Okay, I think I can finally answer the question at the top of this thread.
Collatz is Baker-hard. On another thread, we repeated Tao’s proof that the Collatz conjecture is Baker-hard, meaning a Collatz proof would necessarily imply a very hard Baker-style result of the form 2^l - 3^m > m for all sufficiently large l and m. (Baker’s proved something stronger, but even this Baker-lite result seems very hard to prove.) In other words, solving Collatz would mean doing something akin to Baker’s Fields Medal-worthy research. Tao showed that ruling out non-trivial Collatz cycles is Baker-hard. He did this by assuming 2^l - 3^m \leq m for some large l and m, then showing how to use that assumption to assemble a Collatz cycle. By the contrapositive, if there are no Collatz cycles, then there are no counterexamples to Baker, so Baker’s result is a consequence.
Question. What about the other half of the Collatz conjecture? How hard is it to prove there are no divergent start numbers?
We’re going to show:
Theorem. If no Collatz divergent numbers exist, then no Z_{9/8} numbers exist.
This is formally scarier than Tao’s reduction, because the nonexistence of Z_{9/8} numbers isn’t just hard to show, it’s an open problem.
Background 1. Lagarias’ book mentions “oblique connections” between the Collatz conjecture and Mahler’s Z_{3/2} problem. According to Lagarias, Mahler conjectures that no \xi exists such that:
0 \leq \{ ( \cfrac{3}{2} )^k \xi \} < \cfrac{1}{2}, k = 1, 2, 3 \ldots
where \{ x \} is the fractional part of x. Now, considering the Collatz-like system
W(n) = \frac{3n}{2} if n is even, or \frac{3n+1}{2} if n is odd,
Lagarias relates that if a Z_{3/2}-number lies in [n,n+1), then the associated trajectory of n never hits 3 mod 4. This connects Mahler’s problem to Collatz-like systems, but not to the Collatz rule itself.
Background 2. A Z_{9/8} number \xi, if it existed, would similarly have
0 \leq \left\{ \left(\frac{9}{8}\right)^k \xi \right\} < \frac{1}{8},
\qquad k=1,2,3,\ldots
It should seem very unlikely that the fractional part would stay under 0.125 forever, as you keep multiplying in factors of 9/8. But … the nonexistence of Z_{9/8} numbers is an open problem. Dubickas and Mossinghoff studied Z_{p/q} numbers in 2009, stating that when (2 < q<p<2q) it was widely believed that no such numbers exist. The case (p/q=9/8) lies in this range. Andrieu, Eliahou, and Vivion (2026) formulated a generalized Mahler conjecture that no (Z_{p/q}) numbers exist whenever (p < q^2). So, Z_{9/8}) is an unresolved case.
Main Theorem. If no Collatz divergent numbers exist, then no Z_{9/8} numbers exist.
The idea is to take a Z_{9/8} number and convert it to a Collatz start number n that follows a divergent trajectory whose parity sequence is an infinite concatenation of 110 and 101. Every three steps of the Collatz process guarantees a larger number.
Let
g_k=\left\lfloor \xi\left(\frac{9}{8}\right)^k\right\rfloor
and write
\xi\left(\frac{9}{8}\right)^k=g_k+r_k,
\qquad 0\leq r_k<\frac18.
Since
g_{k+1}+r_{k+1}
=
\frac98(g_k+r_k),
we get
8g_{k+1}-9g_k
=
9r_k-8r_{k+1}.
The left-hand side is an integer, while
-1<9r_k-8r_{k+1}<\frac98.
Therefore
8g_{k+1}-9g_k\in\{0,1\}.
Let
e_k=8g_{k+1}-9g_k.
Then
g_{k+1}=\frac{9g_k+e_k}{8}
=
\left\lceil\frac98g_k\right\rceil.
Reducing modulo 8,
g_k+e_k\equiv0\pmod8.
Thus
g_k\equiv
\begin{cases}
0\pmod8,&e_k=0,\\
7\pmod8,&e_k=1.
\end{cases}
So every g_k is either 0 or 7\pmod8.
Now define
n_k=2g_k-5.
If g_k\equiv0\pmod8, then n_k\equiv11\pmod{16}, so its next three Collatz parities are 110, and
T^3(n_k)=\frac{9n_k+5}{8}
=
2\left(\frac{9g_k}{8}\right)-5
=
2g_{k+1}-5
=
n_{k+1}.
If g_k\equiv7\pmod8, then n_k\equiv9\pmod{16}, so its next three Collatz parities are 101, and
T^3(n_k)=\frac{9n_k+7}{8}
=
2\left(\frac{9g_k+1}{8}\right)-5
=
2g_{k+1}-5
=
n_{k+1}.
Hence in either case,
T^3(n_k)=n_{k+1}.
Since
g_k=
\left\lfloor\xi\left(\frac98\right)^k\right\rfloor
\longrightarrow\infty,
we may choose K large enough that n_K>0. Then
T^{3j}(n_K)=n_{K+j}=2g_{K+j}-5\longrightarrow\infty.
Thus n_K is a divergent positive Collatz starting number.
Therefore, if a Z_{9/8} number exists, then a divergent Collatz number exists. Taking the contrapositive,
If no divergent Collatz numbers exist, then there are no Z_{9/8} numbers.
That ends the proof.
PS. 110 is the parity sequence of a well-known negative Collatz cycle (-5, -7, -10, -5). It’s also possible to take the other negative cycle 11110111000 and prove that ruling out Collatz divergence is alternatively as hard as the nonexistence of Z_{2187/2048} numbers.