I have created an equation that might be able to solve half of the Collatz conjecture, but it requires a proof that is beyond me to make, and so I ask you.
for all positive integers of $y$
$$x = \frac{\displaystyle \sum_{i=0}^{y-1} \left(3^i 2^{\left(\displaystyle\sum_{j=0}^{y-1-i} n_j\right)}\right)}{2^{\left(\displaystyle\sum_{k=1}^y n_k\right)}-3^y}$$
$n_0=0$
$n_{\text{anything else}}=\text{new variable}$
I want someone to prove that for any positive integer $y$ there is only one positive integer solution $x$ where all other entries $(n_1,\space n_2,\space n_3, \cdots)$ are also positive integers.
Notation
If you reformulate the collatz-transformation from one odd element $a_1$ to another odd element $a_2$ as such $$ a_{k+1} = {3a_k+1\over2^{A_k}} \tag 1 $$ then the $N$-fold multiple consecutive transformation can be written as $$a_{N+1} = {3^N a_1+ 3^{N-1}2^{A_1} + 3^{N-2}\cdot 2^{A_1+A_2} + \cdots + 3^1\cdot 2^{A_1+A_2+...A_{N-2}} +2^{A_1+...+A_{N-1}} \over2^S} \tag 2 $$ where $N$ is the number of steps, and $S$ is the sum of the exponents.
Separating the fraction for easier handling and introducing the function $Q(E)$ $$a_{N+1} = {3^N \over 2^S} a_1+ { Q([A_1,A_2,...,A_N]) \over2^S} \tag 3 $$ where $Q(E)$ with the vectorial argument $E=[A_1,...A_N]$ is the long numerator (which is also in the numerator of your sum): $$ Q([A_1,A_2,....,A_N])= 3^{N-1}2^{A_1} + 3^{N-2}\cdot 2^{A_1+A_2} + \cdots + 3^1\cdot 2^{A_1+A_2+...A_{N-2}} +2^{A_1+...+A_{N-1}} \tag 4 $$
For completeness we define $Q([A])=1$ for any $A\ge 1$ if the arguments-vector has length $N=1$.
Cycles
If we want a cycle, then this means that $a_{N+1}=a_1$ and eq (3) must take the form $$a_1 = {3^N \over 2^S} a_1+ { Q([A_1,A_2,...,A_N]) \over2^S} \tag 5 $$ and we can rearrange $$a_1\cdot 2^S-a_1\cdot 3^N = Q([A_1,A_2,...,A_N]) \tag 6 $$ and finally arrive at the formula that you've given in your question: $$ a_1= { Q([A_1,A_2,...,A_N]) \over 2^S-3^N} \tag 7 $$
Simple solutions
Of course the trivial cycle has only one step and looks like $$ 1= { Q([2]) \over 2^2-3^1} = { 1 \over 1}=1 \tag {8.1} $$ or repeated with two steps $$ 1= { Q([2,2]) \over 2^4-3^2} = { 3 + 2^2 \over 7}=1 \tag {8.2} $$ or repeated with three steps $$ 1= { Q([2,2,2]) \over 2^6-3^3} = { 3^2 +3\cdot 2^2 + 2^4 \over 37}=1 \tag {8.3} $$ and so on.
It is not so well known, that this gives quite immediately some solutions (=cycles) in negative integers: $$ -1= { Q([1]) \over 2^1-3^1} = { 2^0 \over -1}=-1 \tag {9.1} $$ $$ -1= { Q([1,1]) \over 2^2-3^2} = { 3+2^1 \over -5}=-1 \tag {9.2} $$ $$ -1= { Q([1,1,1]) \over 2^3-3^3} = { 3^2+3\cdot 2^1 + 2^2 \over -19}=-1 \tag {9.3} $$ with the argument-vector $E=[1,1,1,....,1]$ of arbitrary length.
We even find the cycle $$ -5= { Q([1,2]) \over 2^3-3^2} = { 3+2^1 \over -1}=-5 \tag {10.1} $$ of periodic argument-vector $E$ of arbitrary even length.
One more with $(N,S)=(7,11)$ : $$ -17= { Q([1,1,1,2,1,1,4]) \over 2^{11}-3^7} = { Q(E) \over -139}=-17 \tag {10.2} $$ with argument-vector $E$ which is periodic with period-length $7$.
Open problem
However, nobody has succeeded to either find a (nontrivial) cycle or to prove the non-existence of the structure with $a_1 \gt 1$ being odd integer: $$ a_1 = {Q(E)\over 2^S-3^N } \\ \qquad \text{with } E=[A_1,A_2,...,A_N], A_k \ge 1 , \\ \qquad S=\sum A_k ,\qquad 2^S \gt 3^N \\ \qquad a_1 \text{ is odd integer } \gt 1 \tag {11} $$ In this formula we needed to find some combination of the $A_k$ which allows the result $a_1$ to be positive odd integer greater than $1$. There has been a lot of amateurish as well as professional discussion of this exponential diophantine equation and one can find online a big pile of coverage on this.
Hints towards the problems in eq (11) from other examples
One of the problems for non-fractional divisibility in eq(11) is that the denominator should be small but the distance between perfect powers is generally large. There are few solutions known for the generalized problem $mx+1$ with $m=5$ and $m=181$ each for which the denominator is exceptionally small. In the following I generalize the $Q(E)$ -notation to the $Q_m(E)$-notation where the $m$ replaces the $3$ in eq.(1) and eq.(2) (making it an "$mx+1$-problem") and in the according $Q()$-definition.
For $m=5$ we have $a_1=1 \to 3 \to 1 $ with $N=2$ and $S=5$ such that the formula 11 becomes $$ 1 = {Q_5([1,4])\over 2^5-5^2 } = {Q_5([1,4])\over 5 } \tag {12.1} $$
For $m=5$ we have $a_1=13 \to ... \to 13 $ with $N=3$ and $S=7$ such that the formula 11 becomes $$ 13 = {Q_5([1,1,5])\over 2^7-5^2 } = {Q_5([1,1,5])\over 3 } \tag {12.2} $$
For $m=5$ we have $a_1=17 \to ... \to 17 $ with $N=3$ and $S=7$ such that the formula 11 becomes $$ 17 = {Q_5([1,3,3])\over 2^7-5^2 } = {Q_5([1,3,3])\over 3 } \tag {12.3} $$
For $m=181$ we have $a_1=27 \to ... \to 27 $ with $N=2$ and $S=15$ such that the formula 11 becomes $$ 27 = {Q_{181}([3,12])\over 2^{15}-181^2 } = {Q_{181}([3,12])\over 7 } \tag {12.4} $$
For $m=181$ we have $a_1=35 \to ... \to 35 $ with $N=2$ and $S=15$ such that the formula 11 becomes $$ 35 = {Q_{181}([6,9])\over 2^{15}-181^2 } = {Q_5([6,9])\over 7 } \tag {12.5} $$
(The two first cycles for $m=5$ and $m=181$ have been shown already by R. Crandall in his 1978 article, but no more nontrivial cycles besides the $5$ cycles shown here are found from since until now (see wikipedia which would be aware of significant new results))
"Exceptional" small denominators can be found using the convergents of the continued fraction of the transcendental values $\mu = \log_2(m)$ which for the case of $m=3$ has for instance "small" denominators for $N \in \{1,5,41,306,...\}$ which is fairly small for $N=5$ being $13$ but increases to $1909$ for $N=7$ and so on, and although $N=41$ has some "best" approximation the denominator becomes $420491770248316829$ - which is not really "small" though...
See also the question & discussion at mathoverflow in Nov 2019