Modified central binomial coefficients generating function

Question

Given $n \in \mathbb{N}$, I would like to find the ordinary generating function of the sequence $a_k = \binom{2n-2k}{n-k}$.

If \begin{align} A(x) = \sum_{k = 0}^\infty a_kx^k, \end{align} then I find that \begin{align} A(x) &= \sum_{k = 0}^n \binom{2n-2k}{n-k}x^k \\ &= \sum_{k = 0}^n \binom{2k}{k}x^{n-k} \\ &= x^n\sum_{k = 0}^n \binom{2k}{k} \left(\frac{1}{x}\right)^k \end{align} but I am stuck here. My understanding is that you cannot extend the last finite sum into an infinite series, so I cannot use the generating function of $\binom{2k}{k}$.

I have also tried rewriting $A(x)$ as \begin{align*} A(x) &= [y^n]\left(1 + xy + (xy)^2 + \cdots\right)\left(\sum_{i \ge 0} \binom{2i}{i}y^i\right)\\ &= [y^n] \frac{1}{1-xy}\frac{1}{\sqrt{1-4y}} \end{align*} but I have no idea how to proceed from here.

Any idea is greatly appreciated.

Answer 1

We consider $a_{n,k}=\binom{2n-2k}{n-k}$ with $n,k\geq 0$ non-negative integers.

• Horizontal GF: First of all we note that \begin{align*} A_n(x)=\sum_{k=0}^na_{n,k}x^k=\sum_{k=0}^n\binom{2n-2k}{n-k}x^k\qquad\qquad n\geq 0 \end{align*} is a polynomial in $x$ and as such a perfect ordinary generating function, a so-called horizontal generating function. Since it is a polynomial having a finite number of terms $a_{n,k}x^k$ not equal to zero, we do not expect a representation via $\frac{1}{\sqrt{1-4x}}$ which is an infinite series.

• Vertical GF: On the other hand we can consider the vertical generating function for fixed $k\geq 0$: \begin{align*} B_k(y)&=\sum_{n=k}^\infty a_{n,k}y^n=\sum_{n=k}^\infty\binom{2n-2k}{n-k}y^n\\ &=\sum_{n=0}^\infty\binom{2n}{n}y^{n+k}\\ &=\frac{y^k}{\sqrt{1-4y}} \end{align*}

• Bivariate GF: We have the bivariate generating function $G(x,y)$ with $A_n(x)$ and $B_k(y)$ as horizontal resp. vertical section: \begin{align*} \color{blue}{G(x,y)}&=\sum_{k=0}^\infty\sum_{n=k}^\infty a_{n,k}x^ky^n\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{2n-2k}{n-k}x^ky^n\\ &=\sum_{k=0}^\infty\sum_{n=0}^\infty\binom{2n}{n}x^ky^{n+k}\\ &=\sum_{k=0}^\infty(xy)^k\sum_{n=0}^\infty \binom{2n}{n}y^n\\ &\,\,\color{blue}{=\frac{1}{1-xy}\,\frac{1}{\sqrt{1-4y}}} \end{align*}

Answer 2

Here's the Mathematica result that I mentioned in the comments, in case it's helpful. It uses Gauss's hypergeometric series ${}_2F_1$ (for background, see the beginning of Gasper & Rahman's Basic Hypergeometric Series, Cambridge, 2004).

$$\sum_{=0}^ {2n-2k \choose n-k} ^ = {2n \choose n} \, {}_2F_1\!\left(1,-n;\frac{1}{2}-n;\frac{x}{4}\right)$$