher are also numbers from Pascal’s triangle. If they are, I would guess they might be 55, 91 and 136 respectively for the odd distances 7, 9 and 11. It’s just a guess, but it matches the minimal board sizes of 1×1, 10×10 and 28×28 above. If my assumption is true, then the formula for the minimal chess board size with solutions with odd minimal distance between queens n would be the Binomial number of (((3 * n) + 1) / 2) over 2 (which is equal to (((3 * n) + 1) * ((3 * n) – 1) / 8)), and the number of solutions on the minimal board size would probably also be a number from Pascal’s triangle.

I found out that when using rooks with a minimal distance between rooks n, the minimal chess board size with solutions is n^{2}×n^{2} with 2 solutions, for all n>1 (odd or even). But this needs to be proved.