Questions about n < 2^k

[Collag3n]

In a cycle,

All elements n_i<2^x<2^k
(here x,n,k are used instead of n,k,x)

Elementary proof of x < 2^n in a cycle of size n

Someone asked a question on Math exchange yesterday, and I showed in my reply that the smallest element (a_0 ) of a cycle has the property that a_0<(\frac{3}{2})^k (translated to your notation of k and n which are inverted here ,see https://math.stackexchange.com/questions/5085433/explicit-baker-constants-for-collatz-cycle-constraints/5085630#5085630) This would place the largest element at a_{max}<(\frac{9}{4})^k. If we start using tighter bounds (like for the 2.56^k bound) in the logic, we co…

3^x-2^x \leqslant \beta <x3^{x-1}

How many odds could a Collatz counter-example cycle have, relative to its length?

I see you often use this upper bound \cfrac{3^x\,(x/2)}{2^k-3^x}, but there is a better one: \cfrac{3^x\,(x/3)}{2^k-3^x} It comes from this https://math.meta.stackexchange.com/revisions/4669/655 where you have \frac{2^S}{3^N}<\frac{3a_0+N}{3a_0} leading to a_0<\frac{N3^{N-1}}{2^S-3^N}

I don’t think the truthness of CSTC means no other cycles exists. I saw that the Google search AI seems to imply that from the “paradoxical” paper of @oros, but that seems far-fetched to me.