new_contents: #!/usr/bin/env python
  #-*-coding:utf-8-*-
  
  '''
  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.
  '''
  
  import timeit
  
  
  def b(n):
      return 2*(n/3+1) if n%3 == 0 else 1
  
  def calc(n):
      p = [3, 2]
      q = [1, 1]
      for i in range(n-2):
          p = [b(i)*p[0]+p[1], p[0]]
          q = [b(i)*q[0]+q[1], q[0]]
      return sum(map(int, str(p[0])))
  
  if __name__ == '__main__':
      print calc(100)
      print timeit.Timer('problem_065.calc(100)', 'import problem_065').timeit(1)
  
subject: add problem 065