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Let the Collatz function go $(3x+1)/2$ for odd numbers and $x/2$ for even.

Now running the function, write a $1$ every time you hit an odd number and a $0$ every time you hit an even number. By this means you can record a function $T$ from your domain into binary numbers. $T(x)$ is sometimes called the Parity Vector of a number's Collatz sequence. For example the number $1$ follows the orbit $1,2,1,2,\ldots$ so it writes the binary number $T(x)=\ldots0101_2=-\frac13$

$T$ is a 2-adic isometry $\Bbb Z_2^\times\to\Bbb Z_2^\times$ having the fixed points $0$ and $-1$.

But if we write a $-1$ instead of a $1$, I think by topological conjugacy we still get an isometry and it still has two fixed points. Zero is obviously still a fixed point, but what is the other fixed point of the new function?

My Attempt

Not part of the question - only provided as evidence of my own attempt in line with site policy. I have a hunch this needs to be attacked by somehow writing an equation with two copies of Newton's method, one on each side, and at each step, asing whether $0$ or $-1$ is the solution to the next digit but I'm unfamiliar with even basic applications of Newton's method.

Edit

I did a bit of algebra and I think seek the number such that $T(x)=-x$, i.e. its Collatz Parity Vector is itself negated. Also, heuristically the numbers $n$ among $-\frac13\Bbb N$ are good candidates, in particular ones satisfying $3n\equiv\{1,2\}\pmod 3$

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  • $\begingroup$ By $\Bbb Z_2^\times$, do you mean the 2-adic integers? In the 2-adics, "$-1$" is $\dots 1111$. What does it mean to write this infinite sequence of digits "every time you hit an odd number"? Are you trying to describe an ordinal sequence (i.e. a sequence indexed by some higher ordinals, not just natural numbers)? And what "Newton's method" are you referring to? Usually that name refers to an approximation scheme for finding roots of equations, but I fail to see how it is applicable here. $\endgroup$
    – Paul Sinclair
    Jun 6 at 14:09
  • $\begingroup$ @PaulSinclair actually I mean 2-adic units but it works for all the integers too. What writing the infinite sequence of digits means is: give the p-adic number $-1$ a one-character symbol and write that as if writing a binary number. Or in other words write $\sum_{k=0}^\infty2^ka_k:a:k\in\{0,-1\}$ so the sequence $a_k$ is the sequence you write. The function $T$ is also defined here: en.wikipedia.org/wiki/Collatz_conjecture#2-adic_extension and unfortunately there it is called $Q$. I have the fixed points of $T$ and the question asks for the second fixed point of $-T$. $\endgroup$
    – it's a hire car baby
    Jun 6 at 14:54
  • $\begingroup$ I assume you meant $a_k \in \{0. -1\}$, so you are talking exactly about the function $-T$. $\endgroup$
    – Paul Sinclair
    Jun 6 at 15:33
  • $\begingroup$ @PaulSinclair yes,that's right. Please let me know even if you have a think about it and get nowhere. The bit in my final edit is probably a big clue because it is probably true that if $X=\{x\in-\frac13\Bbb N:3n\equiv\{1,2\}\pmod 3)\}\setminus\{\frac13\}$ then $-T(X)\subset X$. In fact this particular statement would prove the Collatz conjecture if shown to be true. $\endgroup$
    – it's a hire car baby
    Jun 6 at 15:49

1 Answer 1

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Let $C(x) = \begin{cases}\frac x2&x\text{ is even}\\\frac{3x+1}2&x\text{ is odd}\end{cases}$ be the Collatz function. The Collatz sequence of $x$ is then the sequence $x, C(x), C^2(x), C^3(x), \dots$, which is the Collatz sequence of $C(x)$, with $x$ prepended. In particular, the Collatz sequence of $2x$ is the Collatz sequence of $x$ with $2x$ prepended. From this, it follows that $T(2x) = 2T(x)$. Multiplying by $2$ just shifts both the bit sequences of $x$ and $T(x)$ to the left by one bit.

Thus if $-T(x) = x$, then $-T(2x) = -2T(x) = 2x$. That is, if $x$ is a fixed point of $-T$, then so is $2x$. So if $-T$ has any fixed points other than $0$, it has infinitely many of them.

It also goes the other way: if $x$ is an even fixed point of $-T$, then $\frac x2$ is a fixed point as well. Thus if $-T$ has non-zero fixed points, one of them will be odd. So suppose that $x$ is an odd solution to $T(x) = -x$. To negate in the 2-adic integers, you find the rightmost $1$ bit, then flip all bits to its left (the "twos-complement"). Suppose $x$ has only finitely many $0$ bits. I.e., it is an ordinary negative integer. Then $-x = T(x)$ will have only finitely many $1$ bits. This means its Collatz sequence is eventually always even. But if $x \ne 0$, this cannot be, as the first entry after all odd entries are finished, would have to be divisible by $2$ infinitely many times, meaning its 2-adic representation is all $0$ bits, and thus that the number is $0$. But $C(y) = 0$ has only one solution: $y = 0$, so this is impossible.

Next suppose that $x$ has only finitely many $1$ bits. I.e., it is an ordinary positive integer. This means that its Collatz sequence is eventually always odd. Again look at the first entry after the last even entry in the sequence. This Collatz sequence for this number will be strictly increasing. Thus its existence would disprove the conjecture. As the conjecture has not been disproven, we can be sure no such number is known. On the other hand, I am no expert on the Collatz conjecture, but as far as I am aware, it has not been proven no such increasing sequence exists. So it is still a possibility.

So if the conjecture is true, then the only possible non-zero fixed points of $-T$ would have to be strictly 2-adic, not in the real integers.

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  • $\begingroup$ $2x$ is not a unit. Technically, neither is $0$ but I included it for completeness. You are right that for every solution $x\in\Bbb Q_2$, we have that $2^zx:z\in\Bbb Z$ is also a solution. But there is a solution in $\Bbb Z_2^\times$ because $T$ is topologically conjugate to $-T$ over $\Bbb Z_2^\times$. $\endgroup$
    – it's a hire car baby
    Jun 6 at 18:14
  • $\begingroup$ In fact a little more can be said. The periodic points of $-T$ over $\Bbb Z_2^\times\cup\{0\}$ are, like the periodic points of $T$, enumerated by the Lyndon words over an alphabet of two words (given here: oeis.org/A001037 ). This means there are two fixed points. Only I can't find the other one. $\endgroup$
    – it's a hire car baby
    Jun 6 at 18:20
  • $\begingroup$ I haven't fully parsed what you write about finitely many ones and zeroes but at the moment you don't appear to have catered for the case in which there are infinitely many both ones and zeroes, as would be the case for an number such as any element of $X=\{x\in-\frac13\Bbb N:3n\equiv\{1,2\}\pmod 3)\}\setminus\{\frac13\}$ whose tail would end $\overline{01}_2=-\frac13$ $\endgroup$
    – it's a hire car baby
    Jun 6 at 18:35
  • $\begingroup$ I see you write if $T(x)$ ended in infinitely many zeroes or ones, then $x$ would not be a natural number. This is also correct. For $T(x)$ to end in all ones, this is the Collatz sequence of an $x$ which eventually converges to the fixed point $-1$. Only a negative ternary rational can do that (or a multiple of a power of two thereof). Finally, if $T$ ends in all zeroes, again this is a Collatz sequence which converges to $0$ and only a negative ternary rational can do that also. (Although some negative ternary rationals do eventually become positive such as e.g. $-1/9$.) $\endgroup$
    – it's a hire car baby
    Jun 6 at 18:38
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    $\begingroup$ I also didn't mean to sound defensive. I was just trying to explain where I was coming from. I don't know anything about the structure of $\Bbb Z_2^\times$, so I cannot comment on it specifically, I'm a little less surprised by how difficult the matter is. There are other questions relating to $n$-adics that are suprisingly complicated. For instance, when $n$ is composite, it is not too hard to show that there are zero-divisors. But these zero-divisors have very complex expressions. They are not easy to calculate. $\endgroup$
    – Paul Sinclair
    Jun 6 at 19:49

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