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 1-10 of 23 results. Next

A092867 Number of regions in an equilateral triangular figure formed by the straight line segments connecting all vertices and all points that divide the sides into n equal parts.

Original entry on oeis.org

1, 12, 75, 252, 715, 1572, 3109, 5676, 9291, 14556, 22081, 32502, 44935, 62868, 83286, 108384, 140152, 181710, 225565, 282978, 342792, 415614, 502318, 606642, 708505, 839874, 983007, 1141416, 1315102, 1529526, 1733476, 1994550, 2259420, 2559990, 2878053, 3237414, 3593521, 4047906, 4510590, 5002350, 5506918, 6128100, 6704800, 7414518, 8113992, 8858622, 9682927, 10626774, 11478142, 12519492
Offset: 1

Views

Author

Hugo Pfoertner, Mar 15 2004

Keywords

Examples

			a(2)=12 because the 6 line segments mutually connecting the vertices and the mid-side nodes form 12 congruent right triangles of two different sizes.
a(3)=75: 48 triangles, 24 quadrilaterals and 3 pentagons are formed. See pictures at Pfoertner link.
		

Crossrefs

Cf. A092866 (number of intersections), A274585 (number of points both inside and on the triangle sides), A274586 (number of edges), A331911 (number of n-gons).
Cf. A092098 (regions in triangle cut by line segments connecting vertices with subdivision points on opposite side), A006533 (regions formed by all diagonals in regular n-gon), A002717 (triangles in triangular matchstick arrangement).
If the boundary points are in general position, we get A367117, A213827, A367118, A367119. - N. J. A. Sloane, Nov 09 2023

Formula

By the Euler characteristic, a(n) = A274586(n) - A274585(n) + 1 = A274586(n) - A092866(n) - 3n - 1.

Extensions

a(1)=1 prepended by Max Alekseyev, Jun 29 2016
a(6)-a(50) from Cynthia Miaina Rasamimanananivo, Jun 28 2016, Jul 01 2016, Aug 05 2016, Aug 15 2016
Definition edited by N. J. A. Sloane, May 13 2020

A255011 Number of polygons formed by connecting all the 4n points on the perimeter of an n X n square by straight lines; a(0) = 0 by convention.

Original entry on oeis.org

0, 4, 56, 340, 1120, 3264, 6264, 13968, 22904, 38748, 58256, 95656, 120960, 192636, 246824, 323560, 425408, 587964, 682296, 932996, 1061232, 1327524, 1634488, 2049704, 2227672, 2806036, 3275800, 3810088, 4307520, 5298768, 5577096, 6958848, 7586496, 8672520, 9901352
Offset: 0

Views

Author

Johan Westin, Feb 12 2015

Keywords

Comments

There are n+1 points on each side of the square, but that counts the four corners twice, so there are a total of 4n points on the perimeter. - N. J. A. Sloane, Jan 23 2020
a(n) is always divisible by 4, by symmetry. If n is odd, a(n) is divisible by 8.
From Michael De Vlieger, Feb 19-20 2015: (Start)
For n > 0, the vertices of the bounding square generate diametrical bisectors that cross at the center. Thus each diagram has fourfold symmetry.
For n > 0, an orthogonal n X n grid is produced by corresponding horizontal and vertical points on opposite sides.
Terms {1, 3, 9} are not congruent to 0 (mod 8).
Number of edges: {0, 8, 92, 596, 1936, 6020, 11088, 26260, 42144, 72296, 107832, ...}. See A331448. (End)

Examples

			For n = 3, the perimeter of the square contains 12 points:
  * * * *
  *     *
  *     *
  * * * *
Connect each point to every other point with a straight line inside the square. Then count the polygons (or regions) that have formed. There are 340 polygons, so a(3) = 340.
For n = 1, the full picture is:
  *-*
  |X|
  *-*
The lines form four triangular regions, so a(1) = 4.
For n = 0, the square can be regarded as consisting of a single point, producing no lines or polygons, and so a(0) = 0.
		

Crossrefs

Cf. A092098 (triangular analog), A331448 (edges), A331449 (points), A334699 (k-gons).
For the circular analog see A006533, A007678.

Formula

No formula is presently known. - N. J. A. Sloane, Feb 04 2020

Extensions

a(11)-a(29) from Hiroaki Yamanouchi, Feb 23 2015
Offset changed by N. J. A. Sloane, Jan 23 2020

A091908 Number of interior intersection points made by the straight line segments connecting the edges of an equilateral triangle with the n-1 points resulting from a subdivision of the sides into n equal pieces, counting coinciding intersection points only once.

Original entry on oeis.org

0, 1, 12, 13, 48, 49, 108, 109, 192, 193, 300, 301, 432, 433, 576, 589, 768, 769, 972, 961, 1200, 1201, 1452, 1405, 1728, 1729, 2028, 2029, 2352, 2341, 2700, 2701, 3072, 3073, 3444, 3469, 3888, 3889, 4332, 4297, 4800, 4777, 5292, 5293, 5724, 5809, 6348
Offset: 1

Views

Author

Hugo Pfoertner, Feb 19 2004

Keywords

Comments

In a drawing the distinction between simple and multiple intersection points may be difficult due to near-coincidences. E.g. there are no coincident intersections for n=7.
Note that 3 divides a(2k)-1 and a(2k+1). - T. D. Noe, Jun 29 2005
The interior intersection points only can be the result of the concurrency of 2 or 3 segments by construction. It is easy to see that the total number of 2-intersections N2 is 3*(n-1)^2 (which includes every 3-intersection as two 2-intersections) by symmetry. But we are interested in excluding the concurrency of more than 2. By Ceva's theorem necessary and sufficient condition for 3 concurrent segments that connect the edges with the opposite side, the number of 3-intersections N3 is the same as the number of (i,j,k) belonging to [1,n-1]x[1,n-1]x[1,n-1] such that (i/(n-i))*(j/(n-j))*(k/(n-k))=1. Thus the terms a(n)=N2-2*N3. - Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 26 2006
If n is even then a(n) < 3*(n-1)^2; if n is odd then a(n) = 3*(n-1)^2 except for n in A332378. - N. J. A. Sloane, Feb 14 2020

Examples

			a(3)=12 because the 3*2 line segments intersect each other in 12 distinct points (see pictures given at link)
a(4)=13 because the 27 intersections form 6 two line intersection points and 7 three line intersection points.
		

Crossrefs

Cf. A091910 = radial locations of intersection points, A092098 = number of regions that the line segments cut the triangle into, A006561.
For the basic properties of the underlying graph, see A092098 (cells), A331782 (vertices), A331782 (vertices), A332376 & A332377 (edges). - N. J. A. Sloane, Feb 14 2020

Programs

  • PARI
    for(n=1,70,conc=0;for(i=1,n-1,for(j=1,n-1,for(k=1,n-1,if(i*j*k/((n-i)*(n-j)*(n-k))==1,conc++))));print1(3*(n-1)^2-2*conc,",")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 26 2006

Extensions

More terms from T. D. Noe, Jun 29 2005
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Oct 26 2006

A331782 Total number of vertices in graph formed by the straight line segments connecting the edges of an equilateral triangle with the n-1 points resulting from a subdivision of the sides into n equal pieces, counting coinciding intersection points only once.

Original entry on oeis.org

3, 7, 21, 25, 63, 67, 129, 133, 219, 223, 333, 337, 471, 475, 621, 637, 819, 823, 1029, 1021, 1263, 1267, 1521, 1477, 1803, 1807, 2109, 2113, 2439, 2431, 2793, 2797, 3171, 3175, 3549, 3577, 3999, 4003, 4449, 4417, 4923, 4903, 5421, 5425, 5859, 5947, 6489, 6397, 7059, 7063, 7653, 7657, 8271, 8275, 8889
Offset: 1

Views

Author

N. J. A. Sloane, Feb 13 2020

Keywords

Comments

a(n) <= 3*(n^2-n+1), with equality iff n is odd and not a member of A332378. A331423 gives the difference between a(n) and the upper bound.

Crossrefs

Cf. A091908, A092098 (number of cells), A332376 (number of edges), A332378, A331423.

Formula

a(n) = A091908(n) + 3*n.

A108914 Number of regions formed inside square by diagonals and the segments joining the vertices to the points dividing the sides into n equal length segments.

Original entry on oeis.org

4, 32, 96, 188, 332, 460, 712, 916, 1204, 1488, 1904, 2108, 2716, 3080, 3532, 4068, 4772, 5140, 6016, 6392, 7188, 7992, 8936, 9260, 10484, 11312, 12208, 12968, 14396, 14660, 16504, 17220, 18436, 19680, 20756, 21548, 23692, 24728, 25992, 26868, 29204, 29704, 32176, 33068, 34444, 36552, 38552
Offset: 1

Views

Author

Len Smiley and Brian Wick ( mathclub(AT)math.uaa.alaska.edu ), Jul 19 2005

Keywords

Crossrefs

A092098 is the corresponding count for triangles.
A355949 (vertices), A355948 (edges), A355992 (k-gons), A355838, A355798.

Formula

If n=1 or n is prime, a(n)=18*n^2-26*n+12.
If n is composite, vanishing regions from 3- and 4-fold concurrency must be subtracted.
a(n) = A355948(n) - A355949(n) + 1 by Euler's formula.

Extensions

a(23), a(33) corrected, a(41) and above by Scott R. Shannon, Jul 22 2022

A345459 Number of polygons formed when connecting all 4n points on the perimeter of an n X n square by infinite lines.

Original entry on oeis.org

0, 4, 80, 568, 2024, 6052, 12144, 26976, 45024, 76724, 116840, 191128, 245976, 388452, 501888, 661476, 870168, 1199724, 1402096, 1911384, 2188320, 2739280, 3371264, 4224288, 4617224, 5801372, 6780568
Offset: 0

Views

Author

Keywords

Comments

The width/height of the entire figure grows as ~ 2*n^3 for large n. See the Formula section below.

Examples

			a(2) = 80. Connecting the 8 perimeter points results in the creation of forty-eight 3-gons and eight 4-gons inside the square while creating twenty-four 3-gons outside the square, giving eighty polygons in total. See the linked images.
		

Crossrefs

Cf. A255011 (number inside the square), A345648 (number outside the square), A345649 (number of vertices), A345650 (number of edges), A344993, A344857, A092098, A007678.

Formula

a(n) = A345650(n) - A345649(n) + 1.
Assuming the square is of size n x n centered on the origin the x (or y) offset for the eight outermost vertices is n^3 - 2*n^2 + 3*n/2, which have a corresponding y (or x) offset of n^2 - 3*n/2 + 1. The total distance from the origin of these vertices is sqrt(n^6 - 4*n^5 + 8*n^4 - 9*n^3 + 13*n^2/2 - 3*n + 1).

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.

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

A332376 Total number of edges in graph formed by the straight line segments connecting the edges of an equilateral triangle with the n-1 points resulting from a subdivision of the sides into n equal pieces.

Original entry on oeis.org

3, 12, 39, 54, 123, 144, 255, 282, 435, 468, 663, 702, 939, 984, 1245, 1314, 1635, 1692, 2055, 2100, 2523, 2592, 3039, 3042, 3603, 3684, 4215, 4302, 4875, 4950, 5583, 5682, 6339, 6444, 7107, 7254, 7995, 8112, 8895, 8964, 9843, 9936, 10839, 10974
Offset: 1

Views

Author

N. J. A. Sloane, Feb 13 2020

Keywords

Crossrefs

Cf. A091908, A092098 (number of cells), A331782 (number of vertices).
Equals three times A332377.

Formula

a(n) = A092098(n) + A331782(n) - 1 (Euler's formula).

A367117 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 vertices in the resulting planar graph.

Original entry on oeis.org

3, 12, 72, 282, 795, 1818, 3612, 6492, 10827, 17040, 25608, 37062, 51987, 71022, 94860, 124248, 159987, 202932, 253992, 314130, 384363, 465762, 559452, 666612, 788475, 926328, 1081512, 1255422, 1449507, 1665270, 1904268, 2168112, 2458467, 2777052, 3125640, 3506058, 3920187, 4369962
Offset: 0

Views

Author

Keywords

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.
Note that although the number of k-gons in the graph will vary as the edge points change position, the total number of regions will stay constant as long as all internal vertices remain simple.

Crossrefs

Cf. A367118 (regions), A367119 (edges).
If the boundary points are equally spaced, we get A274585, A092866, A274586, A092867. - N. J. A. Sloane, Nov 09 2023

Programs

Formula

Theorem: a(n) = (3/4)*(n+1)*(3*n^3+n^2+4).
a(n) = A367119(n) - A367118(n) + 1 by Euler's formula.

A367118 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.

Original entry on oeis.org

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

Views

Author

Keywords

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.

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

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.
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