<p>The Hailstone sequence of numbers can be generated from a starting positive integer, n by:</p>
If n is 1 then the sequence ends.
If n is even then the next n of the sequence <code> = n/2 </code>
If n is odd then the next n of the sequence <code> = (3 * n) + 1 </code><p>The (unproven) <ahref="https://en.wikipedia.org/wiki/Collatz conjecture"title="wp: Collatz conjecture">Collatz conjecture</a> is that the hailstone sequence for any starting number always terminates.</p>
<p>The hailstone sequence is also known as hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as the Collatz sequence.</p>
Task:
Create a routine to generate the hailstone sequence for a number.
Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with <code>27, 82, 41, 124</code> and ending with <code>8, 4, 2, 1</code>
Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length. (But don't show the actual sequence!)See also: