Mathematical mysteries: Hailstone sequences


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.


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,...

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,...

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

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.

0 is not in N, 0 is in the whole number set but not the natural number set.

But 1x3 is 3, plus 1 is 4 so then you get to 4 again

It's a repeating sequence. The system is correct

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

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

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.

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...

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

Challenge accepted.

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.

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.

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.

i was reading these comments to see if anyone else had seen it but its not always 80 40 20 10 5 16 8 4 2 1 while i found that with alot of numbers there are some cases like 96 that come out as 24 12 6 3 10 5 16 8 4 2 1 always 10, 5, ect but before 10; 20 or 3 least form what ive seen :D

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?

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

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 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

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

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

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.

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

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

I recently came across the Collatz conjecture and have spent some time investigating it. Although I haven’t found a proof of the conjecture I did wonder if a process similar the ‘sieve of Eratosthanes’ (used to isolate the prime numbers) might be used to isolate any integers that didn’t eventually home in on 1. Previous postings have established that we only need be concerned with odd numbers as even numbers divide down to 1 or an odd number so the sieving procedure is as follows:-

Step 1: List all the odd integers from 1 to infinity in column 1.
Step 2: Create a new column, column 2, and record the result of the function 3*n + 1 for each odd integer n in column 1 and reduce the calculated value to its smallest odd number. This obviously takes a long time to do as the column is infinitely long though these are the only calculations that have to be performed! Numbers such as 5, 21, 85 etc. have fallen through the sieve as they go straight to 1 in this step.
Step 3: Create a new column, column 3. Work down column 2 looking at each row in turn. Take the value on row n, column 2 and find the same number in column 1 then record the value found in column 2 at this row, in column 3 of row n. More numbers have now fallen through the sieve leaving fewer odd integers to deal with.
Step 4: Effectively repeat step 3 by forming a fourth column, and look up the values from column 3 in column 1 and record the values found in column 2 in column 3 of the initial row.

I tried to include a table to show how the above process works but, unfortunately, getting it to work in these comments seems to be beyond me at the moment, sorry.

I have tried this process in Excel, with odd integers up to 59995, and the first number which didn’t reach 1 was 703 but this was because it called for a number further down the list than 59995. Just less than 60% of the list went to 1 and this was achieved by column 17 which is much quicker and requires much less computation than the straightforward step by step method.

One thing I think we can conclude is that if there is a smallest number which forms a loop or which never reach 1, then this number must never reach a value in its sequence that is smaller than itself, as this value would, by definition, reduce to 1.

I have also derived formulae which will calculate all the numbers that would eventually lead to any given odd integer. I will post details of this later.

This looks vulnerable to induction.

It looks very much like all sequences starting from an n_odd number contain the sequence for (n-1)_odd in the tail, and n = 3 ends in 4,2,1,4,2,1.

At some point a sequence starting from n_even will either reach an odd number part way through, and hence end in 4,2,1,4,2,1 by the above idea, or it will remain even and hence end in 4,2,1,4,2,1. E.g. n=8. = > 8, 4, 2, 1, 4, 2, 1,...

Oops, I meant to write that it seems n_odd sequence contains (n-2)_odd subsequence.

I worked out new series which always ends with sequence of 9, 4, 6 endlessly , i tried it on 25 numbers it works consistently. I want to take some help to prove sequence i have figured out will work for any number >3 .