# Ordinary generating function for $\binom{3n}{n}$

The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}}$$

follows from the generalized binomial theorem and an application of the duplication formula for the gamma function.

But what about the ordinary generating function for $\displaystyle \binom{3n}{n}$?

According to Wolfram Alpha, $$\sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \big(\frac{1}{3} \arcsin (\frac{3 \sqrt{3x}}{2})\big)}{\sqrt{4-27x}}$$

Any suggestions on how to prove this?

EDIT:

Approaching this problem using the fact that $$\text{Res} \Big[ \frac{(1+z)^{3}}{z^{n+1}},0 \Big] = \binom{3n}{n}$$

you get that $$\sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{-1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1}$$

where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $\displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.

Evaluating that contour integral would appear to be quite tedious.

• Will try and flesh this out into a proper answer when I get the chance, but: it seems like you should be able to do this via the $n=3$ case of the Gauss multiplication formula, generalizing the duplication formula approach? – Steven Stadnicki Nov 8 '13 at 20:19
• @RandomVariable: Another proof variation is based upon the Lagrange inversion formula. See this answer to pretty much the same question. – Markus Scheuer Aug 10 '15 at 21:35
• @MarkusScheuer Thanks for the link. – Random Variable Aug 10 '15 at 22:04

Standard conversion to hypergeometric series and use of the duplication and triplication formulas for the $$\Gamma$$-function yields $$$$\sum_{n\ge 0}\binom{3n}{n}z^n = \sum_{n\ge 0}\frac{\Gamma(3n+1)}{\Gamma(2n+1)}\frac{z^n}{n!} = \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(n+1)} {\Gamma(n+1/2)\Gamma(n+1)}\frac{z^n}{n!} \frac{(2\pi)^{1/2} 3^{3n+1/2}}{2\pi2^{2n+1/2}}$$$$

$$$$= \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(1/2)} {\Gamma(n+1/2)\Gamma(1/3)\Gamma(2/3)}\frac{z^n}{n!} \frac{(2\pi)^{1/2} 3^{3n+1/2}}{2\pi2^{2n+1/2}}$$$$

$$$$= \sqrt{3/(4\pi)} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(1/2)} {\Gamma(n+1/2)\Gamma(1/3)\Gamma(2/3)}\frac{(2^{-2} 3^3z)^n}{n!}$$$$

$$$$= \sqrt{3/(4\pi)} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{(1/3)_n (2/3)_n} {(1/2)_n}\frac{(2^{-2} 3^3z)^n}{n!}$$$$

$$$$= \sqrt{3/(4\pi)} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)}{} _2F_1(1/3, 2/3; 1/2; 2^{-2}3^3z)$$$$ $$$$= \sqrt{3} \frac{\Gamma(1/3)\Gamma(2/3)}{2\pi}{} _2F_1(1/3, 2/3; 1/2; 2^{-2}3^3z)$$$$ Furthermore by equation 15.1.18 of Abramowitz/Stegun this Gaussian Hypergeometric Function can be reduced by $$$$_2F_1(a,1-a;1/2;\sin^2z)=\frac{\cos[(2a-1)z]}{\cos z}$$$$ with parameter $$a=1/3$$. Furthermore $$\Gamma(1/3)\Gamma(2/3) = 2\pi/\sqrt{ 3}$$ according to OEIS sequence A073006.

Proposition : $$f(z,a) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}$$

Proof : Note that,

$$\sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1}$$

where $U_{n} (x)$ is the Chebyshev Polynomial of the second kind.

$$\implies \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{x^2 +2ax +1}$$

Using Ramanujan Master Theorem, we have,

$$f(a,z) = \dfrac{\pi}{\sin \pi z} U_{-s} (a)$$

$$= \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square$$

Now, using Gamma Triplication Formula,

$$(3n)! \; = \; \Gamma(3n+1) \; = \; \dfrac{1}{2\pi} 3^{2n+\frac12} \Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23\right)\Gamma(n+1)$$

and hence

$${3n \choose n} \; = \; \frac{3^{3n+\frac12}}{2\pi} \times \frac{\Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23 \right)}{\Gamma(2n+1)} \; = \;\frac{3^{2n+\frac12}}{2\pi} \operatorname{B} \left(n+\dfrac23,n+\dfrac13\right)$$

Thus,

$$\begin{array}{rcl}\displaystyle \text{S} \; =\; \sum_{n=0}^\infty {3n \choose n}x^n & = & \displaystyle\frac{\sqrt{3}}{2\pi}\sum_{n=0}^\infty 3^{2n} x^n \int_0^1 u^{n-\frac13}(1-u)^{n-\frac23}\,du \\ & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 27xu(1-u)}\,du \; =\; \frac{\sqrt{3}}{2\pi}\int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 4a^2\, u(1-u)}\,du \end{array}$$

where $a = \dfrac{3}{2} \sqrt{3x}$. Substituting $u = \sin^2\theta$ and then $\tan\theta = x$, we have,

$$\text{S} = \dfrac{\sqrt{3}}{2 \pi} \int_0^\infty \frac{x^{\frac{1}{3}}(1 + x^2)}{(x^2 + 2ax + 1)(x^2 - 2ax + 1)} \mathrm{d}x$$

Using Partial Fraction and the Proposition, we have,

$$\text{S} = \dfrac{\sqrt{3}}{2\pi} \left[ \dfrac{1}{2} f \left( \dfrac{1}{3} , -a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{4}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{10}{3} , -a \right) + \dfrac{1}{2} f \left( \dfrac{1}{3} , a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{4}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{10}{3} , a \right)\right]$$

After simplification, we have,

$$\text{S} = \dfrac{2\cos \left(\frac{1}{3} \sin^{-1} \left(\dfrac{3\sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \quad \square$$

• I will always have time to upvote your solutions – Aman Rajput Jul 8 '16 at 8:14

The ordinary generating function of ${kn \choose n}$ is (the derivative of) a power series expansion for the real root near $1$ of a degree $n$ polynomial that looks very much like $$x^k - x - t=0.$$ There is a combinatorial interpretation using $k$-generalizations of Catalan numbers.

The fraction in the question therefore comes from solving a cubic equation, and the trigonometric solution is for the case with three real roots. This is consistent with $x^3 - x = t$ for small $t$.

Unfortunately I don't remember the exact polynomial. "Hypergeometric quintic" at Wikipedia finds