Help me to simplify $\sum\limits_{i=0}^{\lfloor\frac{r}{2}\rfloor}\binom{r}{i}\binom{r-i}{r-2i}$ I tried to simplify $\sum\limits_{i=0}^{\lfloor\frac{r}{2}\rfloor}\binom{r}{i}\binom{r-i}{r-2i}$ by using generating function in the similar way of markus-scheuer's answer to the question How to prove $\sum\limits_{i=0}^{\lfloor\frac{r}{2}\rfloor}\binom{r}{i}\binom{r-i}{r-2i}2^{r-2i}=\binom{2r}{r}$. What I got is $[u^r](1+u+u^2)^r$. I don't know how to compute the coefficient of $u^r$ in $(1+u+u^2)^r$. I tried using by trinomial expasion but I didn't suceed. 
 A: There   is  no closed form available for the  central trinomial coefficients
\begin{align*}
[u^r](1+u+u^2)^r&=\sum_{i=0}^{\lfloor r/2\rfloor}\binom{r}{i}\binom{r-i}{r-2i}
=\sum_{i=0}^{\lfloor r/2\rfloor}\binom{r}{2i}\binom{2i}{i}
\end{align*}

Plausibility check:
We consider the more general expression 
  $$\sum_{i=0}^{\lfloor r/2\rfloor}\binom{r}{i}\binom{r-i}{r-2i}t^{r-2i}$$
   and obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{\lfloor r/2\rfloor}}&\color{blue}{\binom{r}{i}\binom{r-i}{r-2i}t^{r-2i}}\\
&=\sum_{i=0}^\infty[z^{i}](1+z)^{r}[u^{r-2i}](1+tu)^{r-i}\tag{1}\\
&=[u^r](1+tu)^r\sum_{i=0}^\infty\left(\frac{u^2}{1+tu}\right)^i[z^i](1+z)^r\tag{2}\\
&=[u^r](1+tu)^r\left(1+\frac{u^2}{1+tu}\right)^r\tag{3}\\
&\color{blue}{=[u^r](1+tu+u^2)^{r}}\tag{4}\\
\end{align*}
We observe in case of $t=2$ the nice simplification
  \begin{align*}
[u^r](1+2u+u^2)^{r}=[u^r](1+u)^{2r}=\binom{2r}{u}
\end{align*}
  Regrettably there is no obvious similar simplification for other values of $t$.

Comment:


*

*In (1) we apply the coefficient of operator twice and set the upper limit of the sum to $\infty$ without changing anything, since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator, do some rearrangements and use the rule
\begin{align*}
[u^{p-q}]A(u)=[u^p]u^qA(u)
\end{align*}

*In (3) we apply the substitution rule of the coefficient of operator with $z:=\frac{u^2}{1+tu}$
\begin{align*}
A(u)=\sum_{i=0}^\infty a_i u^i=\sum_{i=0}^\infty u^i [z^i]A(z)
\end{align*}

*In (4) we do some simplifications.

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*}
[u^r](a+bu+cu^2)^r=[u^r]\frac{1}{\sqrt{1-2bu+(b^2-4ac)u^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: The central trinomial coefficient (A002426): the coefficient of $x^n$ in $(1+x+x^2)^n$, is given by:
$$\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{i;2n-2i-2\lfloor\frac{n}{2}\rfloor;i}$$
because $i$ represents the number of $2$'s, $2n-2i-2\lfloor\frac{n}{2}\rfloor(=n-2i+n-2\lfloor\frac{n}{2}\rfloor)$ represents the number of $1$'s and $i$ represents the number of $0$'s in a partition of $n$ into $\{0,1,2\}$.
For example, $n=6$, we have $222000,221100,211110,111111$, represented in multinomial format as $303,222,141,6$.
So the sum is:
$$\frac{720}{3!0!3!}+\frac{720}{2!2!2!}+\frac{720}{1!4!1!}+\frac{720}{6!}$$
$$=20+90+30+1$$
$$=141$$
