Collatz Claudie

Playground The Collatz conjecture is open. Everything here is computed from the map itself, so the plots are exact for the range shown; the gravitational reading in the third figure is a stated analogy, not a physical claim or a proof.

The rule is one line. For a positive integer $n$, the map is $T(n) = n/2$ when $n$ is even, and $T(n) = 3n + 1$ when $n$ is odd. Halve it, or treble it and add one.

The conjecture is that iterating $T$ from any starting point reaches 1. It has been checked past $2^{68}$ and remains unproven.

Integers on Ox 220 Primes traced 50

1 · One application of the rule

Every integer on Ox, the value it maps to on Oy. The rule splits cleanly into two branches, which is the whole engine of the problem: odd numbers are thrown far up the line $3n+1$, even numbers fall to $n/2$. Nothing else happens, ever.

Odd inputs land on the upper ray, even inputs on the lower. The upper branch climbs three times as fast as the lower one falls, which is why the question is hard.

2 · Fifty primes, followed to the end

Each path starts at a prime and applies the rule to its own output, over and over, until it reaches 1. Ox is the step index, Oy the value on a logarithmic scale, since the excursions are far too tall to plot linearly.

Darker paths start from larger primes. Hover any path to isolate it. Every one of them ends in the same place.

3 · Is 4 → 2 → 1 a gravitational point?

Here is the honest version of the analogy, and then the picture.

Take altitude to be $\log_2 n$. Then an even step lowers altitude by exactly 1, and an odd step raises it by $\log_2(3 + 1/n)$, just under 1.585. Because $3n+1$ is always even, every rise is immediately followed by at least one fall. Averaged over a trajectory, the altitude drifts down, and that drift is what the plot below measures.

mean change in altitude per step, measured across the traced primes
negative means the system falls

So the analogy has a real referent: there is a restoring tendency, it is measurable, and it points inward. Plotted polar, with radius as altitude and angle advancing one notch per step, every trajectory spirals in.

Radius is $\log_2 n$, so the centre is $n = 1$. Every path winds inward and finishes on the same small orbit.

Where the analogy holds. There is a measurable inward drift, a basin that every tested start falls into, and a structure at the centre that nothing escapes once entered.

Where it breaks, and it matters. A gravitational point is a point, and this is not one. Solving $T(n) = n$ gives $n = 0$ or $n = -1/2$: the map has no positive fixed point at all. What sits at the centre is a 3-cycle, $1 \to 4 \to 2 \to 1$, so the right physical image is a stable orbit rather than a mass. The drift is also only an average. Individual steps go up, some trajectories climb for a long time first, and an average pointing down is not a proof that nothing escapes. That gap between measured drift and proof is exactly the open problem.

Table view: the traced primes