# Convergents of e
The square root of 2 can be written as an infinite continued fraction.
![img](http://img0.ph.126.net/x1Hyc4iHQg0Jz2EInmT3ag==/6597639313681841979.png)  
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.
![img](http://img1.ph.126.net/xme9gNBQdA7bvQkwIftznQ==/6597819633588793645.png)  
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 mth convergent of the continued fraction for e.
- Powered by [Project Euler](https://projecteuler.net/problem=65)
def convergents_of_e(n):
    a0, a1, i = 1, 2, 2
    
    while i <= n: 
        f =  1 if i % 3 else 2 * i // 3
    
        a = a0 + a1 * f
        a0, a1 = a1, a
        i += 1
    
    return sum(map(int, str(a1)))
def convergents_of_e(n):
    a0, a1, i = 1, 2, 2
    
    while i <= n: 
        f =  1 if i % 3 else 2 * i // 3
    
        a = a0 + a1 * f
        a0, a1 = a1, a
        i += 1
    
    return sum(map(int, str(a1)))