Consider this recurrence relation:


$$
\begin{eqnarray\*}
f\_0&=&1\\
 f\_n&=&
\sum\_{m=0}^{n-1} \frac{\left(\frac{m+3}{2}\right)\_{m-1}}{\left(\frac{m+2}{2}\right)\_m} f\_{n-m-1} f\_m\ \ \ \text{for $1\leq n$.}
\end{eqnarray\*}
$$
where the Pochhammer symbol denotes the rising factorial. The generating function $f(z)=\sum\_{n=0}^\infty f\_nz^n$ seems to be a root of
$$
0=12 f^3 z^2- (f-1)^2 (f+2)
$$
I have checked this to be true for the first 600 terms. However, I have been unable to come up with a proof. Do you have any ideas on how I might show this to be true?


Cheers,
Petter



Second order recurrence relation for third order polynomial root
This is sequence [A244038](https://oeis.org/A244038) in OEIS after scaling by $3^n$, so $f\_n=(4/3)^n\binom{3n/2}n$. The fact that it satisfies a cubic equation
is certainly a well-known result in hypergeometric functions.


EDIT: remove the "well-known": set $F(x)=\sum\_{n\ge0}\binom{3n/2}{n}x^n$.
Then $$F(x)={}\_2F\_1(1/3,2/3;1/2;27x^2/4)+(3x/2)\,{}\_2F\_1(5/6,7/6;3/2,27x^2/4)$$
(which can probably be slightly simplified using contiguity relations),
but can a hypergeometric expert explain why $(27x^2/4-1)F^3+3F-2=0$ ?


