
COMMENTS

Only nondegenerate triangles are considered.
Conjecture: sequence is welldefined, i.e., a(n) is an integer for every n > 1.
For n > 1, n odd, a(n) >= 4(n2)^2(n^23n+1)(n^23n+3), with equality for n = 5,7,...,15. The bound is obtained considering the 3 lines passing by the 3 pairs of points {(0,1), (n1, n1)}, {(n2, 0), (1, n2)}, and {((n1)/2, n1), ((n3)/2, 0)}.
We may consider the similar problem of finding the largest triangle. Here, the areas for n>=2 are 1/2, 9/2, 25, 100, 289, 676, 1369,... so it appears that for n >= 4 the maximal area is ((n2)^2+1)^2, (cf. A082044) obtained via the lines passing through the points {(0,2), (1,n1)}, {(n2,0), (n1,n2)}, and {(0,0), (1,n1)}.


EXAMPLE

For n=2, consider the 2 X 2 grid formed by the points with coordinates (0,0), (0,1), (1,0) and (1,1). The two diagonals and the line passing through (0,0) and (1,0) form a triangle whose area is 1/4 and since no smaller triangle can be formed in this way, a(2) = 4.
