Second order recurrence relation for third order polynomial root

Question

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

Answer 1

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

Answer 2

To prove that $F(z)$ is a root of $(27x^2/4-1)F^3+3F-2=0$, we rewrite as:

[1] $x^3+12x/(27z^2-4)-8/(27z^2-4)=0$

Then making use of equation (48) of the roots of the general trinomial https://arxiv.org/pdf/2212.09919.pdf

Letting $I=0$ one can trivially see that it yields the desired binomial expansion like $F(z)$, but it differs slightly because the denominator has $(n+1)$.

After some labor using (48) we have

$2u^{-1}/3-1/3 \sum_{n\ge0}\binom{3n/2}{n}\frac{u^n}{n+1}$

In which $u=(27/(4-27z^2))^{-1/2}$

Which is a root of [1], which we will denote as $G(z)$

$F(z)$ and $G(z)$ are obviously related by differentiating or integrating.

Converting $F(z)$ and $G(z)$ to their trig equivalents (using binomial to -->> hypergeometric --> inverse-trig transformations):

$\sum_{n\ge0}\binom{3n/2}{n}\frac{u^n}{n+1}= (2- \frac{4}{\sqrt{3}}\cos(\frac{1}{3} \arcsin (\frac{3u\sqrt{3}}{2})+\pi/6) )/u $

$\sum_{n\ge0}\binom{3n/2}{n} z^n = ( 4 \sin(\frac{1}{3} \arcsin (\frac{3z\sqrt{3}}{2})+\pi/6) )\frac{1}{\sqrt{4-27z^2}}$

Thus, we have to prove the following is true:

$\sin(\frac{1}{3} \arcsin (\frac{3z\sqrt{3}}{2})+\pi/6)=\cos(\frac{1}{3} \arcsin (\frac{3u\sqrt{3}}{2})+\pi/6)$

Because $G(z)$ converges for $0<z<2z/3\sqrt{2/3} $ we have:

$ \arcsin (\frac{3z\sqrt{3}}{2})+ \arcsin (\frac{3u\sqrt{3}}{2})=\pi/2 $

Using

$\arcsin x + \arcsin y = \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ it is easy to confirm it is true.

Answer 3

Here's a sketch of a proof, using Lagrange inversion, of the equation $(27x^2/4-1)F^3+3F-2=0$, where $$F(x) = \sum_{n=0}^\infty \binom{3n/2}{n}x^n.$$

One form of Lagrange inversion says that if $h(x) = x R(h(x))$ then (writing $h$ for $h(x)$) for any power series $\psi$, $$[x^n] \frac{\psi(h)}{1-xR'(h)} = [t^n]\psi(t) R(t)^n.$$ Since $\binom{3n/2}{n}= [x^n]\bigl( (1+x)^{3/2}\bigr)^n$, taking $R(t) = (1+t)^{3/2}$ and $\psi(t) = 1$ gives $$F =\frac{1}{1-\frac32 x (1+h)^{1/2}}\tag{1}$$ where $h=x(1+h)^{3/2}$ and thus $$h^2 = x^2(1+h)^3\tag{2}.$$ Solving $(1)$ for $h$ in terms of $F$ and $x$, substituting in $(2)$, and factoring gives $$[(27x^2-4)F^3 +12F -9 ][(27x^2-20)F^3+48F^2 -36F]=0\tag{3}.$$ We can check that the second factor on the left side of $(3)$ is not 0 (the coefficient of $x^3$ is $-54$) so the first factor must be 0

It may be noted that if we set $A(x) = F(4x) = \sum_{n=0}^\infty 4^n \binom{3n/2}{n}x^n$, so that the coefficients of $A(x)$ are integers, then in the OEIS page A244038 for the coefficients of $A(x)$, Michael Somos states that $A(x)$ satisfies $A(x)^3 (1 - 108 x^2) = 3A(x) - 2$, which is equivalent to $(27x^2/4-1)F^3+3F-2=0$. Somos also gives a reference to my paper A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, which has some further information about these and related integers.

It remains to prove that proposer's recurrence for the numbers $f_n$ is satisfied by $f_n = (4/3)^n\binom{3n/2}{n}$. This does not seem easy.