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A365929
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Number of intersections formed within a triangle by placing n points "in general position" on each of the three sides and connecting each point to each of the points on the other two sides using straight lines.
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9
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0, 0, 15, 108, 396, 1050, 2295, 4410, 7728, 12636, 19575, 29040, 41580, 57798, 78351, 103950, 135360, 173400, 218943, 272916, 336300, 410130, 495495, 593538, 705456, 832500, 975975, 1137240, 1317708, 1518846, 1742175, 1989270, 2261760, 2561328, 2889711, 3248700, 3640140, 4065930, 4528023
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OFFSET
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0,3
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COMMENTS
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There are n points on each of the three sides (not counting the vertices of the triangle). Each point must be connected to every point on the other two sides. A033428(n) = 3*n^2 gives the number of lines.
"In general position" means that all interior intersection points are simple. No three-way or higher intersections are permitted.
If the 3*n+3 boundary points are included in the count, there are 3*A366478 points.
For the configurations where the boundary points are equally spaced and every pair of boundary points is joined by a chord, see A091908, A092098, A331782.
(End)
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REFERENCES
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Vijay Srinivas Balaji, Formulating A Conjecture For Intersections Created From Crossing Lines Within Different Polygons, International School of Helsingborg, 2023.
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LINKS
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FORMULA
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a(n) = (3/4)*n^2*(n-1)*(3*n-1). [Proof: For intersection points defined by two points on two opposite sides, the number is 3*C(n,2)^2; for intersection points defined by two points on one side and one point on each of the other two sides, the number is 3*C(n,2)*n^2. - N. J. A. Sloane, Nov 07 2023]
G.f.: 3*x^2*(5 + 11*x + 2*x^2)/(1 - x)^5. - Stefano Spezia, Sep 24 2023
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EXAMPLE
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a(5) = (3/4) * 5^2 * (3*5^2 - 4*5 + 1) = 1050.
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MAPLE
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p__3 := n -> 9/4*n^4 - 3*n^3 + 3/4*n^2; for n from 0 to 55 do p__3(n); end do;
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MATHEMATICA
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LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 15, 108, 396}, 50] (* or *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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