24 lines
1.3 KiB
Plaintext
24 lines
1.3 KiB
Plaintext
The Hailstone sequence of numbers can be generated from a starting positive integer, n by:
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* If n is '''1''' then the sequence ends.
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* If n is '''even''' then the next n of the sequence <big><code> = n/2 </code></big>
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* If n is '''odd''' then the next n of the sequence <big><code> = (3 * n) + 1 </code></big>
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The (unproven) [[wp:Collatz conjecture|Collatz conjecture]] is that the hailstone sequence for any starting number always terminates.
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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).
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This sequence is also known as the ''Collatz sequence''.
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;Task:
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# Create a routine to generate the hailstone sequence for a number.
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# 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>
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# Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.<br> (But don't show the actual sequence!)
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;See also:
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* [http://xkcd.com/710 xkcd] (humourous).
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<br><br>
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