TITLE: Second order recurrence relation for third order polynomial root
QUESTION [6 upvotes]: 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

REPLY [9 votes]: This is sequence 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 $(27x^2/4-1)F^3+3F-2=0$ ?