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.

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

Author

Scott R. Shannon and N. J. A. Sloane, Nov 09 2023

Keywords

• nonn

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.

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.

Formula

a(n) = A366485(n) - A366486(n) + 1 (Euler).