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.

The Neighbouring rules view runs three rules side by side. From every start up to 1,000 it lists each cycle that $3n+1$, $3n-1$ and $5n+1$ fall into, and how many starts land in each. Below 1,000, $3n+1$ has the single cycle $1\to 4\to 2$; $3n-1$ has three, and so does $5n+1$. Under $5n+1$ most starts do not repeat a value within 1,000 steps at all. The page calls those starts outran the limit and nothing stronger: whether the orbit from 7 grows without bound is believed, not proved.

The second way is more constrained than it looks. Go once around a hypothetical cycle: each odd step multiplies the value by a little more than 3, each halving divides it by exactly 2, and the loop has to land back where it began. So the number of halvings $m$ and the number of odd steps $k$ must make $2^m$ almost equal to $3^k$, and "almost" is a quantity you can measure: $2^m/3^k$ is exactly the product of the factors $1+1/(3n)$ over the odd values $n$ of the cycle, which is close to 1 when those values are large. A cycle with large values therefore needs $m/k$ to be an unusually good approximation to ${\log }_{2}3$, from above, and such approximations are rare. The "Why cycles are rare" view draws exactly this, and layer 5 proves it.

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.

The cycle equation. Let $q$ be an odd number, $q\ge 3$, and consider the map that sends an odd $n$ to $qn+1$ and halves an even $n$. Take one period $v_0,v_1,\dots ,v_{L-1}$ of a cycle of positive whole numbers, with $k$ odd values and $m$ even values, so $L=k+m$. Each value is the image of the one before it, and the period visits each value once, so the product of the images equals the product of the values:

Cancelling the positive product over the even values leaves

where $n_1,\dots ,n_k$ are the odd values. Every factor exceeds 1, so $2^m>q^k$. ∎ The identity is checked in the test suite on the real cycles: $5\to 14\to 7\to 20\to 10\to 5$ under $3n-1$ has $k=2$, $m=3$ and $(3\cdot 5-1)(3\cdot 7-1)=14\cdot 20=280=2^3\cdot 5\cdot 7$; under $5n+1$ the cycles through 13 and 17 both have $k=3$, $m=7$.

The smallest value of a cycle is at most $\frac{k{2}^m}{q(2^m-q^k)}$. The smallest value $n_{\min }$ of the cycle is odd, because an even value $n$ would have $n/2$ in the cycle too, and $n/2$ is smaller. Write $r=2^m/q^k>1$ and $x=1/(q{n}_{\min })$. Each factor above is at most $1+x$, so $r\le (1+x{)}^k$. Taking logarithms, $\ln r\le k\ln (1+x)\le kx$. Applying $\ln u\le u-1$ to $u=1/r$ gives $\ln r\ge 1-1/r=(r-1)/r$. Together, $x\ge (r-1)/(kr)$, which rearranges to

The right side shrinks as $r$ grows, and $r$ doubles with every extra halving, so for a given $k$ the largest ceiling comes from the least $m$ with $2^m>q^k$. For $3n+1$ and $k=1$ that is $m=2$ and a ceiling of 1, which the cycle $1\to 4\to 2\to 1$ meets exactly. For $5n+1$ and $k=3$ it is $m=7$ and $\lfloor 3\cdot 128/(5\cdot 3)\rfloor =25$, which leaves room for the cycles through 13 and 17.

Why large cycles need convergents of ${\log }_{2}3$. Put $\alpha ={\log }_{2}3$. The distance $d=m-k\alpha$ is positive, and $r=2^d$, so a cycle with large values needs a very small $d$, that is $k\alpha$ just below a whole number $m$. How small $|k\alpha -m|$ can be for a given size of $k$ is the business of continued fractions. Here

with convergents 1, 2, 3/2, 8/5, 19/12, 65/41, 84/53, 485/306, 1054/665, 24727/15601, and so on. Every quotient and convergent printed here was computed twice in exact arithmetic: once from nothing but comparisons of powers of 3 with powers of 2, and once from a pair of rationals that bracket ${\log }_{2}3$ to hundreds of binary places. We use one classical fact, cited rather than reproved: if $p_j/q_j$ is a convergent and $1\le k<q_{j+1}$, then $|k\alpha -m|\ge |q_j\alpha -p_j|$ for every whole number $m$. The test suite checks this exactly for every $k$ below 15,601.

Combining it with the verified bound. Suppose a cycle avoids 1. Any value below $2^{71}$ reaches 1 (the cited verification), and then the cycle would contain 1, so every value of the cycle is at least $2^{71}$, and so is $n_{\min }$. Take a convergent $p_j/q_j$ with ${\epsilon }_j=|q_j\alpha -p_j|$, and suppose $k<q_{j+1}$. Then $d\ge {\epsilon }_j$. Since $2^d\ge 1+d\ln 2$, we get $r-1\ge {\epsilon }_j\ln 2$ and $r/(r-1)\le 1+1/({\epsilon }_j\ln 2)$, so

For the convergent 10,439,860,591 / 6,586,818,670 the next denominator is $q_{j+1}=65,470,613,321$ and, with $\ln 2$ and $\alpha$ replaced by exact rational bounds on the safe side, ${\epsilon }_j>1.46\times 10^{-11}$. The right side is then below $2.16\times 10^{21}$, while $2^{71}$ is about $2.36\times 10^{21}$. That contradicts $n_{\min }\ge 2^{71}$, so such a cycle has at least 65,470,613,321 odd steps. Since $m>k\alpha$, it has more than $2.58k$, so more than 168 billion, values in all. ∎ The final comparison is computed with exact rational arithmetic and reproduced in the test suite; a reader can verify each line but is unlikely to finish the arithmetic by hand. The next convergent only gives a ceiling above $2^{71}$, so this method stops there. For $k$ up to 10,000 the cruder table in the laboratory gives, with no continued fractions at all, a largest ceiling of 7,798,919 at $k=9,616$.

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.

You can run that search yourself. The Record hunt view checks starts from 1 upward on your own device and lists each new delay record and peak record as it finds one, stopping by itself at 100,000,000. Every start it reports was run to 1 in your browser, so what it finds is a record among those starts and no more, and the speed it measures shows how far one browser tab is from the search behind the $2^{71}$ bound.

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.

The cycle bound is conditional, and it uses only one inequality. Everything in the last paragraph of layer 5 stands on the cited verification below $2^{71}$: if the verified limit were smaller, the bound on $k$ would be smaller with it. It uses a single inequality, and sharper tools give stronger results (Eliahou, 1993; Simons and de Weger, 2005; Hercher, 2023 on cycles with few circuits). Those papers are cited here and not reproduced, and none of their numbers is quoted. What is known is that a nontrivial cycle, if there is one, would be extremely long and have extremely large values. What is not known is whether one exists, and nothing in this argument touches the other way the conjecture could fail, an orbit that grows without bound. Under $5n+1$ the same arithmetic leaves room for cycles, and they are real: the argument bites only where a verification and a narrow gap meet.

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.