Cleanup of initial checks

friendly.py

@@ -17,7 +17,7 @@
 import sys, os, re
 
 # SETTINGS
-index_search = 10                       # Get the index of '10'
+friendly_number = 10                    # The integer we want to find friendly numbers for
 where_to_start = 10**12                 # 1 trillion
 where_to_end = 10**13                   # 10 trillion (e.g. search from 1 trillion to 10 trillion)
 checkpoint = 10**4                      # Update the checkpoint after how many checks?
@@ -27,32 +27,32 @@
 if os.path.isfile(progress_file_name):
     # If so, open it...
     progress_file = open(progress_file_name, "r")
-    if progress_file.mode == 'r':
-        # and get the current checkpoint
-        start_num = progress_file.read()
-        
-        # If it contains the text 'FOUND' just print it out and exit
-        match = re.search(r"FOUND:\s(\d{1,})", start_num)
-        if match:
-            print(start_num)
-            print(divisors(int(match.group(1))))
-            sys.exit()
-        # Otherwise we're loading in that checkpoint
-        start_num = int(start_num)
+
+    # and get the current checkpoint
+    last_checked = progress_file.read()
+
+    # If it contains the text 'FOUND' just print it out and exit
+    match = re.search(r"FOUND:\s(\d{1,})", last_checked)
+    if match:
+        print(last_checked)
+        print(divisors(int(match.group(1))))
+        sys.exit()
+    # Otherwise it's just a number and we're starting from that point
+    start_num = int(last_checked)
 else:
     # No progress file found, so just start at the beginning
     start_num = where_to_start
 
 # Find the index of the provided integer
 # (The [0] is because find_index returns 3 values, we only want the first)
-search = find_index(index_search)[0]
+search = find_index(friendly_number)[0]
 
 # Just start looping through each number, one by one
 for num in range(start_num, where_to_end):
     # Destructure find_index into its appropriate variables
     index, sum_of_factors, factors = find_index(num)
 
-    # If it matches index_search, we found one so break out of the loop
+    # If it matches friendly_number, we found one so break out of the loop
     if index == search:
         print(f'{sum_of_factors} / {num} = {index}')
         print(factors)