Understanding the Collatz conjecture — the 3n+1 sequence step by step
The Collatz conjecture — also known as the 3n+1 problem — follows a tiny rule: halve even numbers, and for odd numbers multiply by 3 and add 1. The astonishing part is that no matter where you start, you seemingly always end up at 1. We trace this with the starting number 6 (8 steps) and take a look at the infamous 27, which needs 111 steps. A classic number-theory enrichment topic from Grade 9 onwards.
Quick answer
The Collatz conjecture is one simple rule: if a number is even, halve it (n/2); if it's odd, compute 3n + 1. Repeat, and — so the conjecture claims — you always reach 1. Example: 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1, which is 8 steps. To this day no one has proven it.
At a glance
| Example | 6 → 3 → 10 → … → 1 |
|---|---|
| Rule | even → n/2, odd → 3n+1 |
| Steps | 8 |
| Peak value | 16 |
| Stopping time of 27 | 111 steps (peak 9232) |
| Grade level | Grade 9+ / Enrichment |
Worked example: 6 → 3 → 10 → … → 1
We start at 6 and apply the rule until we reach 1.
How to compute a Collatz sequence
These steps work for any positive whole starting number.
Step 1 · Start
n = 6Starting number. 6 is even, so we halve it.Step 2 · ÷2
6/2 = 3Even → divide by 2. The result 3 is odd.Step 3 · 3n+1
3·3+1 = 10Odd → multiply by 3 and add 1.Step 4 · continue
10 → 5 → 16 → 8 → 4 → 2 → 1Keep applying the rule until 1 appears (peak value 16).Step 5 · Stopping time
= 8 stepsIt takes 8 steps to get from 6 down to 1.
Why the sequence (probably) always ends at 1
Halving shrinks a number, while the 3n + 1 step grows it at first — but the result is always even and is immediately halved again. On average shrinking wins: across many steps each odd number is effectively multiplied by about 3/4 rather than 3, so the sequence tends downward over time. Once you hit a power of two such as 16, 8, 4, 2, everything slides straight down to 1. A complete proof, however, still does not exist.