Project Euler Problem 45

Problem 45

Triangle, 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.

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

三角形数、五边形数和六角形数分别由以下公式给出：

三角形数	Tn=n(n+1)/2	1, 3, 6, 10, 15, …
五边形数	Pn=n(3n−1)/2	1, 5, 12, 22, 35, …
六边形数	Hn=n(2n−1)	1, 6, 15, 28, 45, …

可以验证，T₂₈₅ = P₁₆₅ = H₁₄₃ = 40755。

找出下一个同时是三角形数、五边形数和六角形数的数。

pentagon = {n*(3*n-1)/2 for n in range(1,100000)}
hexagon = {n*(2*n-1) for n in range(1,100000)}
for x in pentagon:
    if x in hexagon:
        print(x)