1)

**You can't have an infinite sequence of "upward steps"**

**If we start from an odd number, we can tell how many upward steps the sequence will take before a downward step. (An upward step is one where (3n+1)/2 is an odd number, for instance 15, 46, 23 where 23>15. A downward step is one where (3n+1)/2 is an even number, for instance, 17, 52, 26, 13 and 13<17).**

If the starting number is X, we can calculate how many upward steps the sequence will take before a downward step by factoring (X+1). For each time 2 is a factor of (X+1), we will get one upward step. For instance, for 7, X+1 is 2^3 so we get three upward steps - 7, 22, 11, 34, 17, 52, 26. For 15 we get 4 upward steps (15, 46, 23, 70, 35, 106, 53, 160, 80).

Note that for the odd numbers in these sequences, the power of 2 in the factorisation of X+1 gradually decreases by 1 each time, and each factor of 2 is replaced by a factor of 3 in the next X+1 - 16 = 2^4, 24 = 2^3*3, 36 = 2^2*3^2, 54 = 2* 3^3, 81 = 3^4 - this process continues until all the factors of 2 have been replaced by a factor of 3 and we reach an odd number (meaning X is even).

This shows that you can't have an infinite sequence of upward steps without intervening downward steps.(I would say it proves it but I think it would need a more rigorous exposition of the reasons for this pattern to call it a proof).

2)

**You can't have an infinite sequence of downward steps**

This is way too obvious to even mention, but of course the number of downward steps from the even number Y also depends on the power of 2 in the factorisation of Y. Each time you halve you lose one of the powers of 2. The only way you can get into a closed loop of numbers which are all pure powers of 2 is when you get to 4, 2, 1, 4, 2, 1 (seeing as how 1 = 2^0).

So we will always be alternating between sequences of upward and downward steps and the number of steps each way is determined by the power of 2 (of X+1 for odd numbers and in Y for evens).

OK, so that's all fun stuff, but unfortunately it doesn't rule out two ways in which you could have a Collatz sequence that never reaches 1 - A) an infinite loop which keeps passing through the same number Z (where Z isn't 1, 2, or 4). B) an infinite run of upward and downward steps which keeps climbing overall.

I think it must be especially hard to prove that the second of those two options isn't possible.

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