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While playing with numbers, I found that every Collatz sequence $n, T(n), T^2(n), \ldots, 1$ can be associated with a strictly decreasing sequence of integers.

The Collatz conjecture asserts that a sequence defined by repeatedly applying the Collatz function \begin{align*} T(n) = \begin{cases} (3n+1)/2 &\text{ if $n \equiv 1 \pmod{2}$, or}\\ n/2 &\text{ if $n \equiv 0 \pmod{2}$} \end{cases} \end{align*} will always converge to the cycle passing through the number 1 for arbitrary positive integer $n$.

Note that multiplying the $n$ by positive odd integer $a$ does not affect the result of the modulo 2 operation. By multiplying the Collatz function by an odd integer $a$, and tracking the $m = an$ rather than $n$, we get \begin{align*} S(m) = \begin{cases} (3m+a)/2 &\text{ if $m \equiv 1 \pmod{2}$, or}\\ m/2 &\text{ if $m \equiv 0 \pmod{2}$,} \end{cases} \end{align*} where each iterate $S^i(m) = a \, T^i(n)$.

Now we can choose a sufficiently large positive integer $A$ and track $m = 3^A n$. But we do a little trick. Instead of multiplying the $m$ by 3 in the "odd" branch, we just replace $3^A$ with $3^{A-1}$, and track the $3^{A-1}$ from that moment on (the effect is the same). We get the following algorithm:

enter image description here

It can be shown that every next $m$ is strictly less than the previous $m$. Since every next $m$ is smaller than its predecessor, we must hit $m = 1$ at the end. Since we track $m = 3^A n$, once the $m = 1$, then the $A = 0$ and $n = 1$. This implies that for arbitrary positive integer $n$, the sequence $n, T(n), T^2(n), \ldots$ leads to one. Note that once the $m = 3^A$, then the $n = 1$.

I am however stuck to show that there is always the sufficiently large $A$ for a given $n$. Is it possible to show this? I found out that the sufficiently large $A$ does not always exist for the $3x-1$ problem.

Example

The trajectory starting at $n=19$ with $A=9$ (termination at $m = 1$): $$\begin{matrix} n & m & A \\ \hline 19 & 373977 & 9 \\ 29 & 190269 & 8 \\ 44 & 96228 & 7 \\ 22 & 48114 & 7 \\ 11 & 24057 & 7 \\ 17 & 12393 & 6 \\ 26 & 6318 & 5 \\ 13 & 3159 & 5 \\ 20 & 1620 & 4 \\ 10 & 810 & 4 \\ 5 & 405 & 4 \\ 8 & 216 & 3 \\ 4 & 108 & 3 \\ 2 & 54 & 3 \\ 1 & 27 & 3 \\ 2 & 18 & 2 \\ 1 & 9 & 2 \\ 2 & 6 & 1 \\ 1 & 3 & 1 \\ 2 & 2 & 0 \\ 1 & 1 & 0 \\ \end{matrix}$$

DaBler
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  • This is a trick, but it won't work in case of cycles or divergent sequences, where $A$ will run out at some point – Collag3n Apr 23 '20 at 08:43
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    @Collag3n So it seems like another hopeless attempt :) – DaBler Apr 23 '20 at 09:25
  • Nice try anyway :) – Collag3n Apr 23 '20 at 12:43
  • @Collag3n I'm still thinking about it... This attempt clearly does not work in the case of a cycle. But is this really also the case of a divergent trajectory? Since the product $n , 3^A$ has to shrink at every step... – DaBler Apr 23 '20 at 14:12
  • The $3^A$ you add just hide the $n$ behind. If $n$ is part of a divergent trajectory, the $3^A$ you add to it will shrink down and at some point will disappear, but your original trajectory won't and will continue....with "naked" $n_i$'s. – Collag3n Apr 23 '20 at 17:29
  • Oh, I see. Okay, totally hopeless... – DaBler Apr 23 '20 at 19:49
  • Yes. $A$ must be at least the number of odd steps (or climb) in the trajectory (6 in your case). But for an infinite trajectory, you can't set $A$ large enough. – Collag3n Apr 23 '20 at 20:12

1 Answers1

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It suffers from the same pitfall as other representations which relies on the fact that the sequence reaches 1.

e.g. in the Collatz tree, you pick a number, and it does not matter if it seems to rise, in the tree it is a step closer to the root.

another one is the "inverse Collatz" representation of a number:

$7 = \frac{2^5}{3^5}\cdot 2^{(3+2+1+0+0)} - \frac{2^4}{3^5}\cdot 2^{(2+1+0+0)} - \frac{2^3}{3^4}\cdot 2^{(1+0+0)} - \frac{2^2}{3^3}\cdot 2^{(0+0)} - \frac{2^1}{3^2}\cdot 2^{(0)} - \frac{2^0}{3^1}\\ 11 = \frac{2^4}{3^4}\cdot 2^{(3+2+1+0)} - \frac{2^3}{3^4}\cdot 2^{(2+1+0)} - \frac{2^2}{3^3}\cdot 2^{(1+0)} - \frac{2^1}{3^2}\cdot 2^{(0)} - \frac{2^0}{3^1}\\ 17 = \frac{2^3}{3^3}\cdot 2^{(3+2+1)} - \frac{2^2}{3^3}\cdot 2^{(2+1)} - \frac{2^1}{3^2}\cdot 2^{(1)} - \frac{2^0}{3^1}\\ 13 = \frac{2^2}{3^2}\cdot 2^{(3+2)} - \frac{2^1}{3^2}\cdot 2^{(2)} - \frac{2^0}{3^1}\\ 5 = \frac{2^1}{3^1}\cdot 2^{(3)} - \frac{2^0}{3^1}\\ 1 = \frac{2^0}{3^0}$

it does not matter if 7 rises to 11. In it's representation, at each step, the exponent decreases, as well as the length of the representation.

Collag3n
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