Mathematical mysteries: Hailstone sequences

Issue 1

Here’s a little game to play. Starting with any positive whole number $n$ form a sequence in the following way:

  • If $n$ is even, divide it by $2$ to give $n^\prime = n/2$.
  • If $n$ is odd, multiply it by $3$ and add $1$ to give $n^\prime = 3n + 1.$

Then take $n^\prime $ as the new starting number and repeat the process. For example, $n = 5$ gives the sequence

  \[ 5, 16, 8, 4, 2, 1, 4, 2, 1,... \]    

and $n = 11 $ gives the sequence

  \[ 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1,... . \]    

Sequences formed in this way are sometimes called hailstone sequences because they go up and down just like a hailstone in a cloud before crashing to Earth. However, it seems that all hailstone sequences eventually end in the endless cycle

  \[ 4, 2, 1, 4, 2, 1. \]    

The ones for $n=5$ and $n=11$ both do, though other values for may $n$ generate a very long sequence before the repeating cycle begins. For example, try starting with $n = 27.$ You can do this by hand, or use our hailstone calculator: enter any positive whole number and the hailstone sequence will be returned.

The question is whether every hailstone sequence eventually settles on the 4, 2, 1 cycle, no matter what starting value you use. Experiments certainly suggest that they all do. Computers have checked all starting values up to 5 x 260, a number that is 19 digits long, and found that the 4, 2, 1 cycle eventually appears. The trouble is that nobody has been able to prove that this is the case for all sequences. This open question is known as the Collatz conjecture after the mathematician Lothar Collatz, who first proposed it in 1937. It's amazing that such an easy recipe for forming sequences leads to a question even the best mathematicians haven't been able to answer yet.


Nice video on this problem

See the video "The Simplest Impossible Problem"
on the channel Tipping Point Math.

If there is a counterexample to Collatz

If there is a counterexample to the Collatz conjecture there is a minimum counterexample and it must be equivalent to 3 mod4.

Goes into an infinite loop

Goes into an infinite loop and hangs when asked to evaluate 0.

My mistake. Zero is not a

My mistake. Zero is not a positive number, so is already excluded from the conjecture. The problem is with validation of the HTML form. It allows any number that can be typed, rounding and making it positive. Seems like it should reject anything but a sequence of digits that are not all zeros.

Yes. The rules says 0' = 0.

Yes. The rules says 0' = 0. So, this needs a special case. And the conjecture should mention the special case! :)

Simple version?

If you change the steps to an much easier version:
if even divide by 2.
if odd add one.
Then you clearly always will end opp with 1. I`m not sure though how to write a proof of this, but you will always end up with the sequence (for n bigger than 2) 4,2,1,
Somehow related?

this is easy to prove

Consider that the n+2 iteration of the function is always less than the nth iteration.

the n+2 iteration is not

the n+2 iteration is not always less than the nth iteration, a mere look at the ending sequence shows that it can at least be equal:
\[ 4, 2, 1, 4, 2, 1. \]

in other sequences, the n+2 is actually bigger than the n.

That's because it has a

That's because it has a different ending cycle. Instead of having a 4,2,1,4,2,1,... cycle, it just has a 2,1,2,1,2,1,... so the original commenter is still correct

Result for any number

Whatever number you enter, before the first 4 appears, the previous 6 numbers are always same (40,20,10,5,16,8). If it is an even number, it is preceded by 26,13 and if it is odd, then 80 and 160.


Not necessarily. Any power of 2 will just be devided by two over and over again and reach 16 without going through 5 first.


The number is divided by 2. If the resulting number is not a whole number, it is multiplied by 3 and 1 is added. Otherwise it is divided by 2 again.

the mother function of 3n+1

there i s function that when you let n=1 the collatz problem appears. but also, when N=-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8...,infinity. as you guys can see, if we solve 3n +1, we will solve all of them. with the function is easy to find and loo of every roblem. why all numbers top at 1, is just because there are not positive numbers less than 0.

Something to think about...

It would be very interesting to know whether or not "stopping length" as a function of n exhibited any regularity, and if so, of what nature? If you plotted the "stopping length" of each sequence against the unique n that generated it, what would the graph look like?

Maybe someone with some programming ability can try this out...

Good idea!

Challenge accepted.

There is an obvious pattern

Try 55, 115, 175, 235. All these have something in common

List of Lengths

My lengths are defined by stopping once a 4 is reached.

n=5 len=3
n=6 len=6
n=7 len=14
n=9 len=17
n=18 len=18
n=25 len=21
n=27 len=109
n=54 len=110
n=73 len=113
n=97 len=116
n=129 len=119
n=171 len=122
n=231 len=125
n=313 len=128
n=327 len=141
n=649 len=142
n=703 len=168
n=871 len=176
n=1161 len=179
n=2223 len=180
n=2463 len=206
n=2919 len=214
n=3711 len=235
n=6171 len=259
n=10971 len=265
n=13255 len=273
n=17647 len=276
n=23529 len=279
n=26623 len=305
n=34239 len=308
n=35655 len=321
n=52527 len=337
n=77031 len=348
n=106239 len=351
n=142587 len=372
n=156159 len=380
n=216367 len=383
n=230631 len=440
n=410011 len=446
n=511935 len=467
n=626331 len=506
n=837799 len=522
n=1117065 len=525
n=1501353 len=528
n=1723519 len=554
n=2298025 len=557
n=3064033 len=560
n=3542887 len=581
n=3732423 len=594
n=5649499 len=610
n=6649279 len=662
n=8400511 len=683
n=11200681 len=686
n=14934241 len=689
n=15733191 len=702
n=31466382 len=703
n=36791535 len=742
n=63728127 len=947
n=127456254 len=948
n=169941673 len=951
n=226588897 len=954
n=268549803 len=962
n=537099606 len=963
n=670617279 len=984

working on it....

Can you find out if the frequence of odd and even "len"s is 50% or not?

Yes you can know! And the

Yes you can know! And the answer is that there are always more even numbers in the sequence. If you assume the 50% hypothesis to be correct, it doesn't hold up. This is because if you have an odd number, the next term will always be even. This is because an odd number times an odd number (odd n times 3) is ALWAYS odd. That is true because if you multiply odd numbers, you can factor each number out, and 2 will not be a factor. This means that the resulting number cannot be divisible by 2 either, meaning it is odd. Anyways, if we assume that the 50% hypothesis is true, we know that an odd number will always be followed by an even one. So if the 50% hypothesis holds up, then every even number would have to precede an odd one. That doesn't hold up because if something is divisible by 4 for instance, you will be able to divide by two twice before you reach an odd number. Since the final cycle is always 4,2,1, and includes 2 even numbers and only one odd, there are always more even numbers than odd ones.

code error

i tested you program and it failed for number 5
the sequence for 5 is 5, 16, 8, 4, 2, 1 definitly not 5, 16, 8, 4, 2, 1, 4, 2, 1,...

Um not wrong

5 DEFINETLY DOES go 5,16,8,4,2,1,4,2,1,4,2,1,4,2,1,4,2,1.... The issue is that there are only 6, 4 or 3 numbers there that matter (depending on how you count)
You are either counting the 5,16,8 (so 3), the 5,16,8,4 (so 4, counting until you reach a 4) or 5,16,8,4,2,1 (so 6, counting until the repetition would start)

The sequence NEVER ENDS, but all (needs proof) eventually repeat 4,2,1,4,2,1 and so each sequence, even for the number 4 is infinitely long, we just don't care about after the repetition.

Perhaps you should re-read the article, and consider the original commentators work. The reason you are disagreeing seems as simple as how are you counting. You each are counting in a different way, yet seem to be saying the exact same thing.

The question isn't whether definitionally if 0 is or is not a natural number, you have to define whether it is and then do the math accordingly. I always asked my professors if they considered 0 in N or not for their classes. The guy you are saying is wrong, just defined his algorithm differently than you did. It isn't that you are right and he is wrong, or vice versa, it is that you both defined what you are looking for differently, but still validly.

Same holds for Fibonacci, does Fib start at 0 or at 1? If 0 is in N than the explicit formula for fibonacci is also different, yet they all are the same if you change the initial definitions to match.

code error

when you get to 1 (which is odd) the next result is 4, 2, 1....

I think you missed the point

Try reading the article next time...
It seems from experiment that such a sequence will always eventually end in this repeating cycle 4, 2, 1, 4, 2, 1,...


This is also called the Collatz conjecture. This is the name of the unsolved problem. Perhaps you should mention this in the article


It's also called: The Ulam sequence