problem65.py

"""
Convergents of e
Project Euler Problem #65
by Muaz Siddiqui

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

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.

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 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued 
fraction for e.
"""
from euler_helpers import timeit

@timeit
def answer():
    """e can be represented by a generalized continued fraction first we generate 
    the continued fraction expansion and then calculate numerators"""
    e = [int(n*(2/3)) if not n%3 else 1 for n in range(1,101)]
    e[0] = 2
    numerators = [2,3]
    for n in range(2, 100):
        numerators.append(e[n]*numerators[n-1] + numerators[n-2])
    return sum(map(int,str(numerators[99])))