Game: divide by 2 or multiply by 3 and add 1

Game
If the number is even, divide by 2. If the number is odd, multiply by 3 and add 1.
When you reach 1, choose another number and restart.

27

This is popularly known as Collatz conjecture:
27 goes through 41 phases
27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

71 * 3 + 1 = 214, not 204

3
10
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16
8
4
2
1

challenge: prove that you are always able to reach 1, regardless of the initial number

Collatz conjectured that every number reaches 1 when following operations are done on it: divide by 2 if it is even or multiply by 3 and add 1 when it is odd. Still any counterexample is not found and computers are running ruthlessly trying to find counterexample. This is one of the most unsolved mathematics problem in the history.
For more, check out the link: en.m.wikipedia.org/wiki/Collatz_conjecture

18
9
28
14
7
22
11
34
17
52
26
13
40
20
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5
16
8
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2
1

@treap_di_dsu #5
A conjecture does not have any proof. For that check out its definition from Internet.

@treap_di_dsu every number when follow you rule it will end in power of 2

that's because the proof hasn't been found yet; but if the conjecture is true, there should be some way to prove it