(b) Do some further experimentation and try to decide whether the 3n + 1 algorithm always terminates and, if so, at what value(s) it terminates.
(c) Assuming that the algorithm terminates at 1, let L(n) be the length of the algorithm for starting value n. For example, L(5) = 6 and L(7) = 17. Show that if n = 8k + 4 with k ≥ 1, then L(n) = L(n + 1). [Hint. What does the algorithm do to the starting values 8k + 4 and 8k + 5?]