A triangular graph $T_n$ is the line graph of the complete graph $K_n$ (see for example here). Can you derive a formula for the number of triangles in the triangular graph $T_n$?
If not, can you at least find an asymptotic expression for it?
Edit: manual calculations lead me to believe that the answer is A002417. Still can't see why.
The Fool's Method: $T_n$ is an strongly regular graph with parameters $({n\choose 2}, 2(n-2),n-2,2)$, so its eigenvalues are $2(n-2)$ (with multiplicity $1$), $n-4$ (with multiplicity $n-1$) and $-2$ (with multiplicity $\frac{n(n-3)}{2}$). So the number of triangles is one sixth of $$tr(A^3) = 8(n-2)^3+(n-4)^3(n-1)-4n(n-3).$$
The KISS Method: Triangles in $T_n$ arise either from a triangle in $K_n$ or from $3$ edges meeting at a vertex. Hence there are ${n\choose 3}+ n{n-1\choose 3}$ triangles.
Both formulas simplify to $(n-2){n\choose 3}$.