new_contents: """level 10
  
  Find len(sequence[30])
  Given the sequence = [1, 11, 21, 1211, 111221, ...]
  This is the 'Look and Say' sequence:
  https://en.wikipedia.org/wiki/Look-and-say_sequence
  """
  
  
  def level_10(num):
      """Return the next number in the 'see-and-say' sequence, given a number.
      
      The next number in the sequence is found by saying the count of each digit,
      in order, in the current number.
      """
      # ensure the input is a number as a string
      num_as_str = str(num)
      
      count = 0
      current_digit = num_as_str[0]
      next_num_as_str = ''
  
      for digit in num_as_str:
          if digit == current_digit:
              count += 1
          else:
              next_num_as_str += '{}{}'.format(count, current_digit)
              count = 1
              current_digit = digit
      else:
          next_num_as_str += '{}{}'.format(count, current_digit)
      
      return next_num_as_str
  
  
  if __name__ == '__main__':
      sequence = ['1']
      for x in range(30):
          sequence.append(level_10(sequence[x]))
      print('\n [+] len(sequence[30]): {}'.format(len(sequence[30])))
  
subject: add solution to level 10