Seeking a closed expression for a combinatorial sum Is there a simple closed expression for the following sum?
$$\sum_{i=0}^{\lfloor\frac n2\rfloor}{n\choose i}{n-i\choose i}$$ 
I can see that this is the constant term in $\big(\frac 1x+1+x\big)^n$. But this is no closer to a simple close expression I am seeking.
 A: We use the coefficient of operator $[z^i]$ to denote the coefficient of $z^i$ of a series. This way we can write for instance
\begin{align*}
[z^i](1+z)^n=\binom{n}{i}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{\left\lfloor \frac{n}{2}\right\rfloor}}&\color{blue}{\binom{n}{i}\binom{n-i}{i}}\tag{2}\\
&=\sum_{i=0}^{n}\binom{n}{i}[z^i](1+z)^{n-i}\tag{3}\\
&=[z^0](1+z)^n\sum_{i=0}^{n}\binom{n}{i}\left(\frac{1}{z(1+z)}\right)^i\tag{4}\\
&=[z^0](1+z)^n\left(1+\frac{1}{z(1+z)}\right)^{n}\tag{5}\\
&\,\,\color{blue}{=[z^{n}](1+z+z^2)^n}\tag{6}
\end{align*}

Comment:


*

*In (3) we apply the coefficient of operator according to (1). We also set the upper limit to $n$ without changing anything, since $\binom{n-i}{i}=0$ if $i>\left\lfloor\frac{n}{2}\right\rfloor$.

*In (4) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (5) we apply the binomial theorem.

*In (6) we do some simplifications and use again the rule as in (3).

We observe the binomial sum (2) represents essentially central trinomial coefficients
\begin{align*}
[z^n](1+z+z^2)^n
\end{align*}
  for which there is no closed form available.

Notes from the experts:

D.E. Knuth gives in Concrete Mathematics, Appendix A 7.56 the following representation of a more general expression
\begin{align*}
[z^n](a+bz+cz^2)^n=[z^n]\frac{1}{\sqrt{1-2bz+(b^2-4ac)z^2}}
\end{align*}
He states that according to the paper Hypergeometric Solutions of Linear Recurrences with Polynomial Coeffcients by Marko Petkovšek there exists a closed form (more precisely: a closed form solution as a finite sum of hypergeometric terms) if and only if
  $$\color{blue}{abc(b^2-4ac)=0}$$
In case of central trinomial coefficients we have $a=b=c=1$. Since then the expression $abc(b^2-4ac)=-3\ne 0$
  there is no such closed form in particular for the central trinomial coefficients.

A: As other people recommended, I computed a few terms by computer and it gives A002426, which seems that there's no exact formula for it. The generating function is 
$$
\frac{1}{\sqrt{1-2x-3x^{2}}}
$$
and we also have an asymptotic formula
$$
a(n) \sim 3^{n}\sqrt{\frac{3}{4\pi n}}
$$
