#### Question

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer N and call it $a_0$. This is the seed of a sequence. The rest of the sequence is generated as follows: For n=0, 1, 2, ...

$a _ { n + 1 } = \left\{ \begin{array} { l l } { a _ { n } / 2 } & { \text { if } a _ { n } \text { is even } } \\ { 3 a _ { n } + 1 } & { \text { if } a _ { n } \text { is odd } } \end{array} \right.$

. However, if $a_n$=1 for any n, then the sequence terminates. a. Compute the sequence that results from the seeds N=2, 3, 4, ...10. You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers N, the sequence terminates after a finite number of terms. b. Now define the hailstone sequence $\left\{ H _ { k } \right\}$, which is the number of terms needed for the sequence $\left\{ a _ { n } \right\}$ to terminate starting with a seed of k. Verify that $H_2$=1, $H_3=7$, and $H_4=2$. c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

#### Solution

Verified#### Step 1

1 of 6Work in Excel is below. The cell $B4$ is open, so that the command is visible. The table shows the sequences $\left\{ a_n\right\}$ for seeds $N=2,3,4,\dots, 10$, until termination. We see that in this example, every sequence has terminated before 21st term (the letter $N$ stays for "Not any more :)" and indicates the termination).