The collatz conjecture states that every number eventually reaches $1$ under the repeated iteration of $$ f(n) = \begin{cases} n/2, & \text{if $n$ is even} \\ 3n+1, & \text{if $n$ is odd} \end{cases}$$

When writing for some number $n$ all the steps required to reach 1 as one big fraction, e.g. $\frac{\frac{3n+1}{2}*3+1}{2}$, simplifying the expression and solving for $n$, one obtains the following representation for a number:

$$ 89 = \frac{2^{21} - 345365}{3^{9}}$$ $$ 90 = \frac{2^{13} - 902}{3^{4}}$$ $$ 91 = \frac{2^{59} - 70586240746870895}{3^{33}}$$ $$\dots$$ $$ 500086 = \frac{2^{62} - 798231474550179022}{3^{26}}$$ $$ 500087 = \frac{2^{62} - 798223848952694035}{3^{26}}$$ $$ 500088 = \frac{2^{62} - 798216223355209048}{3^{26}}$$

In general: $$ n = \frac{2^{r} - s}{3^{t}}$$

Although I haven't tried much yet, I'm quite sure the collatz sequence length can be read out from this representation (because it's literally the whole lengthy operation just written in another way).

My question(s):

  1. Is each representation for some number $n$ unique? Or could one choose different $s$ when also choosing different $r$ and $t$?
  2. For this question I assume three things (if wrong ignore the question): these representation are unique, the cycle length can be read out from this representation, and that there is an algorithm which is not $f$ and results in this representation of a number. Couldn't one than simply prove the collatz conjecture because that cycle length could be calculated this way for any $n$?
  • $\begingroup$ see the response and the last comment on it here: math.stackexchange.com/questions/2711060/… $\endgroup$ – Collag3n Oct 3 '18 at 17:38
  • $\begingroup$ Unsure of what you mean by cycle length? Do you mean potential internal cycles of one number represented by its graph? I have studied some numbers this way and they all seem to be unique. Well there might be a method that looks similar to this: $3n+1+2^x$ I don't remember wether this was the one I studied but anyway, the point is to find out wether the expression reaches pow2 and solve the conjecture that way. But first proving that such an expression is similar to the Collatz. $\endgroup$ – Natural Number Guy Oct 4 '18 at 0:16

With no information about $r,s,t$ the representation is far from unique. Starting from an arbitrary integer $n$, we may choose any power $3^t$, any power $2^r$ greater than $n \cdot 3^t$, and just set $s$ to equal the leftover $2^r - n \cdot 3^t$.

For example, starting with $90$, we may choose $3^3$ and $2^{12}$ to get the representation $$90 = \frac{2^{12} - 1666}{3^3}.$$

Of course, the representation we get from the Collatz calculation does not just have an arbitrary $r$, $s$, $t$: we could be more precise and expand your representation to $$ 90 = \frac{2^{13}-902}{3^4} = \frac{2^{13}-2^9-3^1\cdot 2^6-3^2\cdot2^4-3^3\cdot 2^1}{3^4} $$ from which we may read out the entire Collatz sequence for $90$.

But it is not at all obvious that such a representation exists for any integer...


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