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Number of ways to create an angle excess of n degrees using 3 regular polygons with integral internal angles.
1

%I #24 Oct 02 2022 00:42:29

%S 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,

%T 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,

%U 7,12,6,8,10,12,7,14,6,10,13

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

%C 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.

%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.

%e For n = 1, there are no possible ways to create an angle excess of 1 degree therefore a(1) = 0.

%e 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).

%o (Python)

%o import itertools

%o def subs(l):

%o res = []

%o for combo in itertools.combinations_with_replacement(l, 3):

%o res.append(list(combo))

%o return res

%o 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

%o for n in range(1, 200):

%o k = 0

%o for i in subs(l):

%o if n + 360 == (i[0] - 2)*180/i[0] + (i[1] - 2)*180/i[1] + (i[2] - 2)*180/i[2]:

%o k += 1

%o print(k)

%Y Cf. A356663 (where distinct polygons are allowed).

%K nonn

%O 1,3

%A _Joseph C. Y. Wong_, Aug 21 2022