I was messing around with the sequences of odd numbers in the Collatz conjecture, and (unsurprisingly) found a pattern. Basically, I was calculating the number of steps an odd number takes to reach an even number. Just to give you an example:
For the number 31: 31 -> 47 -> 71 -> 107 -> 161 -> 242 (31 takes 5 steps to reach an even number)
In a table, a pattern appears:
| Number of steps | Odd number |
|---|---|
| 2 | 3 |
| 1 | 5 |
| 3 | 7 |
| 1 | 9 |
| 2 | 11 |
| 1 | 13 |
| 4 | 15 |
| 1 | 17 |
If you were to plot this in a graph, it would look like this:
In which case, the biggest number I have found so far is 131 071 which takes 17 steps to get to an even number.
Doesn't this prove that, in a periodic way, the number of steps per odd number increases as the odd number gets bigger? In which case, if you get an infinitely big number, then it would take an infinite amount of steps to get to zero? Does any of this actually mean anything or is just a random pattern?
The numbers that set records are of the form $2^k-1$. Written in binary they consist of $k\ 1$'s When you do the first iteration you get $\frac {3(2^k-1)+1}2=\frac{3\cdot 2^k-2}2=2^k+2^{k-1}-1$ The last in binary is of the form $10$ followed by $k-1\ 1$'s. It will still be odd if $k \gt 1$. Each iteration on an odd number removes one of the trailing $1$'s, so $k$ iterations will get you to an even. Writing your result from $31$ in binary we get $$11111 \to 101111 \to 1000111 \to 1101011 \to 10100001 \to 11110010$$ You should be able to convince yourself that adding any multiple of $2^{k+1}$ will not impact the run of trailing $1$'s. You have shown that we can have an arbitrarily long run of odd numbers at the start, but it is still a finite run. To get an infinite run of numbers avoiding $1$ you either need to find a loop or find a sequence that never gets below some number.