065.py

#!/usr/bin/python
# -*- coding: utf-8 -*-

#The square root of 2 can be written as an infinite continued fraction.
#√2 = 1 + 	
  	#2 + 	
  	  	#2 + 	
  	  	  	#2 + 	
  	  	  	  	#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.

#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.

#Answer:
	#272

from time import time; t=time()

def e():
    for n in range(33, 0, -1):
        yield 1
        yield 2*n
        yield 1
    yield 2

n, m = 1, 0
for i in e():
    n, m = m+i*n, n

print(sum(int(c) for c in str(n)))#, n*1.0/m, time()-t