cp's OEIS Frontend

A367119 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 edges in the resulting planar graph.

%I A367119 #14 Nov 09 2023 03:28:58
%S A367119 3,24,153,588,1635,3708,7329,13128,21843,34320,51513,74484,104403,
%T A367119 142548,190305,249168,320739,406728,508953,629340,769923,932844,
%U A367119 1120353,1334808,1578675,1854528,2165049,2513028,2901363,3333060,3811233,4339104,4920003,5557368,6254745,7015788
%N A367119 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 edges in the resulting planar graph.
%C A367119 "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 A367119 See A367117 and A367118 for images of the triangle.
%F A367119 Conjecture: a(n) = (3/2)*(3*n^4 + 4*n^3 + 3*n^2 + 4*n + 2).
%F A367119 a(n) = A367117 (n) + A367118 (n) - 1 by Euler's formula.
%Y A367119 Cf. A367117 (vertices), A367118 (regions), A091908, A092098, A331782, A366932.
%Y A367119 If the boundary points are equally spaced, we get A274585, A092866, A274586, A092867. - _N. J. A. Sloane_, Nov 09 2023
%K A367119 nonn
%O A367119 0,1
%A A367119 _Scott R. Shannon_ and _N. J. A. Sloane_, Nov 05 2023