The Collatz Orbit Atlas

The starting value must be a positive whole number. To produce the next value, divide it by two if it is even, or multiply it by three and add one if it is odd. Apply the same choice to the new value, rather than repeatedly using the starting value.

For example, 6 is even, so 6 ÷ 2 = 3. Now 3 is odd, so 3 × 3 + 1 = 10. Continuing gives 10 → 5 → 16 → 8 → 4 → 2 → 1. This list of visited values is called an orbit.

The laboratory stops at the first 1 so that it can count the steps needed to reach it. The mathematical rule itself does not stop: another step gives 4, then 2, then 1 again.

Starting from 27 takes 111 steps and reaches 9,232 along the way, while 26 takes ten steps and 28 takes eighteen. The Collatz conjecture says that every positive start eventually reaches 1. Published computations establish this for enormous finite ranges, but no proof covers all positive whole numbers.

Watch the parity view and one thing becomes obvious: an odd step is always followed by at least one halving. Multiply an odd number by three and you get an odd number; add one and it becomes even. So the climbing move can never happen twice in a row.

For a more careful heuristic, jump from one odd value to the next. Write $3n+1=2^{a}m$, where $m$ is odd and $a\ge 1$. The whole episode takes $n$ to $m$, a change of roughly $3/2^a$. If the powers of two dividing successive values behaved like random residues, then $a$ would equal $1,2,3,\dots$ with probabilities $1/2,1/4,1/8,\dots$, so its mean would be 2. On a logarithmic scale the typical multiplier would therefore be $3/2^2=3/4$.

Because 3/4 is less than one, this random model suggests that values should tend to shrink over many steps. That tendency is consistent with the completed computations that reach 1.

That argument is intuition, not proof. The exponents $a$ are produced by one deterministic orbit, not independent coin flips, and a single starting value is entitled to be atypical forever.

The conjecture can only be false in one of two ways.

1. An unbounded orbit. It reaches arbitrarily large values without reaching 1; this does not mean every step increases, since even values are still halved.
2. A cycle that does not contain 1. It returns to a value it has already visited, and loops forever without ever passing through 1.

There is no third option: an orbit that stays bounded must eventually repeat a value, because there are only finitely many values below any bound - and a repeat is a cycle.

Neither has ever been found for $3n+1$. Both are entirely ordinary for nearby rules. Switch this laboratory to $3n-1$ and start from 5:

A cycle, in five steps, that never touches 1. So "eventually reaches 1" is not something the shape of these rules guarantees. Whatever makes $3n+1$ special - if anything does - is not visible in the statement of the rule.

Define $T:{\mathbb{Z}}_{>0}\to {\mathbb{Z}}_{>0}$ by

The Collatz conjecture. For every $n\ge 1$ there is a $k$ with $T^k(n)=1$.

Three statements to keep apart:

1. Every $n$ below $2^{71}$ has been verified to reach 1 by published computation.
2. Almost all $n$ have orbits attaining almost bounded values (Tao, 2022), in a precise logarithmic-density sense.
3. Every $n$ reaches 1. Open.

The first is a finite check. The second is a theorem, and it is not the third - a set of logarithmic density zero can still be infinite, and "almost bounded" is weaker than "reaches 1". A counterexample is entirely compatible with what is proved.

Every orbit is well defined and stays a positive integer. Halving an even number gives an integer; $3n+1$ for $n\ge 1$ gives an integer at least 4. Neither branch can produce a non-positive value, so the map never leaves ${\mathbb{Z}}_{>0}$. ∎

A bounded orbit must be eventually periodic. If an orbit's values never exceed some $M$, it takes values in a finite set of size $M$, so within $M+1$ steps two of them coincide; the map is deterministic, so everything after the second occurrence repeats what followed the first. ∎

So the conjecture fails only by divergence or by a cycle other than $1\to 4\to 2\to 1$. Immediate from the previous two: an orbit either is bounded, and then cycles, or is unbounded. ∎

An odd step is always followed by an even one. For odd $n$, $3n$ is odd and $3n+1$ is even. ∎ This is why the parity view never shows two vermilion blocks in a row, and it is what the heuristic in layer 2 is built on.

What is not proved. That the heuristic average governs any individual orbit. The gap is not technical: the heuristic treats the parities of successive values as if they were independent coin flips, and they are not - they are determined, by the starting value, in a way nobody knows how to control. Tao's theorem is the strongest thing anyone has extracted from this line of argument, and what it gives is a statement about almost every start rather than about every start.

And note what is not available here. The argument that settles the happy numbers - the map cannot outrun itself, so the space closes and enumeration finishes - has no analogue. $3n+1$ has no bound to close inside.

A finite calculation is not a proof of the conjecture. A completed run establishes that its starting value reaches 1. A completed sweep does the same for the finite range it covers. These are examples consistent with the conjecture, but they cannot rule out a counterexample outside that range. The published verification below $2^{71}$ is a separate computation, not a result established by running this page.

The records are provisional by construction. Below ten thousand the longest run is 261 steps, from 6,171, and the highest excursion is 27,114,424, from 9,663. Both are records only within that range; each held until a wider search found the next. Presented as a sequence of provisional bests rather than as facts about the integers.

This laboratory stops early, on purpose. An individual run gives up after 2,000 steps and says so rather than reporting a result, and any step that would take a value past the range where whole numbers are exact stops the run instead of continuing in rounded arithmetic. A page about the difference between checking and proving cannot afford to print values that are quietly wrong.

Changing the rule changes the question. The alternative rules let you observe cycles or runs that outlast the computation limit. A cycle under $3n-1$ is a result about that variant, not a counterexample to the $3n+1$ conjecture. Each plot and arithmetic transcript should be read with its selected rule and starting value.