Obviously encountering a power of 2 in a Collatz sequence is important as this is what automatically causes the sequence to terminate. If you calculate the first power of 2 that appears in the Collatz sequence for the first 5,000,000 seeds, 93.8% of the time it's 16. The only way for this to happen is if the number in the sequence immediately prior to that is 5 (or if the seed itself is 16). Why do the vast majority of sequences end up at 5? Is it possible to prove that as you test more and more seeds this percentage approaches 100? Maybe! Gonna take a crack at it.

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I understand why only even powers of 2 are ever encountered first by Collatz sequences where the seeds aren't themselves odd powers of 2. It's easy to show that (2\^n-1)/3 is an integer iff n is even. But why 16 all the time? 64 pulls no weight at all and 1024 and 256 are basically on par with each other so it's not like there's some sensible distribution to suggest that the lower the power the more "attractive" it is.

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If I had to guess without analyzing further, I'd say it's because the lowest seed whose sequence can first encounter the powers of 2 rail somewhere higher than 16 is 21 which has the sequence (21, *64*, 32, 16, 8, 4, 2,1). So, given a random seed, if its sequence ever gets below 21, it's going to end up at (..., 5, 16, 8, 4, 2, 1) and there are a lot of ways for a sequence dive below 21, including the 2^k *3 rail, the 2^k *5 rail, 2^k *7, and so on, as well as other paths that don't involve hitting any rail. Even so, I doubt that accounts for the full effect.

Would love to hear anyone's thoughts on this!
Think of it the other way around. Every new branch of the Collatz-tree leads to infinite amount of numbers. Five is the first branch, so it is the largest one (apart from the trunk of the tree, which are the powers of two).

Going from 2 you can create 4, 8 and 16 is the first number when you can subtract one and divide by 3, you get 5.
That's a common feature for functions where everything (or almost everything) ends at one element or one cycle. You can find a similar pattern if you follow the first link of each Wikipedia article, for example. The exact patterns change over time as the articles are not static, but most articles end up in the same loop with a few entry points being the most common ways to get there.

How do you get to 64? From a higher power of 2, but you excluded that from the analysis, or from 21. How do you get to 21? It's divisible by 3, so you can only reach it from 42. You can only reach that from 84, and so on, 21\*2^k is the only way to reach 64 as first power of 2. You should find the same result for all powers of 64.

So let's look at the list:

* 4 - that's a part of 100% of all cycles assuming the conjecture is true, but it's never the entry point
* 16 - by far the most common entry point
* 64 - extremely rare
* 256 - a pretty large number, so most cycles will pass it
* 1024 - even larger
* 4096 - extremely rare

It might be coincidence, but 16 and 256 differ by a factor 16 and 93.8% is almost exactly 15/16 = 93.75%.
by
Wanna know something cool I figured out with the collatz conjecture and the 5 branch? There are infinite odd numbers that have the same branch depth once you find a branch.

For example, 5 goes to 16. Now add those two numbers and you get 21. 21 is the same distance from a power of 2 as 5. 21 becomes 64. Add 21 and 64 and you get 85. 85 becomes 256.

Same works for all odd numbers. 3 becomes 10. Add them, get 13. 13 becomes 40 which becomes 10. The second odd number will always become 4x what the previous odd number became.

I've formalized this behavior a little bit. You can get the first sequence like this:

4⁰=1

4¹+4⁰=5

4²+4¹+4⁰=21.

Etc.
"It's easy to show that (2^n/ -1 )/3 is  an integer if n is even"
Isn't that just a power of 4?
Probably just the strong law of small numbers? Your sample of "the first 5,000,000 seeds" is inherently biased as it specifically considers the *first* 5 million integers out of a set of integers that can be arbitrarily large. What if you instead considered the interval $Grahams number, Grahams number + 5,000,000$?
There needs to be a forum existing solely of posts from varying degrees of mathematicians (from enthusiasts to like hardcore researchers) to just post their thoughts/analyses about the Collatz conjecture.

I don’t think I’ve seen a single problem gain so much notoriety among the general math-enjoyer population, could yield interesting results.
On thing I have been doing while trying to get somewhere with the collatz conjecture is going from the inverse collatz conjecture to figure out the prime factorization of all numbers I solve for, I want to understand how you can start at a prime number and move to another prime and its linear combinations with different primes.

So far I'm almost certain that if you find a prime number on the inverse collatz tree that you will eventually find any of its multiples on the inverse collatz tree, I dont really have a way that I could prove this, but it would basically solve the collatz conjecture if true the multiple of a prime n.

One method I have started down is path adding on the inverse collatz conjecture tree, like if you wanted to multiply 3 and 5 together to get 15 you would add the path from 1 to 3 on the tree to the path 1 to 5 on the tree and you should get to 15, well in practice it gets you to the same neighborhood but it doesn't quite get you to the number you are looking for, but if you allow for a second fractional numbers and follow the paths I like to use even, odd for marking turns on the path and following the rules for each turn like even you multiply by 2 and odd you subtract 1 and divide by 3 you will get a fraction that is very close to what you are looking for,  like I think with the path of 3 + the path of 5 example if i remember right you get something like 15.64... which is pretty close to 15, but it breaks rules because the collatz conjecture deals with whole numbers.

if you also look at the tree you only end up a few spots away from 15, and what is further if you compare the path of 1 to 15 with the paths of 1 to 5 and 1 to 3 you will see the paths of 1 to 5 and the paths of 1 to 3 in the path of 1 to 15. Now I think this is important because while 15 is on the 5 branch, 15 is not on the 3 branch, but it shows that finding a path to any prime number can help you in finding the path on the tree to any linear combination of prime numbers assuming have also found the paths to those.

Edit i realize this might come off as me thinking I have solved the conjecture, but I haven't there are holes in my reasoning, there are terms in the path I dont know where they come from like on the path to 15 there is a section between the path to 5 and the path to 3 that doesn't make sense, or im not completely sure that you can find all primes and all linear combinations of primes on the inverse collatz tree, it just seems to hold true for the few hundred numbers I have calculated by hand.
Of odd integers, multiples of 3, powers of 4, and tree structures...

Even numbers are "less interesting" that the behaviors of one odd integer leading to the next.  Every even gets divided by 2, repeatedly, until no factors of 2 remain, landing you on an odd 'trunk.'

There is an interesting consequence to the rules making multiples of 3 next to useless: Only an initial pick can take you to a multiple of 3.  If that pick is even, the divide-by-2 rule will remove powers of 2 as factors, leaving an odd multiple of 3.  Since the next rule is 3x+1, NO MULTIPLE OF 3 can ever result. (It's always 3x+1, thus 1 mod 3, never 0 mod 3.) No Collatz sequence can contain a multiple of 3 unless that's the initial pick. (Funny the things you *don't* find.)

So 64 & 21 backtrace to nearly nothing, because 3 divides 21, leaving no sources except 21 times a power of 2.  EVERY even is an odd integer times some power of 2: That's the factorization  of everything on the tree, all natural numbers.

Powers of 4 add an interesting 'phantom rule' to the mix: every odd can be run thru 4x+1, giving "the next higher cousin on the tree."  That because it's the simplification of "next in sequence, times 4, and backtraced.")  3x+1 always leads to an even (3×odd is odd, +1 becomes even), so its backtrace opposite is (x-1)/3.  So the backtrace of 'next times 4' is (4*(3x+1) -1)/3 = 12x/3 +4/3 -1/3 = 4x+1.  And since every odd leads to an even, the rules show the maximum growing two-step is just
50% more +1.

That 4x+1 rule it also great for branch detection. Any odd ≡ 5 mod 8 is a branch with a sharp decline in sequence, factoring out 2's three times or more.

I'm solving Collatz by elementary proof as a hobby.  I've identified anatomical features to the tree structure, but had to invent descriptors like "willows, subwillows, and microwillows."  This is the first venue I've found to "talk math" with anyone.  Otherwise, Ijm an agoraphobic disabled shut in, living in the woods under a rock.