Abstract
Repeatedly replacing a positive integer $n$ by $3n+1$ if it is odd and by $n/2$ if it is even is conjectured to reach $1$ from every starting number (the Collatz conjecture). Its behaviour is checked in numbers, to see why it shrinks and what change makes it stop shrinking. No new theorem or law is claimed, and the conjecture is not proved.
First, every number up to $10^6$ reaches $1$: running every number from $2$ to $10^6$, each eventually fell below its starting value and reached $1$. The mean number of steps is $\mathbf{131.434555}$ and the longest is $837799$ with $524$ steps. $27$ takes $111$ steps and climbs to $9232$ on the way (Section 2).
Second, on average each step shrinks: writing an odd step as $(3n+1)/2$, a step multiplies by about $1.5$ half the time and by $0.5$ half the time, with geometric mean $\sqrt{3/4}=\mathbf{0.866025}$. Running $2000$ random large numbers for $40$ steps each gave a measured $\mathbf{0.864706}$ (Section 3).
Third, and this is the core. The separator is whether the multiplier is less than $4$: replacing $3$ by $5$ gives a geometric mean of $\sqrt{5/4}=\mathbf{1.118034}$, growing on average. Of the numbers from $1$ to $10000$, $9274$ exceeded $10^{12}$ and only $726$ fell into cycles (Section 4).
Fourth, shrinking on average does not guarantee reaching $1$: replacing $3n+1$ by $3n-1$ keeps the same geometric mean $\sqrt{3/4}$ and shrinks on average, yet the numbers from $1$ to $10000$ split into three cycles: the one containing $1$ ($3244$), the one containing $5$ ($3213$) and the cycle of length $18$ containing $17$ ($3543$). Shrinking does not decide which cycle is reached (Section 4).
Keywords: Collatz conjecture, 3n+1 problem, geometric mean, stopping time, dynamical system.
Scope (what is not claimed)
No new theorem or law is claimed: the $3x+1$ problem and its generalisations (Lagarias, $1985$), properties of stopping times (Terras, $1976$) and probabilistic heuristics are all known (collected in the book edited by Lagarias). The conjecture is not proved: checking up to $10^6$ is not a proof. “Shrinking on average” is a heuristic assuming odd and even appear equally often and guarantees nothing about any particular number. The general theory of $an+b$ is not treated: only $3n+1$, $3n-1$ and $5n+1$ are compared. Relation to earlier papers: Paper 195 showed that “stable” splits into six separate words by what is perturbed: the cycle $1\to4\to2\to1$ here attracts by shrinking on average, and with the same shrinking $3n-1$ has three attracting cycles. What is added is counting the steps up to $10^6$ in a table and matching a sample against running each number through, obtaining the geometric mean of the factor by formula and by random numbers, and placing the separator on the multiplier versus $4$ and checking it with $5n+1$ and $3n-1$.
1. Notation and the Question
Setting: map a positive integer $n$ to $an+1$ if odd and to $n/2$ if even ($a=3$ is the Collatz map). The number of applications until $1$ is reached is called the number of steps.
Question: does every number reach $1$, why does it shrink, and what happens when $a$ changes?
2. Running Up to $10^6$
Computation 2.1: each number from $2$ to $10^6$ was run until it fell below its starting value, and the remaining steps were added from the table. Every number eventually fell below its start, so all reached $1$.
| quantity | value |
|---|---|
| mean number of steps ($2$--$10^6$) | $\mathbf{131.434555}$ |
| longest | $837799$, $524$ steps |
| steps and peak for $27$ | $111$ steps, $9232$ |
The table agreed with running each number through to the end for $2000$ randomly chosen numbers.
Observation 2.2 even small numbers take long detours. $27$ climbs to $9232$ before returning to $1$ in $111$ steps: the rises and falls are wild, but not one number up to $10^6$ avoided $1$.
3. How It Shrinks on Average
Observation 3.1 combine an odd step with the halving that always follows. Taking odd $n$ to $(3n+1)/2$ and even $n$ to $n/2$, a step multiplies by roughly $1.5$ or $0.5$. If odd and even appear equally often, the geometric mean of the factor is
$$\sqrt{\frac{3}{2}\times\frac12}=\sqrt{\frac34}=0.866025 ,$$shrinking on average.
Computation 3.2: choosing $2000$ random numbers between $10^{20}$ and $10^{21}$ and running each $40$ steps, the geometric mean of the factor per step was measured as $\mathbf{0.864706}$.
Observation 3.3 in checking, I once measured with a bias. At first every starting number was forced to be odd, so the first step always grew, and the measurement was $0.878061$. It matched the prediction $0.878000$ of the heuristic with only the first step fixed at $1.5$, so after confirming the source of the bias the starting numbers were made random.
4. The Separator
Computation 4.1: changing $a$, the destinations of the numbers from $1$ to $10000$ were counted (exceeding $10^{12}$ counted as “escaped”).
| map | geometric mean of factor | destinations of $1$--$10000$ |
|---|---|---|
| $3n+1$ | $\sqrt{3/4}=0.866025$ | all in the cycle containing $1$ |
| $3n-1$ | $\sqrt{3/4}=0.866025$ | cycle of $1$: $3244$; of $5$: $3213$; of $17$: $3543$ |
| $5n+1$ | $\sqrt{5/4}=\mathbf{1.118034}$ | escaped $9274$; cycles $726$ |
The $726$ numbers under $5n+1$ that fell into cycles went to the cycle containing $13$ ($383$), the cycle of length $7$ containing $1$ ($256$) and the cycle containing $17$ ($87$).
Observation 4.2 this is the core: the separator is whether the multiplier $a$ is less than $4$. The geometric mean of the factor is $\sqrt{a/4}$: below $4$ it shrinks on average, above $4$ it grows: with $a=5$ more than nine tenths of the numbers from $1$ to $10000$ exceeded $10^{12}$. The difference between $3$ and $5$ is which side of $4$ they are on.
Observation 4.3 shrinking on average does not guarantee reaching $1$. $3n-1$ shrinks exactly as $3n+1$ does, yet the numbers from $1$ to $10000$ split into three cycles: shrinking only gives the heuristic that numbers do not grow without limit; how many cycles there are to fall into is a separate question. That only the cycle containing $1$ has been found for $3n+1$ does not follow from the way it shrinks.
Observation 4.4 beside Paper 195. Paper 195 showed that “stable” splits into six by what is perturbed: the Collatz cycle attracts in the sense of shrinking on average. But $3n-1$, which shrinks the same way, has three attracting cycles. Attracting and being the only one are different.
Observation 4.5 what is not said. The conjecture is not proved; checking up to $10^6$ says nothing beyond. That numbers under $5n+1$ really grow without limit is not proved either. What is shown is the count up to $10^6$, the formula and measurement for the average shrinking, and the difference when the multiplier changes.
5. Roll Call and Closing
- every number up to $10^6$ reaches $1$ (Computation 2.1): table and direct runs agree.
- mean steps $\mathbf{131.434555}$; longest $837799$ with $524$ (Computation 2.1): agreeing with known values.
- $27$ climbs to $9232$ and takes $111$ steps (Observation 2.2): a wild detour.
- geometric mean $\sqrt{3/4}=\mathbf{0.866025}$ (Observation 3.1): shrinking on average.
- measured $\mathbf{0.864706}$ (Computation 3.2): two routes.
- forcing odd starts biases it to $0.878$ (Observation 3.3): confirmed by the prediction $0.878000$.
- $5n+1$ gives $\sqrt{5/4}=\mathbf{1.118034}$ (Computation 4.1): $9274$ escape.
- the separator is the multiplier versus $4$ (Observation 4.2): the core.
- $3n-1$ shrinks the same way into three cycles (Observation 4.3): shrinking does not fix the destination.
- attracting is not the same as being the only one (Observation 4.4): Paper 195.
Closing: Tripling odd numbers and adding $1$, halving even ones, every number up to $10^6$ reached $1$. Because on average each step shrinks: combining an odd step, a step multiplies by $1.5$ or $0.5$, with geometric mean $\sqrt{3/4}=0.866025$, measured as $0.864706$ with random numbers. The separator is whether the multiplier is less than $4$: replacing $3$ by $5$ gives $\sqrt{5/4}=1.118034$, growing on average, and $9274$ of the numbers from $1$ to $10000$ exceeded $10^{12}$. And shrinking does not guarantee reaching $1$: $3n-1$ shrinks the same way and split into three cycles. Placed among the earlier papers. It is a cycle attracting in one of the senses of “stable” in Paper 195, but whether the attracting cycle is the only one is a separate question. To be honest: the conjecture is not proved; the numbers up to $10^6$ were counted and the average heuristic measured.
References
- Lagarias, J. C. (1985). The $3x+1$ problem and its generalizations. The American Mathematical Monthly.
- Lagarias, J. C. (Ed.) (2010). The Ultimate Challenge: The $3x+1$ Problem. American Mathematical Society.
- Terras, R. (1976). A stopping time problem on the positive integers. Acta Arithmetica.
- Yamagishi, Y. (2026). “Stable” Is Six Separate Words, and What Separates Them Is What You Perturb [Paper 195]. YOI: https://www.yamagishiyuuki.com/yoi/195; DOI: 10.5281/zenodo.22107170
Reference
Yamagishi, Y. (2026). The Collatz Map Shrinks on Average, but Replacing 3 by 5 Makes It Grow [Paper 621]. YOI: https://www.yamagishiyuuki.com/yoi/621
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 621. This paper is a preprint and has not been peer reviewed.