I have found that odd numbers of the form $2^a(2b-1)-1$ will become even at $3^a(2b-1)-1$, after $a$ iterations of $n\to\frac{3n+1}{2}$, then falling to some odd $2^c(2d-1)-1$. If $a\equiv b\pmod 2$, it will be divided by 2 just once. For some combination of $a$ and $b$, $c$ will be even $\frac{1}{3}$ of the time, and odd $\frac{2}{3}$ of the time. I have not been able to prove this mathematically since I am not sure why this is the case, so an intuitive explanation would really be helpful.


There are more than that (see my comment here): How to prove this inequality: $f(2h-1)≤\frac{3h-1}{2}$

Some discussion here too with the same results of yours: http://mathhelpforum.com/number-theory/276791-collatz-conjecture-investigation.html

For the 33%-66% repartition:

As you said, 1 on 2 of them (numbers of the form $4j+3=\{3, 7, 11,...\}$) is only divisible once (They have $c>1$). When you look at numbers expressed as $2^c(2d-1)-1$, you'll see that they are distributed that way:

Odd c: $\frac{1}{4}(c=1)+ \frac{1}{16}(c=3)+ \frac{1}{64}(c = 5)+ \frac{1}{256}(c = 7)...= \sum\limits_{k\geq 2}\frac{4}{4^k}=4\sum\limits_{k\geq 2}\frac{1}{4^k}$

Even c: $\frac{1}{8}(c=2)+ \frac{1}{32}(c = 4)+ \frac{1}{128}(c = 6)...= \sum\limits_{k\geq 2}\frac{2}{4^k} = 2\sum\limits_{k\geq 2}\frac{1}{4^k}$

where the fractions are "fractions of all naturals" (ex: for c=1 which is "all numbers of the form $4j+1$" they obviously represent $\frac{1}{4}$ of "all naturals". for c=2, numbers of the form $8j+3$, this is $\frac{1}{8}$...). Note that $(c=...)$ is only a comment to show which fraction correspond to which $c$.

As you can see, there are 2 times more odd $c$ than even $c$.

Note: if you want to prove the whole thing from start, you'll need to use the fact that $3n+1$ is even and 1 on 2 even numbers (in general but also as a result of apllying that $3n+1$) gives an odd number when divided by 2, so you will apply $3n+1$ again on 1 on two of them...and so on. You'll end up with the fractions above.


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