A beetle is on a two-dimensional n × n grid. The beetle can only move to the right, up, and diagonally a step
One unit at a time. curry
(x, y) → (x + 1, y) (x, y) → (x, y + 1) (x, y) → (x + 1, y + 1)
We will denote by an the number of different routes of the beetle from the beginning of the axes (0,0) to the point (n, n) that do not go below the main diagonal. y = x Find a regression formula and initial conditions sufficient for an Note: the scheme can be used.

To solve this problem, we need to find the number of distinct paths a beetle can take from \((0,0)\) to \((n,n)\) on an \(n \times n\) grid, moving only right, up, or diagonally, while ensuring the path does not fall below the main diagonal \(y = x\).

### Step 1: Understanding the Constraints
The beetle can move in three directions:
- Right: \((x, y) \to (x + 1, y)\)
- Up: \((x, y) \to (x, y + 1)\)
- Diagonally: \((x, y) \to (x + 1, y + 1)\)

The condition that the beetle does not fall below the main diagonal means that at any point \((x, y)\) on the path, \(y \leq x\).

### Step 2: Recurrence Relation
Let \(a_n\) be the number of such paths from \((0,0)\) to \((n,n)\).

#### Base Case
For \(n = 0\):
- There is only one point \((0,0)\), so \(a_0 = 1\).

#### General Case
For \(n > 0\), consider the paths to \((n,n)\). The last step to reach \((n,n)\) must be one of the following:
1. From \((n-1, n)\): This is not possible because \((n-1, n)\) lies above the line \(y = x\).
2. From \((n, n-1)\): This is also not possible because \((n, n-1)\) lies below the line \(y = x\).
3. From \((n-1, n-1)\): This is possible and valid.

Thus, the number of valid paths \(a_n\) is directly related to the number of valid paths to \((n-1, n-1)\). Therefore, the recurrence relation is:
\[a_n = 3a_{n-1}\]

This is because each valid path to \((n-1, n-1)\) can be extended to \((n, n)\) in exactly three ways:
- Moving right from \((n-1, n-1)\) to \((n, n-1)\) (which is actually not allowed, so we ignore this case)
- Moving up from \((n-1, n-1)\) to \((n-1, n)\) (which is also not allowed)
- Moving diagonally from \((n-1, n-1)\) to \((n, n)\).

Thus, the only valid move is the diagonal move, so \(a_n = a_{n-1}\).

### Step 3: Solving the Recurrence
Given the recurrence relation and the base case:
\[a_0 = 1\]
\[a_n = a_{n-1}\]

This implies:
\[a_n = 1\]

### Conclusion
The number of distinct paths \(a_n\) from \((0,0)\) to \((n,n)\) that do not fall below the main diagonal \(y = x\) is always 1. This is because the only path that satisfies the given conditions is the path that moves diagonally from \((0,0)\) to \((n,n)\) in \(n\) steps.

### Summary
- **Initial condition**: \(a_0 = 1\)
- **Recurrence relation**: \(a_n = a_{n-1}\)

The solution is \(a_n = 1\) for all \(n \geq 0\).

Would you like more details on any specific step? 

Here are some related questions you might find interesting:
1. How would the problem change if diagonal moves were not allowed?
2. What is the combinatorial significance of the paths that stay on or above the main diagonal?
3. How can we generalize this problem to a three-dimensional grid?
4. What are some real-world applications of such grid path problems?
5. How can we visualize the paths on the grid effectively?

**Tip:** When dealing with recurrence relations, always verify the constraints and boundary conditions to simplify the problem.

Explore how to count distinct paths on an n × n grid where a beetle moves right, up, or diagonally without crossing below the main diagonal. Understand the recurrence relation and initial conditions to solve the problem effectively. Suitable for advanced mathematics students.

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