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"""

Project Euler Problem 65: https://projecteuler.net/problem=65

The square root of 2 can be written as an infinite continued fraction.

sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / (2 + ...))))

The infinite continued fraction can be written, sqrt(2) = [1;(2)], (2)

indicates that 2 repeats ad infinitum. In a similar way, sqrt(23) =

[4;(1,3,1,8)].

It turns out that the sequence of partial values of continued

fractions for square roots provide the best rational approximations.

Let us consider the convergents for sqrt(2).

1 + 1 / 2 = 3/2

1 + 1 / (2 + 1 / 2) = 7/5

1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17/12

1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/29

Hence the sequence of the first ten convergents for sqrt(2) are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

What is most surprising is that the important mathematical constant,

e = [2;1,2,1,1,4,1,1,6,1,...,1,2k,1,...].

The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10th convergent is

1 + 4 + 5 + 7 = 17.

Find the sum of the digits in the numerator of the 100th convergent

of the continued fraction for e.

-----

The solution mostly comes down to finding an equation that will generate

the numerator of the continued fraction. For the i-th numerator, the

pattern is:

n_i = m_i * n_(i-1) + n_(i-2)

for m_i = the i-th index of the continued fraction representation of e,

n_0 = 1, and n_1 = 2 as the first 2 numbers of the representation.

For example:

n_9 = 6 * 193 + 106 = 1264

1 + 2 + 6 + 4 = 13

n_10 = 1 * 193 + 1264 = 1457

1 + 4 + 5 + 7 = 17

"""

def sum_digits(num: int) -> int:

"""

Returns the sum of every digit in num.

>>> sum_digits(1)

>>> sum_digits(12345)

>>> sum_digits(999001)

"""

digit_sum = 0

while num > 0:

digit_sum += num % 10

num //= 10

return digit_sum

def solution(max: int = 100) -> int:

"""

Returns the sum of the digits in the numerator of the max-th convergent of

the continued fraction for e.

>>> solution(9)

>>> solution(10)

>>> solution(50)

"""

pre_numerator = 1

cur_numerator = 2

for i in range(2, max + 1):

temp = pre_numerator

e_cont = 2 * i // 3 if i % 3 == 0 else 1

pre_numerator = cur_numerator

cur_numerator = e_cont * pre_numerator + temp

return sum_digits(cur_numerator)

if __name__ == "__main__":

print(f"{solution() = }")

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