A356444 Number of ways to create an angle excess of n degrees using 3 regular polygons with integral internal angles.

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.

Links

Table of n, a(n) for n=1..75.

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

Adjacent sequences: A356441 A356442 A356443 * A356445 A356446 A356447

Keyword

nonn

Author

Joseph C. Y. Wong, Aug 21 2022

Status

approved