We are tasked with finding a recursive formula for $Q_n$, the number of ways to place rooks on an $n \times n$ chessboard such that no two rooks can attack each other and the configuration is symmetric about the diagonal from the lower-left to the upper-right.

Step 1: Understanding the Symmetry

The condition that the placement is symmetric about the diagonal means that if a rook is placed at position $(i, j)$, there must also be a rook at $(j, i)$ for the placement to maintain symmetry. This implies that the positions of rooks must be mirrored along the diagonal.

Step 2: Constraints on Placing Rooks

1. Non-attacking rooks: No two rooks can be placed in the same row or column, meaning that each row and each column can contain at most one rook.

2. Symmetry: Since the configuration is symmetric about the diagonal, if a rook is placed at $(i, j)$, then a rook must also be placed at $(j, i)$. This means that we can only place rooks in positions where $i \leq j$ and $i = j$ or $i \neq j$.

Step 3: Recursive Structure of $Q_n$

To find a recursive formula, let's break down the problem by considering the $n \times n$ board:

1. Base cases:

   • $Q_1 = 1$: A 1x1 board has only one position where a rook can be placed, and it trivially satisfies the symmetry condition.
   • $Q_2 = 2$: On a 2x2 board, the two possible placements of rooks are:
     • Both rooks on the diagonal (positions (1,1) and (2,2)).
     • One rook in each of the two symmetric positions (positions (1,2) and (2,1)).

2. Recursive step: For an $n \times n$ board, we can consider the rook placement in two parts:

   • First, place a rook in the first row (and consequently in the first column, by symmetry).
   • After placing the first rook, we have reduced the problem to a smaller $(n-1) \times (n-1)$ board.

Thus, the number of ways to place rooks on an $n \times n$ board symmetrically is related to the number of ways to place rooks on a smaller board, and the recursion follows the pattern:

$Q_n = Q_{n-1} + (n-1) \cdot Q_{n-2}$

Step 4: Recursion Formula

To summarize, the recursive formula for $Q_n$ is:

$Q_n = Q_{n-1} + (n-1) \cdot Q_{n-2}$

This formula accounts for both the placement of a rook on the diagonal and the symmetric placements of rooks in off-diagonal positions.

Step 5: Explanation

• $Q_{n-1}$ corresponds to configurations where one rook is placed on the diagonal and the rest are placed symmetrically about the diagonal.
• $(n-1) \cdot Q_{n-2}$ corresponds to configurations where no rook is placed on the diagonal, and symmetric pairs of rooks are placed in the off-diagonal positions.

Conclusion

The recursive formula for the number of symmetric rook placements on an $n \times n$ chessboard is:

$Q_n = Q_{n-1} + (n-1) \cdot Q_{n-2}$

where the base cases are:

• $Q_1 = 1$
• $Q_2 = 2$.

Would you like more details on this recursive process, or any clarifications on the formula? Here are some related questions for further exploration:

1. How would you extend this formula for larger $n$?
2. How can we calculate $Q_n$ efficiently using dynamic programming?
3. Can this formula be applied to other types of chessboard symmetries, like diagonal or rotational symmetry?
4. How does the concept of symmetry impact other combinatorics problems, like placing queens or knights?
5. What is the relation between $Q_n$ and other chess piece configurations like kings or bishops?

Tip: You can use dynamic programming to compute $Q_n$ for larger values efficiently by storing intermediate results, thus avoiding redundant calculations.