A356663 - OEIS

%I #15 Oct 02 2022 00:43:30

%S 0,1,3,1,3,5,1,3,4,5,2,7,2,5,6,4,2,6,2,4,5,4,2,5,4,4,6,5,2,7,2,5,6,4,

%T 6,7,4,6,9,7,5,9,6,9,9,8,6,10,6,7,8,6,6,8,6,5,7,6,4,10,3,7,7,7,7,10,6,

%U 6,10,9,7,9,6,9,11,10,7,10

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

%C a(n) is the number of positive integer triples (a, b, c) (not including permutations) and with a, b, c distinct that satisfy n+360 = (a-2)*180/a + (b-2)*180/b + (c-2)*180/c.

%C For n >= 175, a(n) = 0. This can be proved. The maximum sum of 3 integral internal angle is of a 360-gon, a 180-gon and a 120-gon with internal angles 179, 178 and 177 degrees respectively. Therefore 179+178+177-360 = 174 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(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. A356444 (where polygons do not have to be distinct).

%K nonn

%O 1,3

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