Triangles within a regular n-gon versus integer triangles with a fixed perimeter of n. Conjecture: The number of incongruent acute triangles formed from the vertices of a regular n-gon is identical to the number of integer triangles formed from a fixed integer perimeter of length n.
Is there a proof? The sequence is give by Alcuin's sequence: https://oeis.org/A005044
 A: Triangles with perimeter $n$ can be characterized by their side lengths $a, b, c$ with $a+b+c=n$. The side lengths $(a,b,c)$ must have the largest side be less than the sum of the two others; when the sum $a+b+c=n$ is known, it's equivalent to ask that $\max\{a,b,c\} < \frac n2$.
Similarly, for a triangle whose vertices are taken from a fixed regular $n$-gon, we can count the number of steps around the $n$-gon between the vertices and get three numbers $a',b',c'$ with $a'+b'+c'=n$. These characterize the triangle: any two triangles with the same triples $(a',b',c')$ are congruent by a symmetry of the $n$-gon. Conversely, to be congruent, two such triangles must be similar, and since $\frac{a'\pi}{n}, \frac{b'\pi}{n}, \frac{c'\pi}{n}$ give the angles of the triangle, they must have the same triple. For all three of these angles to be acute, we must have $\max\{a',b',c'\} < \frac n2$.
So the set of triples $(a,b,c)$ describing triangles with perimeter $n$ is the same as the set of triples $(a',b',c')$ describing triangles whose vertices are vertices of the regular $n$-gon. In particular, when we count the triples, we ge the same number.
