<s>[INST] Waring's problem The first comment on OEIS A002379 states:

It is an important unsolved problem related to Waring's problem to show that a(n) = floor((3^n-1)/(2^n-1)) holds for all n >= 1. This has been checked for 10000 terms and is true for all sufficiently large n, by a theorem of Mahler. [Lichiardopol]

Edit -- a new equality
$$
\frac{2^n( 3^n \mod 2^n)}{4^n-2^n} -\frac{3^n \mod 2^n}{4^n-2^n} -\frac{2^n( (-2+3^n) \mod (-1+2^n))}{4^n-2^n}
-\left(\frac{3}{2}\right)^n +\frac{3^n}{2^{n-1}} -\frac{1}{2^{n-1}} +\frac{2^{2 n}}{4^n-2^n} -\frac{2^{n+1}}{4^n-2^n} = 1
$$
The items with $4^n-2^n$ in the denominators are distances and the items with $2$ or $2^{n-1}$ are locations on the number line.  
An alternate form is:
$$
\left\lfloor\left(\frac{3}{2}\right)^n\right\rfloor=\left\lfloor\left(\frac{3^n}{2^n-1}-\frac{1}{2^{n-1}}\right)\right\rfloor
$$
 [/INST] Note that $$ {3^n-1\over2^n-1} = 3^n{1 \over 2^n-1} -{1 \over 2^n-1} \\
= {3^n \over 2^n} (1+ \frac 1{2^n}+\frac 1{4^n}) -\frac 1{2^n } (1+ \frac 1{2^n}+\frac 1{4^n}) \\ =  {3^n \over 2^n} + \left(\frac {3^n}{4^n}-\frac {2^n}{4^n} +\frac {3^n}{8^n}-\frac {2^n}{8^n}+... \right)   $$
and it is not obvious, that the parenthese is smaller than the difference of $\frac {3^n}{2^n}$ to the next integer above. In the contrary, this is very near the detail in the Waring problem (and only slightly weaker than the (unproven) conjecture) that the parenthese in
$$ {3^n \over 2^n} + \left(\frac {3^n}{4^n} - 0 + 0 ...  \right)   $$ is smaller than that difference.

The following picture is meant to make the problem more visible, that the question is that of the distance of $(3/2)^r$ to the next integer. I've drawn the x-axis of the real numbers, two marks at consecutive integer numbers floor and ceil $(3/2)^N$ and three examples of possible positions of the $(3/2)^N$ in that interval, such that the Collatz/Waring-intervals around $(3/2)^N$ interfer in three different ways with that consecutive integers. Positions 1 and 3 contradict the expectations from the Waring-related and the COllatz-related intervals, and for positions $k$ we see, that the intervals are inside the bounds given by the consecutive integers.
I've also added a further definition of bounds for an interval around $(3/2)^N$, which is always smaller than the unit-interval but includes the Collatz/Waring-intervals, which I've not seen before/elsewhere. It ensures, that at most $1$ integer is in that interval.   

 
The question is: how do we know, that $(3/2)^N$ is at an appropriate position in the unit-interval of the consecutive integers?
</s>