, 2014

We have
$$a_n = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = a_0 + \sum_{k=0}^{n-1} 2 \cdot 3^k = a_0 + 2 \frac{3^n - 1}{3-1} = a_0 - 1 + 3^n.$$
If $a_0 = 1$, then $a_n = 3^n$ and
$$\frac{a_k^j}{a_j^k} = \frac{3^{kj}}{3^{jk}} = 1,$$
so $a_0 = 1$ is clearly a solution.

We now assume that $a_0 \neq 1$ and write $a_0 - 1 = x/y$ with coprime integers $x$ and $y$ with $y > 0$.
We obtain
$$\frac{a_k^j}{a_j^k} = \frac{\left(\frac{x}{y} + 3^k\right)^j}{\left(\frac{x}{y} + 3^j\right)^k} = \frac{(x + y3^k)^j}{(x + y3^j)^k} y^{k-j} \in \mathbb{Z}.$$
We have $\text{gcd}(x + y3^j, y) = \text{gcd}(x, y) = 1$, which implies that
$$\frac{(x + y3^k)^j}{(x + y3^j)^k}$$
is also an integer. This implies that
$$(x + 3^k y)^{k-j} \frac{(x + 3^k y)^j}{(x + 3^j y)^k} = \frac{(x + 3^k y)^k}{(x + 3^j y)^k} = \left(\frac{x + 3^k y}{x + 3^j y}\right)^k$$
is an integer, too. If the $k$th power of a rational number is an integer, then the number itself has to be an integer. Therefore,
$$\frac{x + 3^k y}{x + 3^j y}$$
is an integer. We now set $k = j + 1$ and write
$$\frac{x + 3^{j+1}y}{x + 3^j y} = 3 + \frac{-2x}{x + 3^j y}.$$
As this is an integer by the above considerations, the second summand
$$\frac{-2x}{x + 3^j y}$$
is also an integer. As $y > 0$ by construction, the denominator is unbounded for $j \to \infty$, which results in $x = 0$, i.e., $a_0 = 1$.

We conclude that $a_0 = 1$ is the only suitable initial value. $\square$