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:
$$
\left\lfloor\left(\frac{3}{2}\right)^n\right\rfloor=\left\lfloor\left(\frac{3^n}{2^n-1}-\frac{1}{2^{n-1}}\right)\right\rfloor
$$

$a_n$ 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 $n\ge2$". 
