About the Collatz conjecture:
Let Steps be: Number of steps, S, that a "counting number", n, takes to reach 1 - sometimes referred as the stopping time (of n);
Example:
5 take 5 steps to reach 1 (note: the only number with S = n) Steps(5) = 5
Which is the highest number 'with' S steps? Example: what is the highest number 'with' 10 steps?
The answer is n = 2^S
example: n = 2^10 n = 1024 The highest number 'with' 10 steps is 1024. There are no n > 1024 with 10 steps.
Disprove this. Please? __ {I dare you!} :-)
notes: S = ln(n)/ln(2) ; only gives the steps of powers of 2, obviously.
Example: the number 4194304 'has' 22 steps. It is the highest number 'with' 22 steps.
8388608 has 23. No bigger number has 23. 16777216 has 24. It's the highest number 'with' 24 steps. __ Ln function goes to infinity; For every Counting Number we can find the power of 2 that corresponds to the 'collatz number' that has lm(n)/ln(2) Steps. __ What is the highest number with exactly 23 Steps? j= 1023 2^j= 89884656743115795386465259539451236680898848947115328636715040578866337902750481566354238661203768010560056939935696678829394884407208311246423715319737062188883946712432742638151109800623047059726541476042502884419075341171231440736956555270413618581675255342293149119973622969239858152417678164812112068608
=> the number 89884656743115795386465259539451236680898848947115328636715040578866337902750481566354238661203768010560056939935696678829394884407208311246423715319737062188883946712432742638151109800623047059726541476042502884419075341171231440736956555270413618581675255342293149119973622969239858152417678164812112068608 takes 1023 Steps to reach 1.
If $S(n)$ is the number of steps that $n$ takes to terminate, then it is true (as has been commented) that the maximum $n$ such that $S(n) = k$ is $2^k$.
You can prove this because if $f(n)$ is $n$ after one step,$f(n) \ge \frac{n}{2}$. So if $S(n) = k$, then $f(f(\ldots(f(n)) = 1$ which implies $1 \ge \frac{n}{2^k}$.