pull down to refresh
An excellent exercise—I love this kind of challenge. Here is my answer: You have just described the Collatz conjecture, one of the most famous open problems in mathematics.
The rule is:
[
f(n)=
\begin{cases}
n/2 & \text{if } n \text{ is even}\
3n+1 & \text{if } n \text{ is odd}
\end{cases}
]
The question is precisely:
For every positive natural number (n), does the sequence eventually reach 1?
If it reaches 1, it necessarily enters the loop:
[
1\rightarrow4\rightarrow2\rightarrow1\rightarrow4\rightarrow2\rightarrow1\ldots
]
Does anyone have the answer? 🤔
No. 🧨🧨 As of September 2026, no one has managed to mathematically prove that all sequences reach the (1\to4\to2\to1) cycle, nor has a number been found that escapes it.
Why is it so difficult?
Because there appears to be a sort of balance between two behaviors:
When (n) is even, it decreases: (n\to n/2).
When (n) is odd, it increases: (n\to3n+1).
For example:
[
27\rightarrow82\rightarrow41\rightarrow124\rightarrow62\rightarrow31\rightarrow94\rightarrow47\rightarrow142\rightarrow\cdots
]
The sequence can climb very high before it starts to drop.
And here is the fascinating part: we can experiment with millions, billions, or vastly larger quantities of starting values and observe that they all end at 1, yet we still cannot prove that there isn't some gigantic number whose trajectory never reaches the cycle.
The open question
There are two possibilities:
A. All positive numbers end in:
[
4\rightarrow2\rightarrow1
]
B. There exists a positive number whose sequence:
grows indefinitely, or
enters a cycle other than (4,2,1). So far, we know of no counterexample.
Therefore, if someone were to find a number that did not reach the cycle (4, 2, 1), they would have solved the Collatz conjecture in a spectacular way.
And if someone were to prove that all numbers reach the cycle, they would also have solved one of the great open problems in mathematics.
I started enjoying it after it was recommended to me on Netflix; honestly, it makes me feel like a kid watching cartoons again—or seeing a live-action version of a man who shares my sense of humor, spirit, euphoria, and outlook on life, but in cartoon form. I didn't know there was a museum; I’d give anything to visit it.
Hey, TODAY🫵 I finished one of my surveys. Thanks to everyone for participating⚡⚡
Be free, accumulate ⚡💥 and don't trust anyone.
Hey, TODAY🫵 I finished one of my surveys. Thanks to everyone for participating⚡⚡.
The participants. @Machiavelli23 and @flat24 have earned the reward for this survey.
Be free, accumulate ⚡💥 and don't trust anyone.
Hey, TODAY🫵 I finished one of my surveys. Thanks to everyone for participating⚡⚡
Be free, accumulate ⚡💥 and don't trust anyone.
Thank you for participating ⚡⚡
Be free, accumulate ⚡💥 and don't trust anyone.
Thank you for participating ⚡⚡
Be free, accumulate ⚡💥 and don't trust anyone.
Thank you for participating ⚡⚡
Be free, accumulate ⚡💥 and don't trust anyone.
Thank you for participating ⚡⚡
Be free, accumulate ⚡💥 and don't trust anyone.
Thanks for sharing! Just seeing the picture makes me want to dive right in 🤣. The crystal-clear water looks so delicious 💧.
Now let's look at a step-by-step example using a number that takes quite a while to reach the cycle: 27.
The rule is:
🟢 If it is even → divide by 2.
🔴 If it is odd → multiply by 3 and add 1.
Example: start with 27
Step 0:
[ 27 ]
27 is odd:
[ 27\times3+1=82 ]
Step 1:
[ 82 ]
82 is even:
[ 82\div2=41 ]
Step 2:
[ 41 ]
41 is odd:
[ 41\times3+1=124 ]
Step 3:
[ 124\div2=62 ]
Step 4:
[ 62\div2=31 ]
Step 5:
31 is odd:
[ 31\times3+1=94 ]
Step 6:
[ 94\div2=47 ]
Step 7:
47 is odd:
[ 47\times3+1=142 ]
Step 8:
[ 142\div2=71 ]
Step 9:
71 is odd:
[ 71\times3+1=214 ]
And it continues:
[ 214\rightarrow107\rightarrow322\rightarrow161\rightarrow484\rightarrow242\rightarrow121\rightarrow364\rightarrow182\rightarrow91\rightarrow274\rightarrow137\rightarrow412\righta rrow206\rightarrow103\rightarrow310\rightarrow155\rightarrow466\rightarrow233\rightarrow700\rightarrow350\rightarrow175\rightarrow526\rightarrow263\rightarrow790\rightarrow395\ri ghtarrow1186\rightarrow593\rightarrow1780\rightarrow890\rightarrow445\rightarrow1336\rightarrow668\rightarrow334\rightarrow167\rightarrow502\rightarrow251\rightarrow754\rightarr ow377\rightarrow1132\rightarrow566\rightarrow283\rightarrow850\rightarrow425\rightarrow1276\rightarrow638\rightarrow319\rightarrow958\rightarrow479\rightarrow1438\rightarrow719\r [ 2158\rightarrow1079\rightarrow3238\rightarrow1619\rightarrow4858\rightarrow2429\rightarrow7288\rightarrow3644\rightarrow1822\rightarrow911\rightarrow2734\rightarrow1367\rightarrow4102\rightarrow2051\rightarrow6154\rightarrow3077\rightarrow9232\rightarrow4616\rightarrow2308\rightarrow1154\rightarrow577\rightarrow1732\rightarrow866\rightarrow433\rightarrow1300\rightarrow650\rightarrow325\rightarrow976\rightarrow488\rightarrow244\rightarrow122\rightarrow61\rightarrow184\rightarrow92\rightarrow46\rightarrow23\rightarrow70\rightarrow35\rightarrow106\rightarrow53\rightarrow160\rightarrow80\rightarrow40\rightarrow20\rightarrow10\rightarrow5\rightarrow16\rightarrow8\rightarrow4\rightarrow2\rightarrow1 ]
And finally:
[ \boxed{1\rightarrow4\rightarrow2\rightarrow1} ]
The surprising part 🤯
The starting number was only 27, but before reaching 1, the sequence hits a maximum value of:
[ \boxed{9232} ]
In other words, it starts at 27, climbs to 9232, and eventually ends up at 1.
That behavior is precisely one of the reasons why the Collatz conjecture is so difficult: just because a number eventually goes down doesn't mean it decreases continuously.
And that brings up the big question:
Could there be a number that, unlike 27, never manages to return and enter the (4\rightarrow2\rightarrow1) loop?
So far, we have no way to prove that such a number doesn't exist.