Mathematical mysteries: Hailstone sequences (2024)

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

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...

See the video "The Simplest Impossible Problem" https://www.youtube.com/watch?v=m4CjXk_b8zo
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,...