Pattern concerning the differences between powers of $2$ and $3$

Question

Background
While joyfully wasting my time with the Collatz conjecture, I noticed a pattern concerning the differences (or ratios?) between powers of $2$ and $3$.

Every integer $n$ can be represented as $$n = \frac{2^x-q}{3^b}$$ for some $x,q,b \in \mathbb{N}$.

Now the following seems to hold: $$n = \frac{2^{x+2}-(4q+3^{b}n)}{3^{b+1}}$$

Question
Why does this equivalence (seem to) hold?
I guess it somehow follows trivially from factoring but I can't see it, despite having looked at it really hard.

Examples $$ 1 = \frac{2^2-1}{3} = \frac{2^4-7}{3^2} = \frac{2^6-37}{3^3} \dots$$ $$ 5 = \frac{2^4-1}{3} = \frac{2^6-19}{3^2} = \frac{2^8-121}{3^3} \dots$$ $$ 13 = \frac{2^7-11}{3^2} = \frac{2^9-161}{3^3} = \frac{2^{11}-995}{3^4} \dots$$ $$ 21 = \frac{2^6-1}{3} = \frac{2^8-67}{3^2} = \frac{2^{10}-457}{3^3} \dots$$

Answer 1

You made a slight (but crucial) mistake. The correct formula should be

$$n = \frac{2^{x+2}-(4q+3^{\color{red}b}n)}{3^{b+1}}$$

And then it is indeed a trivial conclusion as

\begin{align*} n&=\frac{2^{x+2}-(4q+3^bn)}{3^{b+1}}\\ n&=4\frac{2^x-q}{3^{b+1}}-\frac n3\\ 3n&=4\frac{2^x-q}{3^b}-n\\ 4n&=4\frac{2^x-q}{3^b}\\ \end{align*}

$$\therefore~n=\frac{2^x-q}{3^b}$$

And the last line is just the original assumption.

Answer 2

We have $$\frac{2^{x+2}-(4q+3^bn)}{3^{b+1}}=\frac{4\cdot 3^bn - 3^bn}{3^{b+1}}=\frac{3\cdot 3^b\cdot n}{3^{b+1}}=n$$

Answer 3

It's not especially deep. Given $$n = \frac{2^x - q}{3^b}$$ it follows that $$ \frac{4n}{3} = \frac{2^{x+2} - 4q}{3^{b+1}}$$ and $$\frac{n}{3} = \frac{3^b n}{3^{b+1}}.$$

Answer 4

https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/ " it was observed by Bohm and Sontacchi that this weak conjecture is equivalent to a divisibility problem involving powers of 2 and 3"(Terence Tao)

$$n\times 3^{k}+3^{k-1}\times 2^{a1}+3^{k-2}\times 2^{a2}+...+2^{ak}=\begin{Bmatrix}n\times 2^{x} \\ \infty \\ 2^{ak+1}\; ,where\: 2^{ak+1}> n\times 3^{k} \end{Bmatrix}$$

"In order to make rigorous progress on this conjecture, it seems that one would need to somehow exploit the structural properties of numbers of the form"(Terence Tao )

$3^{k-1}\times 2^{a1}+3^{k-2}\times 2^{a2}+...+2^{ak}$

"It leaves open the possibility of some very rare exceptional ${n}$ for which the orbit goes to infinity, or gets trapped in a periodic loop." (Terence Tao)

In my question Collatz conjecture, Tao-Collatz remainder and mod n., I call

$3^{k-1}\times 2^{a1}+3^{k-2}\times 2^{a2}+...+2^{ak}$= TCR, you call it q.

When we say that

$n = \frac{2^x-q}{3^b}$

we are assuming the conjecture true. If the series

$n\times 3^{k}+3^{k-1}\times 2^{a1}+3^{k-2}\times 2^{a2}+...+2^{ak}$ converges (or is equal) to a power of 2, It is a sufficient condition for the demonstration of the conjecture. But I understand that this condition has to be extremely difficult to prove. Although exceptionally rare, the circumstance can be given that the conjecture converges to infinity $f^{k}(n)= \infty$ , or enters a periodic loop, $f^{k}(n)=n\times2 ^{x}$

therefore, if we are able to demonstrate that the series does not converge to infinity and does not converge to a multiple of n, the conjecture must converge to a power of 2. In my question, I argued that

$2^{ak+1}\equiv TCR(or\: q) \pmod{n}$

The GCD ($2^{ak+1}$, n) =1 , so there exists just 1 solution for this congruence.And $2^{ak+1}$ – TCR is a multiple of n. Effectively for each power of 2, $2^{ak+1}>n×3^{k}$ exist 1 solution with the form $2^{ak+1}$−TCR (or q).For small TCR(or q) it could be easy that is of the form $3^{k-1}\times 2^{a1}+3^{k-2}\times 2^{a2}+...+2^{ak}$ even being 1 we have that $3^{0}\times2^{0} = 1$, but if the value of q is very large, it seems more difficult to verify if its structure is correct.In any case, we would always arrive at the fact that if the conjecture is true, this implies that it is true, that it does not seem like a demonstration. We could use this argument in the other two possible solutions, although exceptionally rare, not ruled out. If

$f^{k}(n)=n\times2 ^{x}$, then we have that $n\times 3^{k} + TCR = n\times2^{x}$, so $TCR\equiv0(mod\:n)$ and $n\times 3^{k}+ny=nz$, with $z=2^{x}=3^{k}+y$

If

$3^{k-1}\times 2^{a1}+3^{k-2}\times 2^{a2}+...+2^{ak}=ny= TCR\not\equiv0(mod\:n)\rightarrow f^{k}(n)\neq n\times2^{x}$

If it is possible to verify this, then, we can rule out the existence of periodic loops. I have not tried this argument to rule out the possibility of numbers that escape infinity, but if we can rule out the existence of loops, then we can take the problem to the theory of graphs and the Collatz tree. As I think I am extending too much, I will try to raise the question, assuming that the existence of periodic loops is not possible and considering that the validity of the conjecture is independent of the existence of the 1-2-4-1 loop, of whether it is possible to ensure that the tree (or forest) of Collatz, is a disjoint union of graphs, labeled, oriented and weakly connected, equivalent to N, the set of natural numbers.