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A366483
Place n equally spaced points on each side of an equilateral triangle, and join each of these points by a chord to the 2*n new points on the other two sides: sequence gives number of vertices in the resulting planar graph.
5
3, 6, 22, 108, 300, 919, 1626, 3558, 5824, 9843, 14352, 23845, 30951, 47196, 62773, 82488, 104544, 144784, 173694, 230008, 276388, 336927, 403452, 509218, 582417, 702228, 824956, 969387, 1098312, 1321978, 1463580, 1724190, 1952509, 2221497, 2505169, 2846908, 3103788, 3556143, 3978763, 4444003
OFFSET
0,1
COMMENTS
We start with the three corner points of the triangle, and add n further points along each edge. Including the corner points, we end up with n+2 points along each edge, and the edge is divided into n+1 line segments.
Each of the n points added to an edge is joined by 2*n chords to the points that were added to the other two edges. There are 3*n^2 chords.
LINKS
Scott R. Shannon, Image for n = 1.
Scott R. Shannon, Image for n = 2.
Scott R. Shannon, Image for n = 3.
Scott R. Shannon, Image for n = 4.
Scott R. Shannon, Image for n = 5.
Scott R. Shannon, Image for n = 10.
FORMULA
a(n) = A366485(n) - A366486(n) + 1 (Euler).
CROSSREFS
Cf. A366484 (interior vertices), A366485 (edges), A366486 (regions).
If the 3*n points are placed "in general position" instead of uniformly, we get sequences A366478, A365929, A366932, A367015.
If the 3*n points are placed uniformly and we also draw chords from the three corner points of the triangle to these 3*n points, we get A274585, A092866, A274586, A092867.
Sequence in context: A243336 A344605 A029848 * A117850 A090910 A256674
KEYWORD
nonn
AUTHOR
STATUS
approved