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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.

Gil Costa
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    This is true; at best you can divide your number by 2 every step, so after N steps x must be at least x/2^N – jschnei May 25 '17 at 22:36
  • Thanks. We can look at it as a proof that the Collatz conjecture is True for powers of 2, right? – Gil Costa May 25 '17 at 22:59
  • Yes it's true for all powers of 2. And in fact, for any power of 2 multiplied by any odd number for which it is true, multiplied by any powers of 2. This is why people often study only the odd number sequences. – it's a hire car baby May 27 '17 at 13:24

1 Answers1

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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 $1 \ge \frac{n}{2^k}$.

Caleb Stanford
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