def problem065():
    """


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

       √2 = 1 +          1
                2 +        1
                    2 +      1
                        2 +    1
                            2 + ...

       The infinite continued fraction can be written, √2 = [1;(2)], (2)
       indicates that 2 repeats ad infinitum. In a similar way, √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 √2.

       1 + 1 = 3/2
           2

       1 +   1   = 7/5
           2 + 1
               2

       1 +     1     = 17/12
           2 +   1
               2 + 1
                   2

       1 +       1       = 41/29
           2 +     1
               2 +   1
                   2 + 1
                       2

       Hence the sequence of the first ten convergents for √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 10^th convergent is 1+4+5+7=17.

       Find the sum of digits in the numerator of the 100^th convergent of the
       continued fraction for e.
    """
    numer = 1
    denom = 0
    for i in reversed(range(100)):
        numer, denom = e_contfrac_term(i) * numer + denom, numer
    ans = sum(int(c) for c in str(numer))
    return ans


def e_contfrac_term(i):
    if i == 0:
        return 2
    elif i % 3 == 2:
        return i // 3 * 2 + 2
    else:
        return 1


if __name__ == "__main__":
    print(problem065())
