<s>[INST] Conjecture about a continued fraction 

Conjecture:

$$\large 2^{n-1}+\frac{1}{2+\cfrac{1}{2^{n}-1+\cfrac{1}{2+\cfrac{1}{2^{n}-1+\cfrac{1}{2+\ddots}}}}}=\frac{1+\sqrt{3a_n}}{2}\tag*{[1]}$$ such that $a_n=4a_{n-1}+1$ and $a_0=0$.$\quad(n\geqslant 1)$


Ex. If $n=1$, then $a_n=4a_0+1=4\times 0 + 1 = 1$. $$\therefore 1+\frac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\ddots}}}}}=\frac{1+\sqrt{3}}{2}\tag*{[2]}$$ This can be proven using the formula $x=a+\dfrac{1}{b+\dfrac{1}{x}}=a+\dfrac{1}{b+\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{a+\ddots}}}}$
Solving for $x$ results in a quadratic equation for which $x=\dfrac12\left\{a+\sqrt{a\left(a+\dfrac4b\right)}\right\}$.
Substituting $a=1$ and $b=2$ yields $[2]$ as required.

Problem is, I am unsure on how to (dis)prove this conjecture given the recursive sequence involved. How do I appropriately go about this? Any suggestions or counter-examples?
Thanks :)
 [/INST] Claim:
For all $n\ge1$, we have $[2^{n-1},2,2^n-1,2,2^n-1,\cdots]=(1+\sqrt{3a_n})/2$ with $a_n=4a_{n-1}+1$ and $a_0=0$.
Proof:
As $a_n=4a_{n-1}+1=4^2a_{n-2}+4^1+4^0=\cdots=4^ka_{n-k}+\sum\limits_{i=0}^{k-1}4^i$, choosing $k=n$ gives $$a_n=4^na_0+\sum_{i=0}^{n-1}4^i=\frac{4^n-1}3\implies\frac{1+\sqrt{3a_n}}2=\frac{1+\sqrt{4^n-1}}2.$$ We know that the continued fraction converges by the Seidel-Stern theorem, as the sum of the convergents diverges. Thus, writing the continued fraction as $y=2^{n-1}+x=2^{n-1}+[0,2,2^n-1+x]$, $$2+\frac1{2^n-1+x}=\frac1x\implies2x(2^n-1+x)+x=2^n-1+x$$ giving the quadratic $$2x^2+2(2^n-1)x-(2^n-1)=0\implies x=-\frac{2^n-1}2\pm\frac{\sqrt{4^n-1}}2.$$ Evidently, the positive root must be chosen so that $y=(1+\sqrt{4^n-1})/2$ as required.
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