I have a problem in a book I am working in that asks:
Let $c_n = \binom{n}{\lfloor{n/2}\rfloor}$. Prove:
$$\sum_{k=0}^{n} \binom{n}{k}c_kc_{n-k} = c_nc_{n+1}$$
The solution given is some counting argument, but I am interested in a solution involving generating functions since the $c_kc_{n-k}$ term looks really tempting.
On
A partial answer. Let $C(x)=\displaystyle{\sum_{n=0}^{\infty}c_{n}\frac{x^n}{n!}}$ the exponential generating series of $c_{n}$. Then,
$$ C(x)^{2}=\left(\sum_{n=0}^{\infty}c_{n}\frac{x^n}{n!}\right)\left(\sum_{n=0}^{\infty}c_{n}\frac{x^n}{n!}\right)=\sum_{n=0}^{\infty}\bigg(\sum_{k=0}^{n}{n \choose k}c_{n}c_{n-k}\bigg)\frac{x^n}{n!}.$$
and $$C^{\prime}(x)=\sum_{n=1}^{\infty}c_{n}\,n\frac{x^{n-1}}{n!} =\sum_{n=1}^{\infty}c_{n}\,n\frac{x^{n-1}}{n(n-1)!}=\sum_{n=0}^{\infty}c_{n+1}\frac{x^{n}}{n!}.$$
also, $c_{n}c_{n+1}$ is the coefficient of $\frac{x^{n}}{n!}$ in the Hadamard product (or coefficient-wise product) of the generating series of $C(x)$ and $C^{\prime}(x)$:
$$C(x)\times C^{\prime}(x)=\sum_{n=0}^{\infty}c_{n}c_{n+1}\frac{x^{n}}{n!}.$$
Nice temptation. Let us consider the ordinary generating function of the $\{\frac{c_n}{n!}\}_{n\geq 0}$ sequence, which is also the exponential generating function of the $\{c_n\}_{n\geq 0}$ sequence:
$$ f(x)=\sum_{n\geq 0}\frac{x^n}{n!}\binom{n}{\lfloor n/2\rfloor} = \sum_{n\geq 0}\frac{x^n}{\lfloor n/2\rfloor!\lceil n/2\rceil!}=\sum_{m\geq 0}\frac{x^{2m}}{m!^2}+\frac{x^{2m+1}}{m!(m+1)!}=I_0(2x)+I_1(2x).$$ Now
$$\sum_{k=0}^{n}\binom{n}{k}c_k c_{n-k}=n!\sum_{k=0}^{n}\frac{c_k}{k!}\cdot\frac{c_{n-k}}{(n-k)!}=n![x^n]f(x)^2 $$ so the claim is equivalent to $$ n![x^n]\left(I_0(2x)+I_1(2x)\right)^2 = \binom{n}{\lfloor n/2\rfloor}\binom{n+1}{\lfloor (n+1)/2\rfloor} $$ or to
$$ (I_0(2x)+I_1(2x))^2 = \sum_{m\geq 0}\frac{x^{2m}}{(2m)!}\binom{2m}{m}\binom{2m+1}{m}+\frac{x^{2m+1}}{(2m+1)!}\binom{2m+1}{m}\binom{2m+2}{m+1}. $$ Now I can show that both sides are entire functions with the same Laplace transform, which is related to elliptic integrals. But I guess there are simpler ways, currently eluding me.
All right, here they are: $$[x^n]\left(\sum_{m\geq 0}\frac{x^m}{m!^2}\right)^2 = \sum_{k=0}^{n}\frac{1}{k!^2(n-k)!^2}=\frac{1}{n!^2}\sum_{k=0}^{n}\binom{n}{k}^2=\frac{\binom{2n}{n}}{n!^2}=\frac{(2n)!}{n!^4}$$ so the whole exercise can be seen as a consequence of the Chu-Vandermonde identity.