The Simplest Impossible Problem

A 7-year-old can understand this problem which completely baffles mathematicians.

[HeroHero]

Officer: Is there a problem?
Me: *Shows this video*.

[DiodeGoneWild]

After hearing "... INVENTED THE PROBLEM..." I realised what the humankind is all about :).

[MrBeanBag]

me: 6969696969696969696969696969
generator: this is the seventh time this input has been submitted

[theSUICIDEfox]

I got 99n+1 problems but it ain't 4 2 1.

[sebbes]

So all numbers below 2*10^60 has been checked, and all of them eventually reached 1.
That means that all larger numbers only need to reach ANY of the previously tested numbers to prove that that number too eventually will reach 1.
This is not a solution for the problem but it can speed up the testing of numbers (which I'm sure that most mathematicians have already thought of)

[BubuSnow]

Was wondering... What if the conjecture is correct but there is no way to prove it? Could it be possible to prove something is impossible to prove?

[ibrozdemir]

what?.. i was listenning the background music whole time

[faizalntd]

Wow surely everyone in the comment know math more than mathematician mentioned in the video

[michaelalafin]

I could easily write down the proof to this but i really need to go feed my cat.

[ataba]

to anyone that says that: "i don't see the problem" the problem is to prove that n will at some point will hit the cycle 4->2->1 and if you say that is will hit 2^x you also have to prove why is does

[issou]

I wanted to check the comments to see interesting things, what we get :
-Comments saying "how is that a problem".
-People who think they can solve the problem (and do what every mathematician failed to do in the last century) in a 2 lines comment on youtube.

I guess the real mystery is why I expected something from the comments at the first place...

[alvinpalmgren]

2:00 **Immediately checking 5 · 2^60 + 1**

[utetopia]

So you can find the sequences in excel:
Start with a column in a sequence (it could be 30, it could be 300), in column A, so that A1=1, A2=2. This is your reference column. In cell b1, put this formula:

=IFERROR(IF(A1=1, "", IF(ISODD(A1), (A1*3)+1, (A1/2))), "")

This determines if cell A1 =1. If it is, it ends the sequence, and creates a clear cell. The clear cells are important for visual reasons. If its not 1, it determines whether that cell is odd or even, and makes the calculation based on the problem (a1*3)+1 for odd, or (a1/2) for even.

Now fill that down for all your numbered rows, and fill right for all your chosen columns as far as you want.

You end with all the sequences of the numbers for your specified rows. (30 or 300! Whatever you chose)

Then you see some really interesting properties. In some cases, like for the number 31, the next number is 94. 94 is the 6th number in the sequence of 27, so from then on, both 31 and 27 are exactly the same.

The reason that these high sequence numbers dont go off into infinity, is because you end up with a sequence of numbers within that sequence that are even, which halves each time. For example, in the number 31, the 84th number and onward are 976, 488, 244, 122 and then 61. You could then argue that those numbers (27 and 31) become exactly the same as the sequence 61 produces (and even the preceding numbers). So as you can see, the sequence drops dramatically because of the number of even numbers in a row. Because 31 and 27 are the same after 6 numbers (in 27), that sequence also falls into 27, and the reduction occurs quickly.

Each number will create a sequence that falls within the sequence of another number.
So no infinity.

Edit: Check out these sequences from 3377, 3378, and 3379. They align at 1426, and follow the same sequence, making these 3 numbers exactly the same as the sequence 1426 makes. They also have exactly the same amount of numbers in their sequence:

3377 3378 3379
10132 1689 10138
5066 5068 5069
2533 2534 15208
7600 1267 7604
3800 3802 3802
1900 1901 1901
950 5704 5704
475 2852 2852
1426 1426 1426
713 713 713
2140 2140 2140
1070 1070 1070
535 535 535
1606 1606 1606
803 803 803
2410 2410 2410
1205 1205 1205
3616 3616 3616
1808 1808 1808
904 904 904
452 452 452
226 226 226
113 113 113
340 340 340
170 170 170
85 85 85
256 256 256
128 128 128
64 64 64
32 32 32
16 16 16
8 8 8
4 4 4
2 2 2
1 1 1

Edit 2: Maybe the question should be "Would a (virtually) infinitely high number create a finite number of numbers in a sequence?" The numbers will always decrease to 4, 2, 1... but how high does a number have to start to end up, not with the numbers going off into infinity, but having an infinite number of numbers in a sequence?

[illiteratethug]

I honestly can't see how this is a problem.

[chinipapa]

I've edited the comment... Now it'll be complete unexplainable section .

[RemarhBsoul]

I've also invented alot of problems during my life

[ridwaan]

The problem isn't that it happens, it's that we can't prove that it happens.

[tergy]

I know anotherone:
x-x+1, it will always gets to one.

[gilotyna]

Video: The problem is: how can we prove this algorithm will eventually yield infinite sequence 4, 2, 1?
Half the comments section: What is the problem? What is the question? The answer is 0. This is obvious wth.

[swaystar]

Youtube has been recommending me this for 3 years now