A335412 -id:A335412 - cp's OEIS frontend

A335413 Irregular table read by rows: T(n,k) = number of k-sided polygons formed by n-secting the angles of an equilateral triangle for k >= 3.

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, 126, 90, 12, 6, 168, 69, 51, 43, 180, 120, 48, 18, 216, 123, 81, 49, 240, 180, 54, 36, 288, 171, 99, 73, 312, 234, 84, 48, 372, 225, 117, 103, 396, 288, 126, 60

Offset: 1

Lars Blomberg, Jun 08 2020

See A277402 for illustrations.

			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;
126, 90, 12, 6;
168, 69, 51, 43;
180, 120, 48, 18;
216, 123, 81, 49;
240, 180, 54, 36;
288, 171, 99, 73;
312, 234, 84, 48;

Lars Blomberg, Table of n, a(n) for n = 1..1991

Cf. A277402 (regions), A335411 (vertices), A335412 (edges).

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

Ethan Beihl, Oct 13 2016

nonn

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

			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.

Lars Blomberg, Table of n, a(n) for n = 1..500
Lars Blomberg, Coloured illustration for n=3
Lars Blomberg, Coloured illustration for n=4
Lars Blomberg, Coloured illustration for n=19
Lars Blomberg, Coloured illustration for n=20
J. H. Conway and A. J. Jones, Trigonometric diophantine equations (On vanishing sums of roots of unity), Acta Arithmetica 30(3), 229-240 (1976).
Wikipedia, Rhumbline network

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

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&]

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

Lars Blomberg, Jun 08 2020

nonn

See A277402 for illustrations.

Lars Blomberg, Table of n, a(n) for n = 1..500

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

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

cp's OEIS Frontend

A335413 Irregular table read by rows: T(n,k) = number of k-sided polygons formed by n-secting the angles of an equilateral triangle for k >= 3.

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

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

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A335527 a(n) is the number of edges formed by n-secting the angles of a square.

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