Wow, thanks for the deep dive! This is so fascinating that someone would be willing to dive into my ideas. HN rules.
I clearly need a much more clear explanation of things — for example the floor(log2(( part is being misinterpreted here. Thanks for diving in and giving me pointers
1: Derive that, for a cycle with o odds and e evens, that the product of the increases due to odds (3 + 1/x) must exactly equal the product of the decreases by halving, which comes out to 2^e.
The trivial cycle meets this, since (3+1/1) = 2^2.
As another example, for the alternative rule where odd numbers go up by 3*n + 5, there are some cycles, like 19 => 62 => 31 => 98 => 49 => 152 => 76 => 38 => 19. The increases here come out to approximately 3.26, 3.16, and 3.10, the product of which is exactly 32, or 2^5.
2: Observe that each (3 + 1/x) term (except for trivially small x) is greater than 3, but near it. For x=101, (3+1/x) is about 3.01. As a result, the product of the increases is going to be larger than a product of 3, but somewhat close to one.
3: Observe that we need the little error above the power of 3 to be large enough that the LHS exactly equals a power of 2. This is challenging because as the numbers in the cycle get larger, (3 + 1/x) gets closer to 3.0, which makes it harder for the product to be much larger than a power of 3. And, for large powers of 3, the nearest power of 2 is never particularly close — we actually need this error to compound to a product substantially larger than the power of 3. We can attempt to achieve this with longer cycles for more compounding error, or smaller values of x.
4: Observe that, for a given number of odd numbers in a cycle, we can find an upper bound for the lowest value in the cycle. For example, let’s use o=5. We need (3+1/x1)(3+1/x2)(3+1/x3)(3+1/x4)(3+1/x5) = 2^e. 3^5 is 243, so the easiest e value is 8, making the RHS 256.
If we define x1 as the smallest number in the cycle, we can see that x1 can’t be too large. x1=1001 implies that x2, x3 etc are at least 1003 and 1005 etc. Plugging these in, the LHS is 243.4 — not near 256. And making x2, x3 etc larger only makes it worse.
Okay, so try a smaller x1 - you can compute that the largest possible x1 that can possibly make the LHS equal 256 is 31.
However we know that 31 and every number below it does not have a cycle, so we know there are no cycles of length 5.
We can actually define the RHS as the next power of 2 above 3^o, which is floor(log2(3^o))+1.
5: Repeat this process for larger cycle lengths and you see that the largest viable value does not trend larger, it trends smaller.
The powers of 2 and 3 getting further away bit is not correct as you mention but it isn't actually all that important. My Github page is a WIP so bear with me..
If a cycle does exist for some large-ish cycle length, let’s say 100 odd numbers, then it means that (3+1/x1)(3+1/x2)…(3+1/x100) exactly equals some power of 2.
Since (3+1/xn) rapidly approaches 3.0, for any large x1, the product of (3+1/x1)(3+1/x2) etc is going to be pretty close to a power of 3, but it needs to be much larger than that to equal a power of 2, since powers of 2 and 3 are always somewhat far apart. That means that x1, x2 etc must be relatively small so that the product of the (3+1/x) can actually be substantially larger than a power of 3. That puts a hard floor on the possible cycle values, and that floor actually trends smaller as the cycle gets larger since the distance between a power of 3 and the nearest power of 2 also trends larger.
It’s not probabilistic — for any cycle length, you can compute the minimum possible x value that could possibly be the lowest odd number in the cycle. That value trends smaller as the cycle length grows.
A cycle means the product of all of the increases (each of which is a scaling factor of 3x+1 / x) equal the product of the decreases (a power of 2). 3x+1 / x = 3 + 1/x, so except for trivially small x, each increase is just a bit larger than 3. The product of those increases is a small amount larger than a power of 3, and has to exactly equal a power of 2.
Larger cycles allow that little error above a power of 3 to compound, but larger cycles also necessitate larger values of x, which shrink the error.