A365929 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.
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
Offset: 0
Examples
a(5) = (3/4) * 5^2 * (3*5^2 - 4*5 + 1) = 1050.
References
- Vijay Srinivas Balaji, Formulating A Conjecture For Intersections Created From Crossing Lines Within Different Polygons, International School of Helsingborg, 2023.
Links
- Paolo Xausa, Table of n, a(n) for n = 0..10000
- Vijay Srinivas Balaji, Diagram of Intersections for a Triangle.
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
- Scott R. Shannon, Image for n = 2.
- Scott R. Shannon, Image for n = 4.
- Scott R. Shannon, Image for n = 7.
- Scott R. Shannon, Image for n = 10.
Crossrefs
Programs
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Maple
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
LinearRecurrence[{5,-10,10,-5,1},{0,0,15,108,396},50] (* or *) A365929[n_]:=3n^2(n-1)(3n-1)/4;Array[A365929,50,0] (* Paolo Xausa, Nov 07 2023 *)
Formula
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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