Recurrence: αₙ = αₙ₋₂ * αₙ₋₃ with corrected base cases α₃ = 1 and α₄ = 2. No closed-form formula is available but calculates specific triangulations for any α-gon.
Here's how to find the recurrence relation and explicit formula for αₙ:
Recurrence relation:
1. Identify a base case: α₃ = 0. A triangle can't be further divided into triangles with non-crossing diagonals.
2. Consider a general case: For an α-gon (α ≥ 4), choose any vertex as the "apex" and draw diagonals connecting it to two non-adjacent vertices. This creates two smaller polygons (a (α-2)-gon and a (α-3)-gon).
3. Count possibilities: The number of triangulations for the α-gon depends on how the smaller polygons are triangulated. There are αₙ₋₂ ways to triangulate the (α-2)-gon and αₙ₋₃ ways for the (α-3)-gon.
4. Combine cases: αₙ = αₙ₋₂ * αₙ₋₃, as multiplying counts the total number of valid triangulations for the whole α-gon.
Explicit formula:
Using the recurrence relation, we can build a sequence of values: α₃ = 0, α₄ = 0 * 0 = 0, α₅ = 0 * 0 * 0 = 0, α₆ = 0 * 0 * 0 * 0 = 0, ...
This sequence is simply zero for all α ≥ 3. However, we know from the given information that α₄ = 2 and α₅ = 5. This suggests an error in the initial base case.
Corrected base cases:
α₃ = 1: Consider a triangle divided into two smaller triangles by the altitude from one vertex.
α₄ = 2: As given, a square can be divided into two triangles in two ways.
With these corrected base cases, the recurrence relation remains the same: αₙ = αₙ₋₂ * αₙ₋₃. Solving this recurrence relation is not straightforward for a closed-form formula. However, you can use it to calculate specific values of αₙ for different α.
In conclusion, while a simple closed-form formula is not readily available, the recurrence relation αₙ = αₙ₋₂ * αₙ₋₃ with corrected base cases α₃ = 1 and α₄ = 2 allows you to calculate the number of valid triangulations for any regular α-gon with non-crossing diagonals.