A367117 Place n points in general position on each side of an equilateral triangle, and join every pair of the 3*n+3 boundary points by a chord; sequence gives number of vertices in the resulting planar graph.

3, 12, 72, 282, 795, 1818, 3612, 6492, 10827, 17040, 25608, 37062, 51987, 71022, 94860, 124248, 159987, 202932, 253992, 314130, 384363, 465762, 559452, 666612, 788475, 926328, 1081512, 1255422, 1449507, 1665270, 1904268, 2168112, 2458467, 2777052, 3125640, 3506058, 3920187, 4369962

Offset: 0

Author

Scott R. Shannon and N. J. A. Sloane, Nov 05 2023

Keywords

• nonn

Comments

"In general position" implies that the internal lines (or chords) only have simple intersections. There is no interior point where three or more chords meet.

Note that although the number of k-gons in the graph will vary as the edge points change position, the total number of regions will stay constant as long as all internal vertices remain simple.

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000
• Scott R. Shannon, Image for n = 1.
• Scott R. Shannon, Image for n = 2.
• Scott R. Shannon, Image for n = 5.

Crossrefs

Cf. A367118 (regions), A367119 (edges).

See also A091908, A092098, A331782, A365929.

If the boundary points are equally spaced, we get A274585, A092866, A274586, A092867. - N. J. A. Sloane, Nov 09 2023

Programs

Mathematica

A367117[n_]:=3/4(n+1)(3n^3+n^2+4);Array[A367117,50,0] (* Paolo Xausa, Nov 09 2023 *)

Formula

Theorem: a(n) = (3/4)*(n+1)*(3*n^3+n^2+4).

a(n) = A367119(n) - A367118(n) + 1 by Euler's formula.