Is there always one integer between these two rational numbers? It appears that for each integer $k\geq2$, there is always one integer $c$ that satisfies the inequalities below. Can this be proved?
$$\frac{3^k-2^k}{2^k-1}<c\leq \frac{3^k-1}{2^k}.$$
Note that for $k\geq2$ the lower bound is always a proper fraction and will never match an integer.  
Edit 28/1/2018 I have a short proof here on Overleaf. <--- It's done! Edit #28 is perfect!
edit It looks like Waring's problem already has a solution. The sequence: https://oeis.org/A060692 "Corresponds to the only solution of the Diophantine equation 3^n = x*2^n + y*1^n with constraint 0 <= y < 2^n." This is equivalent to our $a+c$ for the Waring if statement. Oops, there is no proof that the diophantine equation is always $\leq 2^n.$ Here is a proof sketch of Waring's
Here is a brute force function:
aplusc[k_] := Module[{c}, c = 1; While[0 < 3^k - 2^k (++c)]; 3^k - 2^k (--c) + c]
where we increment $c$ until the calculation becomes negative, then we decrement by one to get $c$. We recalculate using that $c$ and add them together. A060692(k) equals this value.  
Closed form: here. The brute force function illustrates the sawtooth pattern.  
We can also create a(k), b(k), and c(k) using the same module. And we can use b(k) for the proof. Trivially, we can show b$(k) + 1 \leq 2^k.$
edit. Some more information and references: https://oeis.org/A002379
  PM

We show that $\left\lfloor\frac{3^n-1}{2^n-1}\right\rfloor$ and $\left\lfloor \left(\frac{3}{2} \right)^n\right\rfloor$ have a common floor when $\frac{3^n-1}{2^n-1}-\left(\frac{3}{2} \right)^n =$ $ \left\{\frac{3^n-1}{2^n-1}\right\}-\left\{\left(\frac{3}{2} \right)^n\right\} = \delta(n),$ where $\left\{\cdot\right\}$ are the fractional parts and $\delta(n)$ is the difference function.
We verify that the order of the fractions, $\frac{3^n-1}{2^n-1}>\left(\frac{3}{2} \right)^n,$ is true for $n\not=0$ and set $n\geqslant1$ to constrain to the positive numbers.
Notice that the denominators are consecutive integers and therefore relatively prime to each other. The lower denominator, $2^n-1,$ must be greater than one, so we set $n\geqslant2$ as our final constraint.
By the ratio test, it is easy to show $\frac{3^n-1}{2^n-1}$ and $\left(\frac{3}{2} \right)^n$ are always proper fractions for $n\geqslant2$ and therefore they always have fractional parts.
We define $\delta(n):=\frac{3^n-2^n}{4^n-2^n}$ and its one's complement, $1-\delta(n),$ as $\frac{4^n-3^n}{4^n-2^n}.$ We define the interval which contains the common floor, $x,$ as lower bound, $\frac{3^n-1}{2^n-1}-1$ and upper bound, $\left(\frac{3}{2} \right)^n.$ From this we can show that the upper bound minus the lower bound is equal to $\frac{4^n-3^n}{4^n-2^n}.$
The RHS below is $\left\{\left(\frac{3}{2} \right)^n\right\}+\left(1- \left\{\frac{3^n-1}{2^n-1}\right\}\right).$  
We assume a common floor: Let $(n|x)\in \mathbb{Z},\ x\geqslant 1,\text{ and } n\geqslant 2,$
$\frac{4^n-3^n}{4^n-2^n}=\frac{3^n-2^n x}{2^n}+ \frac{\left(2^n-1\right) x+2^n-3^n}{2^n-1},\ \ \text{expand; subtract LHS from both sides}\Rightarrow$
$\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}+1=\frac{2^n
     x}{2^n-1}-\frac{x}{2^n-1}+\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}-x+1;\ \ \Rightarrow$
$0=\frac{2^n
     x}{2^n-1}-\frac{x}{2^n-1}-x,\ \ \text{simplify}\Rightarrow\text{True.}$     
We assume different floors: Let $(n|w|x)\in \mathbb{Z},\ w\geqslant 1,\ x\geqslant 1,\text{ and } n\geqslant 2,$
$\frac{4^n-3^n}{4^n-2^n}=\frac{3^n-2^n x}{2^n}+ \frac{\left(2^n-1\right) w+2^n-3^n}{2^n-1},\ \ \text{expand; subtract LHS from both sides}\Rightarrow$
$\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}+1=\frac{2^n
 w}{2^n-1}-\frac{w}{2^n-1}+\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}-x+1;\ \ \Rightarrow$
$0=\frac{2^n w}{2^n-1}-\frac{w}{2^n-1}-x,\ \ \text{simplify}\Rightarrow w=x.$
Next, we craft two unit-intervals within which the bounds reside: 
\begin{align}
&\left(\frac{3^n-1}{2^n-1}-1-\left\{\frac{3^n-1}{2^n-1}\right\},\ \frac{3^n-1}{2^n-1}-\left\{\frac{3^n-1}{2^n-1}\right\}\right),\ \ (1)\\ &\left(\left(\frac{3}{2} \right)^n-\left\{\left(\frac{3}{2} \right)^n\right\},\ \left(\frac{3}{2} \right)^n+1-\left\{\left(\frac{3}{2} \right)^n\right\}\right).\ \ (2)
\end{align}
Since lower bound of (1) $<\left(\frac{3}{2} \right)^n-1$, upper bound of (2) $>\frac{3^n-1}{2^n-1},$ and the remaining bounds equal $x,\text{ we have a common floor because }x$ is the greatest integer less than the fraction, which is the floor definition.$\ \ \ \square$
Edit 6/1 I forgot to show that $x$ was the floor.
