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Hailstoning

Joe. The sequence 1,2,4,8,16,5,10,20 is not uniquely arrived at. 20 can only be reached from 40, but 40 can be reached from 80 or 13. Where there is a split possible, then both roads need to be taken into account. Then the conjecture is analogous.

Qualititatively it seems obvious that a hailstone sequence will eventually hit a value of 2^m. However, this is not a proof.

It would be interesting to see the "coverage" of numbers in hailstone sequences. eg

1 - 4 - 2 -1 covers the numbers 1,2, and 4.
3 - 10 - 5 -16 - 8 - 4 - 2 - 1 additionally covers 3,5,8,10, and 16.
7 - 22 - 11 - 34 -17 - 52 - 26 - 13 - 40 - 20 - 10... additionally covers 7,11,13,17,20,22,26,34,40 and 52
9 - 28 - 14 -7... additionally covers 9, 14, 28

I have deliberately omitted even numbers that reduce to already covered numbers and odd numbers that already appear in a hailstone sequence.

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