old_contents: """
  Problem 72 Counting fractions: https://projecteuler.net/problem=72
  
  Description:
  
  Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1,
  it is called a reduced proper fraction.
  If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we
  get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7,
  3/4, 4/5, 5/6, 6/7, 7/8
  It can be seen that there are 21 elements in this set.
  How many elements would be contained in the set of reduced proper fractions for
  d ≤ 1,000,000?
  
  Solution:
  
  Number of numbers between 1 and n that are coprime to n is given by the Euler's Totient
  function, phi(n). So, the answer is simply the sum of phi(n) for 2 <= n <= 1,000,000
  Sum of phi(d), for all d|n = n. This result can be used to find phi(n) using a sieve.
  
  Time: 3.5 sec
  """
  
  
  def solution(limit: int = 1_000_000) -> int:
      """
      Returns an integer, the solution to the problem
      >>> solution(10)
      31
      >>> solution(100)
      3043
      >>> solution(1_000)
      304191
      """
  
      phi = [i - 1 for i in range(limit + 1)]
  
      for i in range(2, limit + 1):
          for j in range(2 * i, limit + 1, i):
              phi[j] -= phi[i]
  
      return sum(phi[2 : limit + 1])
  
  
  if __name__ == "__main__":
      print(solution())
  
new_contents: """
  Problem 72 Counting fractions: https://projecteuler.net/problem=72
  
  Description:
  
  Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1,
  it is called a reduced proper fraction.
  If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we
  get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7,
  3/4, 4/5, 5/6, 6/7, 7/8
  It can be seen that there are 21 elements in this set.
  How many elements would be contained in the set of reduced proper fractions for
  d ≤ 1,000,000?
  
  Solution:
  
  Number of numbers between 1 and n that are coprime to n is given by the Euler's Totient
  function, phi(n). So, the answer is simply the sum of phi(n) for 2 <= n <= 1,000,000
  Sum of phi(d), for all d|n = n. This result can be used to find phi(n) using a sieve.
  
  Time: 1 sec
  """
  
  
  def solution(limit: int = 1_000_000) -> int:
      """
      Returns an integer, the solution to the problem
      >>> solution(10)
      31
      >>> solution(100)
      3043
      >>> solution(1_000)
      304191
      """
  
      phi = [i - 1 for i in range(limit + 1)]
  
      for i in range(2, limit + 1):
          if phi[i] == i - 1:
              for j in range(2 * i, limit + 1, i):
                  phi[j] -= phi[j] // i
  
      return sum(phi[2 : limit + 1])
  
  
  if __name__ == "__main__":
      print(solution())
  
subject: Improve solution (#5705)