In the 14th century ,an Indian mathematician T.V Narayana came up with a sequence now named after him.The sequence satisfies the recurrence $$a_{n}=a_{n-1}+a_{n-3}$$
Starting with $a_{0}=a_{1}=1$, and $a_{2}=2$ hence, $a_n = 1, 1, 2, 3, 4, 6, 9, 13,\dots$
I would like to conjecture a new generating function for narayana's sequence $a_{n}$. Given
$$\psi\Big(q\Big)=\cfrac{1}{1+q-\cfrac{(q^2)}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$
How can we show that
$$\psi\Big(q\Big)= \sum_{n=0}^\infty (-1)^{n} a_{n}q^n = 1 -1q +2q^2-3q^3+4q^4-6q^5+\dots$$
is true,such that the functional equation holds
$$\psi\Big(q\Big)=\frac{1}{q}\psi\Big(\frac{1}{q}\Big)$$
The ordinary generating function for your recurrence is $$ g(x) = \dfrac{1+x^2}{1-x-x^3}$$ Thus $$\sum_{n=0}^\infty (-1)^n a_n q^n = g(-q) = \frac{1+q^2}{1+q+q^3}\tag1$$
If that is $\phi(q)$, then indeed $$ \phi(1/q) = \dfrac{1/q^2+1}{1/q^3 + 1/q + 1} = \dfrac{q (1 + q^2)}{1+q^2 + q^3} = q \phi(q)$$
Now let's try to get your continued fraction. $$ \phi(q) = \dfrac{1}{1+q - \phi_1(q)}$$ where $$ \phi_1(q) = \dfrac{q^2}{q^2+1}$$ $$\phi_1(q) = \dfrac{q^2}{1 + q^3 + \phi_2(q)}$$ where $$ \phi_2(q) = q^2 - q^3$$ $$ \phi_2(q) = \dfrac{q^2 (1-q)(1-q^3)}{1+q^5 - \phi_3(q)}$$ where $$ \phi_3(q) = q^3 + q^5$$ $$\phi_3(q) = \dfrac{q^2(1+q^2)(1+q^4)}{1+q^7 + \phi_4(q)}$$ where $$ \phi_4(q) = q^4 - q^7$$ Hmm. Looks like a pattern developing, and should be possible to prove it by induction.