Expressing the continued fraction $[k,\dots,k]$ as a closed form

Question

For positive integers $a_1,\dots,a_n\geq 2$, let $[a_1,\dots,a_n]$ denote the continued fraction $$ [a_1,\dots,a_n]=a_1-\frac{1}{a_2-\frac{1}{\cdots-\frac{1}{a_n}}}.$$

Then we have $[2,\dots,2]=(n+1)/n$ (with $n$ number of $2$'s) and this can be easily proved by induction. Similarly, can we express $[3,\dots,3]$ (with $n$ number of $3$'s) or more generally $[k,\dots,k]$ $(k\geq 2)$ in closed form?

In the case $k=3$, the first few are given by $3, \frac{8}{3}, \frac{21}{8}, \frac{55}{21},\frac{144}{55}$, so it seems that it is related to Fibonacci numbers, but I can't see how to express this as a closed form.

Answer 1

You can use Fibonacci polynomials of the second kind (alternating signs)

$i$ iterations of $k$ would give $[k,k,k,...,k]=\frac{F_{i+1}(k)}{F_i(k)}$

A closed form for $F_i(k)=\frac{\alpha(k)^i-\beta(k)^i}{\alpha(k)-\beta(k)}$

with $\alpha(k)=\frac{k+\sqrt{k^2-4}}{2}$ and $\beta(k)=\frac{k-\sqrt{k^2-4}}{2}$

so you would end up with $[k,k,k,...,k]=\frac{\alpha(k)^{i+1}-\beta(k)^{i+1}}{\alpha(k)^i-\beta(k)^i}$

Answer 2

$3 - \frac{f_{n+1}}{f_{n-1}} = \frac{f_{n-3}}{f_{n-1}}$

So $\frac{f_{n+1}}{f_{n-1}} = 3 - \frac{f_{n-3}}{f_{n-1}} = 3 - \frac{1}{\frac{f_{n-1}}{f_{n-3}}}= 3 - \frac{1}{3 - \frac{f_{n-5}}{f_{n-3}}}$ and so on.