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 1-5 of 5 results.

A277402 "3-Portolan numbers": number of regions formed by n-secting the angles of an equilateral triangle.

Original entry on oeis.org

1, 6, 19, 30, 61, 78, 127, 150, 217, 234, 331, 366, 469, 510, 631, 678, 817, 870, 1027, 1074, 1261, 1326, 1519, 1590, 1801, 1878, 2107, 2190, 2437, 2514, 2791, 2886, 3169, 3270, 3571, 3678, 3997, 4110, 4447, 4554, 4921, 5046, 5419, 5550, 5941, 6078, 6487, 6630, 7057, 7194
Offset: 1

Views

Author

Ethan Beihl, Oct 13 2016

Keywords

Comments

I like the name "portolan numbers": cf. the rhumbline designs on medieval maps, constructed in a similar way.
The regions can be counted using an adaptation of Smiley and Wick's method in A092098: count regions assuming there are no redundant intersections, then subtract the number of regions that Ceva's Theorem says must vanish.
Off-diagonal redundant intersections occur for triples of integers 1 <= i, j, k < floor(n/2)-1 such that M(i)*M(j) = M(k), where M(x) is the ratio (sin(Pi(n-x)/(3n)))/(sin(Pi*x/(3n))). In the case 10|n, this corresponds to the charming identity (sin(7*Pi/30)*sin(8*Pi/30))/(sin(3*Pi/30)*sin(2*Pi/30)) = sin(9*Pi/30)/sin(1*Pi/30).
Differs from A092098 (which counts regions when *sides*, not angles, are n-sected) for the first time at the tenth term.
The above equation has solutions if and only if 10|n. This can be shown by rewriting the equation in exponential form, and using facts about vanishing sums of roots of unity to narrow the possibilities for n. (See Conway and Jones, 1976.) This is computationally feasible because A273096(6) = 1. - Ethan Beihl, Nov 26 2016

Examples

			For n=3, a(n) gives the 19 regions formed by the intersection of 3*2 lines: 3 pentagons, 3 quadrilaterals, 12 triangles, and 1 big central hexagon.

Links

Crossrefs

Cf. A092098, A335411 (vertices), A335412 (edges), A335413 (ngons).

Programs

  • Mathematica
    regions[n_]:=
If[Mod[n,2]==0, 3n^2-6n+6, 3n^2-3n+1]-
  6*Length@
    Select[
     Flatten@
      With[
       {b=N@
          Table[
             1/2 - (Sqrt[3]/2)Tan[(60Degree/n)(n/2-i)],
             {i, 1, Floor[n/2]- 1}
             ]},
       Table[
        Abs[(1-b[[k]])b[[l]]b[[j]] - b[[k]](1-b[[l]])(1-b[[j]])],
        {j, 1, Floor[n/2]-1},
        {k, 1, Floor[n/2]-1},
        {l, 1, Floor[n/2]-1}]
       ],
     Chop@#==0&]

Formula

Empirical g.f.: x*(1 + 5*x + 12*x^2 + 6*x^3 + 18*x^4 + 6*x^5 + 18*x^6 + 6*x^7 + 18*x^8 - 6*x^9 + 29*x^10 + 13*x^11 - 6*x^12) / ((1-x)^3*(1+x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)). - Colin Barker, Oct 14 2016
Empirically for 12 < n <= 500: a(n) = a(n-2) + a(n-10) - a(n-12) + 120. - Lars Blomberg, Jun 08 2020
Empirical: a(2*k + 1) = 6*k*(2*k + 1) + 1, for k >= 0. - Ivan N. Ianakiev, Jun 27 2020
Empirical: 10*a(n) = 30*n^2 -45*n +23 +13*(-1)^n -15*(-1)^n*n - 24*b(n) where b(n) is the 10-periodic sequence 4, 0, -1, 0, -1, 0, -1, 0, -1, 0, 4, 0 .... of offset 0. - R. J. Mathar, Jul 05 2020

A335411 a(n) is the number of vertices formed by n-secting the angles of an equilateral triangle.

Original entry on oeis.org

3, 7, 21, 25, 63, 67, 129, 133, 219, 199, 333, 337, 471, 475, 633, 637, 819, 823, 1029, 1009, 1263, 1267, 1521, 1525, 1803, 1807, 2109, 2113, 2439, 2419, 2793, 2797, 3171, 3175, 3573, 3577, 3999, 4003, 4449, 4429, 4923, 4927, 5421, 5425, 5943, 5947, 6489
Offset: 1

Views

Author

Lars Blomberg, Jun 08 2020

Keywords

Comments

See A277402 for illustrations.

Links

Crossrefs

Cf. A331782, A277402 (regions), A335412 (edges), A335413 (ngons).

Formula

Empirically for 12 < n < 500: a(n) = a(n-2) + a(n-10) - a(n-12) + 120.
Conjectures from Colin Barker, Jun 08 2020: (Start)
G.f.: x*(3 + 4*x + 11*x^2 + 24*x^4 + 24*x^6 + 24*x^8 - 24*x^9 + 45*x^10 + 20*x^11 - 11*x^12) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-10) - a(n-11) - a(n-12) + a(n-13) for n>13.
(End)
Colin Barker's recurrence conjecture holds for 13 < n <= 500. Lars Blomberg, Jun 12 2020
Empirical: a(2*k - 1) = 3*(4*k^2 - 6*k + 3), for k >= 1. - Ivan N. Ianakiev, Jul 15 2020

A335412 a(n) is the number of edges formed by n-secting the angles of an equilateral triangle.

Original entry on oeis.org

3, 12, 39, 54, 123, 144, 255, 282, 435, 432, 663, 702, 939, 984, 1263, 1314, 1635, 1692, 2055, 2082, 2523, 2592, 3039, 3114, 3603, 3684, 4215, 4302, 4875, 4932, 5583, 5682, 6339, 6444, 7143, 7254, 7995, 8112, 8895, 8982, 9843, 9972, 10839, 10974, 11883, 12024
Offset: 1

Views

Author

Lars Blomberg, Jun 08 2020

Keywords

Comments

See A277402 for illustrations.

Links

Crossrefs

Cf. A332376, A277402 (regions), A335411 (vertices), A335413 (ngons).

Formula

Empirically for 12 < n < 500: a(n) = a(n-2) + a(n-10) - a(n-12) + 240.
Conjectures from Colin Barker, Jun 08 2020: (Start)
G.f.: 3*x*(1 + 3*x + 8*x^2 + 2*x^3 + 14*x^4 + 2*x^5 + 14*x^6 + 2*x^7 + 14*x^8 - 10*x^9 + 25*x^10 + 11*x^11 - 6*x^12) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-10) - a(n-11) - a(n-12) + a(n-13) for n>13.
(End)
Colin Barker's recurrence conjecture holds for 13 < n <= 500. Lars Blomberg, Jun 12 2020

A335528 Irregular table read by rows: T(n,k) = number of k-sided polygons formed by n-secting the angles of a square for k >= 3.

Original entry on oeis.org

0, 1, 4, 12, 16, 0, 0, 0, 1, 24, 4, 4, 48, 28, 8, 4, 4, 1, 52, 32, 80, 68, 28, 12, 0, 1, 96, 68, 20, 4, 128, 132, 36, 16, 8, 1, 156, 120, 24, 16, 192, 196, 60, 36, 0, 5, 200, 220, 28, 12, 244, 312, 80, 56, 0, 1, 296, 276, 60, 44, 360, 372, 112, 76, 12, 1
Offset: 1

Views

Author

Lars Blomberg, Jun 12 2020

Keywords

Comments

See A278823 for illustrations.

Examples

			Table begins:
0, 1;
4;
12, 16, 0, 0, 0, 1;
24, 4, 4;
48, 28, 8, 4, 4, 1;
52, 32;
80, 68, 28, 12, 0, 1;
96, 68, 20, 4;
128, 132, 36, 16, 8, 1;
156, 120, 24, 16;
192, 196, 60, 36, 0, 5;
200, 220, 28, 12;
244, 312, 80, 56, 0, 1;
296, 276, 60, 44;
360, 372, 112, 76, 12, 1;

Crossrefs

Cf. A335413, A278823 (regions), A335526 (vertices), A335527 (edges).

A356119 Irregular table read by rows: T(n,k) is the number of k-sided polygons formed, for k>=3, in an equilateral triangle when straight line segments connect the three corner vertices to the points dividing the sides into n equal parts.

Original entry on oeis.org

1, 6, 12, 3, 3, 1, 24, 6, 36, 9, 9, 7, 48, 24, 6, 72, 21, 15, 19, 84, 48, 12, 6, 108, 51, 33, 25, 132, 78, 18, 18, 168, 69, 51, 43, 180, 120, 48, 18, 216, 135, 57, 61, 252, 156, 66, 36, 294, 159, 105, 67, 312, 234, 84, 48, 372, 225, 117, 103, 408, 264, 138, 60, 456, 291, 159, 121, 486, 372, 138, 84
Offset: 1

Views

Author

Scott R. Shannon, Jul 27 2022

Keywords

Comments

The maximum sided k-gon up to n = 250 is the 6-gon; it is likely this is the maximum sided k-gon for all n.
See A092098 for more images of the triangle.

Examples

			The table begins:
1;
6;
12,  3,   3,   1;
24,  6;
36,  9,   9,   7;
48,  24,  6;
72,  21,  15,  19;
84,  48,  12,  6;
108, 51,  33,  25;
132, 78,  18,  18;
168, 69,  51,  43;
180, 120, 48,  18;
216, 135, 57,  61;
252, 156, 66,  36;
294, 159, 105, 67;
312, 234, 84,  48;
372, 225, 117, 103;
408, 264, 138, 60;
456, 291, 159, 121;
486, 372, 138, 84;
.
.
See the attached text file for more examples.

Links

Crossrefs

Cf. A092098 (regions), vertices (A331782), edges (A332376), A335413.

Formula

Sum of row(n) = A092098(n).
Showing 1-5 of 5 results.

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