From 20d1667a93621ae604891e0e64565735ab261b1d Mon Sep 17 00:00:00 2001
From: Lauchmelder23 <Robert-Altner@t-online.de>
Date: Wed, 10 Oct 2018 17:35:18 +0200
Subject: [PATCH] Added Problem 14

---
 Problem_14.py | 40 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 40 insertions(+)
 create mode 100644 Problem_14.py

diff --git a/Problem_14.py b/Problem_14.py
new file mode 100644
index 0000000..0a70a5b
--- /dev/null
+++ b/Problem_14.py
@@ -0,0 +1,40 @@
+######################################################################
+# The following iterative sequence is defined for the set of positive integers:
+#
+# n → n/2 (n is even)
+# n → 3n + 1 (n is odd)
+#
+# Using the rule above and starting with 13, we generate the following sequence:
+#
+# 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
+# It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms.
+# Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
+#
+# Which starting number, under one million, produces the longest chain?
+#
+# NOTE: Once the chain starts the terms are allowed to go above one million.
+######################################################################
+
+longestChain = []
+currentChain = []
+
+for i in range(2, 1000000):
+    if i not in longestChain:
+        print(i)
+        currentChain.clear()
+        n = i
+        currentChain.append(n)
+
+        while n != 1:
+            if n % 2 == 0:  # even
+                n /= 2
+            else:
+                n = 3*n + 1
+
+            currentChain.append(n)
+
+        if len(currentChain) > len(longestChain):
+            longestChain = currentChain.copy()
+
+print(longestChain[0])
+# Solution: 837799