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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.
+10
9
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
0,3
COMMENTS
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.
Comments from N. J. A. Sloane, Oct 29 2023: (Start)
"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)
REFERENCES
Vijay Srinivas Balaji, Formulating A Conjecture For Intersections Created From Crossing Lines Within Different Polygons, International School of Helsingborg, 2023.
LINKS
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.
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
EXAMPLE
a(5) = (3/4) * 5^2 * (3*5^2 - 4*5 + 1) = 1050.
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;
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 *)
CROSSREFS
Cf. A367015 (number of regions), A366932 (number of edges), A366478 (vertices including boundary points), A033428 (number of chords).
See also A091908, A092098, A331782.
KEYWORD
nonn,easy
AUTHOR
STATUS
approved
Place n points in general position on each side of an equilateral triangle, and join every pair of the 3*n+3 boundary points by a chord; sequence gives number of regions in the resulting planar graph.
+10
8
1, 13, 82, 307, 841, 1891, 3718, 6637, 11017, 17281, 25906, 37423, 52417, 71527, 95446, 124921, 160753, 203797, 254962, 315211, 385561, 467083, 560902, 668197, 790201, 928201, 1083538, 1257607, 1451857, 1667791, 1906966, 2170993, 2461537, 2780317, 3129106, 3509731, 3924073, 4374067
OFFSET
0,2
COMMENTS
"In general position" implies that the internal lines (or chords) only have simple intersections. There is no interior point where three or more chords meet.
LINKS
Scott R. Shannon, Image for n = 1.
Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 5.
FORMULA
Conjecture: a(n) = (1/4)*(9*n^4 + 12*n^3 + 15*n^2 + 12*n + 4).
a(n) = A367119(n) - A367117(n) + 1 by Euler's formula.
CROSSREFS
Cf. A367117 (vertices), A367119 (edges), A091908, A092098, A331782, A367015.
If the boundary points are equally spaced, we get A274585, A092866, A274586, A092867. - N. J. A. Sloane, Nov 09 2023
KEYWORD
nonn
AUTHOR
STATUS
approved
Number of edges formed after n points have been placed in general position on each edge of the triangle in A365929.
+10
7
3, 9, 51, 255, 855, 2193, 4719, 8991, 15675, 25545, 39483, 58479, 83631, 116145, 157335, 208623, 271539, 347721, 438915, 546975, 673863, 821649, 992511, 1188735, 1412715, 1666953, 1954059, 2276751, 2637855, 3040305, 3487143, 3981519, 4526691, 5126025, 5782995, 6501183, 7284279
OFFSET
0,1
COMMENTS
See A365929 for more information. See A365929 and A367015 for images of the triangle.
FORMULA
Conjecture: a(n) = 3*(3/2*n^4 - 2*n^3 + 3/2*n^2 + n + 1).
a(n) = A367015(n) + 3*A366478(n) - 1 by Euler's formula.
CROSSREFS
Cf. A365929 (internal vertices), A367015 (regions), A366478.
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Nov 02 2023
STATUS
approved
Place n equally spaced points on each side of an equilateral triangle, and join each of these points by a chord to the 2*n new points on the other two sides: sequence gives number of vertices in the resulting planar graph.
+10
5
3, 6, 22, 108, 300, 919, 1626, 3558, 5824, 9843, 14352, 23845, 30951, 47196, 62773, 82488, 104544, 144784, 173694, 230008, 276388, 336927, 403452, 509218, 582417, 702228, 824956, 969387, 1098312, 1321978, 1463580, 1724190, 1952509, 2221497, 2505169, 2846908, 3103788, 3556143, 3978763, 4444003
OFFSET
0,1
COMMENTS
We start with the three corner points of the triangle, and add n further points along each edge. Including the corner points, we end up with n+2 points along each edge, and the edge is divided into n+1 line segments.
Each of the n points added to an edge is joined by 2*n chords to the points that were added to the other two edges. There are 3*n^2 chords.
LINKS
Scott R. Shannon, Image for n = 1.
Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 3.
Scott R. Shannon, Image for n = 4.
Scott R. Shannon, Image for n = 5.
Scott R. Shannon, Image for n = 10.
FORMULA
a(n) = A366485(n) - A366486(n) + 1 (Euler).
CROSSREFS
Cf. A366484 (interior vertices), A366485 (edges), A366486 (regions).
If the 3*n points are placed "in general position" instead of uniformly, we get sequences A366478, A365929, A366932, A367015.
If the 3*n points are placed uniformly and we also draw chords from the three corner points of the triangle to these 3*n points, we get A274585, A092866, A274586, A092867.
KEYWORD
nonn
AUTHOR
STATUS
approved
Place n equally spaced points on each side of an equilateral triangle, and join each of these points by a chord to the 2*n new points on the other two sides: sequence gives number of interior vertices in the resulting planar graph.
+10
5
0, 0, 13, 96, 285, 901, 1605, 3534, 5797, 9813, 14319, 23809, 30912, 47154, 62728, 82440, 104493, 144730, 173637, 229948, 276325, 336861, 403383, 509146, 582342, 702150, 824875, 969303, 1098225, 1321888, 1463487, 1724094, 1952410, 2221395, 2505064, 2846800, 3103677, 3556029, 3978646, 4443883
OFFSET
0,3
COMMENTS
See A366483 for further information.
LINKS
Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 3.
Scott R. Shannon, Image for n = 4.
Scott R. Shannon, Image for n = 5.
Scott R. Shannon, Image for n = 10.
FORMULA
a(n) = A366485(n) - A366486(n) - 3*n - 2 (Euler).
CROSSREFS
Cf. A366483 (vertices), A366485 (edges), A366486 (regions).
If the 3*n points are placed "in general position" instead of uniformly, we get sequences A366478, A365929, A366932, A367015.
If the 3*n points are placed uniformly and we also draw chords from the three corner points of the triangle to these 3*n points, we get A274585, A092866, A274586, A092867.
KEYWORD
nonn
AUTHOR
STATUS
approved
Place n equally spaced points on each side of an equilateral triangle, and join each of these points by a chord to the 2*n new points on the other two sides: sequence gives number of regions in the resulting planar graph.
+10
5
1, 4, 27, 130, 385, 1044, 2005, 4060, 6831, 11272, 16819, 26436, 35737, 52147, 69984, 92080, 117952, 157770, 193465, 249219, 302670, 368506, 443026, 546462, 635125, 757978, 890133, 1041775, 1191442, 1407324, 1581058, 1837417, 2085096, 2365657, 2670429, 3018822, 3328351, 3771595, 4213602
OFFSET
0,2
COMMENTS
See A366483 for further information.
LINKS
Scott R. Shannon, Image for n = 1.
Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 3.
Scott R. Shannon, Image for n = 4.
Scott R. Shannon, Image for n = 5.
Scott R. Shannon, Image for n = 10.
FORMULA
a(n) = A366485(n) - A366483(n) + 1 (Euler).
CROSSREFS
Cf. A366483 (vertices), A366484 (interior vertices), A366485 (edges).
If the 3*n points are placed "in general position" instead of uniformly, we get sequences A366478, A365929, A366932, A367015.
If the 3*n points are placed uniformly and we also draw chords from the three corner points of the triangle to these 3*n points, we get A274585, A092866, A274586, A092867.
KEYWORD
nonn
AUTHOR
STATUS
approved
Place n equally spaced points on each side of an equilateral triangle, and join each of these points by a chord to the 2*n new points on the other two sides: sequence gives number of edges in the resulting planar graph.
+10
4
3, 9, 48, 237, 684, 1962, 3630, 7617, 12654, 21114, 31170, 50280, 66687, 99342, 132756, 174567, 222495, 302553, 367158, 479226, 579057, 705432, 846477, 1055679, 1217541, 1460205, 1715088, 2011161, 2289753, 2729301, 3044637, 3561606, 4037604, 4587153, 5175597, 5865729, 6432138, 7327737
OFFSET
0,1
COMMENTS
See A366483 for further information. See A366483 and A366486 for images of the triangle.
FORMULA
a(n) = A366483(n) + A366486(n) - 1 (Euler).
CROSSREFS
Cf. A366483 (vertices), A366484 (interior vertices), A366486 (regions).
If the 3*n points are placed "in general position" instead of uniformly, we get sequences A366478, A365929, A366932, A367015.
If the 3*n points are placed uniformly and we also draw chords from the three corner points of the triangle to these 3*n points, we get A274585, A092866, A274586, A092867.
KEYWORD
nonn
AUTHOR
STATUS
approved

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