cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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.

Original entry on oeis.org

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

Views

Author

Vijay Srinivas Balaji, Sep 23 2023

Keywords

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)

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

Crossrefs

Cf. A367015 (number of regions), A366932 (number of edges), A366478 (vertices including boundary points), A033428 (number of chords).
See also A091908, A092098, A331782.

Programs

  • 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 *)

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

A367015 Number of regions formed after n points have been placed in general position on each edge of the triangle in A365929.

Original entry on oeis.org

1, 4, 28, 136, 445, 1126, 2404, 4558, 7921, 12880, 19876, 29404, 42013, 58306, 78940, 104626, 136129, 174268, 219916, 274000, 337501, 411454, 496948, 595126, 707185, 834376, 978004, 1139428, 1320061, 1521370, 1744876, 1992154, 2264833, 2564596, 2893180, 3252376, 3644029, 4070038
Offset: 0

Views

Author

Scott R. Shannon, Nov 01 2023

Keywords

Comments

See A365929 for more information.

Links

  • Scott R. Shannon, Image for n = 2.
  • Scott R. Shannon, Image for n = 3. Note that although the number of k-gons will vary as the edge points change position the total number of regions will stay constant (at 136 for n = 3) as long as all internal vertices remain simple.

Crossrefs

Cf. A365929 (internal vertices), A366932 (edges), A366478 (vertices/3).

Formula

Conjecture: a(n) = (9*n^4 - 12*n^3 + 15*n^2 + 4)/4.
a(n) = A366932(n) - 3*A366478(n) + 1 by Euler's formula.

A366932 Number of edges formed after n points have been placed in general position on each edge of the triangle in A365929.

Original entry on oeis.org

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

Views

Author

Scott R. Shannon, Nov 02 2023

Keywords

Comments

See A365929 for more information. See A365929 and A367015 for images of the triangle.

Crossrefs

Cf. A365929 (internal vertices), A367015 (regions), A366478.

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.

A366483 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.

Original entry on oeis.org

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

Views

Author

Scott R. Shannon and N. J. A. Sloane, Nov 09 2023

Keywords

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

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.

Formula

a(n) = A366485(n) - A366486(n) + 1 (Euler).

A366484 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.

Original entry on oeis.org

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

Views

Author

Scott R. Shannon and N. J. A. Sloane, Nov 09 2023

Keywords

Comments

See A366483 for further information.

Links

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.

Formula

a(n) = A366485(n) - A366486(n) - 3*n - 2 (Euler).

A366486 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.

Original entry on oeis.org

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

Views

Author

Scott R. Shannon and N. J. A. Sloane, Nov 09 2023

Keywords

Comments

See A366483 for further information.

Links

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.

Formula

a(n) = A366485(n) - A366483(n) + 1 (Euler).

A366485 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.

Original entry on oeis.org

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

Views

Author

Scott R. Shannon and N. J. A. Sloane, Nov 09 2023

Keywords

Comments

See A366483 for further information. See A366483 and A366486 for images of the triangle.

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.

Formula

a(n) = A366483(n) + A366486(n) - 1 (Euler).
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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