OFFSET
2,1
COMMENTS
Only non-degenerate triangles are considered.
Conjecture: sequence is well-defined, i.e., a(n) is an integer for every n > 1.
For n > 1, n odd, a(n) >= 4(n-2)^2(n^2-3n+1)(n^2-3n+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), (n-1, n-1)}, {(n-2, 0), (1, n-2)}, and {((n-1)/2, n-1), ((n-3)/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 ((n-2)^2+1)^2, (cf. A082044) obtained via the lines passing through the points {(0,2), (1,n-1)}, {(n-2,0), (n-1,n-2)}, and {(0,0), (1,n-1)}.
LINKS
Giovanni Resta, Illustration of the first 9 terms
FORMULA
For n>1 odd, a(n) >= 4(n-2)^2 (n^2-3n+1)(n^2-3n+3).
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.
CROSSREFS
KEYWORD
nonn,more,nice
AUTHOR
Giovanni Resta, Dec 28 2013
STATUS
approved