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"""

Calculate the nth Proth number

Source:

https://handwiki.org/wiki/Proth_number

"""

import math

def proth(number: int) -> int:

"""

:param number: nth number to calculate in the sequence

:return: the nth number in Proth number

Note: indexing starts at 1 i.e. proth(1) gives the first Proth number of 3

>>> proth(6)

>>> proth(0)

Traceback (most recent call last):

...

ValueError: Input value of [number=0] must be > 0

>>> proth(-1)

Traceback (most recent call last):

...

ValueError: Input value of [number=-1] must be > 0

>>> proth(6.0)

Traceback (most recent call last):

...

TypeError: Input value of [number=6.0] must be an integer

"""

if not isinstance(number, int):

raise TypeError(f"Input value of [number={number}] must be an integer")

if number < 1:

raise ValueError(f"Input value of [number={number}] must be > 0")

elif number == 1:

return 3

elif number == 2:

return 5

else:

block_index = number // 3

"""

+1 for binary starting at 0 i.e. 2^0, 2^1, etc.

+1 to start the sequence at the 3rd Proth number

Hence, we have a +2 in the below statement

"""

block_index = math.log(block_index, 2) + 2

block_index = int(block_index)

proth_list = [3, 5]

proth_index = 2

increment = 3

for block in range(1, block_index):

for move in range(increment):

proth_list.append(2 ** (block + 1) + proth_list[proth_index - 1])

proth_index += 1

increment *= 2

return proth_list[number - 1]

if __name__ == "__main__":

for number in range(11):

value = 0

try:

value = proth(number)

except ValueError:

print(f"ValueError: there is no {number}th Proth number")

continue

print(f"The {number}th Proth number: {value}")

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