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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.

%I A367117 #36 Jul 09 2025 05:03:12
%S A367117 3,12,72,282,795,1818,3612,6492,10827,17040,25608,37062,51987,71022,
%T A367117 94860,124248,159987,202932,253992,314130,384363,465762,559452,666612,
%U A367117 788475,926328,1081512,1255422,1449507,1665270,1904268,2168112,2458467,2777052,3125640,3506058,3920187,4369962
%N 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.
%C A367117 "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.
%C A367117  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.
%H A367117 Paolo Xausa, <a href="/A367117/b367117.txt">Table of n, a(n) for n = 0..10000</a>
%H A367117 Scott R. Shannon, <a href="/A367117/a367117.png">Image for n = 1</a>.
%H A367117 Scott R. Shannon, <a href="/A367117/a367117_1.png">Image for n = 2</a>.
%H A367117 Scott R. Shannon, <a href="/A367117/a367117_2.png">Image for n = 5</a>.
%F A367117 Theorem: a(n) = (3/4)*(n+1)*(3*n^3+n^2+4).
%F A367117 a(n) = A367119(n) - A367118(n) + 1 by Euler's formula.
%t A367117 A367117[n_]:=3/4(n+1)(3n^3+n^2+4);Array[A367117,50,0] (* _Paolo Xausa_, Nov 09 2023 *)
%Y A367117 Cf. A367118 (regions), A367119 (edges).
%Y A367117 See also A091908, A092098, A331782, A365929.
%Y A367117 If the boundary points are equally spaced, we get A274585, A092866, A274586, A092867. - _N. J. A. Sloane_, Nov 09 2023
%K A367117 nonn
%O A367117 0,1
%A A367117 _Scott R. Shannon_ and _N. J. A. Sloane_, Nov 05 2023