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Let $\Bbb Z[2^{-1},2^{-1/2},2^{-1/4,\ldots}]$ be the ring of dyadic rationals extended to include dyadic powers of $2$.

Then let $2^{\nu_2(x)}$ extend the 2-adic valuation to dyadic powers of $2$ (using the rule given below).

Let $f(x)=\dfrac{3x+2^{\nu_2(x)}}{2\cdot\sqrt{2^{\nu_2(3x+2^{\nu_2(x)})}}}$

This looks like a bit of a handful, but it is easily explained:

Ignore multiples of $3$.

Apply the usual Collatz function $3x+1$ to any odd integer

Then instead of dividing out all of the factors of $2$, square root the power of $2$ and divide by $2$ again. Use this power of $2$ to extend the padic valuation.

The thinking behind this is that it has the same fixed point $f(1)=1$ as the conventional graph, and it also converges to $1$ under infinite iteration if the Collatz conjecture converges.

Is there an argument to show it converges finitely?

There's a quick upper bound on its long-term growth of $\frac{3}{2\sqrt2}$ per iteration (not sure how to convert this to big O notation), and in fact a better estimate limit looks to be something like:

$\dfrac{3x}{2\sqrt{2\sqrt{2\sqrt{\ldots}}}}\to3x/4$ for all values that don't converge (i.e. the odd factors dominate), and approaches $1$ when the even factors dominate.

Robert Frost
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  • @reuns the form I give contains every odd integer subsequence of the form you give, but commutes with $2x$ in the sense $2f(x)=f(2x)$ so it's the same ( but better when you get to know the problem) as the powers of 2 really just generate multiplicative cosets of what's really going on. – Robert Frost Jun 19 '19 at 20:23
  • I don't understand your $f$. The collatz function is $f(n) = \frac{3n+1}{2^{v(3n+1)}}$ can you make your $f$ in this form ? – reuns Jun 19 '19 at 20:24
  • @reuns where you have 3,10,5,16,8,4,2,1, my function just has 3,5,1 or 3,10,32 take your pick. – Robert Frost Jun 19 '19 at 20:24
  • @reuns Actually ignore multiples of 3 as I wrote it to commute with 3x too. So where you have 13,40,20,10,5,16,8,2,1 I have 13,5,1 or 13,40,128 take your pick. Then you can do what you want with the powers of 2, which is what I'm doing here. All you then seek is that the function converges to some power of 2. – Robert Frost Jun 19 '19 at 20:28
  • @reuns I simplified it to remove the commutativity with 3x as that will just cause confusion. – Robert Frost Jun 19 '19 at 20:31
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    I don't understand your denominator. If $n = 3$ then $f(n) = \frac{3n+1}{2^{3/2}}$ why would you want to divide by $2^{3/2}$ ? Also for any $g$, $F(x) = \frac{3x + 2^{v(x)}}{2^{3x + 2^{v(x)}}}$ is equal to the usual Collatz function $f(n) = \frac{3n+1}{2^{v(3n+1)}}$ when quotienting by all the powers of $2$. You can see the usual Collatz function as $f(n) = 3n+2^{v(n)}$ on $\Bbb{Z}[1/2]^* / 2^\Bbb{Z}$ (in which case $f(2n) = 2 f(n)$ by definition) – reuns Jun 19 '19 at 20:43
  • @reuns yes, any Collatz sequence can be recovered from this. So I was trying to pick denominators which give a demonstrably convergent series, e.g. this falls for numbers greater than 1 and has some convergent properties. – Robert Frost Jun 19 '19 at 21:00
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    Let $g(x):=\tfrac{3x+2^{\nu(x)}}{2}$. Then $f(x)=g(x)2^{-g(x)}$. Just for my own sanity. – Servaes Jun 19 '19 at 23:00
  • @Servaes that doesn't look quite right but do you mean it's better not to use the radical symbol? – Robert Frost Jun 20 '19 at 03:56
  • @Servaes did you mean Let $g(x):=\tfrac{3x+2^{\nu(x)}}{2}$. Then $f(x)=g(x)2^{-\nu_2(g(x/2))}$? – Robert Frost Jun 20 '19 at 08:20
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    @user334732 No, though my comment is wrong as well and it's even uglier than I thought at first. What is the relevance of this function? – Servaes Jun 20 '19 at 09:16
  • @Servaes yes, I was aware my comment is also wrong but it was as close as I could quickly get to yours that satisfied me it might clarify what you mean. My thought with this function is that at least it converges without vanishing, while at the same time containing a recoverable Collatz sequence. I think the result that it converges to a valuation of 2 would prove the conjecture. – Robert Frost Jun 20 '19 at 09:22

1 Answers1

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To expand on comments by users reuns and Servaes, this seems rather pointless.

One can define, on any additive subgroup of a field that has a well-defined $2$-adic valuation, a variant of the Collatz function,

$$f(x) = 3x+2^{v_2(x)}$$

which by definition commutes with powers of $2$. We can be generous and look e.g. at the field $K:= \mathbb Q (2^r : r\in \mathbb{Q}$) (viewed as a subfield of $\mathbb R$; if we want to see that as a truly $2$-adic field i.e. contained in something like $\mathbb C_2$, one has to choose a compatible set of $n$-th roots of $2$, but all that's beside the point).

The Collatz conjecture is equivalent to stating that for every positive integer $x$, there is $k\in \mathbb N$ such that $f^{\circ k}(x) \in 2^\mathbb{Q}$. (Notation $^{\circ k}$ meaning $k$-fold composition.) It is well-known that in bigger subsets of our $K$ than just the positive integers, there are non-trivial cycles (already in $\mathbb Z$, namely starting at $-5$ or $-17$), and even though research into the behaviour of $f$ on such bigger rings shows some interesting results (Lagarias!), it seems that these are of rather limited use for the original Collatz problem. But even if we believe they might be, let me now explain why I'm quite sure your adaptation of that map is not, and you might as well stick to the above $f$.

Namely, let $a(x)$ be any (!) map on $K^\times$, or some suitable subset thereof, with range $\subseteq \mathbb Q$. Then the map

$$f_a(x) := f(x) \cdot 2^{a(x)}$$

carries not more or less information than the original $f$ about the Collatz problem. To see this, generalising what reuns writes, $f$ induces a well-defined map $\overline{f}$ on the multiplicative quotient $K^\times/2^{\mathbb Q}$, and $f^{\circ k} (x) \in 2^{\mathbb Q} \Leftrightarrow \overline f^{\circ k} (\overline{x})=\overline 1$. But, punchline, $f$ and $f_a$ induce the same quotient map on $K^\times/2^{\mathbb Q}$, so $$f^{\circ k} (x) \in 2^{\mathbb Q} \Leftrightarrow f_a^{\circ k} (x) \in 2^{\mathbb Q}$$ (this can also be seen directly, without the quotient machinery, just by noting that $f$ commutes with powers of $2$).

Your proposed map is (this is what Servaes was alluding to, but got wrong in a first comment) $f_a(x)$ with $a(x) = \dfrac12 v_2(f(x))-1$. A more standard approach, as noted by reuns, and used in previous questions of yours, is just $a(x) = v_2(f(x))$. And here you tried $a(x) = v_2(f(x))-r$ for $r=0,1,2$ already.

If next time you choose $a \equiv 5000$, you can of course argue heuristically that the real absolute value of iterations of $f_a$ goes to infinity which heuristically might preclude cycles. Or you choose $a(x) := \min(v_2(f(x)), -317, v_2(x+99))$ to argue heuristically that iterations of $f_a$ become smaller, i.e. converge, which heuristically prevents Collatz sequences from going to infinity. The problem would be that in each case, the heuristic is nonsense, and real absolute values of $f_a(x)$, as well as everything else that distinguishes $f_a$ from $f$, say nothing, nothing about the Collatz problem.

Torsten Schoeneberg
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  • I get the issue. One avenue I was contemplating is whether finding some form which converges to a rational without vanishing could lead to an argument there can be no cyclic component to the convergent sequence since this would be a rational function of finitely many irrational terms. But I think I have better leads. – Robert Frost Jun 21 '19 at 19:31
  • I guess one of the questions here is; are there sequences of odd natural numbers which cannot converge to a rational when multiplied by a sequence drawn from $2^r:r\in\Bbb Q$, whereas other sequences do have that ability. – Robert Frost Jun 21 '19 at 21:57