The formula can be rewritten as follows:
$$3^{a-1} \frac{(\frac{2^{b}}{3})^c - 1}{\frac{2^b}{3} - 1}$$
Using the general formula $\sum_{i=0}^{n} x^n = \frac{x^{n+1}-1}{x-1}$, this rewrites to
$$3^{a-1} \sum_{i=0}^{c-1} \left(\frac{2^b}{3}\right)^i$$
When writing the sum as a number in digits with respect to base 3, one can see that the $(c-1)$th digit after the decimal point is non-zero. Thus, $a-1$ has to be at least $c-1$. Hence, the condition that you are looking for is:
$$a\ge c.$$
