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Here is a sort-of-but-not-really explicit answer, following up on Helge's idea. Set $h = e^{\pi^2/(12 \log 2)}$. Recursive define the $b_i$ and $q_i$ as follows: $q_{-2} =1$, $q_{-1}=0$ and $q_i = b_i q_{i-1} + q_{i-2}$. If $q_{i}^{1/i} > h$, then $b_{i+1} = 2$, otherwise, $b_{i+1}=3$. b_{i+1}=4$. This is clearly a well defined recursion; you can decide whether or not you think the result of this process is explicit.

Lets first study, in general, the situation where every $b_i$ is either $2$ or $3$. 4$. Write $r_i = q_i/q_{i-1}$. Then $r_i = b_i + r_{i-1}^{-1}$ so $b_i < r_i < b_i + 1$. So, if every $b_i$ is $2$ or $3$, then every $r_i$ is in $[2,4]$. Let's plug this estimate into itself: If every $b_i$ is $2$ or $3$ then every $r_i$ after the first few is in $[2.25, 3.5]$. Plugging these estimates into themselves once more, we see the following:

Suppose that every $b_i$ is $2$ or $3$. Then, except for the fist few $i$, if $b_i=2$ then $r_i \in [2+1/3.5, 2+1/2.25]$ and, if $b_i$ is $3$, then $r_i \in [3+1/3.5, 3+1/2.25]$.

The key point is Note that $2+1/2.25 3 < h \approx 3.275822 3.28 < 3+1/3.5 \approx 3.285714$4$.

Now,

Set $$\frac{1}{n} s_n = \log q_n = \frac{1}{n} \sum_{i=1}^n sum_{i=0}^{n-1} \log r_i$$Call this quantity $a_n$. r_i$. So, if $a_n > \log h$s_n < (\log h) n$, then $a_{n+1} < a_n$ ands_{n+1} = s_n + \log r_i \in [s_n+\log 4, if s_n + \log 5]$ and we see that $a_n < s_{n+1} - (\log h) (n+1) \in $[(s_n - (\log h) n) + \log h$4 - \log h, (s_n - (\log h) n) + \log 5 - \log h]$. Writing $t_n = s_n - (\log h) n$, we see that

If $t_n <0$, then $a_{n+1} > a_n$t_{n+1} \in [t_n + (\log 4 - \log h), t_n + (\log 5 - \log h)]$.Also

Similarly,

If $$|a_{n+1} - a_n| t_n > 0$, then $t_{n+1} \leq in [t_n + (\log 2 - \sum_{i=1}^n log h), t_n + (\log 3 ) \left| - \frac{1}{n+1} log h)]$

In particular, once $t_n$ gets into the interval $[\log 2 - \frac{1}{n} log h, \right| + log 5 - \frac{\log 3}{n+1} = O(n/n^2) + O(1/n) = O(1/n).$$log h]$, it stays there. I leave it to you to write out the exercise check that these properties force it gets there.

So $t_n$ is bounded, $\log q_n = (\log h) n + O(1)$ and $\lim_{n \to \infty} a_n q_n^{1/n} = \log h$.
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Here is a sort-of-but-not-really explicit answer, following up on Helge's idea. Set $h = e^{\pi^2/(12 \log 2)}$. Recursive define the $b_i$ and $q_i$ as follows: $q_{-2} =1$, $q_{-1}=0$ and $q_i = b_i q_{i-1} + q_{i-2}$. If $q_{i}^{1/i} > h$, then $b_{i+1} = 2$, otherwise, $b_{i+1}=3$. This is clearly a well defined recursion; you can decide whether or not you think the result of this process is explicit.

Lets first study, in general, the situation where every $b_i$ is either $2$ or $3$. Write $r_i = q_i/q_{i-1}$. Then $r_i = b_i + r_{i-1}^{-1}$ so $b_i < r_i < b_i + 1$. So, if every $b_i$ is $2$ or $3$, then every $r_i$ is in $[2,4]$. Let's plug this estimate into itself: If every $b_i$ is $2$ or $3$ then every $r_i$ after the first few is in $[2.25, 3.5]$. Plugging these estimates into themselves once more, we see the following:

Suppose that every $b_i$ is $2$ or $3$. Then, except for the fist few $i$, if $b_i=2$ then $r_i \in [2+1/3.5, 2+1/2.25]$ and, if $b_i$ is $3$, then $r_i \in [3+1/3.5, 3+1/2.25]$.

The key point is that $2+1/2.25 < h \approx 3.275822 < 3+1/3.5 \approx 3.285714$.

So, our algorithm has the property that, if $q_i^{1/i} < h$, then $r_{i+1}>h$ and vice versa.

Now, $$\frac{1}{n} \log q_n = \frac{1}{n} \sum_{i=1}^n \log r_i$$ Call this quantity $a_n$. So, if $a_n > \log h$, then $a_{n+1} < a_n$ and, if $a_n < \log h$, then $a_{n+1} > a_n$. Also, $$|a_{n+1} - a_n| \leq \sum_{i=1}^n (\log 3) \left| \frac{1}{n+1} - \frac{1}{n} \right| + \frac{\log 3}{n+1} = O(n/n^2) + O(1/n) = O(1/n).$$

I leave it to you to write out the exercise that these properties force $\lim_{n \to \infty} a_n = \log h$.