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:
$$
\left\lfloor\left(\frac{3}{2}\right)^n\right\rfloor=\left\lfloor\left(\frac{3^n}{2^n-1}-\frac{1}{2^{n-1}}\right)\right\rfloor
$$
 A: 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?
A: $a_n$ is defined as the floor of $3^n/2^n$. The unsolved problem is whether that is always equal to the floor of ${3^n-1\over2^n-1}$. 
EDIT: where "always" seems to mean "for all $n\ge2$". 
A: This is another answer, motivated by your additional query at MO, where it likely gets/stays closed. Because I understand now your question there in a better way this answer my be useful also for the question here. 
First (the first formula in MO) does nothing else than to express the difference between $ {3^n\over2^n} $ and $ {3^n-1\over2^n-1} $ .Nothing in this to be proved.
Second you ask, whether "a proof of that formula would also prove the formula" $$ \left\lfloor {3^n\over2^n} \right\rfloor = \left\lfloor {3^n-1\over2^n-1} \right\rfloor$$
which indeed is the critical conjecture (and only slightly different from that difficult and unproven conjecture from the Waring problem).
Well, the first formula contains nothing to prove - it is just a statement of the difference, but where an explicite expansion and cancellation can help to express it in a very intuitive way - and only after that it remains to show/prove , that also the expression by the second formula is actually not only empirically but analytically true. So here goes my second thought on it.    
First consider the representation of the numbers in the digit-system of $2^n$. Let's write 
$$ 3^n  = m 2^n + r \qquad \qquad r \in (0..2^n-1) \tag 1 $$
Then in the digit system base $b=2^n$ we have
$$ 3^n = \mathtt{"m r.000"_{|_b}}  \tag 2 $$    and
$$ {3^n \over 2^n} = \mathtt{"m. r000"_{|_b}} \tag 3 $$
Now $ {3^n \over 2^n-1} ={3^n/2^n \over 1-1/2^n}  $  is a geometric series with quotient $2^n$ and occurs as a repunit:
$$ {3^n/2^n \over 1-1/2^n} = \mathtt{" m. r" + "0.mr" +"0.0mr"+ ... _{|_b}} \tag 4$$
which can then be written as
$$ {3^n \over 2^n-1} = \qquad \begin{array} {rcl}
       & &  \mathtt {\; ^"m.m m m m m m \ldots \; ^"} \\
        &+& \mathtt {\; ^"0.r r r r r r \ldots\; ^"}  _{|_b}
 \end{array} \tag 5$$ and if we append the missing $ { -1\over 2^n-1}$ we get
$$ {3^n -1 \over 2^n-1} =  \qquad \begin{array} {rcl}
& & \mathtt{ \; ^"   m.m m m m m m \ldots  \; ^" } \\
&+& \mathtt{ \; ^"   0.r r r r r r \ldots \; ^" }  \\
&-& \mathtt{ \; ^"   0.1 1 1 1 1 1 \ldots \; ^" }  \\
\end{array} _{|_b}  \tag 6 $$ 
In (3) we see, that $m$ is the floor-value for $3^n/2^n$ and what your second question asks, whether (3) and (6) result in the same floor-value.      
With this representation we see easily, that this can only happen, if the expression of the summands after the decimal points do not give a carry on each single "digit", that means, that, the floor-equality is only true if
$$ m + r - 1 \lt b-1 \tag 7$$
or with the reference to a general/indeterminate $n$
$$ m_n + r_n - 1 \lt 2^n-1 \tag 8$$
(This can be reworked to fit into the notations of the Waring-problem.) 
 
A correct handling of your problem is thus:    

Your first question, whether "a proof of the first equation in MO is also a proof of the equality of the floors" is actually a thinking in circles: in this first equation there is nothing to prove, it is just the formulation of the difference between the two expressions  $ {3^n\over2^n} $ and $ {3^n-1\over2^n-1} $ .     
It can be reformulated such that it becomes better visible, what in fact must be proved: and that is that condition (7) or (8) holds not only empirically for a lot of values $n$ but for analytical reasons for all $n$ (perhaps above a certain lower bound). 
