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.

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.

Since the thread is mostly jokes, here's roughly where the actual research stands.

Brute force: every starting number up to 2^71, about 2.4 x 10^21, has been checked and all of them fall into 4-2-1. That's Bařina's 2025 run.

A second loop is nearly ruled out too. Hercher (2023) showed there's no non-trivial cycle built from 91 or fewer up-then-down blocks, and the length floor for any cycle that could still exist is on the order of 100 billion terms. So a counterexample almost certainly isn't a small loop nobody noticed. It would be one trajectory that just keeps climbing forever.

Tao's 2019 paper is the closest anyone has gotten to ruling that out. He proved that almost all starting numbers (in logarithmic density) eventually drop below any function you name that goes to infinity, even something as slow as log log log log N. It doesn't cover every number, but it means a runaway trajectory, if one exists, is extremely rare.

Worth knowing why this is genuinely hard and not just unsolved: Conway proved in 1972 that the generalized version, same setup but other multipliers, is undecidable. There's no single algorithm that settles every problem of this shape. The plain 3x+1 case might still be provable, but that's the reason "just analyze the recursion" keeps stalling out.

So the computational angles are basically spent. What's left is number theory, and the 700k looks safe.