|
| |
|
|
A255011
|
|
Number of polygons formed by connecting all the 4n points on the perimeter of an n X n square by straight lines; a(0) = 0 by convention.
|
|
33
|
|
|
|
0, 4, 56, 340, 1120, 3264, 6264, 13968, 22904, 38748, 58256, 95656, 120960, 192636, 246824, 323560, 425408, 587964, 682296, 932996, 1061232, 1327524, 1634488, 2049704, 2227672, 2806036, 3275800, 3810088, 4307520, 5298768
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
There are n+1 points on each side of the square, but that counts the four corners twice, so there are a total of 4n points on the perimeter. - N. J. A. Sloane, Jan 23 2020
a(n) is always divisible by 4, by symmetry. If n is odd, a(n) is divisible by 8.
For n > 0, the vertices of the bounding square generate diametrical bisectors that cross at the center. Thus each diagram has fourfold symmetry.
For n > 0, an orthogonal n X n grid is produced by corresponding horizontal and vertical points on opposite sides.
Terms {1, 3, 9} are not congruent to 0 (mod 8).
Number of edges: {0, 8, 92, 596, 1936, 6020, 11088, 26260, 42144, 72296, 107832, ...}. See A331448. (End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
EXAMPLE
|
For n = 3, the perimeter of the square contains 12 points:
* * * *
* *
* *
* * * *
Connect each point to every other point with a straight line inside the square. Then count the polygons (or regions) that have formed. There are 340 polygons, so a(3) = 340.
For n = 1, the full picture is:
*-*
|X|
*-*
The lines form four triangular regions, so a(1) = 4.
For n = 0, the square can be regarded as consisting of a single point, producing no lines or polygons, and so a(0) = 0.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,more
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
