Had it run 10 times too many integers woops, but still easy to solve

57/main.py

@@ -0,0 +1,59 @@
+'''
+It is possible to show that the square root of two can be expressed as an infinite continued fraction.
+
+By expanding this for the first four iterations, we get:
+
+The next three expansions are 
+ 
+, 
+ 
+, and 
+ 
+, but the eighth expansion, 
+ 
+, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
+
+In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?
+'''
+
+import math
+import time
+
+def t(n, dDict):
+    assert n >= 0
+
+    if n == 0:
+        return 1, 1
+    
+    num, denom = d(n-1, dDict)
+    return (num + denom, denom)
+
+def d(n, dDict):
+    assert n >= 0
+
+    if n == 0:
+        return 1, 2
+    
+    if n in dDict.keys():
+        return dDict[n]
+    
+    num, denom = d(n-1, dDict)
+    r = (denom, 2 * denom + num)
+    dDict[n] = r
+    return r
+
+def main():
+    print("Hello, this is Patrick")
+    t0 = time.time()
+
+    counter = 0
+    dDict = {}
+    for n in range(1001):
+        num, denom = t(n, dDict)
+        if len(str(num)) > len(str(denom)):
+            counter += 1
+
+    print(counter, time.time() - t0)
+
+if __name__ == "__main__":
+    main()