Python/project_euler/problem_45/sol1.py
Kushagra Bansal 402ba7f49a
Created problem_45 in project_euler and Speed Boost for problem_34/sol1.py (#2349)
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Co-authored-by: Christian Clauss <cclauss@me.com>

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Co-authored-by: Christian Clauss <cclauss@me.com>

Co-authored-by: Christian Clauss <cclauss@me.com>
2020-08-25 13:48:19 +02:00

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"""
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle T(n) = (n * (n + 1)) / 2 1, 3, 6, 10, 15, ...
Pentagonal P(n) = (n * (3 * n 1)) / 2 1, 5, 12, 22, 35, ...
Hexagonal H(n) = n * (2 * n 1) 1, 6, 15, 28, 45, ...
It can be verified that T(285) = P(165) = H(143) = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
All trinagle numbers are hexagonal numbers.
T(2n-1) = n * (2 * n - 1) = H(n)
So we shall check only for hexagonal numbers which are also pentagonal.
"""
def hexagonal_num(n: int) -> int:
"""
Returns nth hexagonal number
>>> hexagonal_num(143)
40755
>>> hexagonal_num(21)
861
>>> hexagonal_num(10)
190
"""
return n * (2 * n - 1)
def is_pentagonal(n: int) -> bool:
"""
Returns True if n is pentagonal, False otherwise.
>>> is_pentagonal(330)
True
>>> is_pentagonal(7683)
False
>>> is_pentagonal(2380)
True
"""
root = (1 + 24 * n) ** 0.5
return ((1 + root) / 6) % 1 == 0
def compute_num(start: int = 144) -> int:
"""
Returns the next number which is traingular, pentagonal and hexagonal.
>>> compute_num(144)
1533776805
"""
n = start
num = hexagonal_num(n)
while not is_pentagonal(num):
n += 1
num = hexagonal_num(n)
return num
if __name__ == "__main__":
print(f"{compute_num(144)} = ")