

A077588


Maximum number of regions into which the plane is divided by n triangles.


7



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


LINKS

Table of n, a(n) for n=0..47.
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.


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(2n1)) 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(n1) + 6*n  6 (with a(1) = 2).  Vincenzo Librandi, Dec 07 2010
For n > 0, a(n) = A002061(n1) + A056220(n); and for n > 1, a(n) = A002061(n+1) + A056220(n1).  Bruce J. Nicholson, Sep 22 2017


EXAMPLE

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.


MATHEMATICA

CoefficientList[Series[(z^3  5*z^2 + z  1)/(z  1)^3, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)


CROSSREFS

Cf. A005448, A077591.
a(n) = A096777(3*n1) for n > 0.  Reinhard Zumkeller, Dec 29 2007
For n > 0, a(n) = 2 * A005448(n).  Jon Perry, Apr 14 2013
a(n) = A242658(n) for n > 0.  Eric W. Weisstein, Nov 29 2017
Sequence in context: A327100 A130238 A038460 * A278212 A025219 A032767
Adjacent sequences: A077585 A077586 A077587 * A077589 A077590 A077591


KEYWORD

easy,nonn


AUTHOR

Joshua Zucker and the Castilleja School MathCounts club, Nov 07 2002


STATUS

approved



