login
A356444
Number of ways to create an angle excess of n degrees using 3 regular polygons with integral internal angles.
1
0, 1, 3, 1, 3, 6, 1, 3, 4, 6, 2, 9, 2, 5, 7, 5, 2, 9, 2, 6, 6, 4, 2, 8, 4, 5, 7, 7, 2, 12, 3, 6, 7, 5, 7, 10, 4, 6, 9, 10, 5, 12, 6, 10, 11, 8, 6, 14, 6, 11, 9, 8, 6, 12, 8, 7, 8, 8, 5, 15, 3, 7, 8, 8, 7, 12, 6, 8, 10, 12, 7, 14, 6, 10, 13
OFFSET
1,3
COMMENTS
a(n) is the number of positive integer triples (a, b, c) (not including permutations) that satisfy n+360 = (a-2)*180/a + (b-2)*180/b + (c-2)*180/c.
For n >= 178, a(n) = 0. This can be proved. The maximum integral internal angle is of a 360-gon with internal angle 179 degrees. Therefore 179*3-360 = 177 degrees is the maximum possible angle excess.
EXAMPLE
For n = 1, there are no possible ways to create an angle excess of 1 degree therefore a(1) = 0.
For n = 3, there are 3 possible ways to create an angle excess of 3 degrees. (3-gon, 8-gon, 30-gon), (4-gon, 5-gon, 24-gon), (5-gon, 6-gon, 8-gon).
PROG
(Python)
import itertools
def subs(l):
res = []
for combo in itertools.combinations_with_replacement(l, 3):
res.append(list(combo))
return res
l = [3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360] # Number of sides of polygons with an integral internal angle
for n in range(1, 200):
k = 0
for i in subs(l):
if n + 360 == (i[0] - 2)*180/i[0] + (i[1] - 2)*180/i[1] + (i[2] - 2)*180/i[2]:
k += 1
print(k)
CROSSREFS
Cf. A356663 (where distinct polygons are allowed).
Sequence in context: A279249 A281893 A049324 * A256095 A181843 A131111
KEYWORD
nonn
AUTHOR
Joseph C. Y. Wong, Aug 21 2022
STATUS
approved