Hello, the following equivalence seems true and is true numerically until big numbers. For all x,p positive integers and x\ge 1,
x \le 2^p/3^{n_p(x)} \iff \exists p’\le p : \mathrm{col}^{p’}(x)=1
with n_p(x) the amount of odd integers in \{x,\mathrm{col}(x),…,\mathrm{col}^{p-1}(x)\} and \mathrm{col}(x)=(3x+1)/2 if x odd and x/2 if x even. The reciprocal sense of this equivalence is easy to proove with the fact that by writing \mathrm{col}^{p’}(x)=1, you have for \mathrm{col}^{p’}(x)=(3^{n_{p’}(x)}/2^{p’})(x+S(x,n_{p’}(x)), x=(2^{p’}/3^{n_{p’}(x)})-S(x,n_{p’}(x)) where S(x,n_{p’}(x)) is a positive remainder sum. But the implication sense seems much more difficult to proove. Indeed, the usual bound (3^{n_p(x)}/2^p)S(x,n_p(x))\le (3^{n_p(x)}/2^{n_p(x)})-1 shown in Theorem 2.4 of Paradoxical behavior in Collatz sequences by Olivier Rozier and Claude Terracol only gives you in the case where x \le 2^{p}/3^{n_p(x)} that \mathrm{col}^p(x) \le 3^{n_p(x)}/2^{n_p(x)} which is not 2 which is what you need to show the assertion. So it seems like you’ve gotta use the fact that x \le 2^p/3^{n_p(x)} to bound (3^{n_p(x)}/2^p)S(x,n_p(x)). But I haven’t found a way of doing this that gives better bounds than the general ones mentioned above.
Make sure to correct me if you spot an error. I will own up to it. I assumed n = n_p(x) (i.e. the number of steps in Collatz for x to go to 1, if possible- not sure 100% if this is what you meant: in fact, what is the difference between n_{p'}(x) and n_{p}(x)? From your definition n_{p}(x)=p which I believe is not the intent.).
\exists p' \le p : col^{p'}(x) = 1 \implies x \le 2^p / 3^n
As you said, this is easy to prove, since if any number reaches 1 (we can safely ignore odd numbers and those that are multiple of 3), then it must be of the form x = (2^{x_1+...+x_n} - \displaystyle\sum_{i=1}^{n}{3^{i-1}2^{x_{i+1}...x_n}}) / 3^n \le 2^p / 3^n where n is the number of Collatz steps and p = x_1 + ... + x_n.
For the other implication
x \le 2^p / 3^n \implies \exists p' \le p : col^{p'}(x) = 1
Collatz is needed, because you assume that for any x there must already exist col^{p'}(x)=1 (and I don’t see a way around it, do you?). Note that if Collatz were to be true, then this implication holds; we can choose n as the number of Collatz steps, then p'=n and clearly p' \le n < p.
My mistake, I guess, I could have been more clear on the definitions and the reciprocal sense is currently wrong, as it is. First of all, n_p(x) is generally defined as the amount of odd numbers in the first p numbers of the trajectory of x for any x,p as defined rigourously in the first message; it is a sequence based on two variables, not the number of steps in Collatz for x to go to 1. n_p(x) and n_p’(x) is this sequence evaluated for p and p for a said x. What I meant in the sketch of proof of the reciprocal sense is that if such p’ exists then for this p’ you have indeed x\le 2^{p’}/3^{n_{p’}(x)} but you cannot obtain x\le 2^{p}/3^{n_{p}(x)} in all generality from here which is why the reciprocal sense of this equivalence is false. The two implications that I meant to talk about were
\mathrm{col}^p(x)=1\implies x \le 2^p/3^{n_p(x)}
which is easy to prove with the same argument that I had in the previous false first sketch proof and
x \le 2^p/3^{n_p(x)} \implies\exists p’\le p : \mathrm{col}^{p’}(x)=1
which is the tricky one to prove true but is true at least numerically until big numbers and to prove it you have to show that x \le 2^p/3^{n_p(x)} \implies \mathrm{col}^{p}(x)\le 2 . The second implication doesn’t seem to need the Collatz Conjecture to be proven true and I don’t think that it shows this implication true as well.
In that case n_p(x) = (\#i \text{ for which } p > \displaystyle\sum_{i=1}^{o(x)}{e_i}) + 1
where o(x) is the number of odd steps before x reaches 1 and e_i is the exponent of two by which we divide after each odd step.
In \operatorname{col}^p(x) = 1 \implies x \le 2^p / 3^{n_p(x)} n_p(x) is the number of odd integers for x before it reaches 1 (as described in my previous response), and therefore n_{p}(x) = o(x).
I still think the second implication requires Collatz, because if any \operatorname{col}^p(x) is to be less or equal than two (it reaches 1), then n_p(x) must be o(x).
However, the second implication in your last post seems as hard to prove as Collatz. At least, it implies the nonexistence of nontrivial cycles. If indeed we assume the existence of a nontrivial cycle , then it would be easy to build a counterexample by choosing x in such a cycle and taking p sufficiently large. The main argument is based on the fact that we have 2^p / 3^{n_p(x)} \rightarrow \infty as p increases whenever x has a periodic trajectory (because the proportion of odd terms is smaller than \log_3 2).
So you should not expect an easy proof of your conjecture
Ok, very interesting. Yeah I was expecting this to be very hard but the fact that it implies the absence of non trivial cycles shows this feeling true. The reason I was trying to show this implication or at least obtain a bound smaller than the one obtained using the bound (3^{n_p}(x)/2^p)S(x,n_p(x))\le (3^{n_p(x)}/2^{n_p(x)})-1 was to have a way of showing that if for the integers x included in the subset of \{1,...,2^p\} having n_p(x)=n for all x in it, you have x smaller than 2^p/3^{n_p(x)} (in this case) then \mathrm{col^p(x)}=1 but I imagine that the problem of this method is exactly that this implication is so hard to show and if you make the bound very small for it to indeed imply \mathrm{col^p(x)}=1 then it will be impossible to show that there are many integers having \mathrm{col^p(x)}=1 because even experimentally you will see that most integers having \mathrm{col^p(x)}=1 are very close to 2^p/3^{n_p(x)} in the sense that they are for most of them greater than c(2^p/3^{n_p(x)}) for a c>0 when p grows to +\infty , at least it seems like. That’s why I think that to bound the counting function of numbers having \mathrm{col^p(x)}=1 as in the papers of Krasikov and more recently Lagarias you have to use the reciprocal image of the \mathrm{col} function to bound this counting function. I would assume that I’m not the first one to think about that and come to this conclusion most likely.
For a heuristic approach regarding the bound of Krasikov and Lagarias on the number of integers reaching 1 below a given value, you should have a look at this preprint of Daudin. It is based on an accurate description of the structure of the inverse tree.
