a conjectured new generating function of narayana's sequence 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 
$$\psi\Big(q\Big)=\frac{1}{q}\psi\Big(\frac{1}{q}\Big)$$
 A: 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.
A: This supplements R. Israel's answer. Given the continued fraction discussed in this post for $|q|<1$,
$$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$
and using a little algebraic manipulation to transform the brown part to the form in the post,
$$\psi\Big(q\Big)=\cfrac{1}{1+q-\cfrac{\color{brown}{q^2}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$
one then gets,
$$\psi\Big(q\Big)= \frac{1+q^2}{1+q+q^3}$$
the same answer $(1)$ given by R. Israel.
