Upper bound on the elements of a Collatz cycle Let $T$ be the (shortcuted) Collatz map: $T(x) = x/2$ when $x$ is even and $T(x) = (3x+1)/2$ when $x$ is odd.
Böhm and Sontacchi, 1978 show that if $x = T^n(x)$ then $x < 3^n$ with $x,n\in\mathbb{N}^+$.
Can this bound be improved to $2^n$?
This question is relevant when analysing Collatz cycles in binary as it asks if iterates in a positive integer cycle of size $n$ can always be represented on $n$ bits or not. We originally asked the question in this paper which studies Collatz in base 2 and 3 -- this is not meant to do self-promotion but merely to give more context on this question.
Arguably, the question is interesting per se as it is generally hard to give any kind of non-trivial results concerning Collatz cycles (such as Eliahou, 1993 or [Steiner, 1977]).
Thank you very much in advance,
 A: I am not aware of any previous answer in a publication, but asymptotically sharper bounds are known. For example, in the context of the $``3x+d"$ generalization, Belaga and Mignotte found that
$$ x < d \cdot k^{C} \cdot \left( \frac{3}{2}\right)^k$$
where $C>0$ is a constant and $k$ is the number of odd terms in the cycle (see  Perigee, Apogee Upper Bounds from Belaga and Mignotte).
In fact, the desired bound can be proved by using Corollary 13 in Simons and de Weger (2005):
$$ x_{min} < A \cdot m \cdot k^{13.3}$$
where $x_{min}$ is the smallest term of the cycle, $A = 457.41 \ldots$ is a constant and  $m$ is the number of sequences of consecutive odd terms in the cycle, making it a $m$-cycle. Note that $m \leq k$.
For a term $x$ of the cycle, we have $x \leq x_{min} \cdot 2^k$ as $T(y) \leq 2y$ for every odd integer $y$ (worst-case scenario). Finally, we get
$$ x < A \cdot k^{14.3} \cdot 2^k$$
and, using the well-known inequality $n/k > \rho = \log_2 3 $ (see, e.g., here) and the obvious one $k \leq n$,
$$ x < A \cdot n^{14.3} \cdot 2^{n/\rho}.$$
To conclude the proof, observe that the right hand side is smaller than $2^n$ for $n \geq 400$, whereas the minimal length of a non-trivial cycle is at least 17087915, according to Eliahou's paper. Moreoever, the bound $x < 2^n$ also holds for the trivial cycle (1,2).
A: $x_{max}<3^k<2^n$ (with $k$ the number of odd steps)
Collatz Conjecture: For a cycle where the maximum odd integer is $x_{max}$, does it follow that $x_{max} < 3^n$
