Squares of triangles of triangles

Numbers of the form n^2/(n^2-1) factor as n^2/(n-1)(n+1), and so it
makes sense they show up on these lists. Triangular numerators are
similar, we have [n(n+1)/2]/[n(n+1)/2 - 1] = n(n+1)/(n-1)(n+1). The
mystery of the squares of triangles of triangles is explained by the
fact that from tt(n) = n(n+1)(n^2+n+2)/8, the triangle of a triangle
function, we get tt(n)^2/(tt(n)^2-1) =
n^2(n+1)^2(n^2+n+2)^2 / (n-1)(n+1)(n^2-n+2)(n^2+n+4)(n^2+3n+4).

I think it would be worthwhile to explore this sort of thing further.