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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. 7
3, 24, 153, 588, 1635, 3708, 7329, 13128, 21843, 34320, 51513, 74484, 104403, 142548, 190305, 249168, 320739, 406728, 508953, 629340, 769923, 932844, 1120353, 1334808, 1578675, 1854528, 2165049, 2513028, 2901363, 3333060, 3811233, 4339104, 4920003, 5557368, 6254745, 7015788 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
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.
See A367117 and A367118 for images of the triangle.
LINKS
FORMULA
Conjecture: a(n) = (3/2)*(3*n^4 + 4*n^3 + 3*n^2 + 4*n + 2).
a(n) = A367117 (n) + A367118 (n) - 1 by Euler's formula.
CROSSREFS
Cf. A367117 (vertices), A367118 (regions), A091908, A092098, A331782, A366932.
If the boundary points are equally spaced, we get A274585, A092866, A274586, A092867. - N. J. A. Sloane, Nov 09 2023
Sequence in context: A043017 A003443 A119581 * A240916 A006292 A067370
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified May 16 08:15 EDT 2024. Contains 372549 sequences. (Running on oeis4.)