This problem's: What is the count of distinct triangulations of a convex octagon?
A convex polygon only has interior angles strictly less than radians. A polygon is a closed set. A polygon boundary has line segments are called sides and the points where the line segments join are called vertices. This means that a straight line segment can be drawn from any point within the polygon to any other point within the polygon without leaving the polygon. The boundary of the polygon is part of the polygon.
For this problem, the polygons must have non-zero area, a triangulation must be made by joining only vertices by line segments called chords, chords must pass into the interior, chords are not permitted to cross each other , two chords are duplicates if they have the same vertices as ends regardless of the vertex order forming the chord, duplicate chords are not permitted, the triangles formed must have non-zero area and the interior of the polygon must be reduced to triangles. Two triangulations are distinct if their sets of chords are different.
An octagon is a polygon with 8 sides and 8 vertices.
A hint: a triangle has one triangulation, a convex quadrilateral has 2 triangulations and a convex pentagon has five triangulations.
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See A000108 re Leonhard Euler used the formula C(n) = Product_{i=3..n} (4*i-10)/(i-1) in his 'Betrachtungen, auf wie vielerley Arten ein gegebenes polygonum durch Diagonallinien in triangula zerschnitten werden könne' and computes by recursion C(n+2) for n = 1..8. (Berlin, 4th September 1751, in a letter to Goldbach.)
The paper's title translates to English as "Considerations on how many ways a given polygon can be divided into triangles by diagonal lines".