Collatz as a Lyapunov stability problem: the β-halving trick

The Collatz conjecture is usually stated as: take any positive integer, if even halve it, if odd do \(3n+1\), repeat — do you always reach 1?

Erdős said "Mathematics is not yet ready for such problems." I'm not sure he was wrong, but I think the framing matters. Here's a framing that makes the problem feel more tractable.

Collatz as dynamical systems

The Syracuse map \(T : \mathbb{N} \to \mathbb{N}\) is defined by \(T(n) = n/2\) if \(n\) is even, \(T(n) = (3n+1)/2\) if \(n\) is odd. Collatz says every orbit of \(T\) eventually reaches 1.

In dynamical systems, the standard tool for proving convergence is a Lyapunov function — a function \(V(n)\) that strictly decreases along orbits and is bounded below. If such a \(V\) exists, convergence is automatic.

The obvious candidate is \(V(n) = \log_2(n)\). An even step decreases \(V\) by exactly 1. An odd step increases \(V\) by \(\log_2(3/2) \approx 0.585\). So on average, if roughly 63% of steps are even, \(V\) decreases.

The β-halving characterization (Theorem D)

Here's the result that surprised me: Collatz is equivalent to the statement that the asymptotic density of even iterates exceeds \(\log_2(3) \approx 1.585\) per odd step.

More precisely, define \(\beta(n)\) as the limiting ratio of even-to-odd steps in the orbit of \(n\). Then:

\[\text{Collatz} \iff \beta(n) > \log_2(3) \text{ for all } n\]

This isn't deep — it's almost a tautology once you write it down. But the consequence is powerful: you've converted a discrete, combinatorial problem into a density estimation problem. The question is no longer "does this specific orbit reach 1?" but "does the average bit-reduction rate exceed a threshold?"

Residue-class coverage (Theorem E)

The second surprise: the density condition can be checked on finite residue classes.

For each modulus \(m\), the first \(k\) iterates of \(T\) modulo \(m\) determine whether those iterates are even or odd. If for every residue class modulo some sufficiently large \(m\), the resulting bit-pattern has density above \(\log_2(3)\), then Collatz holds.

This means: Collatz is equivalent to a finite computation, in principle. The catch is "sufficiently large \(m\)" — we don't know how large is sufficient.

What's verified

All five theorems (A through E) are kernel-verified: 94 declarations, 0 errors. The Lyapunov characterization, the amortized version, the piecewise covering, β-halving, and residue-class reduction are all formally checked.

Where I'm stuck

The greedy selector termination — proving that \(m(x) < \infty\) for the covering strategy — is equivalent to Collatz itself. So the reduction is real but doesn't yet close the problem. The residue-class approach needs either a clever bound on \(m\), or a way to prove the density condition without exhaustive case-checking.

Thoughts welcome. This is genuinely one of those problems where an idea from a different field might unlock everything.