A077588 Maximum number of regions into which the plane is divided by n triangles.
1, 2, 8, 20, 38, 62, 92, 128, 170, 218, 272, 332, 398, 470, 548, 632, 722, 818, 920, 1028, 1142, 1262, 1388, 1520, 1658, 1802, 1952, 2108, 2270, 2438, 2612, 2792, 2978, 3170, 3368, 3572, 3782, 3998, 4220, 4448, 4682, 4922, 5168, 5420, 5678, 5942, 6212, 6488
Offset: 0
Examples
a(2) = 8 because a Star of David divides the plane into 8 regions: 6 triangles at the vertices, the interior hexagon, and the exterior.
Links
- Keyang Li, figures for n=1,2,3,4,5
- Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
Crossrefs
a(n) = A096777(3*n-1) for n > 0. - Reinhard Zumkeller, Dec 29 2007
a(n) = A242658(n) for n > 0. - Eric W. Weisstein, Nov 29 2017
Programs
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Mathematica
CoefficientList[Series[(-z^3 - 5*z^2 + z - 1)/(z - 1)^3, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)
Formula
a(n) = 3n^2 - 3n + 2 for n > 0.
Proof (from Joshua Zucker and N. J. A. Sloane, Dec 01 2017)
Represent the configuration of n triangles by a planar graph with a node for each vertex of the triangles and for each intersection point. Let there be v_n nodes and e_n edges. By classical graph theory, a(n) = e_n - v_n + 2. When we go from n to n+1 triangles, each side of the new triangle can meet each side of the existing triangles at most twice, so Dv_n := v_{n+1}-v_n <= 6n.
Each of these intersection points increases the number of edges in the graph by 2, so De_n := e_{n+1}-e_n = 3 + 2*Dv_n, Da_n := a(n+1)-a(n) = 3 + Dv_n <= 3+6*n.
These upper bounds can be achieved by taking 3n points equally spaced around a circle and drawing n concentric overlapping equilateral triangles in the obvious way, and we achieve a(n) = 3n^2 - 3n + 2 (and v_n = 3n^2, e_n = 3n(2n-1)) for n>0. QED
a(n) is the nearest integer to (Sum_{k>=n} 1/k^2)/(Sum_{k>=n} 1/k^4). - Benoit Cloitre, Jun 12 2003
a(n) = a(n-1) + 6*n - 6 (with a(1) = 2). - Vincenzo Librandi, Dec 07 2010
For n > 0, a(n) = A002061(n-1) + A056220(n); and for n > 1, a(n) = A002061(n+1) + A056220(n-1). - Bruce J. Nicholson, Sep 22 2017