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 A234566 1/a(n) is the area of the smallest triangle delimited by 3 lines each passing through at least 2 points of an n X n unitary spaced grid. 1
 4, 30, 770, 5148, 30566, 89900, 219960, 614460, 1146596, 2624076, 4299916, 8432732, 11016390, 22391148, 28183214 (list; graph; refs; listen; history; text; internal format)
 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 Table of n, a(n) for n=2..16. 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 Cf. A018808, A082044. Sequence in context: A064050 A162128 A337969 * A365366 A201974 A221250 Adjacent sequences: A234563 A234564 A234565 * A234567 A234568 A234569 KEYWORD nonn,more,nice AUTHOR Giovanni Resta, Dec 28 2013 STATUS approved

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Last modified July 19 13:41 EDT 2024. Contains 374394 sequences. (Running on oeis4.)