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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

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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 $(27x^2/4-1)F^3+3F-2=0$ ?

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  • $\begingroup$ Thanks a lot! I don't know how the denominators all being powers of three escaped me. Both the above formulas showed up in some QFT calculations so being able to show they are the same is very nice :). $\endgroup$
    – Petter
    Jan 28 '18 at 0:04

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