Collatz conjecture is equivalent to $n\times 3^{k} = 2^{ak+1} - TCR$ where, for me, $k$=odd steps, and $ak+1 $=even steps. Note that total steps = k +( ak+1) steps. Some numbers have the same total steps, but $k$ steps and $ak+1$ steps are different because it is unique for every n.
Even numbers are $\equiv \pmod{2}$ remainder 0 and $\equiv\pmod{3}$ remainder 0,1,2 . Odd numbers are $ \equiv \pmod{2}$ remainder 1 and $\equiv\pmod{3}$ remainder 0,1,2.
Professor Terence Tao talks about $3^{k-1}*2^{a1} + 3^{k-2}*2^{a2} + ... +2^{ak }$ and the study of his properties to take any advance in the solution on his post about Collatz conjecture. Let´s call it Tao-Collatz remainder, $TCR$ for me, I think that
$2^{ak+1}\equiv TCR \pmod{n}$
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, $2^{x}> n\times 3^{k}$ exist 1 solution with the form $2^{x}-r$ .This simple question based on the Euclidean algorithm of division may not be strong enought for a demostration. We know that
$n^{i}\equiv r_{i}\pmod{n}$
When $3^{0}\rightarrow TCR= 0$ , $3^{1}\rightarrow TCR= 1$ , $3^{k}\geq 2\rightarrow TCR = 3^{k-1}*2^{a1} + 3^{k-2}*2^{a2} + ... +2^{ak }$ And
$3^{0}\equiv 1\pmod{n}$ , $3^{1}\equiv 3\pmod{n}$ , $k\geq 2 ,3^{k}\equiv \chi \pmod{n}$
If $2^{ak+1}\chi \equiv TCR \pmod{n}$ and $3^{k-1}\equiv \chi \pmod{n}$, where $\chi $ is the remainder, is the solution, and the Euclidean algorithm always ends, ¿is it the probe that the Collatz conjecture is true?
What is the solution to this congruence?