problem 45

"""
angle, pentagonal, and hexagonal numbers are generated by the following formulae:

Triangle	 	Tn=n(n+1)/2	 	1, 3, 6, 10, 15, ...
Pentagonal	 	Pn=n(3n−1)/2	 	1, 5, 12, 22, 35, ...
Hexagonal	 	Hn=n(2n−1)	 	1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.

h(n) = n(2n - 1)
n = 2
h(2) = 6

6 = n(3n-1) / 2
12 = 3n^2 - n

3n^2 - n - 2 * h(n) = 0

h(i) = i (2i-1)
p(j) = j (3j-1) / 2
t(k) = k(k+1) / 2

[k(k+1) / 2 ] = k * i

t(i) + p(i+1) = h(i+1)

i(i+1) / 2 + (i+1)(3i + 3 - 1) / 2 = (i+1) (2i+2-1)
left side
i^2 + i  + 3i^2 + 2 + 5i
4i^2 + 6i + 2
2i^2 + 3i + 1

right side

def triangular(num):
  num = n(n+1) / 2
  2 num = n^2 + n
  n^2 + n - 2num = 0
  a = 1
  b = 1
  c = -2num
  -b +- sqrt(-4ac) / 2a

for i in range(inf):
  num = h(i)
  a, b = solve_triangle(num)
  if a is integer or b is integer:
    c , d = solve_pentagonal(num)

Find the next triangle number that is also pentagonal and hexagonal.
"""




def all_formulae(n, tri):
  new_i = tri // n 
  hexa = pow(new_i, 2) * 2 - new_i
  #print(new_i, hexa)
  if tri / n == tri // n:
   if hexa == tri:
     return True
  return False   



pen = set()
for n in  range(286, 100000000):
 tri = (pow(n, 2) + n) // 2
 pen.add((pow(n, 2) * 3 - n) // 2)
 #print(tri, n)
 if all_formulae(n, tri) and tri in  pen:
   print(tri)
   break